Layout

......MAIN SECTION ‫ﻣﻼﺣظﺎت ﻋن وﺿﻊ‬

( 8: 4) ‫ ﻓﻲ اﻻﺗﺟﺎه اﻟﻘﺻﯾر ﺣﯾث ﺗﺗراوح اﻟﻣﺳﺎﻓﮫ ﻣن‬main system ‫ ﯾﺗم وﺿﻊ‬-1
:‫ﻣﺗر‬

𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝐿𝐿
– n.o. frames
4 ∶ 8 𝑚𝑚

A-EX-

Span = 60 m Lat S = 6 m
n.o.frames = 60 / 6 = 10
all frames = 11 frames

B-EX –

span = 60 m lat n.o. all frames = 9

n.o.frames = 9-1 = 8
Spacing the distance between frames = 60/8 = 7.5 m
 N.O.main system

.‫ ( ﻣﺗر‬2: 1.5 ) ‫ ﺗﺗرواح ﻣن‬Burlines ‫ اﻟﻣﺳﺎﻓﺎت ﺑﯾن‬-2

𝐵𝐵
2
-n.o.burlin =
1.5 ∶ 2 𝑚𝑚
EX -a
B = 30 m let( a ) spacing Between burlin = 1.50
N.o.burlin = 15/ 1.50 = 10 burlin
N.o.all Burlin = 10 + 1 = 11 burlin at one side

EX-B
B = 30 M let N.O. all Burlin = 9 at one side
N.o, burlin = 9-1 = 8
a= 15/8= 1.875 m
 ‫‪N.O.Burlin at one side‬‬

‫‪ -3‬ﯾﺗم وﺿﻊ ‪ Bracing‬ﻋﻠﻰ ﻣﺳﺎﻓﺎت ﺗﺗرواح ﻣن ‪ 25‬اﻟﻲ ‪ 30‬ﻣﺗر‬

 ‫‪.‬‬ ‫‪ -4‬ﺗوﺿﻊ ‪Bracing‬ﻋﻠﻲ اﻟﻣﻧﺷﺎ ﺑﺣﯾث ﺗﺗرواح زاوﯾﺗﮭﺎ ‪ 30‬اﻟﻲ ‪60‬‬

‫‪ -5‬ﯾﻔﺿل وﺿﻊ ال ‪ end gabel‬ﻋﻠﻲ ﺗﻘﺎطﻊ ‪ H.bracing‬ﺑﺣﯾث ﺗﺳﺗطﯾﻊ ﻧﻘل‬

‫أﺣﻣﺎل اﻟرﯾﺎح ‪.‬‬
‫‪ – 6‬اذا ﻛﺎن اﻟﻣﻧﺷﺎ ‪ truss‬ﯾﺗم وﺿﻊ ‪longitudinal Bracing‬ﻋﻠﻰ‬
‫ﻣﺳﺎﻓﺎت ﻟا ﺗزﯾد ﻋن ‪ 8‬ﻣﺗر‬
‫‪ – 8‬ﯾﺗم رﺳم ﻟﻛل ‪ side view‬ﻣﻧظورﯾن اﺣدھم ‪ strut‬واﻻﺧر‬
‫‪. burlin‬‬
‫‪ -9‬اي ﺗﻐﯾر ﻓﻰ ‪ in plan‬ﯾﺟب ان ﯾﺗﺑﻌﮫ ﺗﻐﯾر‪ outplan‬ﺑوﺿﻊ‬
‫‪. strut‬‬
 ‫‪ -10‬ﺣﺳﺎب اﻟﻣﯾل واﻟﻣﺳﺎﻓﮫ ‪x‬‬
‫‪At frames .‬‬

‫راﺳﻲ‬ ‫اﻓﻘﻲ‬

‫‪1 : 10‬‬
‫‪X : B/2‬‬
‫𝐵𝐵‬
‫∗‪1‬‬
‫= ‪TO GET X‬‬ ‫‪2‬‬
‫‪10‬‬

‫‪Ex -‬‬
B= 30 m Slop = 1:10
X= 1*15/10 = 1.5 m
AT TRUSS

𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝐵𝐵)
1-calculate H=
12:16
𝐵𝐵 1
2-calculate h = H- [
2
* ]
𝑧𝑧
h(min) 1.25 m

EX

B = 40 M SLOP = 1:10
 ‫‪40‬‬
‫= ‪1- H‬‬ ‫‪= (3.33 : 2.5 ) use 3 m‬‬
‫‪12:16‬‬
‫‪40‬‬ ‫‪1‬‬
‫( – ‪2- h= 3‬‬ ‫∗‬ ‫‪) = 1 m use min = 1.25 m‬‬
‫‪2‬‬ ‫‪10‬‬

‫‪ -11‬ﯾﻔﺿل ﻋﻧد وﺿﻊ ال ‪ members‬ﻓﻲ ‪ truss‬ﺗﻛون ﻣوازﯾﮫ ﻟﻠﻣوﻣﻧت ﺣﺗﻲ‬

‫ﯾﺗوﻟد ‪ tension‬ﻓﻲ ‪-: member‬‬
.main system ‫ أﺳس أﺧﺗﯾﺎر‬-12
EXAMBLE:-
Calculat N.o.frames and burlin
Solution
1- calculate spacing between frames
-let N.o.All frames = 9
- N.o.frames = 9-1 = 8
.
-spacing between frames = 8 =
S= 5.60 m

2-calculate spacing Between burline at frames

/
-N.o.spacing = =6
N.o.burlins = 7 Burlin at one side

3- calculate spacing Between burline at truss

N.o.spacing = =4
.
N.o.burlis = 5 Burlin at one side

4-calculate x=

1:10 1:10
X:B/2 x: 12 x= 1.20 m
5- calculate ɸ
Tan ɸ = 1.20 /12 =0.1 ɸ=5.71
LOADS
 2- Calculate loads

LOADS = 1) Dead load .

2)Live load .
3)Wind load .
4)Temberature .
5)Bracking force .
6)Lateral shock .

1) DEAD LOAD :-
AT FRAMES :-
𝑺
--D.L = (O.W.T) of steel X S + W cover X 𝑪𝑶𝑺 ɸ
1- o.w.t of steel …….. calculat by sap
Or……..let (20:40) kg/m-

3- o.w of single sheet = 5:10 kg/m2

o.w of Double sheet= 10 :18 kg/m2

4- S= Spacing between frames

5- ɸ = slop angel
 AT TRUSS :-
𝑺𝑺𝑺𝑺𝑺𝑺
----D.L = (O.W.T) of steel X S X a + W cover X 𝑪𝑪𝑪𝑪𝑪𝑪 ɸ
1- o.w.t of steel …….. calculat by sap
Or……..let (20:40) kg/m- take W s= B

2-o.w of single sheet = 5:10 kg/m2

o.w of Double sheet = 10 :18 kg/m2

3- S= Spacing between frames

𝟒𝟒 − ɸ = slop angel
5 −a= spacing between burlin

2) LIVE LOAD :-
3) AT FRAMES :-

----Wl.l= L.L X S

1- L.L inaccessible roof= 60 - ( 66.67 tan Φ ) > 20 kg/m2

L.L accessible roof = 200 - ( 250 tan Φ ) > 50kg/m2

2- S= Spacing between frames

AT TRUSS :-

----Wl.l= L.L X S X a

1- L.L inaccessible roof= 60 - ( 66.67 tan Φ ) > 20 kg/m2

L.L accessible roof = 200 - ( 250 tan Φ ) > 50kg/m2

S= Spacing between truss

a= spacing between burlin

3)wind LOAD :-
AT FRAMES
𝑺𝑺
W wind load = ce x k x q x 𝑪𝑪𝑪𝑪𝑪𝑪 ɸ

AT TRUSS
𝑺𝑺𝑺𝑺 𝒂𝒂
W wind load = ce x k x q x 𝑪𝑪𝑪𝑪𝑪𝑪 ɸ
K= Height k
0 -- 10 1.00
10 -- 20 1.15
20 -- 30 1.40
30 -- 50 1.60

q=
place q kg/m2
cairo 68.0
alex 81.0
matroh 110.3
faium 56.3
suze 92.0

-You can calculate (q)_ by this equation

q = 0.5*ρ V2 Ct Cs = …………….. N/m2 (/10) kg/m2

--Wind Speed V =……….. m/s

--Air Density (taken) ρ = 1.25 Kg/m

--Topographic coefficient:- Ct = 1.0 (Flat Area <5.0%)

--Structural coefficient :- Cs = 1.0 (Height <60.0m)

Ce=
 ‫ ﺗﺗوﻗف ﻋﻠﻰ ظل زاوﯾﮫ اﻟﻣﯾل‬ce ‫ﻗﯾم ال‬

If tanα < 0.4 take ce = suction

If tanα > 0.8 take ce = pressure
If tanα =(0.4: 0.8) take ce = pressure and suciton
4)Temberature:-
 5)Bracking force .
At crane girder = 1/7 max load with out imapact factor
6)Lateral shock .
At crane girder =0.1 max load with out impact factor

Cases for design :-

CASE A CASE B case designe if case B/ A <1.2
EX design by (case A)
D.L+L.L D.L+L.L+W.L.L D.L+L.L+W.L.R D.L+W.L.L D.L+W.L.R (MAX case B)/ Case A) if case B/A >1.2
design by case B
MEMBER 1 -12 -9 -13 -2 -6 13/12= 0.9<1.2 case A

‫ ﻓﻲ‬case B ‫طرﯾﻘﮫ اﻓﺿل ﻓﻰ اﻟﺗﺻﻣﯾم ﺑﺄن ﻧﺿرب ﺟﻣﯾﻊ ﻣﻌﺎدﻻت‬

(0.833)

𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐵𝐵
< 1.2 design by case A
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐴𝐴

𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐵𝐵
> 1.2 design by case B
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐴𝐴

case B = case A *1.20

𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐵𝐵
Case A = =( 0.833 * cases B )
1.20
 ‫ﺣﺎﻻت ﺗﺣﻣﯾل اﻟزﻻزل اﻟﻣوﺟوده ﻓﻰ اﻟﻛود ھﻲ ﺣﺎﻻت ‪ UL‬ﻓﺑﺗﺎﻟﻲ ﯾﺗم ﺗﻘﺳﯾم اﻻﺣﻣﺎل‬
‫ﻋﻠﻲ ‪ 1.4‬ﻟﻛﻲ ﯾﺗم ﺗﺣوﯾﻠﮭﺎ ‪ working‬وﯾﺗم ﻗﺳﻣﺗﮭﺎ ‪ 1.2‬ﻟﻛﻲ ﯾﺗم ﻧﺣوﯾﻠﮭﺎ اﻟﻲ ‪case 1‬‬
‫وﻧﻘوم ﺑﺗﺻﻣﯾم ﻣﺑﺎﺷره ‪.‬‬

‫‪Case A = 1/1.4 =0.71428…………….= 0.71428/1.2=0.56 EQ‬‬

‫‪Load combination‬‬
‫ﺣﺎﻻت اﻟﺗﺣﻣﯾل اﻟﻣﺻري ھﻲ ﺣﺎﻻت أﺟﺗﮭﺎدﯾﮫ ﻟﻠﺣﺻول ﻋﻠﻰ اﻗﺻﻲ ‪straing‬‬
‫‪action‬‬
ASD

1- Dead load
2- Dead load+Live load
3- 0.833 dead load + 0.833 Live load +0.833wind load right
4- 0.833 dead load + 0.833 Live load +0.833wind load lift
5- 0.833 dead load+0.833wind load right
6- 0.833 dead load+0.833wind load lift
7- 0.833 dead load+0.56 earth quak+x
8- 0.833 dead load+0.56 earth quak-x
9- 0.833 dead load+0.56 earth quak-Y
10- 0.833 dead load+0.56 earth quak+Y
11- 0.833 dead load+0.833liveload+0.56 earth quak+x
12- 0.833 dead load+0.833liveload+0.56 earth quak-x
13- 0.833 dead load+0.833liveload+0.56 earth quak+y
14- 0.833 dead load+0.833liveload+0.56 earth quak+y
15- 0.833 dead load+0.833Cran
16- 0.833 dead load+0.833liveload+0.833 Cran
17- 0.833 D.L +0.833 W.L.L+0.833cran
18- 0.833 D.L +0.833W.L.R+0.833cran
19- 0.833 D.L + 0.833 L.L +0.833W.L.R+0.833cran
20- 0.833 D.L + 0.833 L.L +0.833W.L.L+0.833cran
21- 0.833 dead load+0.56 Earth quick+x +0.833Cran
22- 0.833 dead load+0.56 Earth quick-x + 0.833Cran
23- 0.833 dead load+0.56 Earth quick+y +0.833Cran
24- 0.833 dead load+0.56 Earth quick -y+ 0.833Cran
25- 0.833 dead load+0.56 Earth quick +0.833L.L+0.833Cran
 Loads
Loads
Loads acting on truss:
1) Dead load:
Weight of corrugated and weight of steel structure
W c = (5 à8) kg / m2 for single layer (default)
= (12 à18) kg / m2 for double layer (if given)
W s = (20 à35) kg / m2/hz proj. We take W s= B
i.e. If the span of truss = 28 m take W s= 28 kg / m2
a
PD = Wc S + Ws aS
cos α

P P P
PDL/2 PDL PDL PDL DL DL DL PDL PDL PDL PDL/2

2) Live load:
W LL = 60 – 66.66 tan α inaccessible roof (default)
= 200 – 300 tan α accessible roof
In kg / m2 /hz proj.
i.e. If the slope 1 : 10
1
∴ W LL = 60 – 66.66 * = 60 – 6.7 = 53.3 kg / m2 /hz proj.
10
PLL = W LL a S
 Loads

P P P
PLL/2 PLL PLL PLL LL LL LL PLL PLL PLL PLL/2

3) Wind load:
W w = C e k q kg / m2
a - K height factor k = 1.0 h 10 m
k = 1.1 10 < h 20
b – q = wind intensity in kg / m2
= 70 kg / m2 Cairo 80 kg / m 2 Alexandria
c – For C e: divide truss into 2 parts

Wind Pressure side Suction side Suction side Pressure side Wind
direction direction

For wind left (WL) For wind right (WR)

For suction side: C e = - 0.5 ( - ve means suction outside truss)

For pressure side: C e is calculated using the following curve
 Loads
Ce

0.8

0.05
0.2 0.4 0.8 tan

- 0.8

ie: For all trusses of slope 1:5 (0.2) to 1:20 (0.05), C e will be = -0.8 (suction)
although it lies in the pressure side.
Only fink truss can have slope more than 0.4
h a
P1 = 0.8 * 70 * 1 * * S P2 = −0.8 * 70 * 1 * *S
2 cos α
a h
P3 = −0.5 * 70 * 1 * *S P4 = −0.5 * 70 * 1 * * S
cos α 2
P3 /2

P2
P2 /2

P2 /2 P2 P2 P2 P3 P3 P3 P3 P3 /2
P1
P1

Note loads at the joint at mid span

 Loads
There are 2 main types of loads
Primary loads: Case I (A): dead and live load
Secondary loads: Case II (B): wind load, braking force & lateral shock
For case II (B) increase the allowable stresses by 20 %

IMP: To determine the design case:

D + L +W D + L +W
If ≤ 1.2 Case I If > 1.2 Case II
D+L D+L
Example:
Determine the design value & case for the following members:
DL LL WL WR
5 7 1 -2
5 7 3 -2
5 7 3 5
5 7 -3 -1
5 7 1 -7
5 7 3 -7
-5 -7 3 -3
-5 -7 3 -1
-5 -7 7 3

Solution:

Case A Case B Comparison Design case

D+L D+L+WL D+L+WR D+WL D+WR and value
12 13 10 6 3 13/12=1.08<1.2 12 T (A)
12 15 10 8 3 15/12=1.25>1.2 15 T (B)
12 15 17 8 10 17/12=1.4>1.2 17 T (B)
12 9 11 2 4 12 T (A)
12 13 5 6 -2 13/12=1.08<1.2 12T (A), 2C (B)
12 15 5 8 -2 15/12=1.25>1.2 15T (B), 2C (B)
-12 -9 -15 -2 -8 15/12=1.25>1.2 15 C (B)
-12 -9 -13 -2 -6 13/12=1.08<1.2 12 C (A)
-12 -5 -9 2 -2 12 C (A), 2 T (B)
DESIGN OF TENSION
MEMBERS
 DESIGN OF TENSION MEMBERS
3.1. GENERAL
This part presents summarized design considerations to design axial tension members. The tensile
resistance is calculated on bases of yielding on the gross area and fracture on the effective net area.

Design considerations:

• Allowable strength:
= 0.58 ( )
= 1.40 / 37
= 2.10 / 52

( )
=
( )

Where,

= = × ( )=

Calculating net area:

1. In case bolts are on the same line:

= × − × ( + 0.20) ×

= ℎ
 =

= ℎ

2. In case of staggered bolts:

In this case, all possible failures shall be studied and in each failure case, the net area is calculated.
The minimum net area will be used in the design of the member.

For example, from the previous figure:

• For section abcdf:

= × − × ( + 2) ×
• For section abekdf:
×
= − ×( × )+
4×
Where,
=
= ℎ ℎ
=
=
=

= ℎ

From the previous general equation, is calculated as follows:

 ×
= × −4× × +2×
4×
• For section abeckdf:
×
= × −5× × +2×
4×
Finally, = ℎ
------------------------------------------------------------------------------------------------------------

3. Calculating of A net for a section composed of two angles connected by a gusset plate and
bolts:

= + 2.00
=2× ( )−2× ×
Where,
= ℎ
------------------------------------------------------------------------------------------------------------

4. Calculating of A net for a section composed of one angle connected by a gusset plate and
bolts:
 3
= +
3 +
Where,
=

= − ×
2

=
2

------------------------------------------------------------------------------------------------------------

Example:

Determine the maximum tensile force (T) carried by 2 channels No. 260 as shown in the given
connection.

Given:

Steel used 52

Bolts are M16 (4.60)

For channel No. 260:

Area of one channel = 48.30 cm2.

tw = 1.00 cm

For M16… = + 2 = 16 + 2 = 18

For section ABCD:

=2× ( ℎ )−4× ×

= 2 × 48.30 − 4 × 1.80 × 1 = 89.40

For section EFGHI:

×
= − ×( × )+
4×
7 ×1
= 2 × 48.30 − 6 × (1.80 × 1) + 4 × = 93.96
4×6
Finally, = 89.40 For St.52, = 2.10 /

Then, = × = 89.40 × 2.10 = 187.74

L / d < 60
Where L: is the length of member and d = depth of member.

λmax < 300

Where:

λmax is the max. Slenderness ratio of λx and λy

λx= Kx . Lx / ix and λy = Ky. Ly / iy

r = radius of gyration = √ (I / A)

I is the inertia and A is the area of cross section of the member.

Buckling length of truss members:

Buckling length in plane:

For all members, the buckling length in plane equal the length of member itself.
i.e. Lbx = L (length of member).

Buckling length out of plane:

For upper chord members: the buckling length out of plane is the distance between two fixed
points. The fixation is either with the purlins, upper bracing for upper chord members, lower
bracing for lower chord members or vertical bracing for the lower chord members.

For vertical and diagonal members, the buckling length in plane equals the length of the member
itself. The buckling length out of plane equals the length of the member itself in case of presence
of upper bracing (general case) and equals 1.2 x the length of the member in case of absence of
upper bracing (special case).

For example:
For the upper figure:
The buckling length for the upper chord members:
• In plane = b
• Out of plane = b
The buckling length for the lower chord members:
• In plane = a
• Out of plane = L1 for member 1 and L2 for member 2.
Calculation of (rx and ry) in case of two angles back to back:
The values (A, Ix, Iy, rx, ry and e) given in the tables are for one angle only. To calculate these
values for two angles:
Area (two angles) = 2 x Area (one angle)

× ( ) ( )
rx (two angles) = = = rx (one angle)
× ( ) ( )

.
ry (two angles) = ( ( )) +( + )
 Design procedure of tension members (2 angles and Bolted
Connections):
1. Select member section.
×
• = = 300
Get a1 = … cm
×
• = = 300
Get a2 = … cm
• >3 +
Get a3 = … cm

• . = … … … …. = 0.58 (0.85 is an assumption of

. ×
the area of holes in the section)

( / )
Steel Grade
≤ 40 > 40
St. 37 1.40 1.30
St. 44 1.60 1.50
St. 52 2.10 2.00

Choose the required angle from tables.

2. Check Stresses:

• Calculate
• Calculate = …≤
3. Check minimum angle:
• >3 +
4. Check slenderness ratio:
• = ≤ 300
• = ≤ 300
5. Check length to depth ratio:
• ≤ 60
 Design procedure of tension members (1 angle and Bolted
Connections):
1. Select member section.
×
• = = 300
Get a1 = … cm
×
• = = 300
Get a2 = … cm
• >3 +
Get a3 = … cm

• . = … … … …. = 0.58 (0.70 is an assumption of

. ×
the area of holes in the section)
( / )
Steel Grade
≤ 40 > 40
St. 37 1.40 1.30
St. 44 1.60 1.50
St. 52 2.10 2.00

Choose the required angle from tables.

2. Check Stresses:

• Calculate
• Calculate = …≤
3. Check minimum angle:
• >3 +
4. Check slenderness ratio:
• = ≤ 300
• = ≤ 300
5. Check length to depth ratio:
• ≤ 60
 Design procedure of tension members (angles and Welded
Connections):
1. Select member section.
×
• = = 300
Get a1 = … cm
×
• = = 300
Get a2 = … cm
• >3 +
Get a3 = … cm

• . = … … … …. = 0.58

Choose the required angle from tables.

2. Check Stresses:

• Calculate
• Calculate = …≤
3. Check minimum angle:

• >3 +

4. Check slenderness ratio:

• = ≤ 300
• = ≤ 300
5. Check length to depth ratio:

• ≤ 60
 Design procedure of tension members (Star Shapes angles):
• Case of bolted connections:
Selection of section:
Assume:

=
0.85 ×

=
2
Choose an angle from the tables.
• Check minimum angle:
>3× +
• Check Stresses:
= ≤
ℎ : =2× −2×∅×
• Check Slenderness:
= ≤ 300
N.B.: ℎ =
……………………………………………………………………………………………....
Example:
For the shown truss, it is required to design the members L5 as 2 angles back to back and
member D5 as on angle given that:
1. The maximum tensile force in member is 25 ton.
2. The maximum tensile force in member is 5.0 ton.
3. Steel used is St. 37
4. Thickness of gusset plate is 10 mm.
5. Bolts used are M16 grade (4.6).

Solution:
Design of member L5:
For steel 37, = 1.40 / , = 2.00 , = 4.00
Choice of section:

. = = 300 , = 0.30

200
300 = ……………. = 2.22
0.30
25
= = = 21
0.85 × 0.84 × 1.40
21
= = = 10.50
2 2
From tables: try 2 angles back to back 75 × 75 × 8
= 11.50
= 2.26
 = 2.13
Check minimum angle:
3 × d + t = 3 × 16 + 8 = 56
= 70 , then >3× + , then ok.
Check stresses:
∅= +2 = 16 + 2 = 18 = 1.80
25
= = = 1.240 /
((2)(11.5)) − ((2)(1.8)(0.80))
= 1.240 / < = 1.40 /
Ok. Safe.
Check slenderness:

= 2.26 ( )= ( ( )) +( + .
)

( ) = (2.26) + (2.13 + 0.50) = 3.46

200
= = = 88.50 < 300 … … … .
2.26
400
= = = 115.60 < 300 … … … .
3.46
Check L/a ratio:
L = 2.00 meters = 75
200
= = 26.60 < 60 … … .
7.50
……………………………………………………………………………………………....
Design of member D5:
For steel 37, = 1.40 / , = 2.76 , = 2.76
Choice of section:

= = 300 , = 0.20

Dr. Amr Mohamed Ibrahim

 276
300 = ……………. = 4.60
0.20
5
= = = 5.10
0.70 × 0.70 × 1.40
From tables: try angle 55 × 55 × 6
= 6.31
= 1.07
Check minimum angle:
3 × d + t = 3 × 16 + 6 = 54
= 55 , then >3× + , then ok.
Check stresses:
∅= +2 = 16 + 2 = 18 = 1.80

Calculate :
6.31
= −∅× = − 1.80 × 0.60 = 2.075
2 2
6.31
= = = 3.15
2 2
3 3 × 2.075
= + = 2.075 + 3.15 = 4.16
3 + 3 × 2.075 + 3.15
5
= = = 1.20 /
4.16
= 1.2 / < = 1.40 /
Ok. Safe.
Check slenderness:
= 1.07
276
= = = 258 < 300 … … … .
1.07
Check L/a ratio:
L = 2.00 meters = 55
200
= = 36.30 < 60 … … .
5.50
Use one angle 55 x 55 x 6
……………………………………………………………………………………………
DESIGN OF COM
MEMBERS
 DESIGN OF COMPRESSION MEMBERS

Allowable stresses in compression members:

1. For symmetric sections:

a) For ≤
= 1.40 − 6.50 × 10 ( ) 37
= 1.60 − 8.50 × 10 ( ) 44
= 2.10 − 13.50 × 10 ( ) 52
b) For >

= 37, 44 52

Where,

= ,

= ≤ 180

= ≤ 180

2. For unsymmetrical sections:

= 0.60 ×

a) For ≤

= 0.60 × 1.40 − 6.50 × 10 ( ) 37

= 0.60 × 1.60 − 8.50 × 10 ( ) 44

= 0.60 × 2.10 − 13.50 × 10 ( ) 52

 b) For >
c) = 0.60 × 37, 44 52
Buckling length of compression members:
1. Chord members:
a) Compression in upper
chord members (Simple
truss):

= =

Where,

= .

-----------------------------------------------------

= =

Where,

= .

----------------------------------------------------

b) Compression in lower
chord members
(cantilever truss):

For member 1:

= =

= = 2( ) = 2 × 3 = 6
To reduce buckling length outside plane
( ), we may use one of the following
steps:

1. Using longitudinal bracing:

For member 1:

= =

= = =3

= Distance between longitudinal bracing

---------------------------------------------------
2. Using lower horizontal bracing:

For member 1:

= =

= =2

= Distance between points of horizontal

bracing

For member 2:

= =

= = ×

= .
 c) Web members (verticals and diagonals):

= = 0.50
= =
d) Subdivided members:

= 1.20

e) K - Truss members:

For Diagonal members:

=

= 1.20
For Vertical members:
= 0.50

= 0.75 + 0.25 × × ……. =

Design of symmetrical sections:
a) Using two angles back to back:
• Given: Design force, length of member.
• Get

Choice of section:
• From stresses:
Assume = 0.75 /

= =⋯
0.75

= =⋯
2
From tables, get angle × ×
• From slenderness:
Assume = 100
= 100 = ......... Get =⋯
.

= 100 = ......... Get =⋯

.

+ max( )
= = ⋯…….
2
Choose 2 angles back to back × ×

Minimum angle 3 × +
Minimum angle 45 × 45 × 5

• Checks:
1. = ≤ 180

= ≤ 180 = ( ) +( + )

2.
3. = ≤
---------------------------------------------------------------------------------------------------
 b) Using two angles star shape:
• Given: Design force, length of member.
• Get

Choice of section:
• From stresses:
Assume = 0.75 /

= =⋯
0.75

= =⋯
2
From tables, get angle × ×
• From slenderness:
Assume = 100
= 100 = ......... Get =⋯
.

+
= = ⋯…….
2
Choose 2 angles star shape × ×

Minimum angle 3 × +
Minimum angle 45 × 45 × 5

• Checks:
4. = ≤ 180
5.
6. = ≤
Design of unsymmetrical sections:

Choice of section:
• From stresses:
Assume = 0.60 × 0.75 / = 0.45 /

= =⋯
0.45
From tables, get angle × ×
• From slenderness:
Assume = 100
= 100 = ......... Get =⋯
.

+
= = ⋯…….
2
Choose 2 angles back to back × ×

Minimum angle 3 × +
Minimum angle 45 × 45 × 5

• Checks:
7. = ≤ 180
8.
9. = ≤

-------------------------------------------------------------------------------------------------------
Design of tie plates:

7
Dr. Amr Mohamed Ibrahim Chapter (4): Design of compression members
 = ⋯ … . . √√
′
=

≤
′
≤

′= ×

Usually ′ >

Then, use one tie plate in the middle of the member.

Note:

We will use one tie plate in the middle of any tension members composed of two elements.

Design of zero members:

Choice of section:

a) Minimum angle L45 × 45 × 5 ( ).

=3× × 1.1 ( )
b) = ≤ 180

For 2 angles back to back

= = 180 = √√ Get =…

= = 180 = √√ Get =…

For one angle

= = 180 = √√ Get =…

For two angles star shape

= = 180 = √√ Get =…
 c) = Get =…

From a, b and c choose the biggest angle.

Example:
For the shown truss, design the three marked members.
Given data:
• Steel used is steel 37.
• tGPL = 10 mm.
• Bolts used are ordinary bolts.

Solution:
For member 1:
F = -17 ton L = 1.80 meters
Lbx = 1.80 meters Lby = 1.80 meters.
Choice of section:
Assume Fc = 0.75 t/cm2 • λmax = 100
• Ag of 2 angles = • λx = Lbx / rx
λx = 180 / 0.30a = 100 a = 6 cm
Ag of 2 angles = = 22.67
.
• λy = Lby / ry
.
Ag of 1 angle = = 11.30
λy = 180 / 0.45a = 100 a = 4 cm
 • From Tables,
Choose 2 angles 80 x 80 x 8 back to back

Aavg = = 7 cm.

Then, choose 2 angles back to back 70 x 70 x 7

Check:

1. Check slenderness
From tables: rx 2 angles = rx 1 angle = 2.12 cm e = 1.97 cm

ry 2 angles = + + = (2.12) + 1.97 + = 3.26

λx = = = 67.92 ˂ 180 safe

.

λy = = = 55.20 ˂ 180 safe

.

λMax = 67.92
2. Check Stresses:
Fc = 1.4 – 6.5x10-5 (λmax) 2 = 1.40 - 6.5x10-5 (67.92)2 = 1.1 t/cm2

Fc = = = 0.90 t/cm2 ˂ Fc = 1.1 t/cm2

.

Design of tie plate:

= 67.92

= = 1.37 ( )

≤
′
≤

′ = 1.37 × 67.92 = 93 cm then, use one tie plate at the middle of the member.
For member 2:

F = -1.80 ton L = (1.8) + (1.6) = 2.41 meters

Lbx = 2.41 meters Lby = 2.41 meters.
Try one angle.
Choice of section:
Assume Fc = 0.75 x 0.60 = 0.45 t/cm2 • λmax = 100
• Ag of 1 angle = • λx = Lbx / rx
. λx = 241 / 0.20a = 100 a = 12.05 cm
Ag of 1 angle = =4
.

• From Tables,
Choose 1 angle 45 x 45 x 5.

. .
Aavg = = 8.2 cm.

Then, choose 1 angle 80 x 80 x 8

Check:

1. Check slenderness
From tables: rv = 1.55 cm

λv = = = 155.48 ˂ 180 safe

.

λMax = 155.48 ˃ 100, then Fc =

( )

2. Check Stresses:

Fc = x 0.60 = x 0.60 = 0.19 t/cm2

( ) ( . )
.
Fc = = = 0.15 t/cm2 ˂ Fc = 0.19 t/cm2
.
For member 3:
F = zero member L = 1.60 meters
∅ = 16 , then, = 3 × ∅ × 1.1 = 3 × 1.6 × 1.1 = 52.8
Use angle 55 x 55 x 5
Due to the presence of longitudinal bracing, 2 angles star shape will be used.
Check:

= = = 67.56 ˂ 180 safe.

.

No need for check L/d for vertical members.

------------------------------------------------------------------------------------------------------------

Design of compression members using unequal angles:

It is recommended to use unequal angles when Lbout ˃˃ 2 x Lbin (For economic design).

= 0.28

= 0.48

Choice of section:
• From stresses:
Assume = 0.75 /

= =⋯
0.75

= =⋯
2

From tables, get angle × × = = √√

• From slenderness:
Assume = 100

= 100 = ......... Get =⋯

.
 = 100 = ......... Get =⋯
.

+
= = ⋯…….
2
Choose 2 angles back to back × ×

• Checks:
1. = ≤ 180

= ≤ 180 = ( ) +( + )

2.
3. = ≤
SHEET 4:
1. For the following figure, design the top compression member (A), if the
design force is -28 tons (case of loading B) and its length is 300 cms, the bolts
used are with diameter 20mm.

2. Design the vertical member (B) if the design force is -3.00 tons (case of
loading Ⅱ) and its length is 400 cms, the bolts used in the connection are with
diameter 20 mms.
3. Design the vertical member (B) if the design force is -3.00 tons (case of
loading Ⅰ) and its length is 400 cms, the connection is welded connection.
DESIGN OF BEAM
 Design of Beams 1

Beams are subjected to moment and shear

We will study how to calculate the actual and allowable stresses:
1- Stress due to moment:
My
fact = ≤ Fbcx
I
Where: fact is the actual stresses due to actual moment affecting the beam. We
calculate "M" using load distribution due to either concrete slab or corrugated
sheets.
Fbcx is: the allowable stress of bending in compression side due to
moment about x-x axis.
For rolled section: we can use "S" in cm3 from tables (in some tables it is called
"Z") instead of (I /Y)

How to calculate F bcx which is the allowable of moment?

In order to calculate the allowable of moment we have to check:
a-Compactness of the section: This is the ability of the section to perform local
buckling for either flange or web only.
We have 3 types of sections:
1- Compact sections:
Are those sections which can achieve the plastic
moment capacity without local buckling of any of its
compression elements.

2- Non-compact sections:
Are those sections which can achieve the yield moment
without local buckling of any of its compression elements.

1/27 Beams
 Design of Beams 1

3- Slender sections:
Are those sections, which undergo local buckling before
achieving the yield moment.

Sections are classified according to their width-thickness ratios. Each

section’s component (flanges and webs) is treated separately as follows:
Look ECP page 9, 10, 11, 12
Rolled I-section Rolled UPN welded I-section

C C C
tf
S
r=

Axis of bending h h h
dw dw dw
tw tw

b b

Rolled I-section Welded I-section

2/27 Beams
 Design of Beams 1

1
c=
1
(b − t w − 2 r ) c = ( b − 2s − t w )
2 2
d w = h − 2(t f + r ) d w = h − 2( t f + s)
 16.9
≤


 15.3 
Compact ≤ Compact 
 Fy   
  Fy
c  23   
≤ Non − compact  c  21 
tf  Fy   ≤ Non − compact 
tf Fy
   
 > 23 Slender   21 
 Fy  > Slender 
   
 Fy 

 127 
 127  ≤ 
≤  Compact

Compact
  Fy 

Fy
  
d w  190  d w  190 
≤ Non − compact  ≤ Non − compact 
tw  Fy  tw  Fy 
   190 
 > 190 

Slender
 > Slender 
 Fy   Fy 
 

From tables pages 9 to 12, we can see that we treat every portion of the
section as a separate part.
Note that in all tables we use the flat portions.

Important note: For rolled section and if "r" is not given, we can take "r" ≈ tf
1 1
So c = (b-tw-2r) = (b-tw-2tf)
2 2
And d w = h-2(tf + r) = h-2(tf + tf) = h-4tf

Important note: If for example, the flange is non-compact and the web is
compact, so we consider the whole section as non-compact (more safe).

3/27 Beams
 Design of Beams 1

Important note:
a) In ECP 9, Web is subjected to moment ( ) in case of beams, to
compression ( ) for columns and to bending and compression (
) in beam-columns.
.

b) In ECP 11, Flange is always subjected to moment ( ) in case

of beams, columns and beam-columns. If there is My, the flange is subjected to
bending and compression ( ).
.

b-Lateral torsional buckling:

We have to know if the section will suffer from lateral torsional buckling or not.
What is L.T.B.?
The COMPRESSION flange of the beam will buckle outside the plan due to
compression forces. At the same time the tension flange will not buckle. This
causes torsion to the section.
Plan of upper flange

Section at Section at
support mid section

How to check whether there is L.T.B. or not?

We have to calculate Lu max and Lu act

4/27 Beams
 Design of Beams 1

Lu max is the maximum unsupported length. This is the maximum length with no
lateral torsional buckling. Lu max calculated from 2 equations in the code.
Lu act is the actual unsupported length. Lu act calculated from the given structure.
To calculate Lu max : ECP page 16
We take the smaller of the following 2 values:
20b f
L u . max =
Fy
OR
1380A f
L u . max = Cb
Fy d

Where: bf is the flange width (from tables), Af = bf x tf (from tables)

Fy = 2.4t/cm 2, d is the beam height= h (for IPE 300, d=30cm)
Lu = the actual unsupported length of the beam
Cb (For beams) = coefficient depending on the type of load and support as
shown in the following table. ECP page 20
Note that:
1- End restrains about Y-axis is always simple
2- If there are end moments, this means that this is an indication for continuous
beams.

5/27 Beams
 Design of Beams 1

For secondary beam:

Shape of moment of section "1" is not in code so we can take it either 1.13 or
1.30 (1.13 is preferable because it is critical)
Shape of moment of section "2" is given in code = 2.1
For main beam: Shape of moment is not in code but it is near to that of
distributed load with cb = 1.13

To calculate Lu act : This depends mainly on the given structure. We have to

study the COMPRESSION flange.

What restrains the compression flange:

1- The connection of beam with column.
2- The tension in flange (if part is compression and then changes to be
tension, Lu act will be from the support to the compression part only)
3- R.C. slabs and steel deck if the flange is completely restrained using shear
connector or the flange embedded in the concrete.

6/27 Beams
 Design of Beams 1
Closed ring
Hooked bar

Spiral
Studs

or
n ect
le con
xib
Friction only Compression flange Fle
embedded in conc

4- Beams perpendicular to the beam under study

But in all cases we have to study the compression flange

1-Floor beams (with concrete slab or steel deck): (Mezzanine floors, tanks)
a- For Simple floor beams:
The system is either main and secondary beams, or secondary, main beams and
main girders as shown in the following figures:
M.B. M.B.
S.B.

S.B.

S.B.

S.B.
Main girder

Main girder
S.B.

S.B.

S.B.

S.B.

S.B.

M.B.
Direction Direction
of load of load
S.B.

S.B.

S.B.

S.B.
M.B. M.B.

System consists of Main and secondary beams System consists of secondary and meain beams
together with main girder

1- If we are using steel decking:

For the shown system, there is a ONE WAY concrete slab regardless of the
dimensions of the slab because we use corrugated sheets in the direction of
secondary beams, which are spaced 1.75m to 2.25m. Since all the beams are
simply supported, therefore the compression flange is the upper flange.
WE DONOT DEPEND ON FRICTION TO RESTRAIN THE
COMPRESSION FLANGE.
We use shear connector to ensure that the concrete restrain the flange.

7/27 Beams
 Design of Beams 1

• If there is shear connector:

The concrete will be supported over the upper compression flange of both
secondary and main beams, So Lu act = zero.

• If there is NO shear connector:

The concrete will Not support the upper compression flange of both
secondary and main beams, So Lu act = length of beam.

If we are using wood shuttering: ( )

• If the flange is embedded:
The upper flange of both the secondary and the main is restraint laterally and if
they are both simple, Lu act for both will be equal to zero.
• If the flange is NOT embedded:
The concrete will Not support the upper compression flange of both
secondary and main beams, So Lu act = length of beam.

b- For Continuous secondary floor beams: (two panels or more):

Lu act = 0.2L for the section at the support (this is an approximate value).
For secondary beams, the upper flange is supported. However, near supports, the
compression flange is the lower flange, which is not supported.

8/27 Beams
 Design of Beams 1

We have to check 2 sections. One at sec1 sec2

mid span where Lu act = zero because
0.2L
the compression flange is the upper
flange, while the other is near the support, where Lu act ≈ 0.2L and calculate the
allowable of moment for each section.

2- For simply supported beams with no slab or steel deck:

The beams carry concentrated loads resulted from machines or hanged
equipments and for beams carry grating or chequered pates
a- Simply supported beams:
The compression flange is the upper flange and not restrained. Lu act will be
equal to distance between secondary beams.
Note that: L.T.B. occurs outside plan, so the members perpendicular to the
member under study support L.T.B. For beams connected to columns, the
column is a support for L.T.B.
i.e. If we have a system of floor beams (main girder and secondary beams),
Always the main girder supports the secondary beams and vise versa.

b- For Continuous beams: (two panels or more):

sec1 sec2 sec3

0.8L 0.2L 0.6L

We have 3 sections.
Section 1, the positive section of the first panel, Lu act is 0.8L because the
compression flange is the upper flange. The upper flange is supported laterally
with the column and tension in the flange at distance 0.2L from support.

9/27 Beams
 Design of Beams 1

Section 2, the negative section, Lu act is 0.2L because the compression

flange is the lower flange. The lower flange is supported laterally with the
column and tension in the lower flange at distance 0.2L from support.
Section 3, the positive section of the intermediate panel, Lu act is 0.6L
because the compression flange is the upper flange. The upper flange is
supported laterally with the tension in the upper flange at distance 0.2L from
supports from each side.

For Cantilever beams: Will be discussed later

Allowable bending stresses for moment about major axis x-x:

Allowable bending stresses for I-sections:
The shear center coincides on the C.G. of the section
The allowable bending stress depends on the section classification and the
presence of L.T.B.
1- If the section is compact and there is no L.T.B., so Fbc = 0.64Fy
2- If the section is non-compact and there is no L.T.B., so Fbc = 0.58Fy
3- If the section is compact or non-compact and there is L.T.B.
Therefore, F bc will be the smaller of F LTB or 0.58Fy.

Important Notes:
• We will know lateral, how to deal with the section if it is slender. So now,
all the sections will be either compact or non-compact.
• We must check compactness of the section to be sure that it is not slender,
this means we cannot begin with the check of L.T.B.

How to calculate F LTB?

Very important note: FLTB is calculated using system of equations, but if at any
step the value exceeds 0.58Fy, so we stop calculation and take Fbcx = 0.58Fy.

10/27 Beams
 Design of Beams 1

i.e. if FLTB 1 is bigger than 0.58Fy, so stop calculations and take Fbcx = 0.58Fy
For the following equations, see page 18 and 19 in the ECP

Fbc = (F 2
ltb1 + F 2 ltb 2 ) ≤ 0.58Fy
800A f C b
Where Fltb1 = ≤ 0.58F
Lud
Lu C
Fltb 2 = 0.58Fy for ≤ 84 b
rt Fy
( L u / rt ) 2 C L C
Fltb 2 = (0.64 − 5
Fy , for 84 b ≤ u ≤ 188 b
1.176 *10 * c b Fy rt Fy
12000 L C
Fltb 2 = 2
* c b , for u > 188 b
( L u / rt ) rt Fy

Where rt is the radius of gyration about Y-Y axis for the compression flange and
(1/6) of the web.
y
I y−y 1
rt = , where A = bf * tf + hw * tw
A 6
1h
6 w
t f * b3f (1 / 6) h w * t 3w
Iy-y = + (the second term can be y
12 12
neglected)
Very important note: Lu in the equations is Lu act NOT Lu max

Allowable bending stresses for Channel sections:

The shear center does not coincide on the C.G. of the section. Therefore, this
section cannot be compact. This section is either non-compact or slender. We
have to check this section by the limits of non-compact only.
When calculating FLTB: There is FLTB 1 only
800A f C b
Fltb1 = ≤ 0.58F
Lud

11/27 Beams
 Design of Beams 1
How to determine allowable compressive bending stress

If compact section, and LU C

≤ 84 b
1380A f  rT Fy
 F d Cc  Fltb2 = 0.58 Fy
 y 
Lu = the least of   ≥ Lu Yes The section is compact
max

 20 b f  Fbt = 0.64Fy
 Fy 
Fbc = 0.64Fy

No Cb LU C
84 ≤ ≤ 188 b
Fy rT Fy
The section is non-compact 2
 ( rT ) × FY

2
Fbt = 0.58Fy Calculate (Lu/rT) LU

Fbc = (F 2
ltb1 )
+ F 2 ltb2 ≤ 0.58Fy Fltb2 = 0.64 −
 1.176 ×105 Cb 
 × Fy ≤ 0.58 Fy
 

800 A f C b
Fltb1 = ≤ 0.58F y
LU d Lu C
≥ 188 b
rt Fy
12000C b
Fltb2 = 2
≤ 0.58F y
 Lu 
 r
 t 

12/27 Beams
 Design of Beams 1

Allowable shear stress:

On the gross effective area-resisting shear, the actual shear stresses must not
exceed the following allowable value:
q all = 0.35Fy

How to calculate the actual stress:

Q Q
q all = =
Area dt w

Allowable deflection of beams:

The calculated deflection due to live load only of any beam shall not be greater
than the values indicated on the table page 132 in the ECP:
• For simple or continuous beams with or without concrete slab
L
all =
300
L
• For cantilever beams: all =
180

How to calculate the actual deflection of beam?

5 wL4
• For simple beam with distributed load: act =
384 EI
• For continuous beam with distributed load: act = 0.8* act of simple beam.

PL3
• For simple beam with one concentrated load at mid span: act=
48EI

wL4
• For cantilever with distributed load: act =
8EI

PL3
• For cantilever with concentrated load at its end: act=
3EI

13/27 Beams
 Design of Beams 1

• If we have 2 of the previous cases, we can make superposition. i.e.

calculate the effect of each load alone and add them together.
• If we do not have any of the previous studied cases, we have to use
conjugate beam or any other method to calculate deflection. In this
method , the elastic moments (the area of the actual bending moment
diagram) are calculated and the supports are changed as following:
Hinged or roller support remains as it is
Free end turns to be fixed, while the fixed end turns to be free

Some examples of how to use conjugate beam to calculate deflection:

1-Beam with 2 concentrated loads:
P P
M = pa, R1 = Ma/2, R2 = 0.5a M
a a a
R = R1 +R2
M = R(a+a/2) – R2 (0.5a+a/3) – R1(a/4)
M1
Mε in m3t (don't forget to R
act =
EI R1 R2

multiply * 106 to change to cm3t)

2-Beam with 3 concentrated loads

P P P
R1 = 0.5aM1, R2 = aM1 a a a a
R3 = 0.5a (M2 – M1)
R = R1 + R2 + R3
M1
M2
M = R*2a – R1 (a/3) – R2 (a/2)
R R1 R2
– R3 (a+a/3). R3

act = Mε in m3t
EI

14/27 Beams
 Design of Beams 1

Estimation of Rolled sections:

In order to make the previous checks, we have to choose a section near to the
safe and economic section.
Nearly all the rolled sections in the tables are compact with respect to
dimensions. Therefore, Lu act and Lu max will be the guide. For this if
• Lu act < 3m , assume that Fbcx = 0.64Fy = 1.536 t/cm2 for steel 37
• Lu act < 6m , assume that Fbcx = 0.58Fy = 1.4 t/cm2 for steel 37
• Lu act > 6m , assume that Fbcx = 1.0 t/cm2 for steel 37
M
• Choose the section from the equation: Sx = and note that M is usually
Fbcx
calculated in mt so we must change it to be in cmt by multiplying it *100.
Choose the section from tables
• If there is Mx only so choose I.P.E. or S.I.B (I.P.E. is preferable)
• If there is Mx and My choose B.F.I.B.

Steps of beam design:

1- Calculate the actual loads acting on the system. Calculate each of
dead and live load separately.
2- Estimate a section by assuming that the section is compact and
L.T.B. is as mentioned before.
3- Calculate the allowable stress for moment Fbcx
4- Check the 3 checks.
My
• ≤ Fbcx
I
Q
• ≤ 0.35Fy
A = dt w
L L
• act = in case of beams or in case of cantilever.
300 180

15/27 Beams
 Design of Beams 1

Example:
These shown beams are used to support a
13 t 13 t
machine (No R.C. slab)

3*2.33=7m
For the given secondary I.P.E. 600, it is
required to: 13 t 13 t

1. Check the safety of the given beam

due to the applied shown loads 3*3=9m

2. If the beam is unsafe, suggest adding

members to make it safe (the members shouldn't carry any vertical loads)
3. Recheck the beam after adding the suggested members
4. Design the main beam
Solution:
1. Check for secondary I.P.E. 600:
• Calculation of loads:
a – Dead load:
o.w. = 122 kg / m\ = 0.122 t / m\

92
WD = 0.122 * = 1.23 m t
8
QD = 0.122 * 9/2 = 0.55 t
b – Live load: 13 t 13 t
3m 3m 3m
M = 39 m t Q = 13 t
• To calculate Fbcx:
1 – Check compactness:
C = 0.5 (22 – 1.2 – 2 * 2.4) = 8 cm, d w = 60 – 2 * 1.9 – 2 * 2.4 = 51.4 cm
C 16.9 dw 127
= 8/1.9 = 4.21 < = 10.9 = 51.4/1.2 = 42.8 < = 82
tf 2.4 tw 2 .4
So the section is compact
2. Check lateral torsional buckling:

16/27 Beams
 Design of Beams 1

20 * 22
Lu act = 900 cm Lu max = = 284 cm < 900
2.4
∴ No need to calculate the other equation L.T.B. will occur and we have to
calculate Fltb . (Cb for simple beam carrying concentrated load = 1.35)
800 * (22 *1.9)
Fltb1 = *1.35 = 0.836 t / cm2 < 1.4 t / cm2
900 * 60
Therefore, we have to calculate Fltb2

I 1.9 * 223 (60 / 6) *1.23

rt = Iy-y = + = 1686 + 1.44 = 1687cm 4
A 12 12
We can neglect the 2nd term
1687
A = 1.9 * 22 + (60/6) * 1.2 = 53.8 cm2 rt = = 5.6 cm
53.8
L 900
∴ u = = 160.7
rt 5.6

Cb 1.35 Cb
84 = 84 = 63 188 = 141
Fy 2.4 Fy

L C
∴ u f 188 b
rt Fy

Fltb2 = (12000/160.72) * 1.35 = 0.63 t / cm2 < 1.4 t / cm2

Fltb = 0.836 2 + 0.632 = 1.04 t / cm2 < 1.4 t/ cm2 ∴ Fbcx = 1.04 t / cm2
Check Stresses:
(39 + 1.23) * 100
1 – Moment : Fact = = 1.31t / cm 2 > 1.04t / cm 2
3070
Unsafe (Don't complete checks)
In order to decrease Lu act, we may add
3*2.33=7m

columns (especially under the loads), but he said

that the members should not carry vertical loads.
Therefore, we want to add members to prevent

3*3=9m
17/27 Beams
 Design of Beams 1

lateral movement & at the same time don't carry vertical loads. These members
are bracing as shown
Lu act = 9/3 = 3m
Lu max = 284 < 300 cm
800 * ( 22 *1.9)
∴ Fltb = *1.35 = 2.51t / cm 2 f 1.4t / m 2
1 300 * 60
So take Fbcx = 1.4 t / cm2
Checks:
(39 + 1.23) *100
1. Moment : Fact = = 1.31t / cm 2 p 1.4t / cm 2 OK
3070
(13 + 0.55)
2. Shear : qact = = 0.19t / cm2 p 0.84t / cm2
60 *1.2
3. Deflection : calculate elastic 13 13

reactions 3 3 3

ME = 117*4.5–58.5*(1.5/2)–
58.5*(1.5+1) = 336.4 m3t 39mt
117
336.4 *106 900 58.5 58.5
∴ δ act = = 1.74cm p
2100 * 92080 300
= 3cm

Design of main beam:

R = 13t, assume o.w. = 100 kg/m\ 13 13
2.33 2.33 2.33
Dead load:
MD = 0.1*(72/8) = 0.61 mt
QD = 0.1*(7/2) = 0.35 t 13

Live load:
ML = 13*2.33 = 30.3 mt
QL = 13 t

Total loads:

18/27 Beams
 Design of Beams 1

MD+ ML = 30.3+0.61 = 30.91 mt

QD+ QL = 0.35+13 = 13.35 t
Estimation of section
Lu act = 2.33 < 3 m, therefore
Assume section is compact and Fbcx = 0.64 Fy = 1.536 t/cm2
30.91*100
Sx = = 2012cm3
1.536
Choose I.P.E. 550
Check compactness:
C = 0.5 (21 - 1.11 - 2 * 2.4) = 7.545 cm
dw = 55 – 2 * 1.72 – 2 * 2.4 = 46.76 cm
C 16.9
= 7.546/1.72 = 4.38 < = 10.9
tf 2.4
dw 127
= 46.76/1.11 = 42.13 < = 82
tw 2 .4
The section is compact
Check lateral torsional buckling:
Lu act = 7/3 = 2.33 m
1380 * ( 21*1.72) 
 *1.35 = 510cm
55 * 2.4 
L u max = the least of   > Lu
 21 * 20 
= 271cm
 2.4 

where cb = 1.35 (simple beam with concentrated loads)

Lu act < Lu max
So No lateral torsional buckling
The section is compact and no lateral torsional buckling
Fbcx = 1.536 t/cm2 (0.64 Fy)
Checks:
30.91*100
1. Moment : Fact = = 1.27t / cm 2 < 1.536t / cm2
2440

19/27 Beams
 Design of Beams 1

2. Shear = 13.35 / (55 * 1.11) = 0.21 t/cm2 < 0.84 t/cm2

3. Deflection: Exactly as before in secondary beam but using the dimensions
of Main beam.

Applications of beams
1- Floor beams
The system consists of secondary and main beams
Calculation of loads for floor beams:
Loads:
Rc slab 10 – 14 cm
F.C. 150 kg / m2
L.L. 300 kg / m2

1 – Design of secondary beam as simply supported on Main beam:

Always distribute loads as one-way slab because of the presence of corrugated
sheets.
a – Dead load:
wd = ts c + F.C. = …….t / m 2 c = 2.5 t/m3
∴ WD = o.w. + (wd)a = …….t / m\ (assume o.w. of secondary beam = 50kg/m')
assume o.w. = 0.05 t / m \ = 50 kg / m\

S2
MD = WD * ……..m t
8
S
Q D = WD * ……….t
2
Where S is the span of secondary beam
b – Live load:
WL = L.L. * a = ………t / m\
Assume L.L. 300 kg / m\

20/27 Beams
 Design of Beams 1

S2 S
ML = W L * ……..m t Q L = WL * ……….t
8 2
c – Total loads:
MTot = MD + ML Q Tot = QD + QL

Design of Main beam:

Concentrated loads = Reaction of secondary beams = WTot * S
Assume o.w. = 0.1 t / m\ = 100 kg / m\
Calculate Mx & Q x
Note that the deflection is calculated due to L.L. only, so we have to RE-
CALCULATE the values of the reaction from the secondary beams.

2 – Design of secondary beam as continuously supported on Main beam:

Calculate loads as before to get Msimple beam
The moment will be as follows:
0.75M 0.75M 0.75M 0.75M
sec1 sec2

0.9M 0.2L 0.8M 0.8M 0.8M

The moment at the end panel is the largest moment, so we estimate the section
from it. We have to design 2 sections. One at the mid of the first panel where
M=0.9Mo and Lu act = zero, while the second is of smaller moment, M=0.75Mo,
but its Lu act = 0.2S
For shear: Q max = 0.6wS as shown in the following figure:
0.4wS 0.5wS 0.5wS 0.5wS

0.6wS 0.5wS 0.5wS

R=0.4wS R=1.1wS R=wS R=wS

Design of Main beam when :

21/27 Beams
 Design of Beams 1

The second main beam is the most critical main beam as it is subjected to the
maximum reaction, which is 1.1wS as shown above.
Concentrated loads = Reaction of secondary beams = 1.1WTot * S
Assume o.w. = 0.1 t / m\ = 100 kg / m\
Calculate Mx & Q x
Note that the deflection is calculated due to L.L. only, so we have to RE-
CALCULATE the values of the reaction from the secondary beams.

Example:
2.0

Main beam

2.5*4m = 10m

For the shown mezzanine floor, it is required to design:

1) Secondary beams as simple beams
2) Secondary beams as continuous beams
3) Main beams in case of secondary beam simple
4) Main beams in case of secondary beam continuous
Given that: ts = 12 cm, F.C. = 175 kg/m2, L.L. = 400 kg/m2
Spacing between trusses is 6m. Assume using steel deck with shear
connector only for secondary beams.

Solution:
1) Secondary beam simple:
a- Dead load:
wD = 0.12 * 2.5 + 0.175 = 0.475 t/m2

22/27 Beams
 Design of Beams 1

Assume o.w. = 50 kg/m2

WD = 0.475 * 2.5 + 0.05 = 1.24 t/m2
MD = 1.24 * 6 2 / 8 = 5.58 mt QD = 1.24 * (6/2) = 3.72 t
b- Live load:
WL = 0.4 * 2.5 = 1 t/m' ML = 1 * 62 / 8 = 4.5 mt
QL = 1 * (6/2) = 3 t
c- Total loads:
MT = 5.58 + 4.5 = 10.08 mt QT = 3.72 + 3 = 6.72 t
Estimation of section:
Simple beams à upper flange is compression.
Therefore Lu act = zero < 3m
Therefore assume Fbcx = 0.64 Fy = 1.536 t/cm2
10.08 *100
Sx = = 656.25 cm3
1.536
Choose Ip E 330
Note the difference between concrete and steel, the beam is of length 600 cm
and depth 33 cm
Span
i.e. d = for secondary beams. (important)
20
- Without calculations of Lu max there is no L.T.B. because Luact = zero
Checks:
C = 0.5 (16 – 0.75 – 2 * 1.8) = 5.825 c m
dw = 33 – 2*1.15 – 2*1.8 = 27.1 cm
C 5.825 dw 27.1
= = 5.06 < 10.9 = = 36.1 < 82
tf 1.15 tw 0.75
The section is compact and No L.T.B.
1- 10.08 *100
fact = = 1.41 t/cm2 < 1.536 t/cm2
713
6.72
qact = = 0.27 t/cm2 < 0.84 t/cm2
33 * 0.75
23/27 Beams
 Design of Beams 1

act due to live load only

WL.L = 1 t/m/

5 1* 64 600
L.L = * * 10 6 = 0.68 < = 2 cm
384 2100 * 11770 300
DONOT forget to multiply *106 to change m4 to cm4 and t/m to t/cm
m4 * t/m' = m 3t = 106 cm3t

b- Design of main girder:

13.44 13.44 13.44
2.5 2.5 2.5 2.5
Rtotal = WD+L * S = (1.24 + 1) o.w.

* 6 = 13.44 t
RL = WL * S = 1 * 6 = 6 t M
20.66
Assume o.w. = 100 kg/m2
MT = 68.45 mt
QT = 20.66 t
There is haunch so Lu act = spacing between secondary beams.
Lu act = 2.5 m < 3 Assume Fbcx = 1.536 t/cm2
68.45 *100
Sx = = 4456 cm3 Max Sx in tables of IPE is 3070 cm3
1.536
So use board flange I-beam HEA
Choose HEA 600
1- Check compactness
C = 0.5 (30 – 1.3 – 2*2.7) = 11.65 cm
dw = 59 -2*2.5 – 2*2.7 = 48.6 cm
C 11.65
= = 4.66 < 10.8
tf 2.5
d w 48.6
= = 37.4 < 82
tw 1.3
Compact sections

24/27 Beams
 Design of Beams 1

 20 * 30 
 2.4 = 387cm 
L u max = the least of  
1380 * (30 * 2.5) *1.35 = 986cm 
 59 * 2.4 

Where cb = 1.35, simple beam with concentrated loads (distributed

o.w. is not effective)
Lu act = 250 < 387 cm Therefore No L.T.B
Fbcx = 1.536 t/cm2
Checks:
1- 68.45 *100
fact = = 1.43 t/cm2 < 1.536 t/cm2
4790
20.66
2- q act = = 0.27 t/cm2 < 0.84 t/cm2
59 *1.3
3- the shown beam is loaded with live load only
R1 = 0.5 * 2.5 (30- 22.5) =
9.375 m2t 6 6 6
2
R2 = 2.5 * 22.5 = 56.25 m t 2.5 2.5 2.5 2.5

R3 = 0.5 * 2.5 * 22.5 = 28.1

m 2t 22.5
30
R = 9.375 + 56.25 + 28.1 = 28.1 25.25
93.75
2
93.75 m t 9.4
M = 93.75*5 – 9.375*2.5/3 –
56.25*2.5/2 – 28.1*(2.5+2.5/3)
= 297 m3t

297 *106 1000

act = = 1 cm < = 3.3 cm
2100 *141200 300

25/27 Beams
 Design of Beams 1

3) Design of secondary beam as Continuous beam:

Same as before in calculating Mo = 10.08mt,
Moment in first panel = 0.9*10.08 = 9.07 mt
M-ve (near support)= 0.75*10.08 = 7.56 mt

7.56mt
sec1 sec2

9.07mt 0.2L 8.06

9.07 *100
Choice of section: S = = 590 cm3
1.536
Choose IPE 330 (same as in case of simple beam)

Checks:
1- Moment:
For sec 1: same as before for compactness
Lu act = zero (compression flange is the upper flange contacted with R.C. slab
Therefore fact = 9.07*100/713 = 1.27 t/cm2 < 1.536 t/cm2
For sec 2: Lu act = 0.2*6 = 1.2 m
Therefore, we have to calculate Lu max
Cb = 1.3 (continuous beam with distributed load)
 20 *16 
 2.4 = 206cm 
Lu = the least of  
max
 1380 * (16 * 0. 75)
*1.3 = 272cm 
 33 * 2.4 

Therefore Lu act < Lu max , NO L.T.B.

7.56 *100
fact = = 1.06 t/cm2 < 1.536 t/cm2
713

26/27 Beams
 Design of Beams 1

2- Shear:
6.72t 6.72t 6.72t

8.06t 6.72t 6.72t

Qmax = 0.6 W S = 0.6 * (1.24 + 1) * 6 = 8.06 t

Note that Qmax = 0.6 wS not 0.5 wS as in case of simple beam.
qact = 8.06 / (33*0.75) = 0.32 t/cm2 < 0.84 t/cm2
3- Deflection:
act of simple beam due to L.L = 0.68 cm (calculated before)
act of continuous beam = 0.8 act of simple beam
= 0.8*0.68 = 0.54 cm < 600/300 =2 cm

4) Design of Main beam (when the secondary is continuous beam)

Note that reaction is (1.1 w*S) not (w*S) as in case of simple beam
RT =1.1 WD+L S =1.1*2.24*6 =
14.78 14.78 14.78
14.78 t
2.5 2.5 2.5 2.5
o.w.
RL = 1.1 WL S = 1.1*1*6 = 6.6 t
Assume o.w. = 100 kg/m2
M
MT = 75.15 mt
20.66
QT = 22.67 t
Sx = 75.15*100 / 1.536 = 4892 cm3 6.6 6.6 6.6
2.5 2.5 2.5 2.5
Use BFIB 650
Exactly as before with the new values of
moment and shear and the new BFIB
(HEA) chosen
Note: use loads 6.6 t in check deflection

27/27 Beams
 Design of beams
Introduction:
Beams are sections subjected to moment and shear only (no
tension or compression forces with the moment and shear). This moment
may be from concrete floor as in case of floor beams, may be from
corrugated sheets as in case of purlins or may be from wind as in case of
side and end girt.

Design of beams:
Beams have same concept of design. We have to choose a section
in which the actual stresses due to moment and shear are less than the
allowable stresses of each of them.
As in case of sections subjected to tension or compression, we have
to estimate an approximate section and check.

In order to check the beam we have to check 2 conditions:

Strength criterion: This requires that the chosen beam cross
section must resist the applied bending moments and shear forces. In
terms of stresses, this means that the actual bending and shear stresses
must not exceed the allowable stresses of each type.
Serviceability criterion: This requires that the beam deflection
under the service loads be within the acceptable range in specifications.
In Classification of section whether compact, non-compact or slender:

Table (2.1a) in page 9, is the limit ratios for stiffened compression

members where stiffened members are webs. This is because the upper
and lower flanges stiffen the web. (The web is stiffened from both
directions by the upper and the lower flange). This table is used for rolled
sections, built-up sections (sections composed of welded plates), box
section or channels.

Table (2.1c) in page 11, is the ratios for un-stiffened compression

members where un-stiffened members are flanges. This is because the
flanges are stiffened from one side only with the web and the other end is
free. This table is used for the flanges of the same section as table (2.1a).

Table (2.1d) in page 12, is used for the ratios of equal and unequal
angles as well as T-sections (T-stub) and circular sections. In case of
unequal angles, we have to make 2 checks (look page 12). Angles and
channels cannot be compact.

Shape of cross section after later torsional buckling

Section at mid section Section at support section

a) Before applying load

c) Due to torsion b) After deflection

To calculate Lu act : This depends mainly on the given structure. We
have to study the COMPRESSION flange.

Support

Support
L

Figure 1 Figure 2

The 2 given beams are simply supported and laterally restrained

at both ends at supports. In figure 1, the upper flange of beam is
restrained at its mid point. While in figure 2, the lower flange of beam is
restrained at its mid point.
The main difference is that, this beam is simply supported, so the
lower flanges subjected to tension while upper flange subjected to
compression.
For beam in figure 1, Lu act = L/2 because the compression flange is
restrained at its mid span. The beam in figure 2, Lu act = L because the
additional support has no effect on the compression flange.

Why to use smaller of FLTB and 0.58Fy?

This is because if FLTB is smaller, so the section will perform
L.T.B. before reach yield. This means that the L.T.B. is guide in design.
While if 0.58Fy is smaller this means that, the section will reach its
maximum stress capacity before maximum stress of L.T.B.
How to calculate the actual stress:
QS
qall = , but since we can neglect
Ib
the contribution of flanges (about 5%), Q QS
so we may consider that the web only ht Ib
carry shear.
We may also assume that the shear
is distributed equally on the web.
Q
Therefore, qall = ,
Area
The effective area in resisting shear is taken as follows: ECP13
Aeff sh = d × tw for rolled section
Aeff sh = hw × tw for built-up section
Where (d) is the total depth of rolled section
 Design of Beams 2

Allowable bending stresses for Channel sections:

The shear center does not coincide on the C.G. of the section. Therefore, this
section cannot be compact. This section is either non-compact or slender. We
have to check this section by the limits of non-compact only.
When calculating FLTB: There is FLTB 1 only
800A f C b
Fltb1 = ≤ 0.58F
Lud

How to calculate Luact for sections of mezzanine floor ?

Sec 1
Case 1

Sec 1 Sec 3
Case 2
Sec 2

For Case 1 :

WL2
a- If there is no shear connector : Luact = L &M=
8

WL2
b- If there is shear connector : Luact = zero &M=
8

1/10 Beams
 Design of Beams 2

For Case 2 :
a -If there is no shear connector:

Sec M Luact
1 0.9M 0 0.8 L
2 0.75 M 0 0.2 L
3 0.8M 0 0.6 L
So only, check Section 1

b -If there is shear connector:

Sec M Luact
1 0.9M 0 zero
2 0.75 M 0 0.2 L
3 0.8M 0 zero

So check Section 1 & 2

How to calculate Luact for section in the cantilever ?

For Cantilever beams ECP page 65 to 67:
Will be later

How and when to calculate F btx?

Fbtx is the allowable of bending moment for the tension
flange due to moment about the major axis x-x. We use this
value only is the section is built-up, un-symmetric as shown.
For symmetric sections, (all table sections are called rolled
sections and are symmetric).
Fbtx = 0.64Fy if the section is compact and no L.T.B. occurs

2/10 Beams
 Design of Beams 2

Or Fbtx = 0.58Fy otherwise (in all other cases).

Allowable bending stresses for moment about minor axis y-y:

:
-1
-2
For doubly symmetric sections (I-shapes) meet the compact requirements,
the allowable for moment when bent about its minor axis is 0.72Fy.
If Fbcx = 0.64Fy then Fbcy = 0.72Fy
Otherwise (in all other cases), the allowable of moment = Fbcy = 0.58 Fy.
If Fbcx = 0.58 Fy or Fltb then Fbcy = 0.58 Fy

For I-beam, box sections

Mx only Un-symmetric Mx and My
section
Compact and Fbcx = 0.64 Fy Fbtx = 0.64 Fy Fbcy = 0.72 Fy
No LTB = 1.54 = 1.54 = 1.73
Otherwise Fbcx = least of Fbtx = 0.58 Fy Fbcy = 0.58 Fy
0.58Fy and Fltb = 1.4 = 1.4

For channel section

Mx only Un-symmetric Mx and My
section
All Cases Fbcx = least of Fbtx = 0.58 Fy Fbcy = 0.58 Fy
0.58Fy and Fltb = 1.4 = 1.4

3/10 Beams
 Design of Beams 2

Example 1:
For the shown plan & elevation of inclined roof

600cm
4*200cm

o
30

It is required to design an intermediate sec. beam twice. First, if there is shear

connector & Second if there is no shear connector
Given that t s = 16 cm , FC = 150 kg / m 2 , LL = 500 kg / m 2
Solution:
Dead load: W DL = (0.160*02.5 + 0.15) 2 + 0.05 = 1.15 t / m/
LL W LL = 0.5 * 2 = 1 t / m/
W T = 1 + 1.15 = 2.15 t / m/
W X = 2.15 * cos 30 = 1.86 t / m/
W y = 2.15 * sin 30 = 1.075 t / m/
wy
1.86 * 6 2
Mx = = 8.37 m t
8 w wx
1.075 * 6 2
My = = 4.84 m t
8
6
Q x = 1.86 * = 5.58 t Q y = 1.075 * 3 = 3.2 t
2

4/10 Beams
 Design of Beams 2

1) If there is shear connector:

M x + 6M y (8.37 + 6 * 4.84) *100
Estimation = Sx = Sx
0.64F y 0.64 * 2.4

SX = 2435 cm3 Choose IPE 550

C dw
Check , ---------- compact
t f tw

Lu act = zero ∴ Fbcx = 0.64F y = 1.536 t / cm 2

& Fb cy = 0.72 * 2.4 = 1.728 t / cm 2

837 / 2440 484 / 254

Check stresses + = 0.22 + 1.1 = 1.32 > 1
1.536 1.728
This is because Stress due to M y is so large because IPE is very weak in
resisting M y
M x + 4M y (8.37 + 4 * 4.84) *100
Estimation = Sx = Sx
0.64 Fy 0.64 * 2.4

SX = 1805 cm 3 Choose HEB 320

C dw
Check , ---------- compact
t f tw

837 / 1930 484 / 616

Check Stresses: + = 0.28 + 0.45 = 0.73<1 O.K
1.536 1.728
5.58
Check vertical shear = = 0.15 t / cm 2 < 0.35 * 2.4 = 0.84 t / cm 2
32 *1.15
3.2
Check horizontal shear = = 0.03 t / cm 2 < 0.84 t / cm 2
2 * 30 * 2.05
Note that the parallel area is the area of flanges
Check deflection: W LL = 1 t / m /
W LLx = 1 * cos 30 = 0.87 t / m /

5 0.87 * 6 4 600
δu = * *106 = 0.23cm < = 2 cm
384 2100 * 30820 300

5/10 Beams
 Design of Beams 2

2) If there is no shear connector

Lu act = 600 cm Assume Fbcx = 1.4 t / cm 2
(8.37 + 4 * 4.84) *100
= S x = 1980 cm 3 Choose HEB 340
1 .4
C dw
Check , ---------- compact
t f tw

20 * 30
Lu max = = 387 cm
2.4
800 * (30 * 2.15)
Fltb1 = *1.13 = 2.86 t / cm 2 > 1.4 t / cm 2
600 * 34
837 / 2160 484 / 646
Check Stresses: f = + = 0.28 + 0.45 = 0.73< 1 O.K
1.4 1.4
Check shear and deflection as before.

Example 2:

B1

B3 B2 1
4
1

1.5 1.5 3.0 1.5 1.5

For the shown plan, A silo. ( ) of weight 40 t & is supported on the 4
shown points. The rest of the ceiling is covered with chequered plate of weight
2
50 kg / m (the chequered plate doesn't support lateral movement) & LL on
ceiling = 300 kg / m 2
It is required to design B1, B2 & B 3 using channel section for B1 & B 3 & I –
section for B2

6/10 Beams
 Design of Beams 2

Solution:
For Beam B1 10t
0.225
50
W DL = 0.05 + * 0.5 = 0.075 t / m /
1000 3m
/
W LL = 0.3 * 0.5 = 0.15 t / m
W T = 0.15 + 0.075 = 0.225 t / m /

3 32
M = 10 * + 0.225 * = 7.75 m t Q = 5 + 0.225 * 1.5 = 5.34 t
4 8
Assume Fb cx = 1.4 t / cm 2 (Channel can not be compact)
775
Sx = = 553 cm 4 Choose UPN (channel) 320
1.4
c 10 − 1.4 23
Check = = 4.9 < = 14.8 (limit of non-compact)
tf 1.75 2.4

d w 32 − 4 *1.75 190
= = 17.8 < = 122.6 (limit of non-compact)
tw 1.4 2.4
20 *10
Lu act = 300 Lu max = =129 cm < 300
2.4
There will be LTB
800 * (10 *1.75)
Fltb1 = *1.35 = 1.97 t / cm 2 > 1.4 t / cm 2
300 * 32
∴ Fbcx = 1.4 t / cm 2
Note: c b = 1.35 because the effective load is conc. Load
775
Check moment stress = = 1.14 t / cm 2 < 1.4 t / cm 2
679
5.34
Shear q = = 0.12 t / cm 2 < 0.84 t / cm 2
32 *1.4

5 0.15 * 34 10 * 33
δu = * *10 6 + *10 6 = 0.25 cm
384 2100 *10870 48 * 2100 *10870
300
< = 1cm O.K
300
7/10 Beams
 Design of Beams 2
0.575
For Beam B 3 :
50 6m
W D = 0.05 + * 1.5 = 0.125 t / m /
1000
W L = 0.3 * 1.5 = 0.45 t / m / W T = 0.125 + 0.45 = 0.275 t / m /

62
M = 0.575 * = 2.6 m t Q = 0.575 * 3 = 1.725 t
8
260
Sx = = 186 cm 3 Choose UPN 200
1.4
c dw
Check , as before
tf tw

20 * 7.5
Lu act = 600 cm Lu max = =96.8 cm < 600 cm
2.4
∴ There is L.T.B.
800 * (7.5 *1.15)
Fltb = *1.13 = 0.65 t / cm 2 < 1.4 t / cm 2
600 * 20
∴ Fbcx = 0.65 t / cm 2
260
Check moment f = = 1.36 t / cm 2 > 0.654 t / cm 2 UNSAFE
191
260
Choose S x = = 400 cm 4 Choose UPN 280
0.65
800 * (9.5 *1.5)
Fltb = *1.13 = 0.77t / cm 2 < 1.4 t / cm 2
600 * 28
260
Check f = = 0.58 t / cm 2 < 0.77 t / cm 2 SAFE
448
1.725
q= = 0.06 t / cm 2 < 0.84 t / cm 2
28 *1

5 0.45 * 6 4 600
δ LL = * *10 6 = < 0.57 cm < = 2 cm
384 2100 * 6280 300

8/10 Beams
 Design of Beams 2

Design of Beam B2: 5.34 10 5.34

Reaction of B1 = 5.34t 0.31
1.5
W D = 0.05 + 0.05 * = 0.09 t / m /
2 1 2 2 1
W L = 0.3 * 1.5 / 2 = 0.225 t / m /
W T = 0.225 + 0.09 = 0.31 t / m /
Q = 11.27t, M = 11.27 * 3 – 0.31 * 3 * 1.5 – 5.34 * 2 = 21.73 m t
Assume Fb cx = 0.64 F y = 1.536 t / cm 2
2173
Sx = = 1415 cm 3 IPE 450
1.536
c dw
Check , compact
tf tw

Lu act = 400 cm (distance between sec. beams)

20 *19
Lu max = = 245 cm < 400 there is L T B
2.4
800 * (19 *1.46)
Fltb1 = *1.13 (For distributed load) = 1.39 ≈ 1.4 t / cm 2
400 * 45
2173
1– f = = 1.45 t / cm 2 > 1.4 t / cm 2 Choose IPE 500
1500
11.27
2- q= = 0.22 t / cm 2 < 0.84 t / cm 2
50 *1.02
3 – Neglect check deflection

Loads on M.B: OW = 0.1 t / m /

1.725 11.27 11.27 1.725

0.275
0.1 0.1

1.5 1.5 3.0 1.5 1.5

w = 0.1 + (0.05 + 0.3) 0.5 = 0.275 t / m 2

9/10 Beams
 Design of Beams 2

Example 3:
For the shown plan.

4*5m

It is required to design a secondary beam if it is continuously supported over

MG without using shear connector
Solution:
W D = (0.2 * 2.5 + 0.15) 2 + 0.05 = 1.35 t / m /
W L = 0.4 * 2 = 0.8 t / m / W T = 1.35 + 0.8 = 2.15 t / m /

52 672
M 0 = 2.15 * = 6.72 m t Sx = = 480 cm 3 Choose IPE 300
8 1.4
Critical section is section 1 with M = 0.9 M 0 (biggest moment) &
Lu act = 0.8 L = 0.8 * 5 = 4 m (biggest Lu act ).
c dw
Check ,
tf tw

20 *15
Lu act = 400 cm & Lu max = =194 cm < 400 There is L T B
2 .4
800 * (15 *1.07)
Fltb1 = *1.3 = 1.39 t / cm 2 ≈ 1.4 t / cm 2
400 * 30
672
f = = 1.21 t / cm 2 < 1.4 t / cm 2
557
0.6 * 2.15 * 5
q= = 0.3 t/cm2 < 0.84 t / cm 2
30 * 0.71

5 0.8 * 5 4 500
δ LL = 0.8 * * *10 6 = < 0.3 cm < = 1.67 cm
384 2100 * 8360 300

10/10 Beams
DESIGN OF BURLIN
 Roof Purlins 1

Purlins are those roof beams used to support the roof covering materials.
Usually, purlins are designed as simply supported beam with span equal to the
spacing between trusses or the main frames.

Loads: a/cos Purlin

1-Dead load:
Own weight of corrugated sheets:
Single layer Wc = (5~8) kg/m2
a
Double layer Wc = (10~15) kg/m2

Own weight of purlin = (15~20) kg/m\. For hot-rolled channels

Own weight of purlin = (5~8) kg/m\. For cold-formed channels
Note that: if he did not mention the type of channel, we take it hot-rolled

WD.L = WC × a + O.W
cos α
Y
Wx D.L = WD.L*cos Corrugated
Wy D.L = WD.L * sin sheets
X
S2
Mx D.L. = Wx D.L * X
W yD . L .
8

S2 WD.L.
WxD . L.
My D.L. = Wy D.L * .
8
Y
Qx D.L. = Wx D.L * S/2
Important note: Neglect shear in y-direction Qy because it is very small.
(Moments are in kg.m, and we want to change it to cm t, so we multiply it
*(100/1000). Shear is in kg so divide by 1000 to change to tons
Where ( α ) is the roof inclination angle with the horizontal direction, and (S) is
the spacing between trusses.

1/15 R.P. 1
 Roof Purlins 1

2-Live load:
The moment due to live load is taken the bigger of uniform load or 100 kg
concentrated load.
a-Due to uniform load:
WL.L

Case of inaccessible roof L.L. = 60-66.67tan (on horizontal projection)

Case of accessible roof L.L. = 200-300tan (on horizontal projection)

L.L. (kg/m2)
WL.L = L.L. * a
Wx L.L = WL.L*cos
Wy L.L = WL.L * sin 200

S2 Accessible
Mx L.L. = Wx L.L * roof
8
Inaccessible
2 60 roof
S
My L.L. = Wy L.L * 20 kg/m2
8
Tan
Qx L.L. = Wx L.L * S/2 0.6

b-Due to one concentrated load P = 100kg.

P = 100 kg Px = 100cosα Py =
100sin α
Mx L.L = Px L.L. × S/4
My L.L = Py L.L. × S/4
Qx L.L. = Px/2

Take bigger of both the 2 previous cases of loading

3-Due to wind load

Wind load affects perpendicular to the surface ⇒ WW .L = W x W .L

2/15 R.P. 1
 Roof Purlins 1

a
Wx L.L = ( C × k × q) ×
cos α
a/cos
Where:

q = wind intensity which depends on

the location of the structure. a

• q = 70 and 80 kg\m2 in ( Cairo

and Alex. respectively)

• k = factor depends on the height of the structure.

Height k
Ce
0 → 10 m 1.0
10 → 20 m 1.1
20 → 30 m 1.3
30 → .... ....
+ 0.8
• Ce = factor depends on the shape of the
0.4
structure and the angle ( α ) Tan
0.8

-0.8
The value of (Ce) is determined from the
shown graph:

Important note:
• When Ce is –ve value (suction), neglect wind load in the design of purlin
as it reduces the positive moment due to dead load and live load. This
case occurs for all trusses when the slope of roof "tan " is less than 0.4
• When Ce is +ve value (pressure), take effect of wind load in the design of
purlin as it increases the positive moment due to dead load and live load.
This case occurs for all trusses when the slope of roof "tan " is more than
0.4. We have to take the effect of wind on the design of roof purlin as
following:

3/15 R.P. 1
 Roof Purlins 1

If Wx W.L. is +ve
S2
∴ Mx W.L. = Wx W.L ×
8
My W.L. = zero
Q x W.L. = WW.L. (S/2)

4-Combinations of loads:
a- If the wind is suction: tan ≤ 0.4, Ce negative
Case (I) ⇒ D.L+L.L
Mx = M x D.L + M x L.L My = M y D.L + M y L.L

Qx = Q x D.L + Q x L.L
Where ML.L. is the bigger of the 2 cases of either distributed or concentrated
load. This case of design is case A

b- If the wind is pressure: tan >0.4, Ce positive

Case (I) ⇒ D.L+L.L "This case of design is case A"
Mx = M x D.L + M x L.L My = M y D.L + M y L.L

Qx = Q x D.L + Q x L.L
Case (II) ⇒ D.L+L.L+W.L "This case of design is case B"
Mx = Mx D.L + Mx L.L + Mx W.L My = My D.L + My L.L

Qx = Qx D.L + Qx L.L + Qx W.L

Roof purlins can be Hot rolled steel section “ Channel section UPN” or
Cold formed steel section “ C, Z, …”

4/15 R.P. 1
 Roof Purlins 1

Design as Hot rolled

Choice of section:
Channel sections are commonly used for purlins.
According to the Egyptian code, channel sections are treated as non-compact.
Since purlins are simple beams, and the corrugated sheets restrains the upper
flange which is the compression flange. The allowable compressive bending
stress is as follows: (Do not forget to check the class of the section
"compactness" to be sure that the used section is not slender.
Fbcx = Fbcy = 0.58Fy.
• If the design is governed by case (I)
Mx My
fb = + = .... ≤ Fbcx = 0.58Fy
Sx Sy

N.B. for C sections assume Sx ≈ 7Sy

M x + 7M y
S xreq = =…cm3
Fbc

• If the design is governed by case (II)

M x + 7M y
S xreq = =…cm3
Fbc × 1.2

Check stresses:
Normal stresses
M xA M yA
Case( I ) fb = + = ... ≤ Fbc = 0.58F y
Sx Sy
M xB M yB
Case( II ) fb = + = ... ≤ 0.58F y × 1.2
Sx Sy

Shear stresses
Q zA
Case( I ) q act = ≤ 0.35Fy Where d = total depth of channel
t wd

Q zB
Case( II ) q act = ≤ 0.35Fy × 1.2
t wd

5/15 R.P. 1
 Roof Purlins 1

Check deflection:
Due to live load only
5W x L.L S 4 span
δ L . L. = = ... ≤
384EI x 300

Using of tie rods in purlins:

Tie rods are used to reduce My in:
1) High slopes (e.g. fink truss)
2) Cold formed sections (Sy is quite small)

We can either use:

1) One tie rod at the mid-span of each purlin
2) Two tie rods at one third of the span.
Very important note:
Adding tie rod will affect only Y-direction, but has no effect on X-direction

Case of using one tie rod at mid span

Truss
S/2 T5

T1 T2 T3 T4 S
S/2
Truss
T5
S/3
T1 T2 T3 T4

S/3 S

S/3
Truss

Case of using two tie rods at the third point

of spacing

6/15 R.P. 1
 Roof Purlins 1

1-Dead load:

WD.L = WC × a + O.W
cos α
Wx D.L = WD.L*cos Wy D.L = WD.L * sin

S2
Mx D.L. = Wx D.L *
8
• Case of using one tie rod at mid span:

(S / 2) 2 1
My D.L. = Wy D.L * = My D.L (case of no tie rods)
8 4
• Case of using 2 tie rods at middle thirds:

(S / 3) 2 1
My D.L. = Wy D.L * = My D.L (case of no tie rods)
8 9
Qx D.L. = Wx D.L * S/2

2-Live load:
a- case of uniform load:
WL.L = L.L. a
Wx L.L = WL.L*cos Wy L.L = WL.L * sin

S2
Mx D.L. = Wx D.L *
8
• Case of using one tie rod at mid span:

(S / 2) 2 1
My L.L. = Wy L.L * = My L.L (case of no tie rods)
8 4
• Case of using 2 tie rods at middle thirds:

(S / 3) 2 1
My L.L. = Wy L.L * = My L.L (case of no tie rods)
8 9
Qx L.L. = Wx L.L * S/2

7/15 R.P. 1
 Roof Purlins 1

b-Due to one concentrated load P = 100kg.

P = 100 kg Px = 100cosα Py = 100sin α
Mx L.L = Px L.L. × S/4
• Case of using one tie rod at mid span:
S/ 2 1
My L.L. = Px L.L * = My L.L (case of no tie rods)
4 2
• Case of using 2 tie rods at middle thirds:
S/3 1
My L.L. = Px L.L * = My L.L (case of no tie rods)
4 3
Qx L.L. = Px/2

Take bigger of both the 2 previous cases of loading

3-Wind load:
The same as before, as in case of purlins without tie rods because the
moments are in direction of X-only and tie rods affect only in Y-direction.

Design of tie rods

The force in the tie rod is calculated from the reaction in y-direction of the
purlin.

Wy = Wy D.L+Wy

Tie rod

8/15 R.P. 1
 Roof Purlins 1

In case of one tie rod in the middle:

1
T1 = Wy × S , T2 = T1 + Wy × S
2 2 2

T3 = T2 + Wy × S , T4 = T3 + Wy × S
2 2

T’5 = T4 + Wy × S
2
Q 2T5 sin θ = T5
'

'
T5
∴ T5 =
2 sin θ

In case of two tie rods:

1
T1 = Wy × S , T2 = T1 + Wy × S
2 3 3

T3 = T2 + Wy × S , T4 = T3 + Wy × S
3 3

T’5 = T4 + Wy × S
3
'
T5
∴ T5 =
sin θ
T’5 in its general form = (0.5wy *S/2) + n (wy *S/2) for using one tie rod
= (0.5wy *S/3) + n (wy *S/3) for using two tie rods
∴ Force = area × stress
π
∴ T5 = 0.7 Aφ × Ft = 0.7 × × φ 2 × Ft
4
Get φ =…
Use minimum rienforcement bar 12 mm

9/15 R.P. 1
 Roof Purlins 1

Example 1:
For a system of trusses with spacing = 6.00 m and panel length =2.00m
Design an intermediate purlin using hot rolled section, Slope 1:10
Assume any missing data.

Solution:
As hot rolled section
Q l (span of purlin ) = 6.00 m
& a (panel length) = 2.00 m
1
& = tan-1 = 5.710
10
1) Dead Load:
Assume wc =6 kg / m2 & o.w. (purlin) = 20 kg / m/
2
∴ wD.L. = 6 * + 20 = 32.06 kg / m/
COS 5.71
wx D.L. = 32.06 * cos5.71 = 31.9 kg / m/
wy D.L. =32.06 * sin 5.71 = 3.19 kg / m/
(31.9)(6) 2
MX D.L. = = 143.55 kg .m.
8

(3.19)(6)2
My D.L. = = 14.355 kg. m
8
(31.9)(6)
QXD.L. = = 95.7 kg
2

2) Live load:
a) Uniform load
L .L . = 60 – 66.67 * (0.1) = 53.33 kg / m2
Wl.l. = 53.33 * 2 = 106.66 kg / m/
wx l.l. = 106.66 * cos5.71 = 106.13 kg / m/
wy l.l. = 106.66 * sin5.71 =10.61 kg / m/

10/15 R.P. 1
 Roof Purlins 1

(106.13)(6)2
∴ Mx L.L.= = 477.59 kg . m
8

(10.61)(6)2
My l.l. = = 47.76 kg .m.
8
(106.13)(6)
Qx L.L. = = 318.39 kg
2
b) 100 kg conc. Load
Px = 100 * cos5.71 = 99.5 Py = 100 * sin 5.71 = 9.95
99.5 * 6 9.95 * 6
Mx l.l. = = 149.3 kg. m My l.l. = = 14.93 kg. m
4 4
Uniform L.L. is more critical

3) Wind Load:.
Q tan = 0.1 < 0.4 wind is suction neglect wind load
4-Combination of loads
Mx = 143.55 + 477.59 = 621.14m. kg = 62.11 cm. t.
My = 14.355 + 47.75 = 62.1m. kg = 6.21 cm. t.
Qx = 95.7 + 318.39 = 414.1kg = 0.41 t
Choice of section
Assume Sx = 7 Sy & Fbc = 0.58Fy =1.4 t / cm2
62.11 + 6.21* 7
Sx = = 75.41 cm3 choose channel 140
1.4
Check compactness: Channel section, non-compact section
c 4.3 23 23
C= 6-0.7-1 = 4.3cm ∴ = = 4.3 < = = 14.84
t f 1.0 Fy 2.4

d w 10 190 190
dw= 14-2*1-2-1 = 10 cm ∴ = = 14.3 < = = 122.6
t w 0.7 Fy 2.4

Note that we used the limits of non-compact because we know that according to the
Egyptian code, the channels are non-compact.

11/15 R.P. 1
 Roof Purlins 1

Simple beam, Compression flange is the upper flange and supported with corrugated
sheets, so Luact = zero (no need to calculate Lu max).
Check stresses:
62.11 6.21
1- fact = + = 1.14 t/cm2 < 1.4 t/cm2
86.4 14.8
0.41
2- q act = = 0.04t / cm 2 <<<< 0.35Fy = 0.84 t/cm2
14 * 0.7
Shear is not critical in purlins at all

5 ( WxLL ) * S 4 5 (106.13) * 6 4 600

3- L.L. = = *103 = 1.4 < =2cm
384 EI 384 2100 * 605 300

Example 2:
For a system of trusses with spacing =
6.00 m
Design an intermediate purlin using hot 3.5m
rolled section and any other additional
member given that the maximum 1.75
9m
available section is channel 100 7m
( ) only available
• Sheets are of weight 10kg/m2
Solution:
3.5
α = tan −1 = 26.60 tan = 0.5
7
a 1.75
1) DL: = = 1.96m, o.w = 20 kg / m/
cos α cos 26.6
WDL = 10 * 1.96 + 20 = 39.6 kg / m/
Wx DL = 39.6 cos = 35.4 kg / m/ , Wy DL = 39.6 sin = 17.7 kg / m/
62
MxDL = 35.4 * = 159.3 kg m
8

12/15 R.P. 1
 Roof Purlins 1

62
MyDL= 17.7 * = 79.6 kg m
8
Shear is not critical
LL à WLL = 60 – 66.66 tan 26.6 = 26.67 > 20 kg / m2
wLL = 26.67 * 1.75 = 46.67 kg /m/
wx LL = 46.67 cos α = 41.73 kg / m / wy LL = 46.67 sin α = 20.9 kg / m /
62 62
Mx LL = 41.73 * = 187.78 kg m My LL = 20.9 * = 94.05 kg m
8 8
Conc. Load ß check 100 kg ( )

3) Wind load:
Q tan = 0.5 > 0.4
There is pressure & suction

0.5 − 0.4 C e
=
0.8 − 0.4 0.8
Ce = 0.2 & and neglect the suction value of Ce
∴ ww= 0.2*1.1(clear height is 9 m)*70*1.96=30.18 kg /m

62
Mx wl = 30.18 * = 135.8 kg m
8
Design loads:
Case I : MxDL + Mx LL = 159.3 + 187.8 = 347.1 kg m =34.7cmt
Case ` : MxD + MXl +Mxw = 347.1 + 135.8 = 482.9 kg m =48.3cmt
In both cases: My = 79.6 + 94.05 = 173.65 kg m =17.4cmt
caseB 482.9
= = 1.39 > 1.2
caseA 347.1
∴ Use case B i.e. D + L +W
48.3 + 17.4 * 7
Estimation of section: Sx = = 101.2 cm3
1.4 *1.2
Choose channel 160 > channel 100 which is available

13/15 R.P. 1
 Roof Purlins 1

∴ Add one tie Rod at mid Span

17.4
∴ Mx = 48.3 cm t & My = = 4.35 cm t
4
48.3 + 4.35 * 7
Sx = = 46.9 cm3
1.4 *1.2
∴ Choose channel120 > channel 100

∴ Add 2 tie Rods at middle thirds

17.4
Mx = 48.3 cm t & My = = 1.93 cm t
9
48.3 + 1.93 * 7
Sx = = 36.8 cm3
1.4 *1.2
Use channel100
Check compactness: Channel section, non-compact section
c 3.55 23 23
C= 5-0.6-0.85=3.55cm ∴ = = 4.17 < = = 14.84
t f 0.85 Fy 2.4

d w 6.6 190 190

dw= 10-2*0.85-2*0.85=6.6cm ∴ = = 11< = = 122.6
t w 0.6 Fy 2.4

Note that we used the limits of non-compact because we know that according to the
Egyptian code, the channels are non-compact.
Simple beam, Compression flange is the upper flange and supported with corrugated
sheets, so Luact = zero (no need to calculate Lu max).
48.3 1.93
Checks: 1) f= + = 1.4 t / cm2 < 1.4*1.2=1.68 t / cm2
41.2 8.49
34.71 1.93
f= + = 1.07 t / cm2 < 1.4 t / cm2
41.2 8.49
(2) (Not critical for shear)
(3) Check deflection

5 41.73 * 6 4 600
δ= * * 103 = 1.62 cm < = 2 cm O.K.
384 2100 * 206 300

Design of tie Rod:

14/15 R.P. 1
 Roof Purlins 1

Wy = wy LL + wy DL = 17.7+20.9 = 38.6 kg / m/
1 6 6
T4/ = * 38.6 * + 3 * (38.6 * ) = 270.4 kg = 0.27 t
2 3 3
T4 cos θ = T4/
Truss
-1 2
θ = tan = 490 T1 T2 T3 T4
1.75
T4 cos 49 = 0.27 T4 = 0.39 t 6m

0.41 = 0.7 A ϕ * 1.4

Truss
2
A ϕ = 0.42 cm 7m

πϕ 2
= 0.4 ϕ = 0.7 cm = 7 mm
4
ϕ min = 12 mm (minimum)

15/15 R.P. 1
DESIGN OF BRACING
 Design of Bracing Members

Bracing is used to carry the loads acting ⊥ to the frames or trusses plane. Such
as wind load, braking force, … etc.
As it is used to carry the wind load, so wind load will be Case A not B.
Wind Load Path:
Wind load à Corrugated sheets à End girts
1/2 base
à End gable columns 1/2 Hz bracing àVL bracing à base

h1 = Clear height + h

h1

h2
h2 = Clear height + H

1) HL bracing: B

R
Vertical
Bracing

W
w

Loads = Ww = (ce + ci )kq * (Hav/2) kg/m'

We divided /2 because half the load is transmitted from the end gable to the
foundation (base), while the other half transmitted to the horizontal bracing.
Where ce = 0.8, ci = 0.3 k =1, q = 70 kg/m2
h1 + h2 B
Hav = R = ww
2 2

2) VL bracing: R
Used to carry the reaction of the Hz bracing system.

.D.B.M-1/8
 Design of Bracing Members

Design of horizontal and vertical bracing:

The horizontal and vertical bracing are subjected to Tension and compression
forces (because wind has no direction), so we have to design the members using
"angles" (single, back to back or star shape). Since force is the same but with
different sign ± , so we have to design as compression and check as tension.
The design of bracing angles, the case of load is considered "case A" because
the wind load is primary (main) load for the bracing.

Steps of design:
1- Calculating of forces and buckling lengths:
a) Hz bracing:
R R R
l

Bracing every 2a Bracing every 3a Bracing every 4a

lin = lout = l/2 lin = lout = l/3 lin = lout = l/4
R R R
F= ± F= ± F= ±
2 cos θ 2 cos θ 2 cos θ

R R
Note that: F= ±
n cosθ
Where:
"n" is the number of members you cut
F
" the angle between member and direction of "R"
F

.D.B.M-2/8
 Design of Bracing Members

b) VL bracing: (Trusses with no crane)

1- V-shape:

R e 3
1
d R 1
h 1 1 Sec 1-1
b 4 a 2 c Sec 1-1
Sec 2-2
2 R
C 1
b a c
2

Sec 2-2
S
Note that: "C" is the clear height, "S" is the spacing and "h" is the distance between
upper and lower chord.
Calculation of F 1:
From summation of forces at joint a: ∑y=0
The forces in member 1 and 1/ are equal and opposite

∑x =0
R
For section 1-1: R = 2 F1 cos 1 So F1= ±
2 cos θ1

Calculation of F 2:
R
For section 2-2: ∑x =0 R = 2 F2 cos 2 So F2= ±
2 cosθ 2
Calculation of F 3:

∑x =0
R
From joint e: R = F1 cos 1 + F3 = cos 1 + F3
2 cosθ1

R
∴F3 =
2
Calculation of F 4:
R
From joint b: ∑x =0 F2 cos 2 = F4
2 cosθ 2
cos 2= F4

R
∴ F4 =
2

.D.B.M-3/8
 Design of Bracing Members

Buckling lengths:
S
For l1 è lin = l, lout = l.2 l l= ( )2 + h2
2

For l2 è lin = 0.5 l2, lout = 0.75 l2 l= S 2 + C2

For l3 è lin = lout = S
For l4 è lin = S/2, lout = S

2- W-shape:
d c 5 b
R 6 1
R
1 Sec 1-1
3 a 4 Sec 1-1
Sec 2-2 R 1
2 b a c
2

Sec 2-2
R
From sec 1: F1= ±
4 cosθ1
R R
From sec 2: F2= ± and F3 = (as before)
2 cosθ 2 2

To calculate F4: From joint a: ∑x =0

R R
F4 = 2F1 cos 1- F3 = 2 cos 1- = zero
4 cosθ1 2

R R
From joint b: ∑x =0 F5 = F1 cos 1= ±
4 cosθ1
cos 1 =
4

From joint c: ∑x =0 F6 = 2F1 cos 1+ F5

R R 3R
F6 = 2 cos 1 + =
4 cosθ1 4 4

.D.B.M-4/8
 Design of Bracing Members

Buckling lengths:
S
For l1 è lin = l, lout = l.2 l l= ( )2 + h2
4

For l2 è lin = 0.5 l2, lout = 0.75 l2 l= S 2 + C2

For l3 è lin = 0.25 S lout = S
For l4 è lin = S/2, lout = S
For l5 è lin = S/2, lout = S
For l6 è lin = S/2, lout = S
Note that:
Member 5 and 6 will have the same section, so we have to design them both
and use the bigger section.
Member 3 and 4 will have the same section, so we have to design them both
and use the bigger section.

• Summary: Calculate the forces in the inclined members using method of

sections and in the horizontal member using method of joints.

2- Estimation of bracing member shape:

For Hl bracing: Design as Single angle ( )
For VL bracing:
• For upper members l1è use single angle à 90*9 if bigger use 2 angles back to back
• For l2è use 2 angles back to back
• For members with no intermediate joints as l3 in the first drawing, use star shape
as l/d is the guide.

.D.B.M-5/8
 Design of Bracing Members

3- Choice of section:
Since the forces are very small (sometimes, the bracing members designed as zero
members), so the guide will be and l/d.
• If the length of the member is smaller than 5m, so will be guide
• If the length of the member is larger than 5m, so use star shape as L/d is the guide.
So we can estimate the section using maximum
l
• If single angle: 200 = out get a
0.2a
l
• If star shape: 200 = out get a
0.38a
• If 2 angles back to back:
l l
200 = in get a 200 = out get a (take bigger a)
0.3a 0.45a
For checks:
Design as Comp member and check as tension member
Steps: 1) Design as comp. member λ ≤ 200 (not 180) ECP 51
F
2) Check ≤ Fc
Agross

F
3) Check ≤ 1.4 t/cm 2
Anet

l
4) Check ≤ 60
d
Don't check 300 as it is smaller than 200
Very Important Notes:
1– For cantilever there is no end gable system, so there is no wind load.
So the members of HL and VL bracing are designed as zero members.
(For zero members, check 2 and 3 are omitted).

L L2
2 – Check L/d for X bracing members = < 60 .
D a
3 – The Connection is always bolted.
.D.B.M-6/8
 Design of Bracing Members

Other shapes and loads on vertical bracing:

1- VL bracing for frames with no crane:

For member 1:
R R 2
F1= ±
2 cos θ1
h
1
Lin = 0.5L, Lout = 0.75L

L
For member 2: C
R
F2 = Lin = Lout = S
2
S

2- VL bracing for trusses with crane:

The braking force is

∑P and distributed on 2 successive columns
7
B = 2P / 7 if there is one crane supported by this bracing

Member Lin Lout Force

1 S S R/2
R 1
1
2 L 1.2L R 2
3
2 cos θ1 4 2

3 0.5S S R/2 B 5 3

4 L 1.2L R 6
2 cosθ 2
L
5 0.5S S (B+R)/2
6 0.5L 0.75L (R + B ) S
2 cos θ3

.D.B.M-7/8
 Design of Bracing Members

3- VL bracing for frames with crane:

Member Lin Lout Force

R 1
1
1 S S R/2 2
B 3 2
2 L 1.2L R
2 cos θ1 4
3 0.5S S (B+R)/2
4 0.5L 0.75L (R + B ) L
2 cos θ 2

4- VL bracing for intermediate columns supporting crane:

Note that: This bracing is not connected to upper horizontal bracing; this means it is
subjected to braking force only.
B = 2P / 7 (the column supports 2 cranes, so the total braking force is 2B

Internal

Member Lin Lout Force 2B 1 1

1 S S 2B/2 2
2 0.5L 0.75L (2B )
2 cos θ1 L

.D.B.M-8/8
DESIGN OF BEAM‐
COL
 Beam-Column I
The beam-column members are subjected to both normal and moment, so their
N Mx My
stress is calculated using the formula: + + ≤ Fallowable . Since allowable of
A Sx Sy

moment is different than allowable of compression, so the equation turns to be

N / A M x /Sx M /S
interaction equation in the following form: + A1 + y y A 2 ≤ 1 ,
Fc Fbcx Fbcy

where A1, A2 are magnification factor and will be explained later.

Steel sections used:

1) Rolled sections:

IPE BFIB

Used for section subjected to N and Used for section subjected to N, Mx

Mx and My

2) Built – upysections: y y

x xx x x x

2 Channels 4 Angles 2 IPE 2 IPE

y y y

Used for section subjected to N and My

1/33-B.C.I
 Beam-Column I
Design steps of columns:
1- Suggest a suitable bracing system
2- Calculate moment and normal and determine the critical sections to be
designed. ( the column may have more than one critical section and the
design is to check the stresses in all the critical sections)
3- Estimation of section.
4- Apply in the interaction equation.
1- If the bracing is not given in the exam, we have to suggest a suitable
bracing system:
Positions of horizontal member in vertical bracing:
1- Level of upper and lower chord of truss.
2- Level of rafter in portal frames.
3- Level of crane girder to avoid braking force.
4- Level of any beam connected to the column especially if this beam carry
load perpendicular to plane of column, (gives My) on column.
5- If hcolumn > 5m and the bracing is not given in the exam.
h > 5m or 6m

post

Beam

End gable section

2/33-B.C.I
 Beam-Column I
For the design of column we have to calculate all the terms of the interaction
equation, so we need to calculate Fc (allowable of compression), which in turns
Lbin Lbout L = k in × L
requires the calculation of both in = and out = . Where bin
rin rout Lbout = k out × L

I- Buckling of Beam-columns
K = is the buckling coefficient, which will be calculated twice, first for in-plane
and second for out-plane.
In order to calculate K for the column, we have to study two points "for in-plane
and for out-plane":-
A-Types of structures according to sway-:
a) Permitted to side sway. (Side movement)
b) Prevented from side sway .
The structure is prevented from sway when we use:
1- Tie member join the structure to the ground
The frame in the figure is prevented from sway
inside plane

2- Structures joined to fixed structure

The frame in the figure is prevented from sway inside
plane
3- Structures with vertical bracing
The frame in the figure is permitted to sway inside plane
while prevented from sway outside of plane

B-End points conditions of the columns

1-Types of bases-:
Bases may be fixed or hinged inside the plane (in the elevation)
We assume all bases are hinged outside plane (in the side view)
EXCEPT: in case of cantilever column out of plane

3/33-B.C.I
 Beam-Column I
2-Types of connection between column and girder-:
a) Fixed:
When the inertia of the girder is too large related to that of the column, the girder
PREVENTS the column from ROTATION.
Example: When the truss is connected to the column with distance "h" min.
1.25m
No rotation (fixed) but sway
Sway Sway

Fixed base Hinged base Fixed bases Hinged bases

b) Free:
Cantilever .

c) Rigid:
When the inertia of the girder is smaller or bigger than that of the column with
small value and the connection is moment connection, the column and the girder
rotate while the angle between them remains CONSTANT.
Example: The column is connected to the girder with head plate while their
inertias are close to each other.
Usually inertia of girder is smaller than that of column (Ic = 2-3 Ig )
This structure is called PORTAL FRAME

Hinged bases Fixed bases

4/33-B.C.I
 Beam-Column I
d) Hinged:
When the connection is hinged (does not translate moment).
Example: The column is connected to the girder using angles connecting the
web of the girder.

Permitted to sway prevented from sway

How to consider hinged or rigid connection:

• If the beam is connected to the column using head plate (connecting
flanges carrying moment and web carrying shear), so the joint is rigid
• If the beam is connected to the column using angles (connecting only web
carrying shear), so the joint is hinged.

Rigid connection Hinged connection

e) Partially fixed ends-:

It occurs when 2 columns of different inertias are connected

5/33-B.C.I
 Beam-Column I
In case the connection between the column and girder is fixed, hinged or
free we use the table ECP 53

Prevented from sway Permitted to sway

Hinged- Hinged- Fixed- Fixed + Fixed + Free
Hinged Fixed Fixed free sway free sway

Buckling
shape

k-value 1.00 0.80 0.65 2.00 1.3 2.10

Applications of buckling lengths Lb

Buckling length of columns in trusses:
i) For hinged base:

h1

h2

hinged hinged

h1
lbin = 2( h2+ ) lbout = h1 or h2 (bigger)
2
ii ) For fixed base:

h1

h2

F ix e d h in g e d
( I n s id e ) ( o u ts id e )

h1
lbin = 1.3 ( h2+ ) lbout = h1 or h2 (bigger)
2

6/33-B.C.I
 Beam-Column I
iii ) For partially fixed base :

h1
roof
column h
2
Partially
fixed
Combined h
3
column

Fixed
base
For roof column:
h1
lbin = 1.5 ( h2+ ) lbout = h1 or h2 (bigger)
2
For combined column:
lbin = 1.5 h3 lbout = h3
iv ) For partially fixed base :
If the bracing is given as shown below:

h4

h3
h2
h1

For roof column:

h4
lbin = 1.5 ( h2+ h3+ ) lbout = h3 or h4 (bigger)
2
For combined column:
lbin = 1.5 h1 lbout = h1 + h2

7/33-B.C.I
 Beam-Column I
v ) For fink truss:

F ix e d
S b ase

lbin = 2.10 h lbout = h

vi ) For cantilever truss:

h
1

h
2

h1
lbin = 2.10 ( h2+ ) lbout = ( h1 or h2 ) (bigger)
2
vii ) For cantilever frame:

h
1

lbin = 2.10 ( h1) lbout = h1

viii ) For cantilever truss and side sway prevented:

h
1

h
2

lbin = ( h2 + h1 ) lbout = ( h1 or h2 ) (bigger)

8/33-B.C.I
 Beam-Column I
In case the connection between the column and girder is rigid we use the
alignment chart ECP page 60 and 61.
Buckling length of columns in rigid frames:

B
I
g

I L
c c

A
L
g

For column AB
lbin = KLc lbout = Lc
: (chart) (k)
- for hinged base GA = 10

- for fixed base GA = 1.0

( rigidity of joint ) G
Σ( I c / Lc )
- GB = ECP page 60 and 62
Σ[( I g / L g ) * F ]

Where F = factor depends on the condition of the far end of girder and on the
case of frame.

C
1) F = 1 for frames (If there is no real support)
( ) Col 2
2) If there is support at the end of the girder: B
• Frame prevented from sway : Col 1
F = 2.0 for fixed far end A
F = 1.5 for hinged far end

9/33-B.C.I
 Beam-Column I

• Frame permitted to sway :

C
F = 0.67 for fixed far end
Col 2
F = 0.50 for hinged far end
B
Col 1
A
Important notes:
1- If the studied column is column 2 in the previous shape, so we have to
calculate Gb and Gc and use the alignment chart.
2- If the column is connected to a cantilever beam it has no effect on the
buckling of the column.
3- Inertia:
For beams and girders :We always use Ix
For columns :We have to check the direction of buckling whether it is around
X-axis or Y-axis of the member under consideration

10/33-B.C.I
 Beam-Column I
Examples for buckling:
Example 1:

IPE 400 IPE 550

IPE 400
IPE 500

Col 2 Col 4

3.0

4.0
IPE 500 IPE 400 IPE 450

IPE 500
IPE 600

Col 3
4.5

5.0
Col 1

8.0 6.0 8.0

Calculate the buckling lengths for columns 1, 2, 3 and 4 in figures 1 2

respectively. Assume there is vertical bracing outside plane connecting columns
at floor levels only.
Solution:
Figure(1): Frame permitted to sway inside plan and
3.0

assume the bracing is as shown:

4.5

Column 1:
Inside plan: GA = 10 (hinged base)
32080 48200
+
ΣI c / Lc 450 300
GB = =
Σ( I g / L g * F ) 48200
800
.·. GB = 6.06
.·. From chart permitted to s way k = 2.65
.·. lbin = 2.65*4.5=11.925 m
Out-plan: lbout = Lc = 4.50 m.

11/33-B.C.I
 Beam-Column I
Column 2:
48200
Inside plan: GB = 6.06 GC = 300 = 5.56
23130
800
From chart k = 2.35
.·. lbin = 2.35*3.0=7.05 m
Out-plan: lbout = 3.00 m

Figure 2:
Column 3: Frame prevented from sway
48200 23130
+
GA = 1.00 (fixed base) GB = 500 400 = 1.10
23130 33740
* 1. 5 + * 2.0
600 800
Note that: although the left support is roller, but this floor level is prevented from
sway, so we use the factor "F" for the prevented from sway frames.
From chart prevented from sway k = 0.78
.·. lb in = 0.78*5.00 = 3.90 m
lb out = 5.00 m

Column 4: Upper part of the frame is allowed to sway

23130
GB = 1.1 (from previous) GC = 400 = 1.03
67120
* 0.67
800
From chart allowed to sway k = 1.3
.·. lb in = 1.3*4.00 = 5.20 m
lb out = 5.00 m

12/33-B.C.I
 Beam-Column I
Example 2:
slope 1:10

IPE 400
col 1 IPE 500 3.00

5.50m

36.0m

Inside: GA (partially fixed) = 3.0

K =2.6
48200 / 300 ∴ lbin = 3x2.6
GB = = 25.0
23130 / 3600 = 7.8m
Outside: 3m (using bracing outside)

Example 3:

3.00 4.00 3.00 3.00 5.00 3.00

col 2 col 1 col 2

col 1

Elevation Side View

All beams are IPE 300.
Columns of first floor are IPE 500, second floor IPE 400
third floor IPE 360.
Take the floor height equals to 3 m.

13/33-B.C.I
 Beam-Column I
Solution:
To solve this problem, we have to draw plan to know exactly the location of the
columns.

3.00 4.00 3.00

Buckling about y-axis of columns

3.00
Side view (Outside plan)

5.00

col 2

3.00
col 1

Elevation (Inside plan)

Buckling about x-axis of columns

Column 1:
48200 / 300 + 23130 / 300
Inside: GA = = 4. 9
8360 / 400 + 8360 / 300

23130 / 300 + 16270 / 300

GB = = 2.7
8360 / 400 + 8360 / 300
K = 2.1 ∴ lb in= 2.1 * 3 = 6.3 m
2140 / 300 + 1320 / 300
Outside: GA = = 0.41
8360 / 300
1320 / 300 + 1040 / 300
GB = = 0.28
8360 / 300
K = 1.1 lbin = 1.1 * 3 = 3.3 m
Column 2:
Inside: lbin =6.3, same as col. no.1
2140 / 300 + 1320 / 300
Outside: GA = = 0.26
8360 / 300 + 8360 / 500
1320 / 300 + 1040 / 300
GB = = 0.18
8360 / 300 + 8360 / 500

∴ K = 1.05 lbin = 3 ×1.05 = 3.15m

14/33-B.C.I
 Beam-Column I
Example 4:
Sec 1 Sec 2
IPE 270
IPE 360 IPE 360
2.0
IPE 400

IPE 500

IPE 400
Col 1

Col 2

Col 3
5.0

28 m 28 m 3m

IPE 330

Sec 1 Sec 2
Portal frame Vertical
bracing bracing
Spacing between frames is 6m
Solution:
Col "1"Inside: → Rigid connection
23130 / 700
GA = = 5. 7
16270 / 2800 Allowed to sway
GB = 1.0 fixed
The crane bracket has no effect on the col., the col. length is 7.0m not 5.0 m
∴ K = 1.7 ∴ lbin = 1.7 * 7 = 11.9 m
lbout : Bracing ∴ Prevented from sway & hinged connection

lbout = 3.5m
Col"2" :
Inside: GA = 1.0 fixed

15/33-B.C.I
 Beam-Column I
48200 / 700
GB = = 5.9
16270 / 2800 * 2
K= 1.7 ∴ lb in = 1.7 * 7 = 11.9 m
Outside: → Rigid connection also, so we have to use the alignment chart
GA = 10.0 (hinged) Bases are hinged outside.
2140 / 700
GB = = 0.15
11770 / 600
∴ K = 1.75 lb in = 1.75 * 7 = 12.25 m
Col 3: Same as col."1" because we neglect effect of cantilever (Its stiffness = zero)

Example 5:
IPE 300 IPE 300
IPE 360
IPE 360

Col 2 Col 2
2.5
2.5

IPE 330 IPE 330

R.C.
IPE 500
IPE 500

Building
3.5
3.5

Col 1 Col 1

4.0 4.0 4.0 4.0

Side View Elevation

For column "1":

Q There is bracing, so the connections in the outside directions are all

hinged.
bracing frame
lbout = 3.5m
Lbin rigid connections, so use chants
GA = 1.0 fixed
48200 / 350 + 16270 / 250
GB = = 6.9
11770 / 400
Q Upper connection is prevented from sway

16/33-B.C.I
 Beam-Column I
The frame is connected to R.C. building which prevent the Sway at the
level of the 1st floor
∴ K = 0.28 from chart of prevented from sway
∴ lbin = 0.85 * 3.5 = 2.97,

Column "2":
Outside is prevented from Sway (bracing) and the connections between
beams & columns are hinged
lbout = 2.5m
Inside: The column is allowed to sway because its top is allowed to Sway
GA (column 2) = GB (column 1) = 6.9

GB (column 2) = 16270 / 250 = 3.1

8360 / 400
∴ K = 2.2 (From chart of allowed to sway)
lbin = 2.2 × 2.5 = 5.5 m
Notes:
1. If the link member is removed from 1st floor and added to second floor.
The 2 columns will be permitted to sway.
2. If 2 link members are added at the levels & 1st & second floor.
The 2 columns will be prevented from Sway .
3. For side view, if the bracing is removed from the 2nd floor:
lout of col"1"= l
lout of col"2"= K l
There is no bracing
Where K is the buckling Col 2 Allowed to sway
Connections are rigid
coefficient calculated using
G A & GB
There is bracing
Col 1 Prevented from sway
Connections are hinged

Side view

17/33-B.C.I
 Beam-Column I
Example 6:
Calculate the buckling lengths of columns 1, 2 and 3.

IPE400

Col 3 Col 4
IPE 500

IPE 500
IPE330
7.0m

Col 1

IPE 400
3.5

Col 2
8.0m
6.0m 6.0m
25.0m
For Col 1,3 and 4 For Col 2

Solution:-
Column 1:
Inside plan: GA = 10 (hinged base)
48200
Σ I c / Lc
GB = = 700 = 7.44
Σ( I g / L g * F ) 23130
2500
.·. From chart permitted to sway k = 2.85
.·. lbin = 2.85*7.0=19.95 m
Out-plan: lbout = Lc = 7.0 m. from bracing system
Column 2:
Inside plan: fixed base, hinged connection with girder and frame is permitted to
sway, as the case of fixed free, K = 2.1
.·. lbin = 2.1*3.5=7.35 m
Out-plan: lbout = Lc = 3.5 m. from bracing system

18/33-B.C.I
 Beam-Column I
Column 3:
Inside plan:
48200 48200
+
GA = 10 (hinged base) GB = 350 350 = 27.08
16270
*0.5
800
.·. From chart permitted to sway k = 3.6
.·. lbin = 3.6*3.5=12.6 m
Out-plan: lbout = Lc = 7.0 m. from bracing system

Column 4:
Inside plan:
48200
GB = 27.08 GB = 350 = 14.88
23130
2500

.·. From chart permitted to sway k = 4.1

.·. lbin = 4.1*3.5=14.35 m
Out-plan: lbout = Lc = 7.0 m. from bracing system

19/33-B.C.I
 Beam-Column I
• To determine Fbcx and Fbcy we have to know whether this section is
compact or non-compact and relation between Lu act and Lu max

To determine compact or non-compact section in case of columns (M,N)

a) Flanges : Subjected to compression as in case of beams
Rolled section built-up section
≤ 10.91 compact ≤ 9.87 compact

C C
≤ 14.8 non-compact ≤ 13.5 non-compact
tf tf

> 14.8 slender > 13.5 slender

b) Web: Subjected to moment and normal (not as case of beams which are
subjected to moment only) ECP 9
Calculate dw = h – 2 [ tf + r ] ≈ h – 4tf (rolled)
= h – 2 [ tf + s ] ( built-up)
Assume compact section:
Fy Fy
-ve
dw

dw
-ve

-ve

2a
+ve

+ve

Fy
Fy Fy
Moment Compression
N
dw = h – 2tf N = 2a x tw x Fy a=
2t w F y

dw dw N 1 N
dw = +a dw = + = x( d w + )
2 2 2t w F y 2 t w Fy

1 N
∴α = ( + 1)
2 d w t w Fy

20/33-B.C.I
 Beam-Column I
Very important note: N is +ve for compression and –ve for tension
> 0.5 (case of compression and moment)
Fy
dw 699
≤
(13α − 1) Fy

dw
tw

-ve
dw
< 0.5 (case of tension and moment)

+ve
dw 63.6
≤
t w α Fy
Fy
Note that: = 0.5 means pure moment, where
dw 63.6 127
≤ =
tw 0.5 * F y Fy

While ≥ 1 means the web is subjected to pure compression, where

dw 699 699 / 12 58
≤ = =
tw (13 * 1 − 1) F y Fy Fy

dw
If is greater than the above ratios
tw

∴ Web may be non-compact or slender

(2) Assume non-compact:

Calculate
Fy
-ve

− N / A + M x / Sx Tension
= =
− N / A − M x / S x Compression
dw
dw 190
>-1 (compression) ≤
t w (2 + ψ ) F y
+

d w 95(1 − ψ ) − ψ Fy
≤ -1 (tension) ≤
tw Fy

dw 190 190
For = -1(pure moment) = =
tw (2 + ( −1)) F y Fy

21/33-B.C.I
 Beam-Column I
dw 190 64
For = 1 (pure compression) = =
tw (2 + 1) F y Fy

dw
If is greater than the above ratios ∴ Web is slender
tw

• To calculate LTB: M M
2 2
Cb = 1.75 + 1.05 + 0.3 2
≤ 2.3
M1 smaller moment
= =
M2 bigger moment
M1 M1

1380 AF 20b f M1 M1
Lumax = Cb or Lumax = = - ve = + ve
F y .d Fy M2 M2

• The maximum value of Cb is 2.3, if Cb calculated from the equation is

larger than 2.3, then take Cb = 2.3
• Smaller α gives smaller Cb which is more critical
• Negative value of α is more critical than positive value.
• Important: When the bending moment at any point within the un-braced
length is larger than the values at both ends of this length, the value of Cb
shall be taken as unity. ECP

22/33-B.C.I
 Beam-Column I
II- We need also to calculate the straining actions on the beam-columns
inplane and outplane “N, Mx and My”:
Edge columns with and without crane
Internal columns with crane

E d g e C o lu m n In tern al C ran e C o lum n

Edge columns without cranes:

N N R wind

M1 1 M1 1

Inplane h Inplane
2
X X M2

Y Y Outplane
N N Rwind
M1 1 M1 1

Inplane Inplane
2
X X M2

Y Y Outplane
Min = M1 = X x h , N1 = Y Min = M1 = given Mout = zero
Critical section is 1 Min = M2 = given or = M1/2
“Max M and N “ N1 = N 2 = N = Y
Critical section:
sec1 - if M1 > M2
sec2 - if M2 > M1

23/33-B.C.I
 Beam-Column I
A- Cantilever column:
The base must be FIXED inside plan to maintain stability

N
M
M

Cantilever Cantilever

N M M

To calculate M and N:
wtot = (wc + ws + wLL) t/m2,
Where wc = (5-8) kg/m2 (corrugated sheets), ws = (20-35) kg/m2
WLL = (60-66.66 tan ) kg/m2 (Note that the value to wtot may be given)
W = wtot x Spacing t/m'

Min = W x L2c /2 N = WLc Mout = zero

Important note: If there is a frame with double cantilever, we have to make
cases of loading to get 2 cases:
1- Case of max. moment and corresponding normal (T, D)
2- Case of max normal and corresponding moment (T, T)
If there is long and short cant, we have to use (for case of max moment) total
on the long cant and dead on the short one. Design always on the case of
maximum moment (it is usually critical) and check on the case of max normal.

Note: Dead Total Total Total

If the load is given in t/m', use this

load to calculate M and N.
If the load is given in t/m2, so
Cantilever Cantilever
multiply load * spacing then calculate M
and N.
Max moment Max normal

24/33-B.C.I
 Beam-Column I
B- Edge columns with cranes:
Cranes give 3 loads in 3 different directions “loads from crane”:
∑P
PY = RDL + RLL+I , PX = 0.1*RLL and B =
7
*-Braking force always distributed on 2 successive columns unless given in
the exam.
The braking force is carried by the bracing if there
is a horizontal member at the same level of the
crane OR cause Mout on the column at section at M
y
B B
the position of crane 2 h
c 2

a- If there is horizontal member in the vertical bracing at the level of crane:

N2 M2 N2 Rwind
M2
2 PY 2 PY
M'1 M1 M'1 B/2
M1 PX PX
1 1
Inplane Inplane
3
X X M3
N1 N1
Y Y Outplane

N2 N2 Rwind

M2
M2 2 2
PY PY
M'1 M1 M'1 B/2
M1 Inplane PX Inplane PX
1 1

3
X X M3
N1 N1
Y Y Outplane
Given X and Y, calculate M1, M1’, M2 and if fixed base M3 Mout = zero
N1 = Y and N2 = Y – PY
Critical Sections:
- If M1 > M2

25/33-B.C.I
 Beam-Column I
Sec1 only (Min=M1, N1 and Mout = zero)
- If M2 > M1
Sec1 (Min=M1, N1 and Mout = zero)
and sec2 (Min=M2, N2 and Mout = zero)

b- If there is NO horizontal member in the vertical bracing at the level of crane:

Assume column is simply supported by bracing

N2 M2 N2 R wind
M2 2
PY PY
M'1 M1 M'1 B/2
M1 PX PX
1 1
Inplane Inplane
3
X X M3
N1 N1
Y Y Outplane

N2 N2 R wind

M2
M2 2 2
PY PY
M'1 M1 M'1 B/2
M1 Inplane PX Inplane PX
1 1

3
X M3
X
N1 N1
Y Y Outplane
Given X and Y, calculate M1, M1’, M2 and if fixed base M3 Mout = …..
N1 = Y and N2 = Y – PY
R
Critical Sections:
B M
y
- If M1 > M2
h
Sec1 only (Min=M1, N1 and Mout = ….) h
c
- If M2 > M1
Sec1 (Min=M1, N1 and Mout = ….)
and sec2 (Min=M2, N2 and Mout = zero)

26/33-B.C.I
 Beam-Column I
C- External combined columns:
The following figures show the BMD and NFD of combined columns and
roof columns, the moment at the partially fixed connection between the 2
columns must be given in the exam.

Roof Roof

Partially Partially
fixed fixed
Combined Combined
Column Column

BM and NF diagrams for

combined and roof columns

Note: If the moment at the partially fixed connection between the 2 columns
is not in the exam, we may consider the connection to be hinged with moment
equals zero as shown:

Roof Roof

Partially Partially
fixed fixed
Combined Combined
Column Column

27/33-B.C.I
 Beam-Column I
D- Internal columns with cranes:

Internal

This short column is used to carry the crane girder, s it will be subjected to loads
from crane “Px , Py and B”
Case 1 “the column carry 1 crane girder“
• Inplane PY
The shown column is cant. Inplane
So Min = Px (h+hI)
PX h I

N = Py
h
Min

28/33-B.C.I
 Beam-Column I
• Outplane
As this column is inside the building, we have three diff ways for
the outside plane.
a- Ordinary bracing system
Very Very important note: This column did not carry any wind load because it
is an internal column.
Since we used bracing, so the bases outside are hinged

B/2

Internal column with bracing

Mout = zero
Lbin = 2.1 h
Lbout = h

29/33-B.C.I
 Beam-Column I
b- Portal frame bracing system.

Internal column with portal

frame bracing
Inside: Cantilever as previous Lbin = 2.1 h
Outside: The column is connected with the beam with rigid connection, outside

(base is hinged) Use GA = 10 & GB =

∑(I y / l)c
∑ (I x / l )b

If section of the beam not given assume Ic= 2Ig

B=
ΣP
Mout =
B
xh
B/2 B/2
7 2
2
M out

1
B/2 B/2
Outplane
So we have to design 2 sections
Section "1" at the base: N & Min , while Mout = zero
Section "2"at the top of the column: N & Mout , while Min = zero

30/33-B.C.I
 Beam-Column I
c- No bracing “cant. Col. In the outside plane”
⇒ This system is unstable outside. So the column must be
Cantilever outside (ie the base is fixed)
⇒ The column is cant. Inside & outside

Internal free column

Here we have 1 critical section “ at the base “
Min = Px (h+hI)
B
Mout = xh
2
N = Py
And Lbin = Lbout = 2.1h

31/33-B.C.I
 Beam-Column I
Case 2 “the column carry 2 crane girders“
We have here 2 cases of loading
1- 2 cranes working at same time.

PY1 PY2

PX1 PX2 h I B total

2
B total
2
d
h
Min
Inplane
Outplane

• Inplane
The shown column is cant. Inplane
Assume crane 1 > crane 2
So Min = Px1 (h+d+hI) + Px2 (h+d+hI) + Py1 (e1) - Py2(e2)
N = Py1 + Py2

• Outplane
We have the same 3 diff bracing systems.
a- Ordinary bracing system
Mout = zero
Lbin = 2.1 h
Lbout = h
ΣP1 ΣP2
Btotal = +
7 7
Same for the other systems with the shown Btotal

32/33-B.C.I
 Beam-Column I
2- The max crane working and the other at rest.

PY1 Pd2

PX1 hI Btotal
2
Btotal
2
d
h
Min
Inplane
Outplane

• Inplane
The shown column is cant. Inplane
So Min = Px1 (h+d+hI) + Py1 (e1) + Pd2 (e1)
N = Py1

• Outplane
We have the same 3 diff bracing systems.
b- Ordinary bracing system
Mout = zero
Lbin = 2.1 h
Lbout = h
ΣP1
Btotal =
7
Same for the other systems with the shown Btotal

33/33-B.C.I
 Beam-Column 2

Design of rolled columns

After calculating the straining actins and choosing the critical sections, we
get Min, Mout and N.
Orientation of the Section:
1- If the column is subjected to Min and N only
Put Min = Mx so My = Mout = zero

N2
M2
2
M'1
M1
1
Inplane

X N1
Outplane
Y Y
X

2- If the column is subjected to Min, Mout and N

Case 1 Min > Mout or Min < Mout
Put Min = Mx so My = Mout

N2
M2
2
M'1
M1
1

Inplane

X N1
Outplane
Y Y
X

1/18-B.C.2
 Beam-Column 2

Case 2 Mout >>>> Min “very rare case”

Put Min = My so MX = Mout
N2
M2 2
M'1 Mout
M1
1
Inplane

Y N1
Outplane
X X
Y

1- Estimation of section (rolled):

a- Section subjected to N only
N
Assume fca = = 0.75 t/cm2
A
Get A from IPE or BFIB tables.
b- Section subjected to Mx and N
Mx
Assume fbcx = = (0.8 to 1.2) t/cm2
Sx

Get S x req. from IPE or BFIB tables.

c- Section subjected to Mx, My and N
Mx
Assume fbcx = = (0.4 to 0.7) t/cm2
Sx

Get S x req. from BFIB tables.

2- Interaction equation of columns and how to use it:

For columns subjected to N, Mx and My, they should be designed to satisfy the
following interaction equation: ECP 25 and 26
f ca f f by
( ) + ( bx ) A1 + ( ) A2 ≤ 1.00 case (A)
Fc Fbcx Fbcy

≤ 1.20 case (B)

Where

2/18-B.C.2
 Beam-Column 2
N
-fca = actual axial compressive stress =
A

-Fc = allowable axial compressive stress = 1.4 – 6.5 × 10-5 λ 2 λ <100

7500
= λ >100
λ2

Mx
-fbcx = actual bending stress @ x – axis =
Sx
My
-fbcy = actual bending stress @ y – axis =
Sy

-Fbcx , Fbcy = allowable compressive bending stress for x , y axes resp.

Where:

Fbcx depends on .Lumax < Luact .

Fbcy = 0.72Fy or 0.58 Fy according to category of the section

C mx C my
A1 = ≥ 1.00 A2 = ≥ 1.00
f ca f
(1 - ) (1 − ca )
FEx FEy
7500
, FEX = (even if x is the , FEy = 7500 (even if is the
λ 2x
y
λ2y
smaller and smaller than 100)
smaller and smaller than 100)
λx
λy

A1 and A2 are modification factors due to p- effect (additional moment)

• To calculate A1 and A2
f ca
For case ≤ 0.15 Take A1 = A2 = 1.00
Fc

- if (fca / Fc ) > 0.15 ⇒ get A1 , A2 as follows:

3/18-B.C.2
 Beam-Column 2

Cmx , Cmy = moment modification factor

a) For frames permitted to sway Cm = 0.85
b) For frames prevented from sway and with transverse loads:
Cm = 0.85 Fixed end

Transverse load
= 1.00 Hinged end

Cmx =1

c) For frames prevented from sway and without transverse load

M1
Cm = 0.60 – 0.40 ( ) > 0.40 M
2
M
2
M2

M 1 smaller moment
α = =
M2 bigger moment
M1 M1
M1 M1
= - ve = + ve
M2 M2

• The minimum value of magnification values A1 and A2 is 1. if A1 or A2

calculated from the equation is smaller than 1, then take them = 1

• To determine Fbcx, we have to know whether this section is compact or

non-compact and relation between Lu act and Lu max
To determine compact or non-compact section in case of columns (M,N)
a) Flanges : Subjected to compression
Rolled section built-up section
≤ 10.91 compact ≤ 9.87 compact

C C
≤ 14.8 non-compact ≤ 13.5 non-compact
tf tf

> 14.8 slender > 13.5 slender

4/18-B.C.2
 Beam-Column 2

b) Web: Subjected to moment and normal ECP 9

Calculate dw = h – 2 [ tf + r ] ≈ h – 4tf (rolled)
= h – 2 [ tf + s ] ( built-up)
Assume compact section:
1 N
∴α = ( + 1)
2 d w t w Fy

Very important note: N is +ve for compression and –ve for tension
dw 699
> 0.5 (case of compression and moment) ≤
tw (13α − 1) Fy

dw 63.6
< 0.5 (case of tension and moment) ≤
t w α Fy

dw
If is greater than the above ratios
tw

∴ Web may be non-compact or slender

(2) Assume non-compact:

Calculate
− N / A + M x / Sx
=
− N / A − M x / Sx

dw 190
>-1 (compression) ≤
t w (2 + ψ ) F y

d w 95(1 − ψ ) − ψ
≤ -1 (tension) ≤
tw Fy

dw
If is greater than the above ratios ∴ Web is slender
tw

5/18-B.C.2
 Beam-Column 2

• To calculate LTB:
Cb = 1.75 + 1.05 + 0.3 2
≤ 2.3
M1 smaller moment
= = (The same sign as before)
M2 bigger moment

1380 AF 20b f
Lumax = Cb or Lumax = Take smaller
F y .d Fy

• The maximum value of Cb is 2.3, if Cb calculated from the equation is

larger than 2.3, then take Cb = 2.3
• Smaller α gives smaller Cb which is more critical
• Negative value of α is more critical than positive value.
• When calculating Cb, we have to consider the Un-braced length
The un-braced length is the distance between the points which are supported
outside using the horizontal member of the vertical bracing.
The column is to be divided into parts, which is braced outside.
Then we design the critical section for each part.

horizontal segments

member of vertical bracing

• Important: When the bending moment at any point within the un-braced
length is larger than the values at both ends of this length, the value of Cb
shall be taken as unity. ECP

6/18-B.C.2
 Beam-Column 2

Some examples to calculate both Cm and Cb :

M M
2

Truss with hinged base Truss with fixed base

= zero =+M /M
1 2 2
C b = 1.75 C b = 1.75+1.05 +0.3
Cmx = 0.85 (Sway) Cmx = 0.85 (Sway)
M
1

Note that: If M1 for the truss with fixed base is not given, it can be assumed to
be 0.5 M2

M M

Cantilever Cantilever
tie

= -1 = zero
Cb = 1 C b = 1.75
Cmx = 0.85 (Sway) Cmx = 0.6-0.4(0)
M
= 0.6

Fink truss
Base must be fixed
= zero
M C b = 1.75
Cmx = 0.85 (Sway)

7/18-B.C.2
 Beam-Column 2

How to determine Lu act and Lout

Lu act. = distance between points, where the comp. flange is laterally braced.
What braces the compression flange? (Lu act)
1- Bracing outside plan (perpendicular to column)
2- Presence of side girts
3- Knee bracing at the level of any side grit
4- If compression in the flange changes to be tension.

Corrugated
sheets

side
girt
IPE

Elevation Plan
Knee bracing

What braces the column? (Lb out)

1- Bracing outside plan
Note that: the side girts and the knee bracing does not affect Lb out

8/18-B.C.2
 Beam-Column 2

Examples:

If h > 6m
h2 h2
M

Truss with hinged base

h1 h1

h1
L =L = L =L =h
u act b out 2 u act b out 1
with no knee bracing
h1
OR L = h AND L =
b out 1 u act 2
Using knee bracing at mid height

If h > 6m
h h
2 2 M
2
Sec1-1 h
1
2 L

h h
1 1
Sec2-2

h1 M
1
L = L = h1 Truss with fixed base
b out 2 b out
Case 2 Case 1

In this example, the buckling outside is h1 or h2 according to shape of bracing,

but Lu act varies according to the location of the compression flange with respect
to side girts.
For section 1-1, the compression flange is the inner flange, while for section 2-2,
the compression flange is the outer flange.
Case 1:
Sec. 1-1: Luact. = L
Sec. 2-2: Lu act. = a (maximum distance between side girts which is mostly the
height of the wall)

9/18-B.C.2
 Beam-Column 2

Case 2:
Sec. 1-1: Luact. = h1 / 2
Sec. 2-2: Lu act. = a (maximum distance between side girts which is mostly the
height of the wall)

(2) If there is no side grits (cantilever sheds)

⇒ We can't use knee bracing

⇒ We can add a hz. Member in the v l bracing at the mid height of the
column
1
⇒ Luact = h1= Lb out
2
If h > 6m
h h
2 2 M

Cantilever

h1 h1

M
h
1 L =L =h
L =L = u act b out 1
u act b out 2

In this example, we can not use knee bracing because there is no side girts

Note that: If the column has 2 different sections as shown, we have to study the
2 sections

N
2 Sec 2
M2
3.0m

N M M3
1 Sec 1 1

5.0m

10/18-B.C.2
 Beam-Column 2

For Section 1: For Section 2:

Luact = 3m Luact = 5.0

For Section 1: For Section 2:

Lb out = 3m Lb out = 5.0
This means that Lu act and Lb out varies according to the place of the section
studied.
M3
For section 1: = zero and Cb = 1.75, while for section 2: =−
M2

11/18-B.C.2
 Beam-Column 2

Summary
• The design of column is the design of compression member subjected to
pure normal force, then design of beam subjected to pure moment except
in 3 differences:
1. There is a magnification factor called A1 due to p – effect
(additional moment). In the direction of Mx, and A2 in direction of
My. (If there is My)

2. The Cb is calculated from the equation Cb = 1.75 + 1.05α + 0.3α 2

instead of using the table of beams.
3. When check the compactness of web, the limit used for beam is
h / t ≤ 82, while for columns the web, is subjected to moment and
normal or normal force only, so the limit is completely changed and
has to be calculated

A) Design of rolled section:

Given or calculated: N=P Mx=P
Choice of section:
Mx
Assume f= = (0.8 ⇒ 1.20) t/cm2 approx 1 t/cm2
Sx

Mx
∴ S x req. = = …………… cm3
f

Choose IPE no
check: Calculate Lbin =P Lbout = P
Lbin Lbout
λx= λy=
rx ry

from λ max ⇒ get Fc =P

N
=
fca A

12/18-B.C.2
 Beam-Column 2

f ca
Get = P ≤ 0.15 ⇒ A1 = A2 = 1.00
Fc

C mx
> 0.15 ⇒ A1 =
F
(1 − ca )
FEx

Mx
fbcx = =P
Sx

Determine compact or non-compact

1380 A f
Lu max = Cb = P >Luact ⇒ Fbcx = 1.536 or 1.4
Fy . d

<Luact ⇒ Fbcx = Fltb

20bf
min =P
Fy

* Finally apply in interaction eq ≅ to check the safety of stresses.

Example 1:
Design the column shown 2.0
in the fig using rolled
5.0
section.
4.0 t

5.0
12.0 t
Solution:
N = 12 ton.
Min = Mx = 4 * 5.00 = 20 m.t.

Choice of section
Assume fbcx = 1.00 t/cm2
20 × 100
∴ Sxreq. = = 2000 cm3
1.00
∴ Choose IPE 500

13/18-B.C.2
 Beam-Column 2

Checks:
C
= 6.25 <10.9
20 / 2
Flange: =
tf 1. 6

1 N co1
Web: = ( + 1) dw = 50 – 4×1.6 = 43.6 cm
2 d w t w Fy

+ 1 ) = 0.556 > 0.5

1 12
∴ = (
2 43.6 × 1.02 × 2.4

d 699
= 42.75] <[
43.6 699 / 2.4
∴[ w = = = 72.4]
tw 1.02 (13α − 1) F y (13 × 0.556 − 1

∴ The section is compact

20 × 20
Luact = 500 cm > Lumax = = 258
2.4

∴ There is LTB
800 A f
∴ Fltb1 = .C b
Lu .d

M1
QFor the lower column segment = Zero ⇒ Cb = 1.75
M2
800 × 20 × 1.6
∴ Fltb1 = × 1.75 = 1.792 > 1.4 t/cm2
500 × 50
2
∴ Fbcx = 1.4 t/cm
2
QLbin = 2(5 + ) = 12 m Lbout = 5.00 m
2

= 58.82 <180 = 116 <180

1200 500
λx= λy=
20.4 4.31

7500
∴ Fc = = 0.557
(116) 2
N 12
fca = = = 0.103 t/cm2
A 116
f ca 0.103
= = 0.186 > 0.15
Fc 0.557

14/18-B.C.2
 Beam-Column 2

C mx
∴ A1 = Cmx = 0.85 (permitted to sway )
f
1 − ca
Fex
7500
fEx = = 2.17 t/cm2
(58.82)2
0.85
∴ A1 = = 0.89 < 1.00 ∴ A1 = 1.00
0.103
(1 − )
2.17
Mx 20 × 100
fbx = = = 1.04 t/cm2
Sx 1930

Applying interaction equation

f ca f
( ) + ( bx ) A1
Fc Fbcx

) * 1.00 = 0.93 < 1.00

1.04
. = 0.186 + (
1.4

∴ Safe and economic

Example 2:

Design the column shown in the figure using rolled section. 2 .0

Spacing between trusses 5.0 m

Y = 12 ton and X = 3.33 ton 6 .0
X
Solution: Y

1- Bracing system:

2 .0
3 .0
3 .0

15/18-B.C.2
 Beam-Column 2

2- Straining action on the column.

In Plane Out of Plane

N = 12ton
M1= 20mt
1

At sec 1 – 1: "critical section"

M = X x h' = 3.33 x 6.0 = 20 mt
N = Y = 12 ton

As the section is subjected to M in plane only, Min = Mx = 20.0 mt

3- Choice of section

Mx 2 20.0 ×100 3
Assume f = = 1.0 t / cm , so Sx = = 2000 cm
Sx 1.0
Choose I.P.E 500

4- Check:

Int. Equ. :

f ca f f bcy
+ A1 ( bcx ) + A 2 ( ) ≤ 1.0
Fc Fbcx Fbcy

As M y = zero

f ca f
+ A1 ( bcx ) ≤ 1.0
Fc Fbcx

N 12 2
* fca = = = 0.103 t/cm
A 116

16/18-B.C.2
 Beam-Column 2

M x 20.0 × 100 2
* fbcx = = = 1.04 t/cm
Zx 1930

1.4 − 6.5 × 10−5 λmax

2
for max < 100

Fc =

7500
for max > 100
λmax
2

So we need for max, so we need Lbin = 2 (6 + 2 / 2) = 14.0 m, Lbout = 3.0 m

Lbin 1400 1400 Lbout 300 300

in = = = = 68.62 < 180 out = = = = 69.6 < 180
rin rx 20.4 rout ry 4.31
max = 69.6 < 100

* Fc =1.4 − 6.5 ×10−5 λmax

2
= 1.4 − 6.5 ×10−5 × 69.62 = 1.085 t/cm2

f ca 0.103
= = 0.095
Fc 1.085

* Fbcx :

1- Study the local buckling " compact or non compact "

C = bf / 2 = 20 / 2 = 10.0 cm, dw = h – 4tf = 50 – 4 x 1.6 = 43.6 cm

C / tf = 10 / 1.6 = 6.25 < 10.9 compact flange

dW / tW = 43.6 / 1.02 = 42.75 < ????

1 N 1 12
We need to calculate α = ( + 1) = ( + 1) = 0.556
2 d w × tw × Fy 2 43.6 × 1.02 × 2.4

699 FY 699 2.4

As > 0.5, so ???? = = = 72.4
(13α − 1) (13 × 0.556 − 1)

dW / tW = 42.75 < 72 .4 compact web

As both are compact, so regarding local buckling the section is a compact

section.
17/18-B.C.2
 Beam-Column 2

2- Study the L.T.B.

From the B M D the compression flange is the inner flange.

From the bracing, at sec 1 – 1 Luact = 3.0 m

20b f 20 × 20
Lumax = = = 258 cm
Fy 2.4

As Luact > Lumax Sec is Non-compact, and LTB may occur

1 + FFltb 2
2 2
Fbcx = Fltb = FFltb
800 × A f 800 × 20 ×1.6
Fltb1 = Cb = Cb =
Lu × d 300 × 50
2
Cb = 1.75 + 1.05 +0.3
M small M small 10.0
= , at sec 1 – 1 = =− = - 0.5
M big M big 20.0
Cb = 1.75 + 1.05 (- 0.5) + 0.3 (- 0.5)2 = 1.3
800 × 20 × 1.6 2 2
Fltb1 = = 1.3 = 1.7 t/cm > 1.4 t/cm
300 × 50
So Fbcx = Fltb = Fltb1 = 1.4 t/cm2 No LTB will occur.

* A1

f ca 0.103
As = = 0.095 < 0.15 so A1 =1.0
Fc 1.085

Applying the interaction Equ:

f ca f 1.04
+ A1 ( bcx ) = 0.096 + 1.0( ) = 0.83 < 1.0 Safe
Fc Fbcx 1.4

18/18-B.C.2
 Beam-Column 3

Design of built-up sections (combined columns)

After calculating the straining action as before, we have Min, Mout and N.
Orientation (Direction) of column:
Always make the axis which resist the moment (or resist the bigger
moment in case of Mx and My) is the bigger inertia axis.
a- if section subjected to Min and N
As for combined col Iy > Ix
So put Min = My
So the in-plane is the Y plane while the out-plane is the X.
b- if the section subjected to Min, Mout and N
• case Min > Mout or Mout > Min
Put Min = My
So the in-plane is the Y plane while the out-plane is the X.
• Case Mout>>>>Min very rare case
Put Mout = My
So the out-plane is the Y plane and the in-plane is the X.

Combined columns are columns composed of rolled sections connected together

with lacing bars or batten plates. There are many shapes of combined columns
as following:
y y y

x x x x x x
h h

y y y
d d d
(1) 4 angles (2) 4 angles (3) 2 channels
and plate

B.C-3-1/49
 Beam-Column 3

y y y

x x

y y y
d d

(4) 2 IPE (5) 4 angles (6) 4 angles

and plate

2 I with batten plates 2 I with lacing bars

Design of section:-
1- Choice of section:- y

For section formed from 2 C or 2I

For section subjected to My and N only x x
h
Assume d if not given, d = (30 → 60) cm
H y
or we may take d =
12 − 15 d
Where "H" is height of column
N M
Calculate compression force on one side: c= +
2 d
Force(C)
∴ A 1[ = = ………. Cm2
Fc

Assume Fc = (1.1 → 1.3) t/cm2, for section subjected to My and N only

B.C-3-2/49
 Beam-Column 3

Assume Fc = (0.5 → 0.8) t/cm2, for section subjected to My, Mx and N only
Choose according to the required section

N.B.:-
For economic section
d = (1.5 → 2.0) h Not a check

2- Checks:-
Calculate A, Ix, Iy, rx and ry for the combined section.
We apply the interaction equation:-
f ca f f by
( ) + ( bx ) A1 + ( ) A2 ≤ 1.00 case (A)
Fc Fbcx Fbcy

≤ 1.20 case (B)

Where,
N N
* fca = =
A 2[ 2A[

Mx Mx h
* fbcx = = y ,y= , I x[ ] = 2(I x[ )
Sx2[ I x2[ 2

My My d d
* fbcy = = x ,x= + e[ , I y[ ] = 2(I y[ + A[ ( )2 )
Sy2[ I y2[ 2 2

* Fc = 1.4 – 6.5 × 10-5 λ max2 λ <100

λ >100
7500
=
λ max 2

So we need to get λ max

B.C-3-3/49
 Beam-Column 3

Global and local buckling:

As this section is composed of several parts so we have 3 buckling modes
Global buckling:
The whole section (all its components) will buckle inside
or outside (as in case of rolled section).
Local buckling:
Each individual section will buckle alone about its minor
axis (local buckling). The local buckling occurs inside
plan or outside plan or in both direction according to the
presence of batten plates or lacing bars as shown.
Overall buckling:
The sum of local and global buckling on each direction.
Calculate Lbin , Lbout (Exactly as before)
Global buckling:-
L bout L bin
λ ’out = and λ ’in =
rx[ ] ry[ ]

Local buckling:-
Lz
λ z if there is lacing bars or batten plate, where λ z =
rmin

Where, Lz is the free distance where each individual part an buckle alone.
rmin is minimum radius of gyration of the part that buckles alone.

rmin = ry[ or ryI

Overall buckling:-
Sum of previous modes

λ in = λ y = (λ 'in ) 2 + (cλ z )2 as both are about y axis

λ out = λ x = λ ’out

Where c = 1 if using lacing bars or 1.25 if using batten plate.

B.C-3-4/49
 Beam-Column 3

Lz 2
Check: λ z = = …< 60 and < λ max (of λ in and λ out)
rmin 3

If z is unsafe, decrease Lz by using double lacing system or hz lacing member

Or increase rmin by increasing the section
How to get Lz?
Different arrangement of lacing bars:

45
°
45
°

Single intesection Single intesection Double intesection Single intesection

Lacing BARS Lacing BARS Lacing BARS Lacing ANGLES
with horizontal bar

Lz is the free distance where each individual part an buckle alone.

For batten plates or channels: Assume Lz = (1 – 2) d = (80-100) cm
For lacing bars: If single intersection system = 500 - 700
If double intersection system = 400 ± 100 See ECP page 133
is the angle of inclination of lacing bar with vertical axis of member.
is the angle of inclination of lacing bar with horizontal axis.
We can take = 450 in all cases ( )
i.e. Lz = 2d in single intersection system and lacing angles
Lz = d in double intersection system and single intersection system with
horizontal bar.

B.C-3-5/49
 Beam-Column 3

Special case for Lz in case of 4L as in figures:-

Lz Lz

C my 7500
*A2 = ≥ 1.00, FEy = , λ in = λ y
f
(1 − ca ) λ2y
FEy

* Fbcy and Fbcx

Always Lu act is neglected, so no need to calculate Lu max
• For Combined sections:
Fbcy = Fbcy = 0.58Fy = 1.4 t/cm2 because these sections are non-compact

B.C-3-6/49
 Beam-Column 3

Different types of built-up sections:

1- Section composed of 2 channels:
y
A2 channels = 2A one channel Ix 2 channels = 2 Ix one channels

⇒ rx[ ] = rx [ (from table) because the axis does not move. x x

h
Ix
Or rx 2 channels = (Ix and A are of the whole section)
A y
d
d Iy
Iy 2 channels = 2[Iy1 channel + A 1 channel ( ) 2] , ry 2 channels =
2 A
rmin = min. radius gyration of one element of column which is ry of single
channel.
My d
fbcy = ( + e[ )
I y[] 2

2- Section composed of 2 IPE :

Exactly as section composed of 2 channels x x

My d b
fbcy = ( + f)
I yII 2 2 y
d
3- Section composed of 4 angels:
2 IPE
y y

x x
h

y y
d d

4 angles 4 angles

N M
Estimation of section: Force on one side = C = + (Force on 2 angels)
2 d
Assume Fc = (1.1 → 1.3) t/cm2

B.C-3-7/49
 Beam-Column 3

Force(C)
∴ Area of each side = = ………. cm2 (Area of 2 angles)
Fc

Area of one angel = A / 2 Get angle from tables

Local buckling will occur in both directions.
L l L l
λ x = λ out = ( bout ) 2 + (C z ) 2 λ y = λ in = ( bin ) 2 + (C z ) 2
rx 4 L rmin ry 4L rmin

Where rmin is the minimum r for single angle = rv

h h d d
rx 4 L = (rx1L ) 2 + ( ) 2 ≈ ry 4 L = ( rx1L ) 2 + ( ) 2 ≈
2 2 2 2
And complete as before
4- Section composed of 4 angles + 2 plates: y
N M
Force on each side = +
2 d
x x
Force
Area of each side = Atot = = (A2L + Apl.)
f
y
Assume the height of the plate = h, and its thickness d

= 10-20mm 4 angles
and plate
Atot − A pl
A1L = Get area of each angle from tables
2
Deal with section exactly like 2 channels
Local buckling will occur inside plan only, so
Lbx L l
λ out = λ in = ( bin ) 2 + (C z ) 2
rx ry rmin

Where rx and ry are of the whole section (4 angles and plate). While rmin is the
minimum radius of gyration of the part buckles alone which is the 2 angles and
I min
the plate of each side. So we have to calculate rmin =
A2 Ls + pl

hp = h
tp as assumed before (10 – 20 mm)
Ap = hp tp
B.C-3-8/49
 Beam-Column 3

Imin is Iy, so we have to calculate X and Iy

A = 2 A1L + Apl
yX
Ap *
hp
2
[
+ 2 AL * (t p + e L ) ]
X = x x
A
2
hp * t p3  tp 
Iy = (can be neglected) + Ap  X −  y
12  2  part behaves
lonely
+ 2  I x1L + A1L * (t p + e L − X ) 
2
 

Iy
rmin = and complete as before
A
5- Section composed of 4 angles and plate:
This section has no local buckling but there is LTB.
H y
Estimation of section: Assume d = , tp = 10mm
12 − 15
N M
C (compression on 2 angles) = + y
2 d
4 angles
2 C/2 and plate
Assume Fc = (1.1 → 1.3) t/cm , A1L =
Fc

Calculate Ix, Iy, Fbcy ( Af = 2area of the angle)

6- Section composed of 3 plates: (Built-up)

Design as Built-up section of crane girder but with equal flanges. y

H dw
Assume dw = , tw is max of 5mm or
12 − 15 190 Fy
y
N M
Force on flange = +
2 d
Assume stress Fc = (1.1 → 1.3) t/cm2 and get area of flange.
Assume bf = 20 tf for both upper and lower flange. Then we have to deal with it
exactly as rolled section but My not Mx

B.C-3-9/49
 Beam-Column 3

Design of Cross column:

flange web portal frame bracing

plan

The cross column can be with the following shapes:

Case 2
Case 1 Economic Case 3
Easiest

Estimation of section:
Here we will deal with each section apart "each IPE resists the moment in its direction.
Mx
S x IPE 1 = =P get IPE no. P
0.9 → 1.3
My
S x IPE 2 = =P get IPE no. P (NOTE: Sx not Sy )
0.9 → 1.3
Checks:
Interaction equ.:
f ca f f 1.0 Case I
+ A1 bcx + A 2 bcy ≤
Fc Fbcx Fbcy 1.2 Case II

* fca = N / Atot
Mx
f bcx = y
I xtot
* Calculate Ix and Iy of the 2 IPE together.
My
f bcy = x
I ytot

Lbin I xtot
λin = , rx =
rx Atot
* Fc
Lbout I ytot
λout = , ry =
ry Atot

* A1 and A2 as before

B.C-3-10/49
 Beam-Column 3

* Calculation of Fbcx and Fbcy:

1- Check compactness:

c
All flanges as before < 10.91
tf

Case 1:
Web of IPE 1 is always compact because
it is stiffened with flange of the other IPE

dw 699 / F y
IPE 2
Web of IPE 2 ⇒ <
tw 13α − 1
1 N IPE 2
Where = *( + 1)
2 dw * t w * Fy

A2 IPE 1
Where NIPE 2 ≈ Ncol
A1 + A2

Case 2:
Web subjected to Web subjected to
bending and comp. IPE 2 comp. only

Web of IPE 1

0.5d w 58/ Fy Compact

∴ <
tw 64/ Fy non - Compact
IPE 1
dw 699 / F y
Web of IPE 2 ⇒ <
tw 13α − 1
1 N IPE 2
Where = *( + 1)
2 dw * t w * Fy

A2
Where NIPE 2 ≈ Ncol
A1 + A2

B.C-3-11/49
 Beam-Column 3

Case 3:
Web subjected to
comp. only
Web subjected to
Web of both IPE 1 and IPE2 bending and comp.
IPE 2
0.5d w 58/ Fy Compact
∴ <
tw 64/ Fy non - Compact

IPE 1

2- There is no LTB because in these columns,

we use stiffener every 1-1.5 m in the vertical
direction, so the LTB is neglected
Stif
f
m
.5
1-1

B.C-3-12/49
 Beam-Column 3

Design of lacing bars and batten plates for combined columns:

Important note: The design of lacing bars and batten plates are not calculated

in the exam except if it is required. ( )

I- Design of lacing bars:

Lacing bars are designed on shear force of column as well as 2% of the normal
force. Q* = Q + 0.02 N (2% of the normal force is due to buckling)

Calculation of force using method of sections and joints:

The lacing bars together with the column act as a truss, where the 2 columns are
the upper and lower chord of the truss (carry moment and normal), while the
lacing bars (inclined and horizontal) act as its diagonals and verticals (carry
shear).
How to calculate force?
1- 2 plans of lacing bars with single intersection system:
F F1
Q /2 Q /2

Q /2 F2
°
4 5
45°

Q /2

Since the given shape has 2 plans of lacing bars, so Q* will be divided to the 2
plans affecting each plan with Q* / 2.
Force in the diagonal: Use method of sections.
Q* Q*
∑x =0 F cos = ± F= ±
2 2 cos α
Force in the horizontal (if any): Use method of joints.
∑x =0 F = ± Q* / 2

B.C-3-13/49
 Beam-Column 3

2- 1 plan of lacing bars with single intersection system:

F F1
Q

Q F2

°
4 5
45°
Q

Since the given shape has 1 plan of lacing bars, so Q* will be carried by one
plan.
Force in the diagonal: Use method of sections.
Q*
∑x =0 F cos = ± Q* F= ±
cos α
Force in the horizontal (if any): Use method of joints.
∑x =0 F = ± Q*

3- 2 plans of lacing bars with double intersection system:

F1 F2
Q /2
Q /2
a F1 F2
°
4 5

Q
45°

Joint a
Q /2

Since the given shape has 2 plans of lacing bars, so Q* will be divided to the 2
plans affecting each plan with Q* / 2.
From the equilibrium of joint "a": F1 equals F2 and in opposite directions
So F1 = F2 = F
Force in the diagonal: Use method of sections.
Q* Q*
∑x =0 2* F cos = ± F= ±
2 4 cos α

B.C-3-14/49
 Beam-Column 3

Summery of previous:
Q*
F= ± (Force in inclined members)
n m cosα
Q*
F= ± (Force in horizontal member if any)
n
Where:
• is the inclination of lacing bars with horizontal
• n is number of lacing plans.
n = 1 if ONE plan of lacing bars, while n = 2 if TWO plans of lacing bars
• m = 1 for single intersection system and m = 2 for double intersection
system.

Dimension of lacing bar:

Note that: The lacing bars may be bolted or welded to the column.
d
For inclined members: Length of lacing bar = L = (Lcos = d)
cos α
For horizontal members: Length of lacing bar = L = d
L
- Width of lacing bar = b = >3 Φ (for bolted connection)
10
L
- Thickness of lacing bar tmin = (ECP 134) 8mm ≤ t ≤ 20mm
50

The lacing bars are designed as compression members and checked as tension
members.

- Check of lacing bar as comp. member:

t

Lbx = Lby = L
y

Buckling outside is more critical because it is about the

b

minor axis of the bar

x

B.C-3-15/49
 Beam-Column 3

bt 3 Ix bt 3 / 12 t
Ix = ⇒ rx = = =
12 A bt 12

L
λx= < 140 ECP 134
rx

Fc = 1.4 – 6.5 × 10-5 λ x2 λ ≤ 100

7500
λ >100
λ2x

F
fact = ≤ Fc
bt

- Check as tension member:

If bolted: Anet = (b - Φ / ) * t Φ / = Φ + 2mm
If welded: Anet = Agross = bt
Force
1) fact = = ………… t/cm2 < Ft = 1.4 t/cm2
Anet

L L
2) = = ………………….. ≤ 60 (b is the depth of bar)
d b

If the previous system of lacing bar is unsafe:

1) Increase (t):
tmax = 20 mm

2) Use lacing angles:

Design of section as un-symmetric section and design it as following:

a) As comp. member:
Lout
λv= = ……………… ≤ 140
rv

Where L = d / cos Lout = L

B.C-3-16/49
 Beam-Column 3

Fc = 0.60 1.4 – 6.5 ×10-5 λ v2 λ v ≤ 100

7500
λ v >100
λ2v

Note that: we multiplied *0.6 because it is un-symmetric compression member

Force
fact = = ……………….. ≤ Fc
AL

b) As tension member:
3 A1
Anet = A1 + A2 ( )
3 A1 + A2

Where A1 = at in case of welded connection

And A1 = (a - Φ / ) t in case of bolted connection

While A2 = ( a – t ) * t for bolted and welded
Force 2
fact = ≤ 1.4 t/cm
Anet

L
≤ 60
d

3) Use double lacing bar system:

Q*
Force in lacing bar: F = ± (i.e the force is decreased to its half)
4 cos α
L
tmin =
50
Lbx = Lbout = 0.7L ECP 134
Note that: In case of using double lacing bars, we must use packing under one of
the lacing bars with thickness equals to thickness of lacing bar.

B.C-3-17/49
 Beam-Column 3

Design of connection:
a) If bolted connection:
Force < Rs.s.

πΦ 2 0.25 Fu
∴ Force = ( )* ( 0.2 Fu )
4
⇒ Φ = ……………..mm ≤ 20 mm.
For big Φ ⇒ change system or use welded connection

b) If welded connection:
Assume size of weld ≤ thickness of lacing bar
S = (5 or 6) mm
Force
Get Lact = + 2S
(2 S ) * (0.2 Fu )

Smin = 4mm (building) Smin = 6mm (dynamic loads)

Very important note for drawing:

Batten plates must be used at both ends of combined columns connected
by lacing bars.
Also, Batten plate must be used at any connection of vertical bracing.
Depth of batten plates is taken ≈ (1 → 1.25) d
If it is required to draw the whole column, arrangement of lacing bars
and end batten plates must be calculated before drawing.

(1-1.25)d (1-1.25)d

(1-1.25)d

(1-1.25)d (1-1.25)d
d

B.C-3-18/49
 Beam-Column 3

II- Design of batten plates:

The system consists of batten plate together with column acts as multi-storey
frame.

d d d

a L a a
z

a a

Dimensions of batten plates

See ECP 136 and 137
(1-1.25)d
The height of the 2 end batten plates ranges from
(1-2)d
(1-1.25) d (minimum "d"). (0.75-1)d
The height of the intermediate batten plates (1-2)d
(0.75-1)d
ranges from (0.75-1) d (minimum "0.75d").
(1-2)d
The distance between batten plates Lz = (1-2) d
(1-1.25)d

Calculation of forces on batten plates:

The force affecting batten plate = Q* = Q + 0.02N (as case of lacing bars)
If the system has 2 plans of batten plates, F = Q* / 2
If the system has one plan of batten plates, F = Q*

B.C-3-19/49
 Beam-Column 3

For system with 2 plans of batten plates:

Q /4
2
Q /2
d
2 3
a a
d
2 R
1
Q /4 Q /4 Q /4
Q*/2 Q*
The system is said to be anti-symmetric, so reaction at joint 1 is =
2 4
x = 0, so reaction at joint 2 = Q* / 4
d Q* Q * .a
M1 = 0, R* = *a R=
2 4 2d
Where: tmin = d /50 Ecp 136 ( as lacing bars)
d is the distance between centerlines of column sections
"a" is distance between assumed joints = Lz + h batten pl d
The forces affecting batten plate are 2
M = R*d/2 and Q = R
R

The batten plate may be welded or bolted to column

Checks:
M h
1- Check flexure stress: ≤ 0.72 F y
Ix 2
h h
3
t.h
Where: Ix = for welded batten plate
12
3
t t
t.h
Or Ix = - I φ for bolted batten plate
12
3Q
2- Check shear: q = (we use total area for shear as given in ECP13
2 h.t

B.C-3-20/49
 Beam-Column 3

Design of connection between batten plate and column:

This connection is subjected to torsion and shear
Qy = R and Mt = R* d / 2

1- Bolted connection:
The height of the batten plate is taken (0.75-1) d
Mt
So we arrange bolts with edge 2d and pitch (3-4) d M r
t 2
r
The critical bolt (the upper bolt) is subjected to: Q Q
n
Vertical force V= Qy /n
Mt
Horizontal force = H= * y1
∑r2
Where ∑ r 2 = ∑ y 2 and y1 is the ordinate of the critical bolt.

H 2 + V 2 ≤ Rleast Where R least is the smaller of Rb and RS.S

2- Welded connection:
The height of the weld = h batten plate Mt Q
1 1
b = flange of column x y
Assume size = 10 mm Mt
Important: assume C.G. is the same as that of
Q
the column.
Calculate Ip = Ix + Iy, also calculate Avl
Point 2 Point 1
Q Mt Mt Mt Mt
V= + *x2 H= * y2 V= *x1 H= * y1
Avl I p Ip Ip Ip

H 2 + V 2 ≤ 0.2 Fu H 2 + V 2 ≤ 0.2 Fu

B.C-3-21/49
 Beam-Column 3

Solved examples:
Example 1:
Design the shown roof column (N=7.00 t., M = 4.00 m.t.). Design also the
combined column using welded batten plates (N = 25.00 t., M = 29.00m.t. Q =
3.00 t.). Design also the batten plate and its welded connection to the column
section. Recheck the batten plate if it is bolted to the column and check the M16
grade 4.6 bolts

1.5
roof
column 2.5

Combined 6.5
column

Fixed
base

Solution:
1.5
Q Moment at the partially fixed connection is not
2.5
given, so we can assume moment = zero (hinged
connection) and the bending moment is as shown
6.5

) Roof column:
400
Assume f = 1.00 t/cm2 .·. Sx = = 400cm 3
1.00
Choose IPE 270
Check:
1.5
lbin= 1.5(2.5 + ) = 4.875m lbout=2.5m
2

487.5 250
x= = 43.5 < 180 y= = 82.8 < 180
11.2 3.02

Fc =1.4-6.5*10-5(82.8)2=0.95 t/cm2

B.C-3-22/49
 Beam-Column 3

N 7
Fca= = = 0.153t / cm 2
A 45.9
f ca 0.153 7500
= = 0.16 > 0.15 FEX = = 3.96t / cm 2
fc 0.95 2
(43.5)
0.85
A1 = = 0.88 < 1.00 Take A1 = 1
0.153
1−
3.96
400
Fbx = = 0.932t / cm 2
429
c 13.5 / 2
Check compactness: Flanges = = 6.6 < 10.9 compact
tf 1.02

web assume compact . dw = 27 – 4*1.02 = 23.46 cm

1 7
= ( + 1) = 0.59
2 23.46 * 0.66 * 2.4
d w 23.46 699 / 2.4
.·. [ = = 34.73 ] < = 67.6 compact
tw 0.66 13 * 0.59 − 1

20b f 20 * 13.5
Lu max = = = 174.3cm < Luact = 250cm
Fy 2.4

.·. The cross Section may be non-compact.

800 AF 800 * 13.5 * 1.02
F1tb1= .cb = * 1.0 = 1.632 > 1.4 t/cm2
Lu d 250 * 27

.·. Fbcx = 1.4 t/cm2 (no lateral torsional buckling)

0.932
0.16+ *1.00 = 0.83 < 1.00
1.4
.·. Safe and economic.

) Design of combined column:

650
Assume 2 channel spaced d = = 50 cm.
12 − 15
N M 25 29
.·. Force on 1channel = + = + = 70.5ton
2 d 2 0.5

B.C-3-23/49
 Beam-Column 3

70.5
Assume f = 1.2 t/cm2 .·. A1 [= = 58.75cm 2
1.2
d 50
choose 2 channels 300 = = 1.67 .·. o.k.
h 30
Check:

lbin = 1.5*6.5 = 9.75 m lbout = 6.50 m

Iy[ ] = 2[495+58.8(25)2] = 74490 cm4 A[ ] = 2*58.8 =117.6 cm2

Iy 74490 d
rx[ ]= rx[ = 11.7 cm ry[ ]= = = 25.17 ~ = 25cm
A 117.6 2

650 975
out = x = = 55.55 /
in = /
y = = 38.7
11.7 25.17

Assume Lz = 2d=100 cm rZ= r y of single channel = 2.9 cm

L z 100
z = = = 34.5 < 60 ∴ in = (38.7) 2 + (1.25 * 34.5) 2 = 57.94
rZ 2.9

2
z = 34.5 < * 57.94 = 38.62 ok
3
Fc = 1.4-6.5*10-5(57.94)2 = 1.18 t/cm2
25
fca = = 0.213t / cm 2
117.6
f 0.213 7500
( ca ) = = 0.18 > 0.15 FEy = 2
= 2.23t / cm 2
tc 1.18 (57.94)

C my 0.85
A2 = = = 0.94 < 1.00 Take A2 = 1
f 0.213
1 − ca 1−
t Ey 2.23

My d 2900
fbcy = ( + e[ ) = (25 + 2.7) = 1.08t / cm 2
I y[] 2 74490

Fbcy = 1.4 t/cm2

1.08
Applying in the interaction equation: (0.18) + ( )(1.00) = 0.95 < 1.00
1.4
.·. Safe & economic.

B.C-3-24/49
 Beam-Column 3

) Design of batten plate:

2
N = 25 t. Q=3t .·. Q = 3 + (25) = 3.5t.
100

For using 2 channels, there is 2 plans of batten plate

Dimension of batten plate:
d 50
h = (0.75 1) d = 40 cm. Assume t= = = 1cm.
50 50
a = Lz + h = 100+40 = 140 cm
M1 = zero Q /4
.·. R*25 = (3.5/4)*140 .·. R = 4.90 ton. 2

Q * .a 3.5 * 140
Or R = = = 4.9 t ECP 136 3
nd 2 * 50 a
d
M = R * d/2 = 4.9*50/2 = 122.5 cmt 2 R
Q * a 3.5 * 140 1
Or M = = = 122.5 cmt ECP 136
2n 2*2 Q /4

Very important note: If you want to use the equations of the ECP directly, you

must note that, the definition of "a" and "d" is the opposite ( ). This

means the code says:

Qd Q *a
The longitudinal shear force = , so R =
na nd
Qd Qa
The moment = , so M = (n is the number of batten plate plans)
2n 2n
Straining action on batten plate:
Q = R = 4.90 t. M = 122.5 cmt
(1.0)(40) 3
If the batten plate is welded: Ix = = 5333.3cm 4
12
M h 122.5
f= = * 20 = 0.46t / cm 2 < 0.72 * 2.4 = 1.73t / cm 2
I x 2 5333.3

B.C-3-25/49
 Beam-Column 3

3Q 3 4.90
q= = * = 0.12t / cm 2 < = 0.84t / cm
2 A 2 1.0 * 40

Check on weld: Qy = 4.90 t. Mt = 122.5 cm.t

Assume size of weld = 10 mm

Avl = 40*1 = 40 cm2
10
1
3
(1)(40)
Ix = + 2(1)(10)( 20.5) 2 = 13738cm 4
12 40
(1)(10) 3 e=2.7
Iy = 2[ + (1)(10)(5 − 2.7) 2 ] + (1)(40)( 2.7 − 0.5) 2 = 466cm 4
12
IP = Ix + Iy = 14204 cm4
122.5 40
Point (1): q1x = 0 + * ( + 1) = 0.18t / cm2
14204 2
122.5
q1y = 0 + * (10 − 2.7) = 0.063t / cm 2
14204

q1 max = (0.063)2 + (0.18)2 = 0.19t / cm2 < 0.2Fu = 0.72t / cm

122.5 40
Point (2): q2x = 0 + * ( + 1) = 0.18t / cm 2
14204 2
4.90 122.5
q2y = + * (2.7) = 0.146t / cm 2
40 14204

q2 max = (0.146)2 + (0.18)2 = 0.23t / cm 2 < 0.2Fu = 0.72t / cm

0.23
Snew = 10* = 3.2 take s = 5mm
0.72

If the batten plate is bolted: Dim of plate – as before.

Assume e = 2*1.6 = 3.2 cm
40 − 2 * 3.2
40
Number of pitches = = 5.25 Taken 5 bolts (6 bolts)
4 * 1.6
40 − 2 * 3.2
Pitch = ≈ 6.6 cm and e= 3.5 cm
5
/ 2
1
Area of hole = d t= 1.8*1 = 1.8 cm

B.C-3-26/49
 Beam-Column 3

(1.0)(40) 3
Ix = − 2 * 1.8(3.3 2 + 9.9 2 + 16.5 2 ) = 4157cm 4
12
M h 122.5
f= = * 20 = 0.59t / cm 2 < 0.72 * 2.4 = 1.73t / cm 2
I x 2 4157
3Q 3 4.90
q= = * = 0.12t / cm 2 < = 0.84t / cm (use gross area)
2 A 2 1.0 * 40

Check on bolt: Qy = 4.90 t. Mt = 122.5 cm.t

∑ r 2 = 2 (3.3 + 9.9 + 16.5 ) = 762.3 cm

2 2 2 2
1
Bolt (1): V= 4.9 / 6 = 0.82 t M
t
122.5 Q
H= * 16.5 = 2.65 t
762.3

2 2 π * 1.6 2
R= 2.65 + 0.82 = 2.77 t > Rss = 0.25*4 = 2.01t
4
So we have to increase dimensions of the plate to be 50 cm instead of 40 cm and
increase number of bolts, and then recheck.

B.C-3-27/49
 Beam-Column 3

Example 2:
For the shown main system with spacing = 7.00 m
It is required to:
1- Design a built-up section for column 1using section composed of 4 angles.
2- Design also its welded lacing bars and its connection to the column.
3- Redesign the lacing bars if it is bolted to the column as well as its connection.
4- Redesign bolted lacing bars with horizontal member and its connection.
5- Redesign the lacing bars using double intersection system assuming welded
connection.
6- Redesign the lacing system using lacing angles assuming bolted connection.
IPE400

10 10
Col 1
7.0m

1.5
5.0

25.0m

Solution:
To get reactions from crane girder
Dead load: Assume o.w. of crane girder = 0.2 t /m'
RD.L. = 0.7*2 = 1.40 ton
10 10
5.50
Live load: RL.L. = 10 + 10 * = 17.86ton
7.50
1.5 5.5
* Assume impact factor I = 25% R

.·. RL.L.+ I = RL.L. (1+I) = 17.86*1.25 = 22.32 t.

1
* Lateral shock = of wheel load without impact
10
1
(Px) = RL.L. =1.79 ton.
10

B.C-3-28/49
 Beam-Column 3

1 1
* Total braking force = vertical loads without impact = (2 *10) = 2.86ton.
7 7
Since bracing is not given, so we can assume a suitable bracing system that the
braking force is resisted by the column and the column is not subjected to My

1- Straining actions on column (1):

B
Py = RD.L. + RL.L. +I = 1.40 + 22.32 = 24.22 ton 2 B
2 =1.43

Px = 1.79 ton, M = Px * h = 1.79*5.00 = 8.93 m.t.

Choice of section:
500
Assume d = = 35 cm.
12 − 15
35
Assume h = = 25 cm
1.5
N M 24.22 8.93
.·. F one side (on 2 angles) = + = + = 37.62t
2 d 2 0.35
Force on one angle = 37.62 / 2 = 18.81t
Assume f = 1.20 t/cm2 A1L = 18.81 / 1.2 = 15.7 cm2
Choose angles 90*9 too large for h = 25 cm.
Take d = 40 cm and h = 30 cm and use angles 90*9 y
or any other estimation
Check: x x 30 cm

A4Ls = 4AL = 4*15.5 = 62 cm2 y

40 cm
I y = 4 [116 + 15.5 * (20)2] = 25264 cm4
4 angles 90*9
I x = 4 [116 + 15.5 * (15)2] = 14414 cm4
25264 14414
ry = = 20.19 ≈ 20 cm rx = = 15.25 ≈ 15 cm
62 62
24.22
* f ca = = 0.39 t / cm2
62
893
* f bx = * ( 20 + 2.54) = 0.8 t / cm 2
25264
B.C-3-29/49
 Beam-Column 3

* Fc :
Lbin = 2.10*5.00 = 10.5 , Lbout = 5.00 m
1- Global buckling:-
1050 500
λ/y = = 52 λ/x = = 32.8
20.19 15.25
2- Local buckling:-
Shape of lacing bars:
70
So L Z = 35 cm ≈ d ≈ h 35 cm
70
r min = r v of single angle = 1.76 cm
35
λz = = 19.9 < 60 40
1.76 30
Elevation Side view
3- Overall buckling:-

λ y = 52 2 + 19.9 2 = 55.67

λ x = 32.8 2 + 19.9 2 = 38.36

2
λZ = 19.9 < * 55.67 = 37.1 O.K.
3
F C = 1.4 – 6.5 * 10-5 * (55.67)2 = 1.2 t / cm2
* A2:
f ca 0.39
= = 0.32 > 0.25
FC 1.2

7500
FEX = = 2.42 t / cm2
55.67 2
0.85
A2 = = 1.01
0.39
1−
2.42
* Fbcy = 1.4 t/cm2
Applying in the interaction equation:
0.8
0.32 + * 1.01 = 0.9 < 1.2 (case of lateral shock)
1.4
So we may try 4 angles 80 * 8 to get an economic sec.

B.C-3-30/49
 Beam-Column 3

2- Design of lacing bar as single intersection system:

2
Q = 1.786 t , N = 24.22 Q* = Q + N = 2.27t
100

Q*
Force in lacing bar F = + = tan-1 (35/40) = 410
2 cosα
2.27
.·. F = + = 1.5t
2 * 1 * cos 41
Dimension of lacing bar:
d 40 l
l= = = 53cm Assume b = = 5.00cm.
cos 45 cos 41 10
l 53
Thickness of lacing bar t. = = = 1.06cm Taken 12mm
50 50
Check of lacing bar as comp. member:
t 1.2
lbx = l = 53 cm rx = = = 0.35cm
12 12
lbx 53
.·. x = = = 151.5 > 140
rx 0.35

Increase (t) t = 1.4 cm.

1.4 53
rx = = 0.4 x = 132.5 < 140
12 0.4

7500 1.5
Fc = = 0.43 t/cm2 fact = = 0.21t / cm 2 < 0.43 t/cm2
132.5 2 5.00 * 1.40

Check as tension member:

1.5 2
fact = = 0.21t / cm 2 < 1.4 t/cm no need for it in case of welded lacing
5.00 *1.40
l 53
= = 10.6 << 60
b 5.00
Design of welded connection:
Assume size of weld s = 6 mm (min. size in case of dynamic loads)

F 1.61
lact = + 2S = + 2 * 0.6 = 3.06cm take lact = 5.00 cm.
2 * S (0.2 Fu ) 2 * 0.6 * 0.72

B.C-3-31/49
 Beam-Column 3

3- If lacing bars are bolted to the column: Assume use M16 grade 4.6
For dimensions, the same as in case of welded except b= 10cm > 3*1.6= 4.8 cm
because this is the minimum edge from both sides.
Check compression is the same as in welded
1.5
Check tension: f = = 0.34t / cm 2 < 1.4 t/cm2
5 * 1.4 − (1.8 * 1.4)

l 53
= = 10.6 << 60 .·. o.k. The same as in welded
b 5.00

Design of the bolted connection:

tmin = 14mm of lacing or 9mm of angle 90 Rb = 0.8*3.6*1.6*0.9=4.1t

π * 1.6 2
Rss = 0.25*4 = 2.01t < 4.01t, so R least =2.01t > 1.5t (force on lacing)
4
Ok safe use one M16 bolt

4- If lacing bars are bolted to the column and have horizontal member:
The diagonal will be the same as previous
The horizontal member:

Q * 2.27
F= = = 1.135 t
2 2
lin = lout = d = 40 cm
Dimensions of horizontal bar:
l = 40 cm, b= 40/10 = 4 cm, tmin= 40/50 = 0.8 cm
Check as compressions member:
0.8
r= = 0.23 cm
12
40 1
= = 173 > 140 Take t = 1 cm, r= = 0.29 cm
0.23 12
40
= = 138 < 140
0.29

B.C-3-32/49
 Beam-Column 3

7500 1.135
Fc = = 0.39 t/ cm2 fca = = 0.28 t/ cm2 < 0.39 t/ cm2
138 2 4 *1

Check as tension member (bolted)

l 40
= =10 < 60
b 4
1.135
f= = 0.51 t/ cm2 < 1.4 t/ cm2 o.k.
4 * 1 − (1.8 * 1)

5- Design of lacing bar as double intersection system :( welded connection)

Q*
Force in lacing bar F = + = tan-1 (35/40) = 410
2 * 2 cos α
2.27
.·. F = + = 0.75t
2 * 2 * cos 41
Dimension of lacing bar:
d 40 l
l= = = 53cm Assume b = = 5.00cm.
cos 45 cos 41 10
l 53
Thickness of lacing bar t. = = = 1.06cm Taken 12mm
50 50
Check of lacing bar as comp. member:
t 1.2
lbx =0.7 l = 0.7*53 = 37.1cm rx = = = 0.35cm
12 12
lbx 37.1
.·. x = = = 106 < 140
rx 0.35

7500 0.75
Fc = = 0.67 t/cm2 fact = = 0.125t / cm 2 < 0.67 t/cm2
106 2 5.00 * 1.20

Check as tension member welded:

0.75 2
fact = = 0.125t / cm 2 < 1.4 t/cm
5.00 *1.20
l 53
= = 10.6 << 60
b 5.00

B.C-3-33/49
 Beam-Column 3

6- Design of lacing angles: (bolted connection)

2
Q = 1.786 t , N = 24.22 Q* = Q + N = 2.27t
100

Q*
Force in lacing bar F = + = tan-1 (35/40) = 410
2 cosα
2.27
.·. F = + = 1.5t
2 * 1 * cos 41
d 40
Dimension of lacing angle: l= = = 53cm = Lb
cos 45 cos 41
Assume minimum angle 55*5 (using bolt M16) or choose as comp member
Check as compression member:
53
v = =49.53 < 140
1.07
Fc = 0.6 * [1.4 – 6.5*10-5(49.53)2] = 0.74 t/ cm2
1.5
fca = = 0.28 t/ cm2 < 0.74 t/ cm2
5.32

Check as tension member:

A1 = (5.5 – 1.8) 0.5 = 1.85 cm2 A2 = (5.5 – 0.5) 0.5 = 2.5 cm2
3 * 1.85
Anet = 1.85 + *2.5 = 3.6 cm2
3 * 1.85 + 2.5
1.5
f= = 0.42 t/cm2 < 1.4 t/ cm2
3.6
l 53
= = 9.6 < 60
d 5.5

B.C-3-34/49
 Beam-Column 3

Example 3:
For the shown figure, it is required to design the B.U. column using the given
section (4 angles and 2 plates). The straining actions are M=30mt and N=40t.

1.5
roof
column d
1 2.5

d
2
Combined 6.0
column
Proposed
Fixed section
base
Solution:
6.0 d1
Assume d1 = = 40 cm Take d2 = = 25 cm
12 − 15 1.5 → 2
d1
Note economic section = 1.5 2
d2

You can assume any other reasonable dimensions estimation of section:

Assume F c = 1.2 t / cm2
3000 40 95
Force on one side: + = 95 t Area of one side: = 79 cm2
40 2 1.2
Assume h p ≈ 25 cm & t p = 1 cm
Actual h of plate = 25 + 2 e (but we don't know the angles yet)
79 − 25
Area of one angle = = 27 cm2 use angle120 Too large
2
Spacing is 25 cm & angle is 120 mm is not logic
So we have to increase the estimated dimensions & take thicker plate
Assume d 1= 50 cm & d 2 = 30 cm
3000 40 80
Force on one side = + = 80 t Area = = 66.7 cm2
50 2 1.2
h p ≈ 30 & t p = 15 mm
66.7 − 30 * 1.5
∴ Area of one angle = = 10.85 cm
2

B.C-3-35/49
 Beam-Column 3

Use angle 75 * 7 (we may use 80*8)

h plate = 30 + 2 * 2.03 = 34 cm
1.5 2
I in = 4 [52.4 + 10.1* 252] + 2 *(34*1.5) (25 + 2.03 + ) = 104176 cm4
2

1.5 * 34 3
I out = 4 [52.4 + 10.1* 152] + 2 * = 19126 cm4
12
A = 4 * 10.1 + 2 * 34 * 1.5 = 142.4 cm2
104176 19126
rin = = 27 cm rout = = 11.6 cm
142.4 142.4

/ 1.5 * 600 600

λin = = 33.3 λout = = 51.7
27 11.6
Assume using lacing bar with angle 450
L z = 2 * 50 cm = 100 cm (2 d)
To calculate rmin for the part buckles alone:
yX
1.5
34 * 1.5 * + 2 * 10.1 * (1.5 + 2.03)
x= 2 = 1.54 cm
2 * 10.1 + 34 * 1.5 x x

34 * 1.53 1.5
Iy= + 34 * 1.5 (1.54 - ) 2 + 2 [52.4 y
12 2
part behaves
+ 10.1 * (1.5 + 2.03 – 1.54)2] = 226.19 cm4 lonely

226.19 100
rmin = = 1.78 cm λz = = 56.2 < 60
2 * 10.1 + 34 * 1.5 1.78

λin = 56.2 2 + 33.3 2 = 65.3

2
λ z = 56.2 > * 65.3 = 43.5 Unsafe
3
So we can use horizontal member. Lz = d = 50 cm
50
λz = = 28.1 λin = 28.12 + 33.3 2 = 43.6
1.78
FC = 1.4 – 6.5 * 10-5 * 51.72 = 1.23 t / cm2
40
f ca = = 0.28 t / cm2
142.4

B.C-3-36/49
 Beam-Column 3

f ca 0.28
= = 0.23 > 0.15
FC 1.23

0.85
A2 = = 0.91 < 1 take A2 = 1
0.28
1−
7500 / 43.6 2
Applying in the interaction equation:
3000
* ( 25 + 2.03 + 1.5)
0.23 + 104176 = 0.82 < 1 O. K.
1.4

Example 4: L3
Calculate Lz for the shown section. The section
L1 30
consists of 4 angles as shown. The distance is 60
and 30 cm. L2
60

Solution:
L1 L2 L3

60 cm
Lz = 60 cm
120 cm

Point 1 is fixed by the lacing bar in the side

Pt1 Pt1
view
60 cm

Always for 4 angles we can use Lz = d

60 cm 30 cm
Elevation Side view

B.C-3-37/49
 Beam-Column 3

Example 5:
For the shown frame it is required to:
1. Suggest a suitable bracing system

7m
2. Design column using the shown section
using batten plate 70cm
5t
3. Calculate force affecting batten plate 25 m
20 t
4. If lacing bars are used, calculate the
maximum force.
Solution:
1. Bracing system is as shown

3.5m
2. M = 5 * 7 = 35 mt N = 20 t

3.5m
C = (20/2) + (35/0.7) = 60 t
Area of 2 angles = 60 / 1.2 = 50 cm2
6.0m
Area of 1 angle = 50/2 = 25 cm2
Use 4 angels 120 * 12
A = 4 * 27.5 = 110 cm2
Iy = 4 (368 + 27.5 * 352) = 136222 cm4
Assume thickness of batten plate = 70/50 = 1.4 cm (tmin)
  1.4  
2
Ix = 4368 + 27.5 *  3.4 +   = 3321 cm4
  2  

136222 3321
ry = = 35.2 cm rx = = 5.5 cm
110 110
To calculate buckling lengths:
Q It's a frame, so we have to use alignment chart
Assume Ic = 2-3 Ig
So GA = 10 (hinged base)
Ic / 7
GB = = 7.14 → K = 2.8
0.5 I c / 25

lb in = 2.8 * 7 = 19.6 m , lb out = 3.5 m

B.C-3-38/49
 Beam-Column 3

1960 350
λ'in = = 55.7 , '
λout = = 63.6
35.2 5.5
Assume Lz = d = 70 cm (use batten plate each 70 cm)
rmin = rv = 2.35 cm
70
λz = = 29.8 < 60
2.35

λin = 55.7 2 + (1.25 * 29.8) 2 = 67

λin = 63.6 2 + (1.25 * 29.8) 2 = 73.7

λ z = 29.8 < (2/3)* 73.7 = 49.1

Fc = 1.4 – 6.5 * 10-5 (73.7)2 = 1.05 t/cm2

fca = 20/110 = 0.18 t/cm2
f ca
= 0.18/1.05 = 0.17 > 0.15
Fc

FEX = 7500/672 = 1.67 t/cm2

0.85
A1 = = 0.95 < 1 take A1 = 1
0.18
1−
1.67
3500
fbx = * (35 + 3.4) = 0.99 t/cm2
136222
Applying in the interaction equation
0.17 + (0.99 / 1.4) = 0.88 < 1 ok

3. There is one plan of batten plates n=1

Assume hpl = (0.75 → 1) d = 60 cm
Q
a = lz + hpl = 70 + 60 = 130 cm d
2
d = 70 cm a
Q* = 5 + 0.02 * 20 = 5.4 t
5.4 * 130
Q=R= = 10 t
1 * 70
Q /2 Q /2

B.C-3-39/49
 Beam-Column 3

5.4 * 130
M= = 351 cmt
2 *1

F1
4. If using lacing bars:

Q* = 5.4 t , α = 45 o Q F2

45°
Q
5.4
F1 = = 7.64 t
1 * 1 cos 45
F2 = Q* = 5.4 t

B.C-3-40/49
 Beam-Column 3

Example 6:
The given figure shows the main structural system of an industrial building:
1:10
IPE 400
3.0 m

shape 1

7.0 m col 1
H
V
30 shape 2

Design column "1" in the following cases:

1- Design a BU section for column "1" if we use portal frame bracing outside.
Use shape 1 for design.
2- Redesign a BU section for column "1" if we use portal frame bracing outside.
Use shape 2 for design.
Data :Crane girder wheel loads are 15t each and spaced 2m.
Spacing between frames = 6 m

B.C-3-41/49
 Beam-Column 3

Solution
The crane is simply supported over columns
Assume ow = 0.15 t/m 15 15
RLL = 15+15*4/6 = 25 t
2 4
Py = 25 ×1.25 + 0.15 * 6 = 32.15 t (max) R

Px = 0.1× 25 = 2.5 t PD = 0.15 × 6 = 0.9t (min)

For column "1" Inside plan [cantilever]

32.15 t 32.15 t 0.9 t 32.15 t

0.4m 0.4m 0.4m 0.4m
2.5 t 2.5 t 2.5 t

BFIB 360 BFIB 360

IPE 500 IPE 500

5.0m 5.0

Case of max normal Case of max moment

Case I : Max moment & corresponding normal
N = 32.15 + 0.9 = 33.05 t
Mx = (32.15 - .09) 0.4 + 2.5 (5 + 0.5 + 0.36) = 27.15 mt
Case II: Max normal & corresponding moment
N = 32.15 * 2 = 64.3 t

Mx = 2 ×2.5 (5 + 0.5 + 0.36) = 29.3 mt

Note that we have to design the column twice for each case BUT
Q For case II Mx2 >Mx1 & N2 >N1 ∴ Case II is more critical

42/49-B.C-3
2007-2008
 Beam-Column 3

1) BU of shape 1 using portal frame bracing outside:

2 × 15
B= = 4.28t
7

2B = 4.3 x 2 = 8.6 ton , will be carried by 2 columns.

Inplan:
From case of studying the column inside plan (cantilever) same as previous
case.
N = 32.15 * 2 = 64.3 t
4.3
4.3
Mx = 2 ×2.5 (5 + 0.5 + 0.36) = 29.3 mt My

Outplan:

My = 4.3 ×5 = 21.5 mt (outside plan)

4.3 4.3
6.0 m
So we must design 2 sections or design one & check the other
The 2 sections are:
(1) one at the Base Mx & N
(2) at top of column My & N
Estimation of section:
Fbcx = Fbcy = 1 t/cm2
2930 2150
Sx IPE 1 = = 2930 cm3 Sx IPE 2 = = 2150 cm3
1 1
Use IPE 600 Use IPE 550
Properties of section:
A = 134 + 156 = 290 cm2
IY (inside) = 92080 + 2670 = 94750 cm4 IPE500
55 y
134 *
IX (outside) needs y y= 2 = 12.7 cm IPE600
290
IX (outside) = 3390 + 156 * 12.72 + 67120 + 134 (55/2 – 12.7)2 = 125023 cm4
B.C-3-43/49
 Beam-Column 3

94750 125023
rYin = = 18 cm , rxout = = 20.7 cm
290 290
lb in = 2.1*5 = 10.5 cm (cantilever)
Outside → Allowed to sway with Rigid connections because it's frame
∴ We must use GA & GB

QInertia of the girder is not given we have to assume it reasonably

Assume Ic = 2Ig
∴ GA = 10 (hinged outside)

I c 15 (2 I g ) / 5
GB = = = 2.4
I g 16 Ig /6

From chart (Allowed to sway)

K =2.2 ∴ Lout = 2.2 ×5 = 11m

1050 1100
in = = 58.3 out = = 53.14
18 20.7
∴ Fc = 1.4 – 6*10-5 (58.3)2 = 1.18 t/cm2
fca = 64.3 / 290 = 0.22 t/cm2
f ca 0.22
= = 0.19 > 0.15
Fc 1.18

Use hl stiffener every 1m or 1.5

Lu act = 1 m or 1.5
20 * b Stiff
Calculate Lu max =
2.4
Check for section 1: (at base)
Mx = 29.3 mt, N = 64.3 t and My = zero M
x
Where b = 22 cm (inside for section at base)
20 * 22
Lu max = = 284 cm > 100 or 150 cm
2.4
∴ NO LTB (We can neglect this check as it is always no LTB)

B.C-3-44/49
 Beam-Column 3

Check compactness:
IPE600
C 0.5(22 − 1.2 − 2 * 1.9)
Check = = 4.47 <10.9 compact
tf 1.9

Web of IPE 600 is stiffened by the flange of IPE 550, so it is compact

IPE550
C 0.5(21 − 1.11 − 2 * 1.72)
= = 4.78 <10.9 compact
tf 1.72

1 N 
For web of IPE 550: Get α =  + 1
2  d w * t w * F y 
134 M Stiff
N IPE 500 = 64.3 * = 29.7 t y
134 + 156

dw = 0.5* (55 – 4×1.72) = 24.06cm.

1 29.7  699 / 2.4

α=  + 1 = 0.73 Limit = = 53.1
2  24.06 * 1.11 * 2.4  13 * 0.73 − 1
d w 24.06
= = 21.7 < 53.1
tw 1.11

Wes and flanges are compact, so Fbcx =Fbcy = 0.64Fy

2930
fbcx = * 30 = 0.93 t/cm2
94750
7500
FEx = = 2.2
58.32
0.85
A1 = = 0.94 A1 = 1
0.22
1−
2.2
Applying in the interaction equation:
0.19 + 0.93/1.536 = 0.79 < 1.2 (section at base) ok
If unsafe: use IPE 600 in both directions

B.C-3-45/49
 Beam-Column 3

Check for section 2: (at base) Mx = zero, N = 64.3 t and My = 21.5mt

2150 2 7500
f bcy = * (55 − 12.7) = 0.73 t/cm FEy = = 2.65
125023 53.142
0.85
A2 = = 0.93 A2 = 1
0.22
1−
2.65
Applying in the interaction equation:
0.19 + 0.73/1.536 = 0.66 < 1.2 (section at top) ok

2- If we use "shape 2":

Estimation of section:
Fbcx = Fbcy = 1 t/cm2
2930 2150
Sx IPE 1 = = 2930 cm3 Sx IPE 2 = = 2150 cm3
1 1
Use IPE 600 Use half IPE 550

Properties of section:
A = (1/2) * 134 + 156 = 223 cm2
Mx
I xin = 92080 + 0.5 * 2670 = 93415 cm4
Stiff
25.8
IPE 600

25.8 * * 1.11 + 1.72 * 21 * 26.6

y= 2 = 5.97cm
223 1 IPE 550
2
25.8 3
Iy = 3390 + 1.11 * + 1.11 * 25.8 (25.8 – 5.97)2 +
12
21*1.72 (26.2)2 = …..

B.C-3-46/49
 Beam-Column 3

Head PL. Crane girder

Head PL.

Stiff PL. Packing Stiff.Pl

Bracket

Portal Frame Bracing Rafter

Stiff PL.

Stiff PL.

Stiff PL.

B.C-3-47/49
 Beam-Column 3

Head PL. Crane girder

Head PL.

Stiff PL. Packing Stiff.Pl

Bracket

Portal Frame Bracing Rafter

Stiff PL.

Stiff PL. Stiff PL.

B.C-3-48/49
 Beam-Column 3

Head PL. Crane girder

Head PL.

Stiff PL. Packing Stiff.Pl

Bracket

Portal Frame Bracing Rafter

Stiff PL.

Stiff PL.
Stiff PL.

B.C-3-49/49
 Examples-Column 4

Very important solved example:

The given figure shows the main structural system of an industrial building:

1:10
IPE 400
Vc
3.0 m
H
c

col 2 5.0 m col 1

H
V
30

It is required to:
a) Design column "1" in the following cases:
1- Design a rolled section for column "1" in the following cases using bracing
system and calculate the loads on the bracing system you suggested.
2- Redesign a rolled section for column "1" using portal frame bracing. If
unsafe, try BU section.
3- Redesign a BU section for column "1" (2 IPE with batten plates) using no
bracing at all (Cantilever)

b) Design the welded and bolted connection between the double crane bracket
and column "1" using pretensioned bolts M24 grade 8.8 for the case of using
portal frame bracing outside.

c) Design column 2 (the column is braced outside) in the following cases

1/24-B.C.4
 Examples-Column 4

1- Using horizontal member at the level of crane girder.

2- Without using horizontal member at the level of crane girder.
Data: :
Crane girder wheel loads are 15t each and spaced 2m.
Spacing between frames = 6 m
Assume brackets are IPE 500, the crane girder BFIB360, distance between
centerline of crane girder and centerline of column is 40 cm. Assume braking
force is to be carried by 2 columns.
The total reactions due to roof loads and crane are as follows
H = 6t and V = 38 t

Solution:
The crane is simply supported over columns
Assume ow = 0.15 t/m 15 15
RLL = 15+15*4/6 = 25 t
2 4
Vc = 25 ×1.25 + 0.15 * 6 = 32.15 t (max) R

Hc = 0.1× 25 = 2.5 t QD = 0.15 × 6 = 0.9t (min)

For column "1" Inside plan [cantilever]

To calculate Mx:

2/24-B.C.4
 Examples-Column 4

32.15 t 32.15 t 0.9 t 32.15 t

0.4m 0.4m 0.4m 0.4m
2.5 t 2.5 t 2.5 t

BFIB 360 BFIB 360

IPE 500 IPE 500

5.0m 5.0

Case of max normal Case of max moment

Case I : Max moment & corresponding normal

N = 32.15 + 0.9 = 33.05 t
Mx = (32.15 - .09) 0.4 + 2.5 (5 + 0.5 + 0.36) = 27.15 mt
Case II: Max normal & corresponding moment
N = 32.15 * 2 = 64.3 t

Mx = 2 ×2.5 (5 + 0.5 + 0.36) = 29.3 mt

Note that we have to design the column twice for each case BUT
Q For case II Mx2 >Mx1 & N2 >N1 ∴ Case II is more critical

a) 1) using Bracing system :

15
B= = 2.14t
7

2B * 2 = 4B = 2.14 × 4 = 8.6 t

For h ≥ 5m use horizontal member at mid height

3/24-B.C.4
 Examples-Column 4

2B 4B = 8.6 t F1
2B
F2
F3
5.0

6.0

-1 5
θ = tan = 39.8o
6
2F2 cos θ = 8.6 F2 = 5.6 t
F1 = 5.6 cos θ = 4.3t F3 = zero

Design the 3 previous members as Tension & compression

M1 → Star shape Lin = Lout = 6m

M2 → 2 angles back to back Lin = 0.5l Lout = 0.75l

2 2
Where l= 5 + 6 = 7.81m

Design of column:
Estimation of section Assume Fbcx = 1 t/cm2
29.3 × 100 3
=1 ⇒ Sx = 2930 cm Use IPE 600
Sx
64.3 2
fca = = 0.41t / cm
156

Fc ⇒ Lin = 2.1×5 = 10.5m Lout = 2.5 m

1050 250
λ in = = 43 λ out = = 54
24.3 4.66

Fc = 1.4 – 6.5 ×10-5 * 542 = 1.2 t/cm2

4/24-B.C.4
 Examples-Column 4

f ca 0.41 2930
= = 0.34 > 0.15 fbx = = 0.95 t/cm2
Fc 1.2 3070
1
11 − × 1.2 − 1.9
C 2
Fbcx ⇒ = = 4.47 <10.91
tf 1.9

dw = 60 – 4 ×1.9 = 52.4 cm

1 64.3
= [ + 1 ] = 0.71 > 0.5 (compression member)
2 52.4 × 1.2 × 2.4
699 d w 52.4
= 54.6 = = 44 < 54.6
(13α − 1) 2.4 tw 1.2

∴ The section is compact

Luact = 2.5 m
20 × 22
Lumax = = 284cm
2.4
= -0.5 (for the suggested bracing system)

Cb = 1.75 – 1.05 ×0.5 + 0.3(0.5)2 = 1.3

0.5M
1380 × (22 × 1.9)
Lu max = × 1.3 = 520cm
60 × 2.4
2
∴ No LTB Fbcx = 1.536 t/cm M

0.85
A1 = = 0.94 → A1 = 1
0.41
1−
4.06
7500
Where FEx = 2
= 4.06 t/cm2
43

Applying Interaction Equation

0.41 0.95
+ * 1 = 0.96 < 1.2 (Because presence of lateral shock)
1.2 1.536
This is the most economic solution. So if we can use Bracing, we must use
bracing.

bracing

5/24-B.C.4
 Examples-Column 4

2) Using portal frame bracing:

p 15
Bracing force of 2 cranes are * 2 each side = 2× = 4.3t
7 7

My = 4.3 ×5 = 21.5 mt (outside plan)

4.3
From case of studying the column inside 4.3
My
plan (cantilever)
N = 32.15 * 2 = 64.3 t

Mx = 2 ×2.5 (5 + 0.5 + 0.36) = 29.3 mt

4.3 4.3
6.0 m
So we must design 2 sections or design one & check the other
The 2 sections are:
(1) One at the Base Mx & N
(2) At top of column My & N

Estimation of section:
Assume f = 0.8 t/cm2 [for there is My]
2930
= 0.8 ⇒ Sx = 3660 cm3
Sx

→ There is no IPE of Sx = 3660

→ There is Mx & My So use BFIB Use HEB 500

Lbin = 2.1 ×5 = 10.5m

Outside → Allowed to sway with Rigid connections because it's frame

∴ We must use GA & GB

QInertia of the girder is not given we have to assume it reasonably

Assume Ic = 2Ig
∴ GA = 10 (hinged outside)

6/24-B.C.4
 Examples-Column 4

I c 15 (2 I g ) / 5
GB = = = 2.4
I g 16 Ig /6

"N " :

"N "

∴ We assume Ic = (2 → 3) Ig (if it's not given)

From chart (Allowed to sway)

K =2.2 ∴ Lout = 2.2 ×5 = 11m

1050 1100
λ in = = 49.5 λ out = = 151.3 < 180
21.2 7.27
64.3 2 7500
fca = = 0.27 t/cm Fc = 2
= 0.33 t/cm2
239 151.3

f ca 0.27
= = 0.82 Too large
Fc 0.33

∴ Use BFIB 600

1050 1100
λ in = = 41.7 λ out = = 156
25.2 7.07

7500 2 64.3 2
Fc = 2
= 0.31 t/cm Fca = = 0.24 t/cm
156 270

∴ f ca =
0.24
= 0.77
Fc 0.31

1
15 − * 1.55 − 3
C 2
Fbcx ⇒ = = 3.74 <10.9
tf 3

dw = 60 – 4×3 = 48 cm

1 64.3
= [ + 1 ] = 0.68 >0.5
2 48 × 1.55 * 2.4

699 dw 48
= 57.5 = = 31 < 57.5
(13α − 1) 2.4 t w 1.55

7/24-B.C.4
 Examples-Column 4

Compact
Lu act = 5m
20 × 30
Lu max = = 387 cm
2.4
∴ There is L.T.B

= zero Cb = 1.75
800 × 30 × 3 2 2
fltb 1 = *1.75 = 4.2 t/cm ∴ Fbcx = Fbcy =1.4 t/cm
500 × 60
7500 2
FEx = 2
= 4.31 t/cm
41.7

0.85
A1 = = 0.9 taken =1
0.24
1−
4.31
→ Check for section at base Mx = 29.3 mt
2930 2
fbx = = 0.51 t/cm
5700
0.24 0.51
+ *1 = 1.14 <1.2
0.31 1.4
Check for section at top of column
My = 21.5 mt
7500 2 0.85
A2 ⇒ FEy = 2
= 0.31 t/cm A2 = = 3.84
0.24
156 1−
0.31
2150
Fby = = 2.7 >> 1.4 t/cm2
902
2150 2
∴ We must use BU.S.[ = 1.97 t/cm >> 1.4]
1090
Note that: 1090 is the Sy of HEB 1000 (biggest section is the tables)
So if we have portal frame outside we have to use the cross
column as shown:

2 IPE

8/24-B.C.4
 Examples-Column 4

The previous 2 pages are only to prove that we cannot use single I-beam if we
have portal frame outside. In the exam, if there is portal frame outside, use
cross column directly.

Estimation of section: Assuming Mx is carried by IPE in direction x and My is

carried by IPE in direction y. Assume Fbcx = Fbcy = 1 t/cm2
2930 2150
Sx IPE 1 = = 2930 cm3 Sx IPE 2 = = 2150 cm3
1 1
Use IPE 600 Use IPE 550
We have to treat the whole section as one part:
Properties of area:

IPE 600
A = A1 + A2 = 156 + 134 = 290 cm2 Stiff

Ix = Ix 1 + Iy 2 = 92080 + 2670 = 94750 cm4

Iy = Iy 1 + Ix 2 = 3390 + 67120 = 70510 cm4 IPE 550
94750 70510
rx = = 18.07 cm ry = = 15.6 cm
290 290
Fc: Lb in = 2.1*5 = 10.5 m
I c 15 (2 I g ) / 5
Lb out → GA = 10 (hinged base) GB = = = 2.4
I g 16 Ig /6

K =2.2 ∴ Lout = 2.2 ×5 = 11m

1050 1100
λ in = = 58.1 < 180 λ out = = 70.51 < 180
18.07 15.6
Fc = 1.4 – 6.5*10-5 *(70.51)2 = 1.076 t/cm2

Check for section 1: (at base) Mx = 29.3 mt, N = 64.3 t and My = zero

fca =
64.3
= 0.22 t/cm2 ∴ f ca =
0.22
= 0.205 > 0.15
290 Fc 1.076

7500
FEx = = 2.22 t/cm2
2
58.1

9/24-B.C.4
 Examples-Column 4

0.85
A1 = = 0.9 4 taken =1
0.205
1−
2.22
29.3 * 100
fbx = * 30 = 0.927 t/cm2
94750
C 0.5(22 − 1.2 − 2 * 1.9)
Fbcx ⇒ = = 4.47 <10.9
tf 1.9 Mx

58
Limit of compact: = 37.44 Stiff

IPE 600
2.4

Web of IPE 600: dw = 0.5* (60 – 4×1.9) = 26.2 cm. IPE 550

d w 26.5
= = 22.1 < 37.44
tw 1.2

Lu act = distance between stiffeners (Take it 1m)

20 × 22 2
Lu max = = 284 cm ∴ There is no L.T.B ∴ Fbcx =1.536 t/cm
2.4
Applying in the interaction equation:
0.927
0.205 + = 0.8 < 1.2 Case "B" lateral shock is included
1.536

Check for section 2: (at base) Mx = zero, N = 64.3 t and My = 21.5mt

fca =
64.3
= 0.22 t/cm2 ∴ f ca =
0.22
= 0.205 > 0.15
290 Fc 1.076

7500 0.85
FEy = = 1.51 t/cm2 A2 = = 0.98 Taken =1
70.512 1−
0.205
1.51
29.3 * 100
fby = * 30 = 0.927 t/cm2
94750
C 0.5(21 − 1.11 − 2 * 1.72) Stiff
IPE 600

= = 4.78 <10.9
My

tf 1.72

dw = 0.5* (55 – 4×1.72) = 24.06cm. IPE 550

10/24-B.C.4
 Examples-Column 4

d w 24.06
= = 21.7 < 37.44 Compact
tw 1.11

Lu act = distance between stiffeners (Take it 1m)

20 × 21 2
Lu max = = 271 cm ∴ There is no L.T.B ∴ Fbcx =1.536 t/cm
2.4
Applying in the interaction equation:
0.927
0.205 + = 0.8 < 1.2 Case "B" lateral shock is included
1.536

3) Using No bracing at all:

The braking force of the 2 cranes is distributed on 2
successive columns 2B

∴ Each carries B

But we have 2 cranes

∴ Each carries 2B

My = 4.3 ×5 = 21.5 mt
15
∴ 2B = 2× = 4.3t
7
My
From case of studying the column inside plan (cantilever)
N = 32.15 * 2 = 64.3 t Outside

Mx = 2 ×2.5 (5 + 0.5 + 0.36) = 29.3 mt

Section 1 is subjected to Mx My & N

Lin = 2.1*5 = 10.5m = Lout

Note: The base is fixed in both directions (inside & outside)
Use 2 IPE. Assume spacing is 60cm
64.3 2930
F= + = 81 t
2 60
Assume f = 0.8 t/cm2 ∴ A = 101 cm
2

(Presence of My)
Use 2 IPE 450 spaced 60 cm

11/24-B.C.4
 Examples-Column 4

d 60
= = 1.3 <1.5
h 45
d
But QLin = Lout Min ≈ Mout ∴ may be = 1
h
Use batten plate (given) each 1.5d = 1m

A = 2×98.8 = 197.6 cm2 30 30

rx = 18.5 (outside)
2 2
d
ryII = (4.12) + (30) = 30.28 ≈
2
Ix = 2* 33740 = 67480 cm4
Iy = 2[1680 + 302 * 98.8] = 181200 cm4
1050 1050
λ out = = 57 λ in = = 35
18.5 30.28
Lz
λz= rmin of IPE is ry =4.12 cm
rmin

Lz assumed ≈ 1.5 × 60 = 100 cm

100
∴λz= = 24.3 < 60
4.12
2
< × 57 = 38
3
2 2
in = (35) + (1.25 × 24.3) = 46.3

Fc = 1.4 – 65×10-5 ×572 = 1.19 t/cm2

fca = 64.3/197.6 = 0.32 t/cm2

f ca
= 0.27 > 0.15
Fc

For Built up closed section: Fbcx = Fbcy = 1.4 t/cm2

2150 45 2
Outside fbx = * = 0.72 t/cm
67480 2
2930 19 2
Inside fbx = * (30 + ) = 0.64 t/cm
181200 2

12/24-B.C.4
 Examples-Column 4

7500 7500
FE in = 2
= 3.5 t/cm2 FE out = 2
2
= 2.31 t/cm
46.3 57

0.85 0.85
Ainside = = 0.98 → 1 Aoutside = = 0.98 → 1
0.32 0.32
1− 1−
3.5 2.31
Applying Interaction Eqn.
0.72 0.64
0.27 + *1 + * 1 = 1.24 > 1.2
1.4 1.4
∴ We may increase distance between the 2IPE to be 70 cm

d 70
= = 1.56 (1.5 → 2) ok
h 45
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
(b) Design of welded & bolted connection for Bolt & weld.
⇒ Case of max moment & corresponding normal

N= 33.05 t M = (32.15 – 0.9) ×0.4 + 2.5(0.5+0.36) = 14.15 mt

Q = 2.5 t
For case of max normal [weld only]

N = 64.3t M = 2.5×2×(0.5 + 0.36) = 4.3 mt

Q= 5t This is a solved example in Eccentric connections 3

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

13/24-B.C.4
 Examples-Column 4

C) Design of column "2": ( External column with crane):

1) Using horizontal member at level of crane:

41

5.85
Vc
15.6 3.0
x Hc 38
30

5.0

6t

38t

x=36+25 = 61cm N crane = V c = 32.15t H c = 2.5t

M cant = 0.4×32.15 + 2.5×0.61 =14.4 mt

M upper = 6×8-32.15×0.4 + 2.5×(3-0.61) = 41 mt

M1' = 30 - 14.4 = 15.6 mt

We have to check 2 sections

(1) Mx = 30 mt N= 38t

(2) Mx = 41 mt N= 5.85t

1) Estimation of section
Use bigger moment for estimation
Assume f= 1.0 t/cm2
4100
∴ = Sx ⇒ Sx = 4100 cm2
1.0
For IPE Max Sx = 3070 cm3
∴ Use HEB 500

Check for section 1: N = 38t and Mx = 30mt

107200 / 8
Lin ⇒ GA = 10 GB = = 17.4
23130 / 30

14/24-B.C.4
 Examples-Column 4

Kin = 3.8 Lin = 3.8 ×8 = 30.4m

Lbont = 5m
3040 500
λ in = = 143.4 < 180 λ out = = 68.8
21.2 7.27

7500 3000
∴ Fc = = 0.36 t/cm2 fbx = = 0.7 t/cm2
143.4 2 4290

C 0.5 * (30 − 1.45 − 2 * 2.8)

Fbcx ⇒ = = 4.1 < 10.91
tf 2.8

dw = 50- 4×2.8 = 38.8 cm

1 38
α= [ + 1] = 0.64 > 0.5
2 38.8 × 1.45 × 2.4
699 d w 38.8
= 61.6 = = 26.75 < 61.6
(13 * 0.64 − 1) 2.4 t w 1.45

∴ The section is compact

Lu act =5m
= zero → Cb = 1.75 (This column has no segments because there is no
horizontal member at mid height. If there is horizontal member at mid height,
will be equal -0.5)
20 × 30
Lu max = = 387cm < 500cm ∴ There is LTB
2.4
800 * (30 × 2.8)
filb1 = × 1.75 = 4.7 t/cm2 >1.4 t/cm2 2
∴ Fbcx = 1.4 t/cm
500 × 50
38
fca = = 0.16 t/cm2
239
f ca 0.16 7500
= = 0.44 > 0.15 FEx = = 0.36 t/cm2
Fc 0.36 143.4 2
0.85
A1 = = 1.53
0.16
1−
0.36

15/24-B.C.4
 Examples-Column 4

Applying in the interaction equation:

0.7
0.44 + × 1.53 = 1.205 ≈ 1.2
1.4
Check using BFIB "450" for section 2: Section 2 is in the upper segment of
column with M = 41mt and N = 5.85 t
4100
fbx = = 0.95 t/cm2
4290
C
Fbcx ⇒ = 4.1 < 10.91 as before
tf

dw = 38.8 cm
1 5.85
α= [ + 1] = 0.52 > 0.5
2 38.8 × 1.45 × 2.4
699 dw
= 78.3 = 26.75 < 78.3
(13 * 0.52 − 1)( 2.4) tw

Lb out = 3m Lb in = 30.4 (as before)

300 3040
out = = 41.3 out = = 143.4
7.27 21.2
7500
Fc = = 0.36 t/cm2
143.4 2
15.6
Luact = 3m α =− = −0.38
41

Cb = 1.75 – 0.38×1.05+0.3(0.38)2 = 1.39

20 × 30
Lu max = = 387cm
2.4
1380 * 30 × 2.8
= × 1.39 = 1342cm
2.4 × 50
∴ The section is compact & No LTB

Fbcx = 1.536 t/cm2

16/24-B.C.4
 Examples-Column 4

5.85
f ca = = 0.024 t/cm2
239
f ca 0.024
= = 0.07 < 0.15 ∴ A1 = 1
Fc 0.36

Applying in the interaction equation:

4100 / 4290
0.07 + × 1 = 0.82 < 1.2
1.536
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Forces on bracing:
15 F3
RB = 2 × = 4.3 t
7 Rw
15 F1 3.0
h1= 8m h2 = 8 + = 9.5m
F4
10
RB
8 + 9.5 F2
hav = = 8.75m 5.0
2
8.75
ww = 0.8×1×70× = 245 kg/m
2
6.0
245
Rw = *15 = 3.675t
1000
2 2
For M1 ⇒ Lin =L Lout = 1.2×L where L= 3 + 3 = 4.24m

F1 ⇒ 2F1 cos 45 = 3 F1= 2.6t

2 2
For M2 ⇒ L in = 0.5L Lout= 0.75L where L= 5 + 6 = 7.81 m
5
2F2 cos θ = (3.675 + 4.3) where θ = tan-1 = 40
6
F2 = 5.2t
M3 ⇒ Lin = Lout = 6m use star shape ( L > 6m)
F3 = F1 cos θ = 1.84t

M4 ⇒ Lin = 0.5L Lout = L where L =6m

F4 = F2 cos θ = 5.2 cos 40 = 4t

17/24-B.C.4
 Examples-Column 4

L
Design all previous members as compression members & check
d

(2) Without horizontal member at the level of crane girder

Sec 2
Rw
41
5.85

15.6 3.0
38
8.06 30 Sec 1 R
B

5.0

M outside 6.0
y M inside
x

Choose BFIB because of double moment

Assume f=0.5 t/cm2 (Neglecting of My and N)
3000 3
⇒ Sx = = 6000 cm Use HEB 650
0.5

107200 / 8
Lin ⇒ GA = 10 GB = = 17.4
23130 / 30

Kin = 3.8 Lin = 3.8 ×8 = 30.4m as before

Lout = 8m

3040 800
λ in = = 112.2 λ out = = 114
27.1 6.99

7500 2
Fc = 2
= 0.57t / cm
114

Check for section 1:

38 2 f ca 0.13
fca = = 0.13t / cm = = 0.23 > 0.15
286 fc 0.57

18/24-B.C.4
 Examples-Column 4

3000 806 2
fbx = = 0.46t / cm 2 fby = = 0.84t / cm
6480 932
1
15 − × 1.6 − 3.1
C 2
Fbcx ⇒ = = 3.58 < 10.9
tf 3.1

dw = 65 – 4×3.1 = 52.6 cm

1 32
α= [ + 1] = 0.59 > 0.5
2 52.6 × 1.6 × 2.4
699 d w 52.6
= 67 = = 17 < 67 Ok compact
(13 * 0.59 − 1) 2.4 tw 3.1

20 × 30
Luact = 8m Lu max = =387 cm
2.4
∴ There is LTB

800 × 30 × 3.1
fltb1 = * C b = 1.43Cb > 1.4t / cm 2
800 × 65
Since the segment of the column for the points braced outside is 8m, and since
the moment is not straight, so take Cb as unity
2
∴ Fbcx = 1.4 t/cm ∴ Fbcy = 1.4 t/cm

7500 7500 2
FEX = = 0.6t / cm 2 FEy = 2
= 0.57t / cm
112.2 2 114

0.85
The frame is allowed to sway inside plan (Mx) A1 = = 1.09
0.13
1−
0.6
The frame is not allowed to sway outside plan (My) because of the presence of
bracing. RB causes transverse load and with hinged base (outside). Cmy=1.0
C my = 1
A2 = = 1.3
0.13
1−
0.57
Applying in the interaction equation:
0.46 0.84
0.23 + × 1.09 + × 1.3 = 1.37 > 1.2 Use HEB 700 & Recheck
1.4 1.4

19/24-B.C.4
 Examples-Column 4

We have to check section "2" with Mx = 41mt, N=5.85t and My = zero

Lu act = 8m, Lb in = 30.4m, Lb out = 8m, Cmx = 0.85
Cb = 1 (moment is not straight in the segment between the points braced outside)

Note: For bracing it carries Rw only

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Solved example "2":
P
(1) Design as hot rolled column if P=2t
(2) Design as cold formed box section (square) if P=2t
(3) Calculate Pmax if we use cold formed rectangular box section 4.0 m

300×200×6.3

Solution:
(1) Assume Fc = 0.75t/cm2 in = out = 100
2 2
∴A = = 2.7cm Lin = Lout = 4m (hinged – hinged)
0.75
400
IPE 80 = 100 ry = 4cm
ry

IPE 450
80 + 450
∴ IPE = = 265 Choose IPE 270
2
2 2
fca = = 0.04t / cm
45.9
7500 2
Fc = 2
= 0.43t / cm ∴ fca << Fc
(400 / 3.02)

400
Use smaller IPE & Recheck 180 = ⇒ r y = 2.22
ry

7500
Use IPE 200 Fc = = 0.235t / cm 2
2
(400 / 2.24)
2
fca = = 0.07 t/cm2 < 0.235 t/cm2
28.5

20/24-B.C.4
 Examples-Column 4

(2) As cold formed:

λ = 100

2 2 400
A= = 2.7cm = 100
0.75 rx

40×40×4 rx = 4 cm = 40

100×100×4

40 + 100
∴ = 70
2

Use 70×70×36

2
Fca = = 0.21 t/cm 2
9.5
400
λx= = 148
2.7
7500
∴ Fc = = 0.34t / cm 2 we may check 60×60×3.2
2
148
400
(3)Max "P" → x = = 48.2 A= 61.2 cm2
8.3

Fc = 1.4 – 6.5 ×10-5 ×48.22 = 1.25t/cm2

P
= 1.25 → P = 76.4t
61.2
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

21/24-B.C.4
 Examples-Column 4

Solved example "3": 1:10

For the shown cantilever, it is required to:
1 – Suggest as suitable bracing system and design it Spacing
6m 8m
2 – Design a rolled section for the column
Take w corrugated sheets = 6 kg/m2 wsteel = 20 kg/m2

2m 8m
Solution:
Since it is double cantilever so we have to make cases of loading
W corr = 6 kg / m2 , W steel = 20 kg / m2
WLL = 60 – 66.66 * 0.1 = 53.3 kg / m2
6
WDL= ( + 20) * 6 = 156 kg / m/ = 0.160 t / m/
cos α
WLL =53.3 * 6 = 320 kg / m/ = 0.32 t / m/
WTotal = 0.16 + 0.32 = 0.48 t / m/ 6.0

Case 1: Maximum normal and corresponding moment

N = 0.48 * 8 + 0.48 * 2 = 4.8 t
M = 0.48 * 8 * 4 - 0.48 * 2 * 1 = 14.4 m t
Case 2: Maximum moment and corresponding normal
N = 0.48 * 8 + 0.16 * 2 = 4.16 t
M = 0.48 * 8 * 4 - 0.16 * 2 * 1 = 15.04 m t
T D
T T

M M

Case 1 Case 2

22/24-B.C.4
 Examples-Column 4

15.04 * 100
Assume f = 1.2 t / cm2 Sx= = 1253 cm3
1.2
Use I PE 450
Lb out = 4 m , L b in = 2.1 * 8 = 16.8 m
400 1680
λout = = 97 , λin = = 90.8
4.12 18.5
Fc = 1.4 – 6.5*10-5 (90.8)2 = 0.86 t / cm2
Calculation of Fbcx: Check compactness:
C 0.5 * (18 − 0.86 − 2 * 1.35)
For flange: = = 5.3 < 10.9
tf 1.35

For web: dw = h – 2c = 37.8 cm

1 4.16
= *( + 1) = 0.52 > 0.5
2 37.8 * 0.94 * 2.4

699 / 2.4 d w 37.8

Limit = = 78 = = 40 < 78 Compact
13 * 0.52 − 1 t w 0.94

Note that: we used the larger normal force, so the smaller will be compact also.
Segment between braced points
α =-1 , C b = 1.75 + 0.5 (-1) + 0.3 (- 1)2 = 1
LU act = 4 m
4.0 m
20 *19
Lu max = = 245 cm < 400 cm
2.4
Segment
SO LT B will occur y
18*1.35
800 *19 *1.46
F1tb =
1
*1 = 1.23 t / cm2 < 1.4 t / cm2
400 * 45
6.22*0.86 h
1
h = (45 – 2 * 1.46) = 7 cm
6
y
3
1.46 * 19
I= = 834.5 cm4
12
A = 19 * 1.46 + 7 * 0.94 = 34.32 cm2
834.5
rt = = 4.9 cm
34.32

23/24-B.C.4
 Examples-Column 4

Lu 400
= = 82
rr 4.9

1 1
84 = 54.2 , 188 = 121.35
2.4 2.4

cb Lu c
84 < < 188 b
Fy rt Fy

82 2 * 2.4 2 2
Fltb2 = (0.64 − )2.4 = 1.18 t / cm < 1.4 t / cm
1.176 * 105 *1

Fltb = 1.182 + 1.232 = 1.7 t / cm > 1.4 t / cm

2 2

Take F bcx = 1.4 t / cm2

Check for case 1:

4.8 f ca 0.05
f ca = = 0.05 t / cm2 = = 0.06 < 0.15
98.8 F 0.86
C

14.4 *100
A1 = 1 , Fbx = = 0.96 t / cm2
1500
0.96
0.06 + * 1 = 0.74 < 1
1. 4

Check for case 2:

4.16 f ca 0.042
f ca = = 0.042 t / cm2 = = 0.05 < 0.15
98.8 F 0.86
C

15.04 * 100
A1 = 1 , Fbx = = 1 t / cm2
1500
1
0.05 + * 1 = 0.76 < 1 Waste try IPE 400
1. 4

24/24-B.C.4
COMBOSIT
SECTIONES
 Composite sections

Composite Bridges
Explain the main concept of composite action in roadway bridges: ( )
Beam Theory for Composite Beams:
a) If the friction between the slab and beam is neglected. The slab
lower surface is in tension and elongates while the upper surface of the
beam is in compression and shortens. There are 2 neutral axes, one at mid
height of concrete slab and the other at mid height of steel beam. This is
shown in fig 1.
b) Partial interaction, the neutral axis of the slab is closer to the beam
and that of the beam closer to the slab; the horizontal slip has now
decreased. This is shown in fig 2.
c) When a system acts compositely no relative slip occurs between the
slab and beam. Horizontal forces (shears) are developed that act on the lower
surface of the slab to compress and shorten it, while simultaneously they act
on the upper surface of the beam to elongate it. There is 1 neutral axis for the
whole section. This is shown in fig 3.

N.A.
N.A.(Steel)

Fig (1) Fig (2) Fig (3)

No interaction Partial sliding Full contact
Non-composite Composite

For design of composite section, we assume the shear connection will be

enough for transmitting the required shear forces.

1/17
 Composite sections

State advantages and disadvantages of using Composite action: ( )

Advantages:
1- A saving in the weight of steel required about 20 to 30 %.
2- A reduction in the depth of members which in turn reduce the height of a
building.
3- Increased stiffness of the floor system.
4- Can achieve long spans that cannot be achieved with non-composite
members.

Disadvantage:
1) In continuous construction, the negative moment region will have a
different stiffness because the concrete slab in tension is cracked and not
working.
2) Long-term deflection caused by concrete creep and shrinkage

Methods of construction: ( ) ECP page 154

Two different methods of construction are to be considered:
1- Without shoring (case 1)
When no intermediate shoring is used under the steel beams or the
concrete slab during casting, the steel section alone supports the dead and
construction loads. The composite section supports the live loads and the
dead loads (flooring, walls, -- etc.) after the slab has reached 75 % of its
required characteristic strength.

[live load] [ floor cover]

composite

2/17
 Composite sections

2- With shoring (case II)

When an effective intermediate shoring system is used during casting ,
the composite section supports both the dead and live loads. Shoring shall
not be removed until the concrete has attained 75 % of its required
characteristic strength, f cu.

composite

Transformed area method: ECP page 158

The concrete area is reduced dividing by the factor n, where n is defined as the
modulus of elasticity of steel divided by the modulus of concrete or
n = ES/ EC E S = 2100 t / cm2
The value of n to be used in design usually is given in the applicable
Table (1) Recommended Values of the Modular Radio (n)
Concrete Modulus of
Characteristic Elasticity of Modular Radio, n
Cube Strength, Concrete, E C
F cu (kg / cm2) (t / cm2)
250 220 10
300 240 9
400 280 8
>500 310 7

Important note:
If the neutral axis of the member falls below the bottom of the concrete slab,
the entire concrete slab is in compression.
If the neutral axis of a composite section falls within the concrete slab, only
the portion of the concrete slab above the neutral axis should be included in
the calculation of section properties.

3/17
 Composite sections

VERY IMPORTANT NOTE: ECP page 162

If the system is shored and because the effect of creep and shrinkage affects
the composite section, use:
Modular ratios = "2n" instead of "n" when design composite in building
Modular ratios = "3n" instead of "n" when design composite in bridges

Why do we use "3n" in shored systems instead of "n" in un-shored:

The creep is due to long-term load, so the creep is a secondary effect of dead
load only (may be live load also in case of garages)
In un-shored systems: the dead load is loaded on steel section, so no
effect of creep on composite section.
In shored systems: the dead load is loaded on composite section, so
we have to take the effect of creep on composite by reducing the area of
concrete slab working "be" (i.e. using "3n" instead of "n" for dead load and
use "n" for live load because live load has no creep effect.

What do u know about effective width concept? "be"

steel flange max.
non-linearly &
&
(b e)

The total compression force

2b
e

carried by the equivalent system

must be the same as that carried
by the real system. (Same area
under the curve)

4/17
 Composite sections

Effective width of concrete slab: ECP page 156

For bridges:
Least of (L / 4), or 12 t slab + b flange (neglecting the thickness of the
haunch if any)
Where L is the actual span of designed beam between the supports.

Minimum thickness of concrete slab: ECP page 155

Bridges: t 16 cm
If not given, take ts = 20cm always

Design procedure:
a) Calculation of straining actions:
1- Calculate MDL-1 [Dead load of concrete slab and O.w. of steel beam]
2- Calculate MDL-2 [Floor cover]
3- Calculate ML+I
Important note: Equivalent WL+I may be given in the form of t/m2 or t/m\
If "w" is given in t/ m\ ML+I = w *L2 / 8
If "w" is given in t/m2 ML+I = (a*w) *L2 / 8 where "a" is spacing
between main girders.

b) Estimation of section:
Span
1- hw = ECP page 155
16 → 22
2- tw : For hw < 1, no vl or hz "longitudinal" stiff are required
For 2.0 > hw > 1, only vl stiff are required @ spacing 1.5 → 2m
For 2.8 > hw > 2.0, vl stiff, hz "longitudinal "stiff @ d/5
For hw > 2.8, vl stiff, hz "longitudinal "stiff @ d/5 and d/2
Then we calculate tw from the following checks:
1- tw min = 8 mm

5/17
 Composite sections

Q (d + L + I )
2- Check shear: q = ≤ 0.35 Fy ,
hw × tw

Q (d + L + I )
tw ≥ ECP 2.6.3.1 page 13
hw × 0.35Fy

hw 830
3- Flange to web buckling ≤ ECP 7.1 page 122
tw Fy

4- Web buckling limits

hw 190
For web not stiffened longitudinally : ≤ ECP 7.5 page 123
tw Fy

hw hw 320
For web longitudinally stiffened @ : ≤ ECP 7.7 page 124
5 tw Fy

hw hw hw 365
For web longitudinally stiffened @ , ≤ Euro code
5 2 tw Fy

We take the max tw of the previous cases

tw must be in mm ,an even number or devisable by 5
Example "8,10,12,14, 15 or 16…"
M D+ L+ I
3- Flanges: Assume F=0.58Fy T=C=
0.97hw
T 1
Area of flange = Af = - hw tw
0.58 Fy 6

We take economic factor 20% → 40% assume 30%

2 4
So Aupper = Af *0.7 Alower = Af *0.7
3 3
Assume b=20t and get b, t for both upper & lower flanges.
4- be: as mentioned in ECP 156
5- Get modular ratio "n" from table ECP page 158
6- Check compactness of both flanges (The flange must be non-compact)
If the flange is in the slender limit, increase thickness.
We may check upper flange only (compression flange)

6/17
 Composite sections

Checks:
2
1) If the system is Un-shored:
1- The moment due to O.W. and concrete slab will
be loaded on steel section only. MDL-1
N.A.
Calculate y s and Is
ys
1
2- The moment due to F.C. and L+I will be loaded
on composite section
MDL-2 + ML+I
A f2
Anv = As + conc f2
n
N.A.
AC t
AS y s + * (hs + s )
n 2 y
y nv = y nv
A s
AS + S
n
AC t
I nv = I S + A S ( y nv − y s )2 + (hs + s − y nv ) 2
n 2
Note that: f2\ = f2 / n
Case Upper steel fiber Lower steel fiber Upper concrete fiber
DL1 f US (upper steel) f LS (lower steel)
M DL −1 M DL −1
= * (hs − y s ) = * ys
IS IS

DL2 f UC (upper composite) f LC (lower composite) =f u-conc (upper concrete) =

L+I M DL −2 + M L + I M DL−2 + M L+ I
= (hs − y nv ) M DL −2 + M L + I * y nv (hs + t s − y nv ) / n
I nv I nv I nv

Total
Checks:

7/17
 Composite sections

1) For total stresses (for tension or compression flange) < 0.58 F y

L u act = zero because the flange supports the comp.- flange)
2) Check compressive stresses for concrete ( )
f u-conc. 75 kg / cm2 (allowable of concrete 300 kg/cm2)
0.6M L + I
4) Check fatigue in tension flange: y nv ≤ Fsr
I nv
Note: detail 'B " and longitudinal member (if M.G.)

1) If the system is Shored:

Very Important note:
Due to the effect of creep & shrinkage we take modular ratio = 3 n for
dead load which makes creep and "n" for live load.
1- Calculate M DL-1 & M DL-2
2- Calculate M L+I f2
f2
N.A.
M DL-1 and MDL-2 affects the composite
section with modular ratio = 3n y
nv
MLL+I affects the composite section
with modular ratio = n
Aconc. A
Calculate y nv and I nv twice, using for D.L. and conc. for L.L. +I
3n n

Ac /3n Ac /n
3 f 3-D 3 f 3-L
f2 f2
2 f 2-D 2 f 2-L
N.A. N.A.
h h
y y
nv-1 nv-2
1 1
f 1-D f 1-L
For DL & DL For LL+I
1 2

8/17
 Composite sections

Case Upper steel fiber(Pt2) Lower steel fiber(Pt1) Upper concrete fiber(Pt3)
DL1 f US (upper steel) = f LS (lower steel) f u-conc(upper concrete) =
+DL2 M DL −1 + M DL −2 M DL −1 + M DL −2 M DL −1 + M DL− 2
I nv−1 I nv−1 I nv−1
* (h − y nv−1 ) * ( y nv−1 ) * (h − y nv −1 − t s ) / 3n

L+I f UC (upper composite) f LC (lower composite) f u-conc (upper concrete) =

M L+ I M L+ I M L+I
= * (h − y nv−2 ) = * ( y nv−2 ) * (h − y nv− 2 − t s ) / n
I nv −2 I nv−2 I nv −2

Total

1- Lower steel fiber: fLC ≤ 0.58Fy (Tension side)

2- Upper steel fiber: fUC ≤ 0.58Fy (Lu act = zero)
3- Upper concrete fiber: f u-conc ≤ Fall conc
0.6M L + I
4- Check fatigue in tension flange: y nv−2 ≤ Fsr
I nv−2

9/17
 Composite sections

Solved example:
The main girder of span 40m and
spacing between main girders 2.70
40m
m will be designed in this example
assuming composite action between
2.70m
the deck slab and the steel section.
Steel used is st.52 and
RC. slab 22 cm with Fcu=300kg/cm2 (Fall = 75 kg/cm2)
Assume F.C. = 175 kg/m2.
Equivalent L.L. = 1.7 t/m2 Distance between X.G. = 5m
It is required to design an intermediate main girder twice: using
a- Un-shored system b- Shored system.

Solution:
1. Staining Actions:
The staining actions for composite girder design are taken as follows:
1.1 Dead load (DL1):
R.C. slab = 22 cm
Assume own weight of steel girder = 0.35 t/m`
WDL1 = ts c x spacing + o.w. = 0.22 x 2.5 x 2.7 + 0.35 = 1.835 t/m`
MDL1 = 1.835x402 / 8 = 367 m.t
QDL1 = 1.835x40 / 2 = 36.7 t
1.2 Dead load (DL2):
Flooring = 0.175 t/m2
WDL2 = flooring x spacing = 0.175 x 2.7 = 0.47 t/m`
MDL2 = 0.47x402 / 8 = 94 mt QDL2 = 0.47x40 / 2 = 9.4 t
1.3 Live load + Impact (L+I):
W eq = 1.7 t / m2 (given) ∴ w = 1.7 * 2.7 = 4.6 t / m/
M L+I = 4.6 * 402 / 8 = 920 m t Q L+I = 4.6 * 40 / 2 = 92 t

10/17
 Composite sections

Q T = 36.7 + 9.4 + 92 = 138.1 t M T = 367 + 94 + 920 = 1381 m t

Estimation of section:
4000
hw = = 180 à 250 cm
16 → 22
500
Take h w = 200 cm with vertical stiffener each = 167 cm (5m is the
3
h
distance between X.G.) and with horizontal stiffener at
5
138.1
tw :1− = 0.35 * 3.6 t w = 0.54 cm
200t w
200 830
2- = t w = 0.87 cm
tw 3.6
200 320 h
3- = (Use one horizontal stiffener at ) t w = 1.18 cm
tw 3.6 5
Take t w = 12 mm > 8 mm (min)
Assume F b cx = 0.58 F y
(1381 * 100) /(0.97 * 200) 1
Af = − * 200 * 1.2 = 300 cm2
0.58 * 3.6 6
Assume economic factor = 30 %
2 4
Auf = 300 * * 0.7 = 140 cm2 ALf = 300 * * 0.7 = 280 cm2
3 3
Assume for both flanges b = 20 t
For upper flange: 20 t2 = 140, t = 26 mm, b = 55 cm
For lower flange: 20 t2 = 280, t = 38 mm, b = 75 cm
Note that: b> 0.25h, which is not economic
We can take h w = 250 cm instead of 200 cm

b e : Intermediate M.G.
270
1- C.L. to C.L. : bE = 2 * = 270 cm
2

11/17
 Composite sections

L 4000
2- bE = 2 * = 2* = 1000 cm
8 8
1
3- b E = 2 (6 t s + b f ) = 2 (6 * 22 + 0.5 * 55) = 319 cm
2
SO b e = 270 cm t slab = 22 cm

a- Design as Shored:
For F CU = 300 kg / cm2 à n = 9 ECP 158

To calculate the section carrying DL-1 and DL-2 USE "3n" = 27

A S = 75*3.8 + 200*1.2 + 55*2.6 f2
2
= 668 cm f2
N.A. 90.85
A C = 270*22 = 5940 cm2
MDL-1+MDL-2 = 367+94 = 461mt
115.55

3.8 200 2.6

75 * 3.8 * + 200 * 1.2 * (3.8 + ) + 55 * 2.6 * (3.8 + 200 + )
y nv−1 = 2 2 2
5940
668 +
27
5940 22
(3.8 + 200 + 2.6 + )
+ 27 2 = 115.55 cm
5940
668 +
27
3.8 2 1.2 * 200 3
I n v-1 = 75 * 3.8 *(115.55 - ) + + 1.2 * 200 (115.55 – 103.8)2
2 12
2.6 2 5940
+ 55 * 2.6 * (90.85 - ) + *(90.85 + 11)2 = 7'943'186 cm4
2 27

270
To calculate the section carrying LL+I f2
f2
USE "n" = 9 57.1
N.A.

149.3 12/17
 Composite sections

b e = 270 cm t slab = 22 cm
h s = 3.8 + 2.6 + 200 = 206.4 cm
A C = 270 * 22 = 5940 cm2

3.8 200 2.6

75 * 3.8 * + 200 *1.2 * (3.8 + ) + 55 * 2.6 * (3.8 + 200 + )
y nv−2 = 2 2 2
5940
668 +
9
5940 22
(3.8 + 200 + 2.6 + )
+ 9 2 = 149.3 cm
5940
668 +
9
3.8 2 1.2 * 200 3
I nv = Inv-2 = 75 * 3.8 *(149.3 - ) + + 1.2 * 200 (149.3 – 103.8)2
2 12
2.6 2 5940
+ 55 * 2.6 * (57.1 - ) + *(57.1 + 11)2 = 10995975 cm4
2 9

Case Upper steel fiber(Pt2) Lower steel fiber(Pt1) Upper concrete fiber(Pt3)
DL1 f US (upper steel) f LS (lower steel) f u-conc(upper concrete) =
+DL2 461 * 100 461 * 100 461 * 100
= * 90.85 = * 115.55 = * (90.85 + 22) / 27
7943186 7943186 7943186
= 0.53 t/cm2 = 0.67 t/cm2 = 0.024 t/cm2

L+I f UC (lower composite) f LC (upper composite) f u-conc (upper concrete)

920 * 100 920 * 100 920 *100
= * 57.1 = *149.3 = * (57.1 + 22) / 9
10995975 10995975 10995975
= 0.48 t/cm2 = 1.25 t/cm2 = 0.073 t/cm2

Total = 1.01 t/cm2 = 1.92 t/cm2 = 0.097 t/cm2

13/17
 Composite sections

1- Check stresses:
Lower steel fiber: f 1 = 1.92 t / cm2 < 0.58 *3.6 = 2.1 t / cm2
Upper steel fiber: f 2 = 1.01 t / cm2 < 0.58 *3.6 = 2.1 t / cm2
Upper concrete fiber= 0.097 t/cm2 = 97 kg / cm2 > 75 kg / cm2 (Unsafe)
2- Check fatigue: detail B, n = 2000000 (longitudinal member)
0.6 * 920 *100
àF sr = 1.26 t/cm2 *149.3 = 0.74 t / cm2 < 1.26 t / cm2
10995975

b- Design If un- Shored:

55*2.6
Check compactness:
55 1.2
− 200*1.2
c 2 = 10.1 < 21 = 11.1
= 2 non-compact
tf 2. 6 3. 6

75*3.8
For steel section:
A S = 75 * 3.8 + 200 * 1.2 + 55 * 2.6 = 668 cm2
3.8 200 2.6
75 * 3.8 * + 200 * 1.2(3.8 + ) + 55 * 2.6(3.8 + 200 + )
y= 2 2 2 = 82 cm
668
3.8 2 200 3
I S = 75 * 3.8 * (82 - ) + 1.2 * + 1.2 * 200 *(103.8 - 82)2
2 12 124.4
2.6 2
+ 55 * 2.6 (124.4 - ) = 4'909'587 cm4
2 82
For composite section:
b e = 270 cm t slab = 22 cm h s = 3.8 + 2.6 + 200 = 206.4 cm
A C = 270 * 22 = 5940 cm2
5940
668 * 82 + * (206.4 + 11)
y nv = 9 = 149.3 cm
5940
668 +
9

14/17
 Composite sections

270
f2
f2
N.A. 57.1

149.3

5940
I n v = 4909587 + 668 * (149.3 – 82)2 + (57.1+11)2 = 10995975 cm4
9
Lower steel fiber Upper steel fiber Upper con. fiber
367 * 100 367 * 100 --------
f1 = * 82 f2 = * 124.4
4909587 4909587
= 0.61 t / cm2 = 0.93 t / cm2
(94 + 920) *100 (94 + 920) *100 (94 + 920) *100 * 79.1
f1 = *149.2 f 2 = * 57.1 f 3 = /9
10995975 10995975 10995975
= 1.38 t / cm2 = 0.53 t / cm2 = 0.081 t / cm2
= 1.99 t / cm2 = 1.46 t / cm2 = 81 kg / cm2 > 75 kg / cm2
< 2.1 t / cm2 < 2.1 t / cm2

Checks:
1- For Upper & lower fiber f < 0.58 F y = 2.1 t / cm2
2- For concrete f con. = 81 kg / cm2 > 75 kg / cm2 Unsafe

15/17
 Composite sections

Shear connectors

Connector type Way to Way to design Example

transmit stress the connector
to concrete
Non- rigid Bond only Connector cross Studs and spiral
(flexible) Or bond+ section
bearing

Semi- rigid Bearing Connector cross Channels and

section & its angles
connection
Rigid Bearing Concrete I – section
bearing& channel
Connector cross T - section
section

T-sec
Closed ring
I-sec Hooked bar
channel
Spiral
Studs

r r
ect
o ecto
n n
on con
gi dc xibl
e
Ri Fle

16/17
 Composite sections

Angle
channel

ct or
ne
c on
i gid
-r
mi
Se

17/17
DESIGN OF STAIRS
Stairs

1/7-Stairs
 Stairs
Design of step:
Steps are made of checkered plate Checkered plate
6/8 as shown

5-7
Assume o.w. of step = 20 kg / m\
L.L = 300 → 500 kg / m2 b = 30 cm
F.C. may be used
20 Beam at 1/2 floor level
W= + (F.C. + L.L.) b
1000
(4.50)
wa 2 wa

a
Calculate M = Q=
8 2
Fbcx = 1.4 t/cm2 (lip is small, so neglect L.T.B.)
L
Check f, q, δ ≤ →L=a
300

a
(6.00) (3.00)
Width of stairs is about 1 → 1.5 m a a
Slope is 1:2 in most cases.

Design of stair beam: (Usually used as channel)

θ ≈ 26.5
assume o.w. = 25 kg/m\
w = ...... t/m'
weight of step, handrail, (all steel structure)
ws = 70 kg/m2/hz proj
L.L = 300 → 500 kg / m2 1
2
F.C. = …….. if any
25
W= + (70 + F.C. + L.L.) (a/2) M2
1000
M1
Draw M, Q, (neglect normal) R

wL2 wL
M= Q=
8 2
Lu act = zero

2/7-Stairs
 Stairs
Fbcx = 1.4 t/cm2
dmin of channel → h-22
t
1
4
Check f, q, δ 2
0 3
26.5
L70*7

Design of connection using butt weld:

hw= hC

hchannel
hweld
°31

= 2
cos 2

Calculate Ix
M
Check y = …..< 0.7 ft (allowable of good butt weld subjected to tension due
Ix
to moment or due to tension). Look ECP page

Design of beam at 1/2 floor height: (Usually IPE)

R = Reaction of stair beam R RR R
Calculate M, Q

a a
If channel is flushed & checkered plate is
welded, so Lu act = zero
If we use grating Lu act = a

Design of connection between channel & I beam:

Force = Rt → Rss, Rb as before

3/7-Stairs
 Stairs
Solved example: Beam B
b c
It is required a complete design of stairs (4.50)

1.25
(steps, beam of stairs with its connection &

3.0
beam at 1/2 floor height)
Given that: L.L. = 400 kg/m2, F.C. =150
kg/m2

1.25
(6.00) (3.00)
a d
1.0 1.25 1.25 1.0
Solution: 0.5

Design of step:
Assume o.w. of step = 20 kg/m\
Take shape of step as Z – section (assume b = 30cm) (checkered plate 6/8)
WDL = 20 + 150 * 0.3 = 65 kg/m\
WLL = 400 * 0.3 = 120 kg/m\
WT = 120 + 65 = 185 kg/m\

1.25 2 Lip = 5cm

M = 0.185 * = 0.036 mt = 3.6 cmt t = 0.6 4.4
8
Q = 0.185 * (1.25/2) = 0.11 t 30

30 * 0.6 3  0.6 * 4.4 3 

Ix = (neglected) + 2  + 0.6 * 4.4 * 2.5 2  = 41.52 cm4
12  12 
 
Checks:
3.6 * 4.7
1. f = = 0.41 t/cm2 < 1.4 t/cm2
41.52
(Lip is small, so we can neglect L.T.B)
0.11
2. q = = 0.02 t/cm2 <<< 0.84 t/cm2
2 * 5 * 0.6

5 120 * 1.25 4 125

3. δ L.L. = * *10 3 = 0.04 cm < = 0.42 cm
384 2100 * 41.52 300

4/7-Stairs
 Stairs
Design of stair beam:
Assume o.w. = 25 kg/m\
Weight of other steel structure is 70 kg/m\
WDL = 25 + (70 + 150) * (1.25 / 2)
= 162.5 kg/m\
WLL = 400 * (1.25 / 2) = 250 kg/m\
Wtotal = 250 + 162.5 = 412.5 kg/m\ w = 0.41 t/m'
0.41 * 5.5 2
M= = 1.56 mt
8
Q = 0.41 * (5.5 / 2) = 1.13 t 1.56
1.41
Assume Fbcx = 1.4 t/cm2 1.13
Sx = 156 / 1.4 = 111 cm3
Choose channel 160
Checks:
Lu act = zero So No L.T.B.
c 6.5 − 0.75 − 1.05
= = 4.48 < 10.9
tf 1.05

dw
= 11.6 / 0.75 = 15.5 < 82
tw
Therefore, the section is not slender (simply symmetric channel)
1. f = 156 / 116 = 1.34 t/cm2 < 1.4 t/cm2
1.13
2. q = = 0.09 t/cm2 < 0.84 t/cm2
16 * 0.75

5 0.25 * 5.5 4
3. δ L.L. = * *10 6 = 1.53 cm < 550/300 = 1.83cm
384 2100 * 925

5/7-Stairs
 Stairs
Min channel: t 9 . 94
30cm h-2 =
h − 2t 2 5cm
= 9.94 (From drawing)
2
t ≈ 1cm h = 22 cm 0
26.5
Choose channel minimum 220
L70*7

Design of connection butt weld:

θ
M connection = 1.13 * 1.25 = 1.41 mt = 13.25 ο 8
2 t=1.25
22
hweld = = 22.6 cm 22.6
t=0.9
cos13.25

0.9 * 22.6 3 22.6 1.25 2

Ix = + 2 *1.25 * (8 − 0.9) * ( − ) = 2888.4 cm4
12 2 2
141 22.6
f= * = 0.55 t/cm2 < 0.7*1.4 = 0.98 t/cm2
2888.4 2
Design of connection between channel & beam at 1/2 floor beam:
R = 1.13 t (use only one angle)
π
Rss = *1.6 2 * 0.25 * 4 = 2.01t
4
tmin is smaller of 0.7 or 0.9 (tw of channel)
Rb = (0.8 * 3.6)*1.6*0.7 = 3.22 t
n = 1.13 / 2.01 = 0.56 use 2 bolts min

Chequerd L70*7
plate

6/7-Stairs
 Stairs
Design of beam B (at mid floor level)
Rtotal stair beam = 1.13 t, O.W. = 50 kg/m' 1.13 1.13 1.13
RLL stair beam = 0.25 * 5.5 / 2 = 0.69 t 0.05
(To be used in check of deflection)
1.25 1.25
Mconc = 2.385 x 2.25 – 1.13 x 1.25 – 0.05 x 1.00 0.5 1.00
2
1.25 /2= 3.83 mt
Q = 0.03 x 5 / 2 + 1.13 x 4 / 2 = 2.385 mt
Assume Fbcx = 0.64 Fy = 1.54 t/cm2
Sx = 383/ 1.54 = 259 cm3 choose IPE 220
Checks:
c 11 − 0.59 − 2 * 1.2 dw
= = 8.7 < 10.9 , = 21.2 / 0.59 =35.9 < 82
tf 0.92 tw

Assume using checkered plate welded to upper flange of beam

So Lu act = zero Fbcx = 0.64 Fy
1. f = 383 / 252 = 1.52 t/cm2 < 1.54 t/cm2
2.38
2. q = = 0.18 t/cm2 < 0.84 t/cm2
22 * 0.59

0.69 0.69 0.69

1.25 1.25

3.51
0.69 1.725
0.54 0.56

3. deflection due to live load only " RLL stair beam "
M ε = 3.51 * 2.5 – 0.56 * (0.25/2) – 0.54 * (0.5 + 1.25/3) – 1.725 * (0.25 +
1.25/2) – 0.69 * (0.25 + 1.25 + 1/3) = 5.43 m3t

5.43 * 10 6
δ L.L. = = 0.93 cm < 550/300 = 1.67cm
2100 * 2770

7/7-Stairs
PROJECTS
 Portal Frame I

General Layout

vertical bracing
horizontal bracing ? = 30 - 60

Portal frame ( main system)

B

purlins

spacing
(4-8m)
S

end gable
4 - 8m
h

L
h > 6.0 m
Used if

1/26 P.F.I
2007-2008
 Portal Frame I

1.5 - 2.0 m purlin

10:1

L/10 - L/20
1.5 - 2.0 1.0 m
1.0 - 1.5

2.0 - 3.0

span = L

Main frame

corrugated steel purlin

10:1
side girts

window

end girts gate

4.0-8.0 m
end gable column

End Gable Section

2/26 P.F.I
2007-2008
 Portal Frame I

Design Of Portal Frame

Loaded area on each

Frame

h S

1-Loads on portal frame:

1) Dead load :
• O.W. of the steel structure which include ( purlins, rafters, bracings)
WS = (20 – 35) kg/m2 "depend on span L ".

• Weight of steel cover

WC = (5 – 8) kg/m2 for single layer.
= (12 – 18) kg/m2 for double layer.

Wd.L. = ( (WC / cos ) + WS ) x S

2) Live load :
• L.L. = 60 – 66.67 tan "for inaccessible roofs"
= 200 – 300 tan "for accessible roofs"

WL.L. = L.L. x S

3/26 P.F.I
2007-2008
 Portal Frame I

3) Wind load :
Wwind = ( Ce x K x q ) x S
W W
2 3
K= 1.0 for h 10 m
= 1.1 for h 20 m
q = 70 kg/m2 in Cairo C
2
C
3

Pressure side
W C Pressure Suction C W
o Vertical surface " surface 1" 1 1 4 4
side side
Ce = +0.8
W1 = 0.8 x 1.0 x 70 x S

o Inclined Surface " surface 2"

Ce = depends on inclination of surface
For tan < 0.4 - Ce = -0.8
For tan > 0.8 - Ce = +0.8
For 0.8 < tan < 0.4 - Ce = we have 2 values

For slopes 5:1 to 20:1 tan < 0.4 Ce = -0.8

W2 = -0.8 x 1.0 x 70 x S
Ce

Suction side +0.8

o Vertical surface " surface 3"
Ce = -0.5
W3 = -0.5 x 1.0 x 70 x S
0.4 tan
0.8
o Inclined Surface " surface 4"
Ce = -0.5 -0.8
W4 = -0.5 x 1.0 x 70 x S

2-Straining actions in portal frames

We solve the frame as 2 hinged frame using computer program or by using virtual work
method.
"In case of using computer program take Icol = (2 – 3) Igirder"

* In case of neglecting Wind Loads we can calculate approximate value of B.M.

M-ve = WT x L2 / (13 – 15)

M -ve 2 M -ve

1 M +ve 3
M +ve = (0.55 – 0.60) x M-ve 4
Ncol = WT x L / 2
Case (A) D.L. + L.L.

4/26 P.F.I
2007-2008
 Portal Frame I

Wind effect

3
2 2
3
1 1
4

Case of wind Case of wind

Neglected May be critical

For all sections 1, 2&3 For sections 1&2

Taking wind load will decrease the Taking wind load will decrease the
moment moment
So neglect the wind effect So neglect the wind effect

For sections 3&4

Taking wind load will increase the
moment
So we must study the effect of that
increase

MA = MD + ML Case A
MB = MD + M L + M W Case B

If (MB – MA) / MA < 20% neglect

wind

Design on MA case A, else design on

MB case B

5/26 P.F.I
2007-2008
 Portal Frame I

Design of rafter

a
M 3
Mh
M 2

* We will divide the rafter in

M1
3 sections , M2 max +ve
M3 max –ve X
h
Mh at end of haunch X
o

* The design of rafter is based

on the Mmax of M2 or Mh ,
we will use haunch to safeguard
section M3.

* The rafter is subjected to

both M , N
but we can neglect N as it is very small.
WT
Calculation of haunch length and zero moment point
x tan
Sec 5-5: End of the haunch
To get the position of sec 5-5 ( position of the end of the haunch )
If not given, we have to calculate the M-ve at any point on the rafter. x
h
From the figure by taking a section at distance X from column

Mx = Q ( h + X tan ) + WT X2 / 2 – Y x X Q

1 to get Xh put Mx = M 2 Y
2 to get the point of Zero Moment “beginning of the –ve moment zone " Xo
put Mx = zero

• L / 20 Xh L /10

Mmax in the max of Mh and M2

1- Choice of Section:

M max
fb = ≤ Fbcx
Sx
Assume fb = 1.536 t/cm2 " compact section "
M
and get the Sx required = max cm3
1.536
and from the tables we choose the appropriate IPE .

6/26 P.F.I
2007-2008
 Portal Frame I
2- Check

1- Check for compact Section

2- Lumax = min of 20 b f / fy

1380 Af Cb / ( fy d ) " Cb = 1.3 "

Sec 2-2: Max +ve

Luact = a " at positive moment zone,compression is on upper flange purlins are
laterally supporting the upper flange "

Luact = a < Lumax so Fbcx = ( 1.4 if non compact

1.536 if compact sec ).
> Lumax so Fbcx = Fltb

Fbx = M2 / Sx < Fbcx

Sec 5-5: End of the haunch

Luact = Xo " at negative moment zone,compression is on lower flange "

Luact = Xo < Lumax so sec 5-5 is safe as sec 2-2

> Lu max so use knee bracing at every purlin to decrease Luact = a

and so Fbcx = ( 1.4 if non compact sec

1.536 if compact sec
Fltb if Luact = a > Lu max ).

L 80X80X8

Purlin

L 60X60X6
PL 10MM Y

h
Sec 3-3: Max -ve
Ix = tw x (2h)3 / 12 + 2 ( bf x tf )( h – tf / 2 )2
Sx = Ix / y = Ix / h X X
Fbcx = ( 1.4 if non compact
h
1.536 if compact sec ).
tf
Fbx = M3 / Sx < Fbcx
bf
Y
7/26 P.F.I
2007-2008
 Portal Frame I

Design of column

Choose the critical section 1 " without wind " or 4 " with wind "

1-Choice of Section:

Mx
fb = ≤ Fbcx
Sx
Mx
Assume fb = (0.8 to 1.2) t/cm2 and get the Sx required = cm3
fb
and from the tables we choose the appropriate IPE .

2-Check

f ca f bcx
Finally get fca, Fc, fbcx, Fbcx and A1 and check + × A1 ≤ 1.00 as previously described
Fc Fbcx
in design of rolled columns.

8/26 P.F.I
2007-2008
 Portal Frame I

End Gable Column

The end gable column is the supporting element of the end girts, its statical system is hinged
base and roller connection with the rafter as shown in figure, so there is no normal force transmitted
from the rafter to the end gable.

Design Procedure:

1- The acting loads:

a. Dead Load:
i. Own weight (40 – 60) kg/m’.
ii. Weight of the steel sheets wc = (5-8) kg/m2 for single layer.
iii. Weight of the end girts wg = (10 – 20) kg/m’.

b. Wind load:
i. In this case the wind load will be a main load so it will be a case A
ii. Wwind = ((Ce + Ci ) x K x q) x S1

Ce = 0.8 for pressure case, 0.5 for suction case

Ci = 0.3 for pressure case
K= 1.0 for h ≤ 10 m
= 1.1 for h ≤ 20 m
q = 70 kg/m2 in Cairo

2- The straining actions:

c. Wx = due to wind load only

Wx = Wwind

WX × h2
MX =
8

9/26 P.F.I
2007-2008
 Portal Frame I

d. N = due to dead loads only

N = wc x (S1 x (h - hwall)) + o.w. x h + wg x ( number of girts ) x S1
Where
S1 = the span of end girts = spacing between end gables columns.
h = the end gable height.

3- Choice of Section:
M
f b = x ≤ Fbcx
Sx
Mx
Assume fb = 1.2 t/cm2 and get the Sx required = cm3
1.2
And from the tables we choose the appropriate IPE.

4- Check

1- lbin = lby = the bigger of ( distance between end girts or height of wall ) .
2- lbout = lbx = h

3- Check for compact Section

4- Luact = distance between end girts or height of wall for pressure case.
= h "for case of suction (compression flange is the inner flange), we can use knee
bracing to reduce it."

20 × b f
Fy
5- Calculate Lu max = Cb = 1.13
1380 × A f
× Cb
Fy × d
f ca f bcx
6- And finaly get fca ,Fc , fbcx , Fbcx and A1 and check + × A1 ≤ 1.00 as previously
Fc Fbcx
described in design of rolled columns.

10/26 P.F.I
2007-2008
 Portal Frame I

Example:

It is required for the shown industrial building to:-

42 m

24 m

a. Draw with a suitable scale a complete layout for the main system "2 hinged steel frame" of
clear height 8 m.

b. Calculate the loads on the frame for the diff. cases of loading.

c. Design the Col, Rafter.

d. Design the end gable column and draw its connection with the rafter.

e. Design and draw the connection between the rafter and the column
Use M24 grade (10.9) bolts, T = 22.23 ton Ps = 7.11 ton.

f. Design and draw the connection between the two rafters

Use M24 grade (10.9) bolts, T = 22.23 ton Ps = 7.11 ton.

11/26 P.F.I
2007-2008
 Portal Frame I
Solution:-

A ) The general layout.

2.0 m

8
7
6
5
42.0 m
4
3
2
6.0 m

6.0 m
1

6.0 m
4.0 m 4.0 m
24.0 m

Vertical Bracing Plan

12/26 P.F.I
2007-2008
 Portal Frame I
2.0 m

10:1

1.5 m 1.0 m
8.0 m
1.0 m

2.5 m

24.0 m

Elevation
@ axis 2 to 7
10:1

1.5 m
8.0 m
1.0 m

2.5 m

6.0 m
24.0 m

End Gable Section

@ axis 1

1.5 m
8.0 m
1.0 m

2.5 m

6.0 m
24.0 m

End Gable Section

@ axis 8

13/26 P.F.I
2007-2008
 Portal Frame I
B – Loads on frame:
tan = 0.1 = 5.71o
1) Dead load :
• WS = 25 kg/m2
• WC = 5 kg/m2

Wd = ( (WC / cos ) + WS ) x S = ( (5 / cos 5.71) + 25 ) x 6 = 180 kg/m

2) Live load :
• L.L. = 60 – 66.67 tan = 60 – 66.67 x 0.1 = 53.33 kg/m2
WL = L.L. x S = 53.33 x 6 = 320 kg/m

3) Wind load :
Wwind = ( Ce x K x q ) x S
W W
2 3
K= 1.0 for h 10 m
= 1.1 for h 20 m
q = 70 kg/m2 in Cairo C
2
C
3

Pressure side
W C Pressure Suction C W
o Vertical surface " surface 1" 1 1 4 4
side side
Ce = +0.8
W1 = 0.8 x 1.0 x 70 x 6 = 340 kg/m

o Inclined Surface " surface 2"

tan = 0.1 < 0.4 Ce = -0.8
W2 = -0.8 x 1.0 x 70 x 6 = -340 kg/m

Suction side
o Vertical surface " surface 3"
Ce = -0.5
W3 = -0.5 x 1.0 x 70 x 6 = - 200 kg/m

o Inclined Surface " surface 4"

Ce = -0.5
W4 = -0.5 x 1.0 x 70 x 6 = -200 kg/m

WT = Wd + WL = 0.32 + 0.18 = 0.5 t/m

14/26 P.F.I
2007-2008
 Portal Frame I

Original system
W2 W
3
WT

W Pressure Suction W
1 side side 4

Case (A) D.L. + L.L. Case (B) W.L.

Modified system
W2 W
3
WT

W W
1 4

4.32 ton

6 ton 6 ton 3.53 ton 1.84 ton

Case (A) D.L. + L.L. Case (B) W.L.

23.7 mt 0.64 mt 6.4 mt

9 mt 36 mt

Mo Total load Mo Wind load

9.2 mt
6.12 mt 3.6 mt
8 mt 8 mt 23.7 mt 0.64 mt 6.4 mt

2.72 mt 1.6 mt

1 ton
M1 Mo Wind load

15/26 P.F.I
2007-2008
 Portal Frame I

From case A .( Dead + Live Only )

10 = 2 ( 36 x 12 / 2 x –( 8 + 2/3 x 1.2 ) + ( 2/3 x 9 x 12 ) x –( 8/2 + 9.2/2 )) = -5040

11 = 2 ( 8 x 8 / 2 x 2/3 x 8 + 8 x 12 x ( 8/2 + 9.2/2 ) + 1.2 x 12/2 x (8 + 2/3 x 1.2 )) = 2119.25

10 + X1 x 11 =0
X1 = 2.378

From case B .( Wind Load)

10 = - 2/3 x 2.72 x 8 x 4 – 23.7 x 8/2 x 2/3 x 8

+ 2/3 x 6.12 x12 x 8.6 – 23.06 x 12/2 x 8.4 – 0.64 x 12 x 2 x 8.6
+ 2/3 x 3.6 x 12 x 8.6 – 5.75 x 12/2 x 8.4
+ 2/3 x 8 x 1.6 x 4 – 6.4 x 8/2 x 2/3 x 8 = -1581.4

11 = 2 ( 8 x 8 / 2 x 2/3 x 8 + 8 x 12 x ( 8/2 + 9.2/2 ) + 1.2 x 12/2 x (8 + 2/3 x 1.2 )) = 2119.25

10 + X1 x 11 =0
X1 = 0.746

Final Moment

6.22 mt
20mt 2 20 mt 3
2

1 13 mt 3 1
4
17.7 mt 0.432 mt
Case (A) D.L. + L.L.
Case of wind
neglected

16/26 P.F.I
2007-2008
 Portal Frame I
Design of rafter:

M2 = 13 mt M3 = 20 mt N = neglected 0.5

Calculation of haunch length and zero moment point x tan

Sec 5-5: End of the haunch

x
2 8
Mx = Q ( h + X tan ) + WT X / 2 – Y x X

20/8 = 2.5
Mx = 2.5 ( 8 + X x 0.1 ) + 0.5 X2 / 2 – 9 x X

= 0.25 X2 – 5.75 X + 20 6
1 to get Xh put Mx = M2
13 = 0.25 Xh2 – 5.75 Xh + 20
Xh = 1.29 take length of haunch = 1.5 m

2 to get the point of Zero Shear "beginning of the –ve moment zone" Xo
Put Mx = zero
0 = 0.25 Xo2 – 5.75 Xo + 20
Xo = 4.27 m

24 / 20 = 1.2 m < Xh = 1.5 m < 24 / 10 = 2.4 m so Mmax = M2 = Mh = 13 mt

Choice of sec:

Assume f = 1.536 t/cm2 Sx = 13 x 100 / 1.536 = 850 cm3

Choose I.P.E 360

Check:

Dw / tw = ( 36 – 4 x 1.27 ) / 0.8 = 38.6 < 82

C / tf = 8.5 / 1.27 = 6.7 < 10.9

Lumax = 20 bf / Fy = 20 x 17 / 2.4 = 219

Sec 2-2: M+ve

Luact = a = 200 cm < Lumax = 219

Sec is compact

Fbcx = 1.536 t / cm2

Fbx = Mx / Sx = 1300 / 904 = 1.438 t / cm2 < Fbcx = 1.536 t / cm2 Safe

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Sec 5-5: End of the haunch

Y
Luact = Xo = 4.27 m > Lu max
so use knee bracing at each purlin Luact = 2 m < 2.19 m

and so Fbcx = 1.536 t/cm2 safe as sec 2-2 36

X X
Sec 3-3: Max -ve
Ix = tw x (2h)3 / 12 + 2 ( bf x tf )( h – tf / 2 )2 36
Ix = 0.8 x (72)3 / 12 + 2 ( 17 x 1.27 )( 36 – 1.27 / 2 )2
= 78887 cm4 1.27
Sx = Ix / y = 78887 / 36 = 2191.6 cm3 17
Fbcx = 1.536 t/cm2 Y

Fbx = M3 / Sx = 2000 / 2191.6 = 0.91 t/cm2 < Fbcx

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Design of column
Choice of sec 1 – 1 :

Mx = 20 mt N = 6 ton

Assume f = 1.2 t/cm2 Sx = 20 x 100 / 1.2 = 1666.6 cm3

Choose I.P.E 500

Check:

Fca / Fc + A1 fbcx / Fbcx 1 case A

Fca = 6 / 116 = .052 t/cm2

fbcx = Mx / Sx = 2000 / 1930 = 1.036 t / cm2

Fc:
Lbin = K x 8

GA = 10 hinged base
GB = ( Ic / Lc ) / ( Ig / Lg ) = ( 48200 / 8 ) / ( 16270 / 24 ) = 8.8 K = 2.90

Lbin = 2.90 x 8 = 23.20 m, Lbout = 4.0 m

in =Lbx / rx = 2330 / 20.4 = 113.7 < 180

out =Lby / ry = 400 / 4.31 = 92.8 < 180
max = 113.7 > 100

Fc = 7500 /(113.7)2 = 0.58 t/cm2

Fca / Fc = .09

Fbcx :
1- Local buckling "compact and non-compact"
C = 10 cm , dw = 50 – 4 x 1.6 = 43.6

For flange C / tf = 10 / 1.6 = 6.25 < 10.9 compact

1 6 699 2.4
For Web , = ( + 1) =0.528 > 0.5 compare with = 76.6
2 43.6 × 1.02 × 2.4 (13α − 1)

Dw / tw = 43.6 / 1.02 = 42.7 < 76.6 compact

Regarding local buckling section is compact

2- L.T.B.

Luact = 400 cm > Lumax = 20 bf / Fy = 20 x 20 / 2.4 = 258.2

Section is non compact

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Column height is devided out plane in 2 parts by the hz member, critical one is sec 1 .

= Msmall / Mbig = - 10 / 20 = -.5

20
2 1
Cb = 1.75 + 1.05 + .3 = 1.3

Fltb = 800 Af Cb / ( Lu d ) = 800 x 20 x 1.6 x 1.3 / ( 400 x 50 ) = 1.66 > 1.4

10
2
Fbcx = 1.4 t / cm

A1:

Fca / Fc = .09 < .15 A1 = 1.0

Applying interaction equation :

.09 + 1.0 x 1.036 / 1.4 = 0.83 < 1 "case A" safe.

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D ) End gable colum:

1- The acting loads:

h = 8 + 12 x 0.1 = 9.2 m
a. Dead Load:
i. Own weight 50 kg/m’.
ii. Weight of the steel sheets wc = 6 kg/m2.
iii. Weight of the end girts wg = 20 kg/m’.

b. Wind load:
i. In this case the wind load will be a main load so it will be a case A
ii. Wwind = ((Ce + Ci ) x K x q) x S1

Ce = 0.8, Ci = 0.3
K= 1.0 for h ≤ 10 m
q = 70 kg/m2 in Cairo

2- The straining actions:

a. Case of wind pressure

Wx = due to wind load only
Wx = Wwind = 1.1 x 1 x 70 x 6 = 0.462 t/m'

WX × h2
MX = = 0.462 x 9.22 / 8 = 4.88 mt
8
b. Case of wind suction
Wx = due to wind load only
Wx = Wwind = 0.5 x 1 x 70 x 6 = 0.21 t/m'

WX × h2
MX = = 0.21 x 9.22 / 8 = 2.22 mt
8

c. N = due to dead loads only

N = wc x (S1 x ( h – hwall)) + o.w. x h + wg x ( number of girts ) x S1
N = (6 x (6 x 4.2) + 50 x 9.2 + 20 x ( 5 ) x 6)/1000 = 1.21 ton

Choice of sec:
From wind pressure case
Mx = 4.88 mt N = 1.21 ton

Assume f = 1.2 t/cm2 Sx = 4.88 x 100 / 1.2 = 406 cm3

Choose I.P.E 270

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Check:

1- Wind pressure case:

Fca / Fc + A1 fbx / Fbcx 1

fca = 1.21 / 45.9 = .026 t/cm2

fbcx = Mx / Sx = 488 / 429 = 1.13 t / cm2

Fc:
Lbin = 2.5 m "wall height" , Lbout = 9.2 m
in = Lby / ry = 250 / 3.02 = 82.8 < 180
out = Lbx / rx = 920 / 11.2 = 82.14 < 180
max = 83 < 100

Fc = 1.4 – 6.5 x 10 -5 (83)2 = 0.95 t/cm2

Fca / Fc = .027

Fbcx :
1- Local buckling "compact and non-compact"
C = 6.4 cm , dw = 27 – 4 x 1.02 = 22.92

For flange C / tf = 6.4 / 1.02 = 6.27 < 10.9 compact

1 1.21 699 2.4

For Web, = ( + 1) =0.51 > 0.5, compare with = 78.33
2 22.92 × 0.66 × 2.4 (13α − 1)

d w / tw = 22.92 / 0.66 = 34.72 < 78.33 compact

Regarding local buckling section is compact

2- L.T.B.

Luact = 250 cm > Lumax = 20 bf / Fy = 20 x 13.5 / 2.4 = 174.3

Section is non compact

Cb = 1.13

Fltb = 800 Af Cb / ( Lu d ) = 800 x 13.5 x 1.02 x 1.13 / ( 250 x 27 ) = 1.84 > 1.4

Fbcx = 1.4 t / cm2

A1:
Fca / Fc = .027 < .15 A1 = 1.0

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Applying interaction equation :

.027+ 1.0 x 1.13 / 1.4 = 0.84 < 1 "case A" safe.

2- Wind suction case:

fbcx = Mx / Sx = 222 / 429 = 0.517 t / cm2

Fbcx :

Regarding local buckling section is compact as before

1- L.T.B.

Luact = 920 cm > Lumax = 20 bf / Fy = 20 x 13.5 / 2.4 = 174.3 cm

Section is non compact ,Cb = 1.13

Fltb = 800 Af Cb / ( Lu d ) = 800 x 13.5 x 1.02 x 1.13 / ( 920 x 27 ) = 0.5 > 1.4

Try to reduce Luact by using knee bracing each side girt.

Luact = max of (wall height or distance between side girts) = 250

Fltb = 800 Af Cb / (Lu d) = 800 x 13.5 x 1.02 x 1.13 / ( 250 x 27 ) = 1.84 > 1.4
Fbcx = 1.4 t / cm2

A1:
Fca / Fc = .027 < .15 A1 = 1.0

Applying interaction equation :

.027 + 1.0 x 0.517 / 1.4 = 0.39 < 1 "case A" safe.

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E )connection between rafter and column:

Mact = 20 mt , Qact = 6 ton

PL. 10 MM
1
STEEL SHEET
RAIN GUTTER .7MM

IPE 360

PART OF IPE 360

17.0
1
1.27 2.0 fb
4.0 11.27
8.0 f1
36.0
26.73
H = 2 + 36 + 36 + 2 = 76 cm

Ix = B H3 / 12 = 17 x 763 / 12 = 621882.667 cm4

Y = H / 2 = 38 cm 36.0
X1 = 2 + tf + e + P /2 = 2 + 1.27 + 4 + 4 = 11.27 cm
X2 = H / 2 – X1 = 38 – 11.27 = 26.73 cm
2.0
fb = Mmax Y / Ix = 2000 x 38 / 621882.6 = 0.12 t/cm2
f1 = Mmax X2 / Ix = 2000 x 26.73 / 621882.6 = 0.086 t/cm2 SEC 1-1
Check bending stress on bolts :
Text,b1,M = 1 / 2 ( ( 0.12 + 0.08 ) /2 x ( 17 x 11.27 ) ) = 9.886 < 0.8 T = 17.84 safe

Text,b2,M = 1 / 2 ( ( 0.08 + 0 ) /2 x ( 17 x 26.73 ) ) = 9.76 < 0.8 T safe

Check shear stress on bolts :

Qact / n = 6 / 6 = 1 ton < Ps = 7.11 ton safe

Thickness of head plate:

As before

Check on weld between head plate and rafter section:

Properties of the weld :

Assume size of weld S = 8 mm for web
and S flange = 10 mm

Awvl = 4 x 28 x 0.8 = 89.6 cm2

Ix = 2 (17 x 1 x 36.52 + 2 x 6 x 1.0 x (36 – 1.27 – 0.5)2

+ 2 x 0.8 x 283 / 12 + 2 x 0.8 x 28 x 182)
= 108301.4 cm4

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17.0
Checks:

At point 1-

q1 = 0 t/cm2
f1 = 2000 x ( 36 + 1 ) / 108301.4 = 0.69 t/cm2 < 0.72 t/cm2 6.0

At point 2-

q2 = 6 / 89.6 = 0.067 t/cm2

f2 = 2000 x ( 18 + 28 / 2 ) / 108301.4 = 0.59 t/cm2 28.0
R1 = ( 0.592 + 3x0.0672 ) = 0.6 t/cm2 < 0.72 t/cm2 x 1.1

F )connection between2 rafter:

Mact = 13 mt , Qact = 0

1
RIDGE CAP .7MM

L 80X80X8 C 140X65X4

IPE 360 IPE 360

17.0

1.27 2.0
4.0
H = 2 + 36 + 36 + 2 = 76 cm
36.0
Ix = B H3 / 12 = 17 x 763 / 12 = 621882.667 cm4
Y = H / 2 = 38 cm
X1 = 2 + tf + e + P /2 = 2 + 1.27 + 4 + 4 = 11.27 cm
X2 = H / 2 – X1 = 38 – 11.27 = 26.73 cm 26.73
fb = Mmax Y / Ix = 1300 x 38 / 621882.6 = 0.079 t/cm2 36.0
f1 = Mmax X2 / Ix = 1300 x 26.73 / 621882.6 = 0.056 t/cm2 8.0 f1
4.0 11.27
1.27 2.0 fb

Check bending stress on bolts :

SEC 1-1
Text,b1,M = 1 / 2 ( ( 0.079 + 0.056 ) /2 x ( 17 x 11.27 ) ) = 6.46 < 0.8 T = 17.84 safe

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Text,b2,M = 1 / 2 ( ( 0.056 + 0 ) /2 x ( 17 x 26.73 ) ) = 6.36 < 0.8 T safe

Check shear stress on bolts :

Qact / n = 0 / 6 = 0 ton < Ps = 7.11 ton safe

Thickness of head plate:

As before

Check on weld between head plate and rafter section:

Properties of the weld : 17.0

Assume size of weld S = 6 mm

Ix = 2 ( 17 x 0.6 x 36.32 + 2 x 6 x 0.6 x ( 36 – 1.27 – 0.3 )2

+ 2 x 0.6 x 283 / 12 + 2 x 0.6 x 28 x 182 )
= 70213.472 cm4
6.0
Checks:
No shear so only point 1 is critical
At point 1-

q1 = 0 t/cm2 28.0
f1 = 1300 x ( 36 + 0.6 ) / 70213.472 = 0.67 t/cm2 < 0.72 t/cm2

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Side and End grits

Both side and end girts are used to mount the side or end cladding, the difference between
the two is that the side girts are supported on the main column and the end girts are supported on the
end gables column.

Design Procedure:

1- The acting loads:

• Dead Load:
- Own weight (10 – 20) kg/m.
- Weight of the steel sheets wc = (5-8) kg/m2 for single layer.

• Live Load :

- For case of maintenance there is a concentrated load of 100 kg

• Wind load:
- In this case the wind load will be a main load so it will be a case A
- Wwind = ( (Ce + Ci ) x K x q ) x a

Ce = 0.8 , Ci = 0.3
K= 1.0 for h ≤ 10 m
= 1.1 for h ≤ 20 m
q = 70 kg/m2 in Cairo

2- The straining actions:

• Wx = due to wind load only
Wx = Wwind

WX × S 2 W ×S
MX = , QX = X
8 2

• Wy = due to dead loads and live load

Wy = wc x a + o.w.
Py = 100 kg
W × S 2 PY × S W × S PY
MY = Y + , QY = Y +
8 4 2 2

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Where S is:
The span of side girts = spacing between main frames columns.
The span of end girts = spacing between end gables columns.
And a is the spacing between side or end girts.

3- Choice of Section:

Mx My
fb = + ≤ Fbcx
Sx Sy

As the channels are non-compact or slender sections so Fbcx = 1.4 t/cm2

Assume Sx = 7 Sy for hot rolled C sec. or Sx = 6 Sy for cold formed sec.

M x + (6or 7)M y
By solving the first equation we get the Sx required = cm3
1.4
And from the tables we choose the appropriate channel.

4- Checks
1- Bending stress:
M My
fb = x + ≤ Fbcx = 1.4t / cm 2
Sx Sy
2- Shear stress:
Q
q x = x ≤ 0.35Fy
A web
Qy
qy = ≤ 0.35Fy
A flanges

2- Deflection "due to live load only":

P ×S 3 span
δ act = <
48E × IY 300
N.B.:
In case of not satisfied we increase the channel section or use tie rod to reduce S in the Y
direction, and so we decrease My and the deflection.

MY with 1 tie = 1 / 4 MY without tie rod

MY with 2 tie = 1 / 9 MY without tie rod

For case of using cold formed section we must use tie rod, to safeguard the deflection

N.B.
We have to check another case of wind, if wind is suction, Ce = 0.5
Wx = Wwind = 0.5 x K x q x a
W ×S2
MX = X And Luact = span
8

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Rafter Splices

- What is the splice?

It is a connection joining 2 parts of the rafter together.

- Why and where we put the splice?

The splice is put to divide the rafter in parts so the max part length is less than 12 m
to be able to transport it to site.
It is preferred to be as near as possible from the point of zero moment.

- Design straining actions :

As the rafter section is subjected to M , N , Q so the splice will be subjected to the
same straining actions.

• N is small and can be neglected

• Q is the actual shear force at the position of the splice. Qact = Y -WT . X of splice
• M is the max moment capacity of the section. Mmax = Sx x Fbcx "at splice section"

Splice
WT
x tan?
?
< 12 m < 12 m
x of splice
h

L / 2 > 12 m
X
span = L

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2 head plate splice. " using Pretensioned bolts "

M max

Q act
B
2 fb
tf
X1 e
p f1
X
2 h H
I

Pretension bolts: “as before page 9 to 16 E.C.2”

Arrange bolts
As the connection is designed on max moment, so it must be symmetric.

By using pretensioned bolts the head plate is fully effective, so Ix = B H3 / 12, Y = H / 2

fb = Mmax Y / Ix = 6 Mmax / B H2

f1 = Mmax X2 / Ix X1 = 2 + tf + e + P /2
X2 = H / 2 – X 1

Check bolts : " Always N is neglected for bolts as it decreases tension on bolts"
Text,b1,M = 1 / 2 ( ( fb + f1 ) /2 x ( B x X1 ) ) 0.8 T

Text,b2,M = 1 / 2 ( ( f1 + 0 ) /2 x ( B x X2 ) ) 0.8 T

Check shear stress on bolts :

Qact / n Ps n is the total number of bolts.

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Check on weld between head plate and rafter section:

Properties of the weld :

b f -2S
Awhz = ( bf – 2S ) S x 2 + 4 x 0.4 x bf x S
Awvl = 2 x 0.8 x h x S
Awtot = Awhz + Awvl 0.4 b f

Ix = 2 ( S x (0.8h)3 / 12 ) + 2 ( b f – 2S ) S x (( h + S ) / 2)2 0.8h

+ 4 ( 0.4 bf x S ) x ( h / 2 – S / 2 – tf )2

Checks:

At point 1-

q1 = 0
f1 = N / Awtot + Mmax x ( h /2 + S ) / Ix 0.72 t/cm2

At point 2-

q2 = Qact / Awvl
f2 = N / Awtot + Mmax x ( 0.8 h / 2 ) / Ix
R2 = ( f22 + 3q22 ) 0.72 t/cm2 x 1.1

N.B.: if N is neglected put N=0

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Example:

It is required for the previous example:

a. Design and draw the rafter splice at 4m from column

Use M24 grade (10.9) bolts.

b. Design Side girts as : 1 – Hot rolled C section

2 – Cold formed C section

Solution:

a. Rafter splice:

At splice sec:-
Qact = Y -WT . X of splice = 6 – 0.5 x 4 = 4 ton
Mmax = Sx x Fbcx = 1.536 x 904 = 1388 cmt

Splice Type I

17.0
1.27 2 fb
X1 4.0
8.0 f1 M
X2
36 40

H = 36 + 2 + 2 = 40 cm
Q act
Ix = B H3 / 12 = 17 x 403 / 12 = 90666.6 cm4
Y = H / 2 = 20 cm
X1 = 2 + tf + e + P /2 = 2 + 1.27 + 4 + 4 = 11.27 cm
X2 = H / 2 – X1 = 20 – 11.27 = 8.73 cm
fb = Mmax Y / Ix = 1388 x 20 / 90666.66 = 0.3 t/cm2
f1 = Mmax X2 / Ix = 1388 x 8.73 / 90666.66 = 0.13 t/cm2

Check bending stress on bolts :

Text,b1,M = 1 / 2 ( ( 0.3 + 0.13 ) /2 x ( 17 x 11.27 ) ) = 20.6 > 0.8 T = 17.84 unsafe

Text,b2,M = 1 / 2 ( ( 0.13 + 0 ) /2 x ( 17 x 8.73 ) ) = 4.82 < 0.8 T safe

Try to safeguard the first row by putting 4 bolts / row ,

but B / 2 = 17 / 2 = 8.5 cm < tw / 2 + 2 e + p = 0.4 + 2 x 4 + 8 = 16.4 we cant use 4 bolts / row.
Increase H and move second row of bolts up the flange

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4.0
fb
H = 36 + 8 + 8 = 52 cm 8.0
4.0
1.2 7
4.0 f1
Ix = B H3 / 12 = 17 x 523 / 12 = 199194.66 cm4
Y = H / 2 = 26 cm 52.0 36.0

X1 = tf / 2 + 2e = 1.27 / 2 + 2 x 4 = 8.63 cm
X2 = H / 2 – X1 = 26 – 8.63 = 17.37 cm 8.0
fb = Mmax Y / Ix = 1388 x 26 / 199194.66 = 0.181 t/cm2
f1 = Mmax X2 / Ix = 1388 x 17.37 / 199194.66= 0.12 t/cm2
S T . P L 1 0m m
Check bolts :
Text,b1,M = 1 / 2 ( ( 0.181 + 0.12 ) /2 x ( 17 x 8.63 ) ) = 11 < 0.8 T = 17.84 safe

Text,b2,M = 1 / 2 ( ( 0.12 + 0 ) /2 x ( 17 x 17.37 ) ) = 8.8 < 0.8 T safe

Check shear on bolts :

Qact / n = 4 / 8 = 0.5 ton < Ps = 7.11 ton safe

Thickness of head plate:

We take a strip of width b/2 = 17/2 = 8.5cm, we calculate the

moment from this strip Mp = force in bolt x e = 11 x 4 = 44cmt
MP t 44 t
fb = ×( P ) = × ( P ) = Fb = 0.72Fy t/cm2
h ×t 3
2 8 ×t P
3
2
( 1 P) ( )
12 12
so tp = 4.4 cm very big

Check on weld between head plate and rafter section:

Properties of the weld: 0.8x8

Assume size of weld S = 8 mm for web
and S flange and stiff = 12 mm
Awhz = 8 x 0.4 x 17 x 1.2 = 65.28 cm2
Awvl = 2 x 0.8 x 36 x 0.8 + 0.8 x 8 x 4 x 1.2 = 76.8 cm2
Awtot = Awhz + Awvl = 142.08 cm2 0.8x36

Ix = 2 (0.8 x (0.8 x 36)3 / 12) + 4 (0.4 x 17 x 1.2) x ((36 + 1.2) / 2)2

+ 4 (0.4 x 17 x 1.2) x ( 36 / 2 – 1.2 / 2 – 1.27)2
+ 4 (1.2 x (0.8 x 8)3 / 12) + 4 (0.8 x 8 x 1.2) x (36 / 2 + 0.5 x 8)2
= 37942.7 cm4

Checks:
At point 1- 0.4x17

q1 = 4 / 76.8 = 0.052 t/cm2

f1 = 1388 x (18 + 8) / 37942.7 = 0.95 t/cm2
R1 = (0.952 + 3x0.0522) = 0.96 t/cm2 > 0.72 t/cm2 x 1.1 try to increase stiff length and recheck

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b. Design of Side girts

1- As hot rolled C section

1- The acting loads:

• Dead Load:
- Own weight 20 kg/m.
- Weight of the steel sheets wc = 6 kg/m2 .

• Live Load :

- Py = 100 kg

• Wind load:
- Wwind = ( (Ce + Ci) x K x q ) x a

Ce = 0.8, Ci = 0.3
K= 1.0 for h ≤ 10 m
q = 70 kg/m2 in Cairo

2- The straining actions:

• Wx = due to wind load only

Wx = Wwind = 1.1 x 1 x 70 x 1.5 = 115.5 kg /m = 0.115 t/m'

WX × S 2 0.115 × 62
MX = = = 0.517 mt
8 8
W ×S 0.115 × 6
QX = X = = 0.345 t
2 2

• Wy = due to dead loads and live load

Wy = wc x a + o.w. = 6 x 1.5 + 20 = 29 kg/m = 0.029 t/m'
Py = 100 kg = 0.1 t
W × S 2 PY × S 0.029 × 62 0.1× 6
MY = Y + = + = 0.28 mt
8 4 8 4
W × S Py 0.029 × 6 0.1
Qy = y + = + = 0.137 t
2 2 2 2

3- Choice of Section:

Mx My
fb = + ≤ Fbcx
Sx Sy

As the channels are non compact sections so Fbcx = 1.4 t/cm2

Assume Sx = 7 Sy for hot rolled sec.

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M x + 7M y
Sx required = = (51.7 + 7 x 28) / 1.4 = 176.9 cm3
1.4
Choose C 200

4- Check
1- Bending stress:
Luact = zero as the compression flange is fully laterally supported by the corrugated sheets.
Section is non-compact as we are using channels
Fbcx = 1.4 t/cm2
M M 51.7 28
fb = x + y = + = 1.3 < Fbcx = 1.4t / cm 2
Sx Sy 191 27

2- Shear stress:
Q 0.345
qx = x = = 0.02 ≤ 0.35Fy
A web 20 × 0.85
Qy 0.137
qy = = = 0.008 ≤ 0.35Fy
A flanges 2 × 7.5 × 1.15

3- Deflection "due to live load only":

P ×S 3 0.1× 6003 span 600

δ act = = = 1.44cm < = = 2cm
48E × IY 48 × 2100 ×148 300 300

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2- As cold formed C section

1- The acting loads:

• Dead Load:
- Own weight 10 kg/m.
- Weight of the steel sheets wc = 6 kg/m2.

• Live Load :

- Py = 100 kg

• Wind load:
- Wwind = (( (Ce + Ci) x K x q ) x a

Ce = 0.8, Ci = 0.3
K= 1.0 for h ≤ 10 m
q = 70 kg/m2 in Cairo

2- The straining actions:

We will use 2 tie rods
• Wx = due to wind load only
Wx = Wwind = 1.1 x 1 x 70 x 1.5 = 115.5 kg /m = 0.115 t/m'

WX × S 2 0.115 × 62
MX = = = 0.517 mt
8 8
W ×S 0.115 × 6
QX = X = = 0.345 t
2 2

• Wy = due to dead loads and live load

Wy = wc x a + o.w. = 6 x 1.5 + 10 = 19 kg/m = 0.019 t/m'
Py = 100 kg = 0.1 t
As there is 2 tie rods S = 6/3 = 2.0 m
W × S 2 PY × S 0.019 × 22 0.1× 2
MY = Y + = + = 0.0595 mt
8 4 8 4
W × S Py 0.019 × 2 0.1
Qy = y + = + = 0.07 t
2 2 2 2
3- Choice of Section:

Mx My
fb = + ≤ Fbcx
Sx Sy

As the channels are non compact sections so Fbcx = 1.4 t/cm2

Assume Sx = 6 Sy for cold formed sec.
M x + 6 M y 51.7 + 6 × 5.95
Sx required = = = 62.5 cm3
1.4 1.4
Choose C180x75x4

10/12-P.F.II
2007-2008
 Portal Frame II
4- Checks:-
a- Code limits for slender sections.

For web subjected to moment

h = H – 2r – 2 t = 180 – 2(6)-2(4) = 160 mm
h 160
= = 40 < 200 Ok
t 4

For Unstiff. flange

b = B – r – t = 75 – 6 – 4 = 65 mm
b 65
= = 16.25 < 40 Ok
t 4

Section satisfies code limits.

b- Determine the effective parts of the section.

1- Flange
Unstiffened flange subjected to compression = 1 , K = 0.43

b = 65 mm
b t Fy 65 4 2.4
λP = = = 0.87
44 K σ 44 0.43
λ − 0.15 − 0.05ψ 0.87 − 0.15 − 0.05 ×1
ρ= P = = 0.885 < 1
λP2 0.87 2

be = b = 0.885 x 65 = 57.53 mm

2- Web
Stiff. Web subjected to moment = -1 , K = 23.9

h = 160 mm
b t Fy 160 4 2.4
λP = = = 0.29
44 K σ 44 23.9
λP − 0.15 − 0.05ψ 0.29 − 0.15 − 0.05 × −1
ρ= = = 2.25 > 1
λP2 0.292

The web is fully effective.

As the flange is not fully effective so we must calculate the new

properties of the section.

IX eff = IX table – IX of the reduced part = 606.25 – 0.747x0.4 (9-0.2)2 = 583.11 cm4

0.4×0.7473 0.747 2
IY eff = IY table – IY of the reduced part = 67.2 – ( + 0.4×0.747×(7.5 −1.9 − ) ) = 59cm4
12 2

11/12-P.F.II
2007-2008
 Portal Frame II

a- Check bending stress.

Luact = zero as the compression flange is fully laterally supported by
the corrugated sheets.
Fbcx = 1.4 t/cm2

MX MY 51.7 5.95
fbc = y + x = 9+ (7.5 − 1.9) =1.36 < 1.4 t/cm2
IX IY 583.11 59
b- Check shear stress.

Very small shear force, shear stress will be safe

Q 0.345
qx = x = = 0.05 ≤ 0.35Fy
A web 18 × 0.4
Qy 0.07
qy = = = 0.012 ≤ 0.35Fy
A flanges 2 × 7.5 × 0.4 − 0.7 × 0.4

c- Check deflection. "due to live load only"

P ×S 3 0.1× 2003 span 200

δ act = = = 0.13cm < = = 0.66cm
48E × IY 48 × 2100 × 59 300 300

12/12-P.F.II
2007-2008
DRAWING
 Drawing and remarks

The following frame is key elevation:

1/12
 Drawing and remarks

Detail A:

2/12
Drawing and remarks

3/12
 Drawing and remarks

Detail B:

4/12
Drawing and remarks

5/12
 Drawing and remarks

Detail C:

6/12
Drawing and remarks

7/12
 Drawing and remarks

General remarks:
1) For fixed base:
Take effect of horizontal shear on the vertical weld
Shear on weld is bigger of:
−Y M Y M
Q= + or 0.6 ( + )
2 d 2 d
On 2 welds from one side
Normal on weld X
N=X on 4 welds Y
Q N M
q= f=
2 sh 4 sh
"h" is the height of side plate.

M
- Y+ M Y+ M
2 d 2 d

h
x

8/12
 Drawing and remarks

2) Hinged base:

Hinged base
Gusset plate connected to the web of the I-beam

9/12
Drawing and remarks

Hinged base

10/12
 Drawing and remarks

Fixed base:

Fixed base
Gusset plate connected to the side plate

11/12
Drawing and remarks

Fixed base

12/12
 SHEAR
CONNECTION
 LECTURES IN STEEL STRUCTURES DESIGN – 1.

DESIGN OF STEEL CONNECTIONS

Types of steel connections:

1. Riveted connections.
2. Bolted connections:
3. Welded connections.

Bolted Connections:

 Bolts Diameter:

Bolts used in steel structures are with diameters ranges from 12 mm to 36 mm.

 Bolts Grades:

Bolts grades are:

Bolt Grade 4.6 4.8 5.6 5.8 6.8 8.8 10.9

Fyb (t/cm2) 2.40 3.20 3.00 4.00 4.80 6.40 9.00
Fub (t/cm2) 4.00 4.00 5.00 5.00 6.00 8.00 10.00

What is meant by a bolt has a grade 4.60?

 This mean that this bolt has ultimate strength = 4.0 t/cm2, and the ratio between
the yield stress to the ultimate stress (Yield stress/ultimate stress) = 0.60

What is meant by bolt M20 (4.6)?

 This means that this bolt is with diameter 20 mm and its ultimate strength is 4.0
t/cm2 and the ratio between the yield strength to the ultimate strength is 0.60.
 ‫‪LECTURES IN STEEL STRUCTURES DESIGN – 1.‬‬

‫‪Bolts used in structures can be classified into two classes according to their grades:‬‬

‫‪(,‬المسامير العادية) ‪a) ordinary bolts‬‬

‫‪ ‬هي المسامير ذات الرتب ‪ 6.4 ,6.4 ,6.4 ،6.4‬و ‪.4.4‬‬
‫‪ ‬يتم ربط المسامير بمفتاح عادي وال يتولد شد سابق في المسمار اثناء ربطة وتنتقل القوة من عضو آلخر من‬
‫خالل القص‪.‬‬
‫‪ ‬يسمى هذا النوع من الوصالت )‪.(Bearing Type‬‬
‫‪(.‬المسامير عالية المقاومة) ‪b) High strength bolts HSB‬‬
‫‪ ‬هي المسامير ذات المقاومة العالية‪ ،‬وذات الرتب ‪ 4.4‬و ‪9..1‬‬
‫‪ ‬يتم ربط المسامير بمفتاح مخصوص به مؤشر لقيام العزم (‪ )Ma‬الالزم لربط المسمار‪.‬‬
‫‪ ‬أثناء ربط المسمار يتولد شد سابق (‪ )Pre tension force‬في المسمار قيمته ‪ T‬وينتج عن ذلك مقاومة إحتكاك‬
‫قيمتها ‪ Ps‬بين األلواح المربوطة تقاوم القوة الخارجية‪.‬‬
‫‪ ‬تسمى هذه الوصالت (‪.)Friction Type‬‬

‫‪Bolts subjected to shear force:‬‬

‫‪I. Single Shear:‬‬

‫‪Failure in connection will be due to:‬‬

‫‪ Shear in bolts:‬‬

‫𝑏𝑞 × ∅𝐴 = 𝑠𝑠𝑅‬

‫‪𝜋 × ∅2‬‬
‫= ∅𝐴‬
‫‪4‬‬
‫𝑏𝑢𝐹 × ‪𝑞𝑏 = 0.25‬‬ ‫)‪𝑓𝑜𝑟 𝑏𝑜𝑙𝑡𝑠 (4.6, 5.6 𝑎𝑛𝑑 8.8‬‬
‫𝑏𝑢𝐹 × ‪= 0.20‬‬ ‫)‪𝑓𝑜𝑟 𝑏𝑜𝑙𝑡𝑠 (4.8, 5.8, 6.8 𝑎𝑛𝑑 10.9‬‬
‫‪ Bearing in plates:‬‬
‫) 𝑏𝐹( × ) 𝑛𝑖𝑚𝑡 × ∅( = 𝑏𝑅‬
 LECTURES IN STEEL STRUCTURES DESIGN – 1.

𝑊ℎ𝑒𝑟𝑒,
𝐹𝑏 = 𝛼 × 𝐹𝑢 𝐹𝑢 𝑖𝑠 𝑓𝑜𝑟 𝑠𝑡𝑒𝑒𝑙 𝑝𝑙𝑎𝑡𝑒.
𝑡𝑚𝑖𝑛 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑜𝑓 𝑡1 𝑜𝑟 𝑡2
Edge Distance (e) ≥3ϕ ≥ 2.5 ϕ ≥2ϕ ≥ 1.5 ϕ
α 1.20 1.00 0.80 0.60

𝐹𝑜𝑟𝑐𝑒
𝑇ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑛𝑜. 𝑜𝑓 𝑏𝑜𝑙𝑡𝑠 = 𝑛 =
𝑅𝑙𝑒𝑎𝑠𝑡
𝑅𝑙𝑒𝑎𝑠𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑓𝑟𝑜𝑚 𝑅𝑠𝑠 𝑜𝑟 𝑅𝑏

II. Double Shear:

𝑅𝐷𝑆 = 2 × 𝑅𝑆𝑆

𝜋 × ∅2
𝐴∅ =
4
𝜋 × ∅2
𝑅𝐷𝑆 =2× × 𝑞𝑏
4

𝑞𝑏 = 0.25 × 𝐹𝑢𝑏 𝑓𝑜𝑟 𝑏𝑜𝑙𝑡𝑠 (4.6, 5.6 𝑎𝑛𝑑 8.8)

= 0.20 × 𝐹𝑢𝑏 𝑓𝑜𝑟 𝑏𝑜𝑙𝑡𝑠 (4.8, 5.8, 6.8 𝑎𝑛𝑑 10.9)
 Bearing in plates:
𝑅𝑏 = (∅ × 𝑡𝑚𝑖𝑛 ) × (𝐹𝑏 )
𝑊ℎ𝑒𝑟𝑒,
𝐹𝑏 = 𝛼 × 𝐹𝑢
𝐹𝑢 𝑖𝑠 𝑓𝑜𝑟 𝑠𝑡𝑒𝑒𝑙 𝑝𝑙𝑎𝑡𝑒.
𝑡𝑚𝑖𝑛 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑜𝑓 (𝑡1 + 𝑡3 ) 𝑜𝑟 𝑡2
 LECTURES IN STEEL STRUCTURES DESIGN – 1.

𝐹𝑜𝑟𝑐𝑒
𝑇ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑛𝑜. 𝑜𝑓 𝑏𝑜𝑙𝑡𝑠 = 𝑛 =
𝑅𝑙𝑒𝑎𝑠𝑡
𝑅𝑙𝑒𝑎𝑠𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑓𝑟𝑜𝑚 𝑅𝐷𝑆 𝑜𝑟 𝑅𝑏

𝑅𝑠𝑠 = 𝐴∅ × 𝑞𝑏
𝜋 × ∅2
𝑅𝑠𝑠 = × 𝑞𝑏
4
𝑅𝑏 = (∅ × 𝑡𝑚𝑖𝑛 ) × (𝐹𝑏 )
𝑊ℎ𝑒𝑟𝑒,
𝑞𝑏 = 0.25 × 𝐹𝑢𝑏 𝑓𝑜𝑟 𝑏𝑜𝑙𝑡𝑠 (4.6, 5.6 𝑎𝑛𝑑 8.8)
= 0.20 × 𝐹𝑢𝑏 𝑓𝑜𝑟 𝑏𝑜𝑙𝑡𝑠 (4.8, 5.8, 6.8 𝑎𝑛𝑑 10.9)
𝐹𝑏 = 𝛼 × 𝐹𝑢 𝐹𝑢 𝑖𝑠 𝑓𝑜𝑟 𝑠𝑡𝑒𝑒𝑙 𝑝𝑙𝑎𝑡𝑒.
𝑡𝑚𝑖𝑛 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑜𝑓 (𝑡1 + 𝑡3 ) 𝑜𝑟 𝑡2
𝐹𝑜𝑟𝑐𝑒
𝑇ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑛𝑜. 𝑜𝑓 𝑏𝑜𝑙𝑡𝑠 = 𝑛 =
𝑅𝑙𝑒𝑎𝑠𝑡

𝑅3𝑆 = 3 × 𝐴∅ × 𝑞𝑏
𝜋 × ∅2
𝑅3𝑆 =3× × 𝑞𝑏
4
𝑅𝑏 = (∅ × 𝑡𝑚𝑖𝑛 ) × (𝐹𝑏 )
𝑊ℎ𝑒𝑟𝑒,
𝑞𝑏 = 0.25 × 𝐹𝑢𝑏 𝑓𝑜𝑟 𝑏𝑜𝑙𝑡𝑠 (4.6, 5.6 𝑎𝑛𝑑 8.8)
= 0.20 × 𝐹𝑢𝑏 𝑓𝑜𝑟 𝑏𝑜𝑙𝑡𝑠 (4.8, 5.8, 6.8 𝑎𝑛𝑑 10.9)
𝐹𝑏 = 𝛼 × 𝐹𝑢
𝐹𝑢 𝑖𝑠 𝑓𝑜𝑟 𝑠𝑡𝑒𝑒𝑙 𝑝𝑙𝑎𝑡𝑒.
𝑡𝑚𝑖𝑛 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑜𝑓 (𝑡1 + 𝑡3 ) 𝑜𝑟 (𝑡2 + 𝑡4 )
 LECTURES IN STEEL STRUCTURES DESIGN – 1.

𝐹𝑜𝑟𝑐𝑒
𝑇ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑛𝑜. 𝑜𝑓 𝑏𝑜𝑙𝑡𝑠 = 𝑛 =
𝑅𝑙𝑒𝑎𝑠𝑡

Arrangement of bolts:
I. One gauge line:

1.50 ∅ ≤ 𝐸𝑑𝑔𝑒 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑒) ≤ 12 𝑡𝑚𝑖𝑛

3 ∅ ≤ 𝑃𝑖𝑡𝑐ℎ 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑝) ≤ 14 𝑡𝑚𝑖𝑛 𝑜𝑟 200 𝑚𝑚
𝑡𝑚𝑖𝑛 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑎𝑙𝑙 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑒𝑑 𝑝𝑎𝑟𝑡𝑠.
𝑎𝑚𝑖𝑛 = 3 × ∅ + 𝑡
𝑎 =𝑒+𝑒+𝑡
II. Two gauge lines:

1.50 ∅ ≤ 𝑒 ≤ 12 𝑡𝑚𝑖𝑛

3 ∅ ≤ 𝑝 ≤ 14 𝑡𝑚𝑖𝑛 𝑜𝑟 20 𝑚𝑚

𝑎𝑚𝑖𝑛 = (6 × ∅) + 𝑡 ≅ (6 × ∅) × 1.1
𝑎 = 2𝑒 + 𝑝 + 𝑡
 LECTURES IN STEEL STRUCTURES DESIGN – 1.

III. Two gauge lines with staggered arrangement:

This way is preferred in case
of tension members in order to
increase the value of Anet.
𝑎𝑚𝑖𝑛 = (6 × ∅) + 𝑡
𝑎 =2𝑒+𝑡

Notes:
 Minimum number of bolts /one gauge line is two bolts.

IV. Long Joints:

After arranging all the bolts, the distance Li is measured and to be checked with the
following:

𝐿𝑖 > 15 ∅, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑖𝑠 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑙𝑜𝑛𝑔 𝑗𝑜𝑖𝑛𝑡.

𝐿𝑖 − 15∅
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝐵𝐿 = 1 −
200 ∅

0.75 < 𝐵𝐿 ≤ 1.0

Then, Number of bolts to be recalculated:

𝐹𝑜𝑟𝑐𝑒
𝑛=
𝐵𝐿 × 𝑅𝑙𝑒𝑎𝑠𝑡
 LECTURES IN STEEL STRUCTURES DESIGN – 1.

For Chord members, we have two different types of joints:

a) Discontinuous joint (Simple Joint):

When the two chord members 1 & 2 are of different sections

Member 1:

𝐹1
n1 = = 5 𝑏𝑜𝑙𝑡𝑠
𝑅𝑙𝑒𝑎𝑠𝑡

Member 2:

𝐹2
n2 = 𝑅 = 4 𝑏𝑜𝑙𝑡𝑠
𝑙𝑒𝑎𝑠𝑡

Member 3:

𝐹3
n3 = 𝑅 = 3 𝑏𝑜𝑙𝑡𝑠
𝑙𝑒𝑎𝑠𝑡

Member 4:

𝐹4
n4 =𝑅 = 2.2 𝑏𝑜𝑙𝑡𝑠, take n4 = 4 bolts, 2 in each angle.
𝑙𝑒𝑎𝑠𝑡

b) Continuous joint:

When the two chord members are of the same sections.

For members 1 and 2, the design force will
be the greater from (F1-F2) or F3 cos (α).
Example:

 Design the connection shown in the figure using non – pre tensioned bolts M16
(4.6).
 Draw the connection to scale 1:10
 Given Data:
1. Steel used is 37.
2. Thickness of gusset plate is 10 mm.

Solution:

𝜋 × ∅2 𝜋 × 1.62
𝑅𝑠𝑠 = × 0.25 × 𝐹𝑢𝑏 = × 0.25 × 4.00 = 2.01 𝑡𝑜𝑛
4 4
𝑅𝐷𝑆 = 2 × 𝑅𝑆𝑆 = 2 × 2.01 = 4.02 𝑡𝑜𝑛
Member 1:

2 angles back to back 80 x 80 x 8 𝑡𝑚𝑖𝑛 = 8 𝑚𝑚

𝑅𝑏 = (∅ × 𝑡𝑚𝑖𝑛 ) × (𝛼𝐹𝑈 ) = (1.6 × 0.80) × (0.80 × 3.6) = 3.69 𝑡𝑜𝑛

𝐹𝑜𝑟𝑐𝑒 21
𝑛1 = = = 5.69 Take 𝑛1 = 6 𝑏𝑜𝑙𝑡𝑠
𝑅𝑙𝑒𝑎𝑠𝑡 3.69

Member 2:

2 angles back to back 60 x 60 x 6 𝑡𝑚𝑖𝑛 = 6 𝑚𝑚

𝑅𝑏 = (∅ × 𝑡𝑚𝑖𝑛 ) × (𝛼𝐹𝑈 ) = (1.6 × 0.60) × (0.80 × 3.6) = 2.76 𝑡𝑜𝑛

𝐹𝑜𝑟𝑐𝑒 15
𝑛2 = = = 5.43 Take 𝑛2 = 6 𝑏𝑜𝑙𝑡𝑠
𝑅𝑙𝑒𝑎𝑠𝑡 2.76

Member 3:

1 angle 80 x 80 x 8 𝑡𝑚𝑖𝑛 = 8 𝑚𝑚

𝑅𝑏 = (∅ × 𝑡𝑚𝑖𝑛 ) × (𝛼𝐹𝑈 ) = (1.6 × 0.80) × (0.80 × 3.6) = 3.69 𝑡𝑜𝑛

 𝐹𝑜𝑟𝑐𝑒 9
𝑛3 = = = 4.50 Take 𝑛3 = 5 𝑏𝑜𝑙𝑡𝑠
𝑅𝑙𝑒𝑎𝑠𝑡 2.01

Member 4:

2 angles Star shape 60 x 60 x 6 𝑡𝑚𝑖𝑛 = 6 𝑚𝑚

𝑅𝑏 = (∅ × 𝑡𝑚𝑖𝑛 ) × (𝛼𝐹𝑈 ) = (1.6 × 0.60) × (0.80 × 3.6) = 2.76 𝑡𝑜𝑛

𝐹𝑜𝑟𝑐𝑒 9
𝑛4 = = = 4.50 Take 𝑛4 = 5 𝑏𝑜𝑙𝑡𝑠
𝑅𝑙𝑒𝑎𝑠𝑡 2.01
 Welded Connections:

Types of welds:

1. Fillet Welds:

2. Butt welds:

S= size of weld
Aweld = Leff. X S
Lact. = Leff + (2 x S)

Size of weld:
Minimum size of weld Smin = 4 mm (Buildings)
= 6 mm (Bridges and dynamic loads)
Maximum size of weld Smax ≤ Thickness of thinner plate
 Length of weld:
Minimum length of weld Min. Leffec = 4 S or 5 cm (the Bigger)
Maximum length of weld Max. Leffec = 70 S

Allowable stress in fillet weld:

For all kinds of stresses (Tension, compression, Shear)

FPW = 0.20 Fu Where, Fu = 3.60 t/cm2 for steel 37.

i.e. FPW = 0.72 t/cm2 (For Steel 37)
Design of welded connection:
Weld is subjected to shear force.
1. Simple Connection:

Assume size of weld = thickness of minimum angle.

Memebr 1:
Force = F
𝑎−𝑒
𝐹1 = 𝐹 × ( )
𝑎
𝑒
𝐹2 = 𝐹 × ( ) Or 𝐹2 = 𝐹 − 𝐹1
𝑎

𝐹𝑜𝑟𝑐𝑒 = 𝐴𝑟𝑒𝑎 × 𝑆𝑡𝑟𝑒𝑠𝑠

𝐹1 = (2 × 𝑆 × 𝐿1 ) × F𝑃𝑊
𝐹1
𝐿1 𝑒𝑓𝑓𝑒. =
(𝑛 × 𝑆)(0.2 𝐹𝑢 )
𝐿1 𝑎𝑐𝑡. = 𝐿1 𝑒𝑓𝑓𝑒 + (2 × 𝑆)
Then, General Formula:
𝐹1
𝐿1 𝑎𝑐𝑡. = + (2 × 𝑆)
(𝑛 × 𝑆)(0.2 𝐹𝑢 )
𝐹2
𝐿2 𝑎𝑐𝑡. = + (2 × 𝑆)
(𝑛 × 𝑆)(0.2 𝐹𝑢 )
𝑛 = 1 𝑖𝑛 𝑐𝑎𝑠𝑒 𝑜𝑓 𝑜𝑛𝑒 𝑎𝑛𝑔𝑙𝑒
𝑛 = 2 𝑖𝑛 𝑐𝑎𝑠𝑒 𝑜𝑓 𝑡𝑤𝑜 𝑎𝑛𝑔𝑙𝑒𝑠
2. Continuous Connection:

For Chord members:

𝐹𝑑𝑒𝑠 = 𝐹𝑢 1 − 𝐹𝑢 2
Or
𝐹𝑑𝑒𝑠 = 𝐹𝐷 × cos 𝛼
𝑎−𝑒
𝐹1 = 𝐹 × ( )
𝑎
𝑒
𝐹2 = 𝐹 × ( ) Or 𝐹2 = 𝐹 − 𝐹1
𝑎

𝐹1
𝐿1 𝑎𝑐𝑡. = + (2 × 𝑆)
(𝑛 × 𝑆)(0.2 𝐹𝑢 )
𝐹2
𝐿2 𝑎𝑐𝑡. = + (2 × 𝑆)
(𝑛 × 𝑆)(0.2 𝐹𝑢 )
Example:
Design the two connections shown in the figure as welded connections.
Given that:
 Steel used is St. 37
 Thickness of gusset plate = 10 mm.
Solution:
Joint (A) will be designed as a simple
joint.
Assume size of weld (S) = 5 mm.

Member 1:
𝐹 = 23 𝑡 2 angles back to back 80 × 80 × 8 𝑒 = 2.26 𝑐𝑚

8 − 2.26
𝐹1 = 23 × ( ) = 16.50 𝑡𝑜𝑛.
8
𝐹2 = 23 − 16.50 = 6.50 𝑡𝑜𝑛
16.50
𝐿1 𝑎𝑐𝑡. = + (2 × 0.50) = 23.90 𝑐𝑚
(2 × 0.50)(0.2 × 3.60)
Take 𝐿1 𝑎𝑐𝑡. = 24.0 𝑐𝑚
6.50
𝐿2 𝑎𝑐𝑡. = + (2 × 0.50) = 10.03𝑐𝑚
(2 × 0.50)(0.2 × 3.60)
Take 𝐿2 𝑎𝑐𝑡. = 11.0 𝑐𝑚
Member 2:
𝐹 = 19 𝑡 2 angles back to back 60 × 60 × 6 𝑒 = 1.69 𝑐𝑚

6 − 1.69
𝐹1 = 19 × ( ) = 13.65 𝑡𝑜𝑛.
6
𝐹2 = 19 − 13.65 = 5.35 𝑡𝑜𝑛
13.65
𝐿1 𝑎𝑐𝑡. = + (2 × 0.50) = 19.96 𝑐𝑚
(2 × 0.50)(0.2 × 3.60)
Take 𝐿1 𝑎𝑐𝑡. = 20.0 𝑐𝑚
5.35
𝐿2 𝑎𝑐𝑡. = + (2 × 0.50) = 8.40 𝑐𝑚
(2 × 0.50)(0.2 × 3.60)
Take 𝐿2 𝑎𝑐𝑡. = 9.0 𝑐𝑚
Member 3:
𝐹 = 10 𝑡 2 angles back to back 55 × 55 × 5 𝑒 = 1.52𝑐𝑚
5.5 − 1.52
𝐹1 = 10 × ( ) = 7.24 𝑡𝑜𝑛.
5.5
𝐹2 = 10 − 7.24 = 2.76 𝑡𝑜𝑛
7.24
𝐿1 𝑎𝑐𝑡. = + (2 × 0.50) = 11.06 𝑐𝑚
(2 × 0.50)(0.2 × 3.60)
Take 𝐿1 𝑎𝑐𝑡. = 12.0 𝑐𝑚
2.76
𝐿2 𝑎𝑐𝑡. = + (2 × 0.50) = 4.80 𝑐𝑚
(2 × 0.50)(0.2 × 3.60)
𝐿𝑚𝑖𝑛 = 4𝑆 = 4 × 0.50 = 2.00𝑐𝑚 Or 5cm
𝐿2 𝑎𝑐𝑡 ˂ 𝐿𝑚𝑖𝑛
Take 𝐿2 𝑎𝑐𝑡. = 5 + 2𝑆 = 5 + 2 × 0.50 = 6.00 𝑐𝑚
Member 4:
𝐹 = 8.50 𝑡 2 angles star shape 60 × 60 × 6 𝑒 = 1.69 𝑐𝑚

6 − 1.69
𝐹1 = 8.50 × ( ) = 6.10 𝑡𝑜𝑛.
6
𝐹2 = 8.50 − 6.10 = 2.40 𝑡𝑜𝑛
6.10
𝐿1 𝑎𝑐𝑡. = + (2 × 0.50) = 9.47 𝑐𝑚
(2 × 0.50)(0.2 × 3.60)
Take 𝐿1 𝑎𝑐𝑡. = 10.0 𝑐𝑚
2.40
𝐿2 𝑎𝑐𝑡. = + (2 × 0.50) = 4.30 𝑐𝑚
(2 × 0.50)(0.2 × 3.60)
Take 𝐿2 𝑎𝑐𝑡. = 6.00 𝑐𝑚 = 𝐿𝑚𝑖𝑛
 Beam connections

Design of beam connections:

1. We always design FROM ONE SIDE ONLY (take the reaction and
moment of one side to calculate the number of bolts from one side).
2. Always design on the reaction of the smaller element. If we are designing
the connection between secondary and main beam, so use reactions of
secondary beam. If we are designing the connection between main beam
and column, so use reactions of main beam. If we are designing the
connection between secondary beam and column, so use reactions of
secondary beam.
3. Beams are always flushed (their upper flanges are on the same level)
except otherwise indicated.

1- Beam to beam connection:

I- Design of connection between secondary beam and main beam:
a- The secondary beam is simply supported over main beam:
Connection "A": Secondary beams from 2 sides:
R sec
1- n1 =
Rl
Where: Rsec is the reaction of secondary beam from one side only = 0.5*w*S
Rl is the least of RD.S. or Rb where tmin is the least of 2tL (2*0.8) or tw of
secondary beam.
R sec
2- n2 =
Rl
Where: Rsec is the reaction of secondary beam from one side only = 0.5*w*S

1/20C
 Beam connections

Rl is the least of RS.S. Or Rb where tmin is the least of tL (0.8) or tw of main

beam. n2 must be even number ( 2 sides)

Connection "B": Secondary beam from 1 side: (Must be simple

connection)
Same design in calculating n1 and n2 as if the secondary beam from 2 sides
because we always design on one side only.

2/20C
 Beam connections

b- The secondary beam is continuously supported over main beam:

0.4wS 0.5wS 0.5wS 0.5wS

0.6wS 0.5wS 0.5wS

0.4wS 0.6wS 0.5wS

0.5wS

0.5wS 0.5wS
0.5wS
R=0.4wS R=1.1wS R=wS R=wS

If we have secondary beams with different reactions (0.5wS, 0.6wS) at the

same connection as shown in the above figure, we choose the largest value.
R sec
1- n1 = (Same as before)
Rl
Where: Rsec is the reaction of secondary beam from one side only = 0.6*w*S
Rl is the least of RD.S. or Rb where tmin is the least of 2tL (2*0.8) or tw of
secondary beam.

R sec
2- n2 =
Rl
Where: Rsec is the reaction of secondary beam from one side only = 0.6*w*S

3/20C
 Beam connections

Rl is the least of RS.S. Or Rb where tmin is the least of tL (0.8) or tw of main

beam. n2 must be even number ( 2 sides)
T M
3- n3 = Where T = connection and M connection = 0.75Mo
Rl hsec
Rl is the least of RS.S. Or Rb where tmin is the least of tupper tie plate (assumed
1cm) or tf of secondary beam. n3 must be even number ( 2 sides)
0.6C M
4- n4 = Where C = connection and M connection = 0.75Mo
Rl hsec
We assume that 40% of the compression is transferred by bearing.
Rl is the least of RS.S. Or Rb where tmin is the least of lower tie plate (assumed 1cm)
or tf of secondary beam. n3 must be even number ( 2 sides)

4/20C
 Beam connections

Design of upper tie plate:

The plate is subjected to tension so we have to calculate Anet
We have 2 sections as shown:
We have to measure the lengths "L1", "a" and "b" from drawing
L1 ( )

L2 can be calculated by 2 methods:

2b
(1) L2= (a+ ) from drawing ( ) Approximate method
1.1
(Length "b" is decreased by 10% because it is inclined on the direction of
tension)
(2) We can exactly calculate the length using the projection perpendicular
S2 S2
"g" to force + . L2 = a+2g+2
4g 4g

And assume tpl = 1 cm and check as following:

Take smaller of A1 = (L1-4d')*tpl & A2 = (L2-2d')* tpl

5/20C
 Beam connections

T
= .....< 1.4 t/cm2
(A1orA 2 )

Design of lower tie plate:

The plate is subjected to compression 0.6C, = zero, so Fc = 1.4t/cm2
0.6C
= .... < 1.4t / cm 2 Where bfl is the flange width of the secondary beam.
b fl * t

Design of connection "C":

Connection C is cantilever so it must be continuous
Using moment and shear diagrams

Q
Q 3
M 1
cant

M1

Q2

Straining actions used in design are:

1- For shear: The greater of Q 2 and Q3
2- For moment: Mconnection = Mcant (not 0.75Mo)
For drawing: Same as secondary beam continuously supported over main
beam.

6/20C
 Beam connections

2. Beam to column connection:

I- Design of connection between secondary beam and column:
This connection is designed on the reaction of secondary beam.
R sec
1- n1 = (Same as before)
Rl
Where: Rsec is the reaction of secondary beam from one side only = 0.5*w*S
Rl is the least of RD.S. or Rb where tmin is the least of 2tL (2*0.8) or tw of
secondary beam.

R sec
2- n2 =
Rl
Where: Rsec is the reaction of secondary beam from one side only = 0.5*w*S
Rl is the least of RS.S. Or Rb where tmin is the least of tL (0.8) or tw of column.
n2 must be even number ( 2 sides)

II- Design of connection between main beam and column:

R main
1- n1 = (Same as before)
Rl
Where: Rmain is the reaction of main beam from one side only
Rl is the least of RD.S. or Rb where tmin is the least of 2tL (2*0.8) or tw of main
beam.

R main
2- n2 =
Rl
Where: Rmain is the reaction of main beam from one side only

7/20C
 Beam connections

Rl is the least of RS.S. Or Rb where tmin is the least of tL (0.8) or tf (connected to

flange) of column. n2 must be even number (2 sides).

40
80
80
80
40

8/20C
 Beam connections

Example:
Design the shown connections using non-
pretensioned M18 grade 5.6. Given that total
dead load (slab + ow) is 0.5t/m2 and live load
is 0.35t/m2. Assume that secondary beam is
IPE 270, main beam is IPE 400 and column
is IPE 600.
Solution:
W = (0.5 + 0.35)1.5 = 1.275 t / m /
π
RSS = * 1.82 * 0.25 * 5 = 3.18 t
4
RDS = 2*3.18 = 6.36 t
Assume e = 2 d, = 0.8
Rb = 0.8 * 3.6 * 1.8 * tmin = 5.18 tmin
Connection A:
• If secondary beam is simply supported over M.B.
R = 0.5 w s = 0.5 * 1.27 * 6 = 3.83 t
n1 : tmin = least of (2 * 0.8) or tw of secondary beam = 0.66 cm
Rb = 5.18 * 0.66 = 3.42 t < RDS
3.83
n1 = = 1.12 use 2 bolts
3.42
n2 : tmin = least of (0.8) or tw of main beam = 0.86 cm
Rb = 5.18 * 0.86 = 4.45 t > RSS
3.83
n2 = = 1.2 taken 4 bolts, 2 each side
3.18
• If secondary beam is continuously supported over M.B.
R = 0.6 w s = 0.6 * 1.275 * 6 = 4.59 t
4.59
n1 = = 1.34 taken 2 bolts
3.42

9/20C
 Beam connections

4.59
n2 = = 1.44 taken 4 bolts, 2 each side
3.18

1.275 * 62
M0 = = 5.74 m t
8
M connection = 0.75 * 5.74 * 100 = 430 cm t
430
T=C= = 15.9 t assume t plate = 1 cm
27
n3 and n4 : t min =1 cm or t f of secondary beam = 1.02 cm
R b = 5.18 * 1 = 5.18 t > RSS
15.9
n3 = = 5 bolts taken 6 bolts, 3 each side
3.18
(0.6 *15.9)
n4 = = 3 bolts taken 4 bolts, 2 each side
3.18
• Design of tension tie plate:
From drawing:
L1 = 7.1+2*6.8/1.1=19.46 cm
(approximate)
OR L1 = 7.1+2*6.57 +
1.76 2
2* = 20.47 cm
4 * 6.57
(Exact as ECP)
L2 = 24.2 cm
d' = 1.8 + 0.2 = 2cm
A1 = (19.46 - 2*2)*1
= 15.46 cm2
A2 = (24.2 – 4*2)*1
= 16.2 cm2
Take smaller area (critical)
ft = 15.9 / 15.46 = 1.03t/cm2
We may decrease thickness of the plate to be 8mm and recheck

10/20C
 Beam connections

• Design of compression tie plate

A = 13.5 * 1 = 13.5 cm2
0.6 *15.9
fc = = 0.71 t / cm2 < 1.4 t / cm2
13.5
We may use plate of thickness 8 mm and recheck

Connection B:
If the secondary beam is simply supported over M.B.
This connection is designed same as before
R= 0.5 w s = 3.83 t
3.83
n1 = = 2 bolts
3.42
3.83
n2 = = 4 bolts, 2 each side
3.18
If secondary beam is continuously supported
R = 0.4 w s = 0.4 * 1.275 * 6 = 3.06 t
3.06
n1 = = 2 b0lts
3.42
3.06
n2 = = 4 bolts, 2 each side
3.18

Design of connection C
If secondary beam is simply supported over
1.275 t/m'
M.G, This connection must be continuous.
Max. Shear = 4.25 t 3.4 2.55

22
M connection = 1.275 * = 2.55 m t
2
4.25
4.25
n1 = taken 2 bolts
3.42
4.25
n2 = Taken 4 bolts, 2 each side
3.18
11/20C
 Beam connections

Design of connection D
I - Secondary beam with column
Assume all beams as simply supported
1.275
R = 0.5 w s = 0.5 * 6 * ( ) = 1.91 t
2
Note that: The beam connected to column carry half weight
n1: t min is the least of (2 * 0.8) or t w secondary = 0.66 cm
Rb = 5.18 * 0.66 = 3.42 < RDS
1.91
n1 = = 0.5 use 2 bolts
3.42
n2 = tmin is the least of (0.8) or tw of column = 1.2 cm
Rb = 5.18 * 0.8 = 4.14 t > RSS
1.91
n2 = = 0.46 taken 4 bolts, 2 each side
4.14
II – Main beam with column
R sec = 1.275 * 6 = 7.65 t
n1= t min is the least of (2 * 0.8) or t w 7.65 7.65 7.65 7.65 7.65 7.65

main = 0.86 cm
Rb = 5.18 * 0.86 = 4.45 < RDS
22.95
n1 = = 5.16 bolts, taken 6 bolts 22.95
4.45
We may increase edge distance to be 2.5d and = 1.0
Rb = 1 * 3.6 * 1.8 * 0.86 = 5.57t
n1 = 22.95 / 5.57 = 4.12 taken 5 bolts
n2 = tmin is the least of (0.8) or tf of column = 1.9 cm
Rb = 5.18 * 0.8 = 4.14 t > RSS
22.95
n2 = = 7.2 taken 8 bolts, 4 each side
3.18
Detailed drawing with enlarged scale:
Simple beam: Connection "A"

12/20C
Beam connections

13/20C
Beam connections

14/20C
 Beam connections

Continuous connection: Connection "A" and "C"

15/20C
Beam connections

16/20C
 Beam connections

Connection of beam with column: Connection "D"

40
80
80
80
40

17/20C
 Beam connections

Isometric:
(1) IF the main beam is simply supported over column:

Column

S.B.

M.B.

18/20C
 Beam connections

(2) IF the main beam is rigidly connected with column:

Column

M.B.

(3) Connection between main beam and secondary beam: (Simple)

19/20C
 Beam connections

(4) Connection between main beam and secondary beam: (Continuous)

20/20C
TABELS
TABLES FOR STEEL CONSTRUCTIONS
TABLES FOR STEEL CONSTRUCTIONS
 CONTENTS

List of symbols i

Introduction ii

Hot Rolled Sections 1

Combined Hot Rolled Sections 26

Cold Formed Sections 46

Combined Cold Formed Sections 50

Bolts 55

Sample of Corrugated Sheets 66

Accessories 76

Cranes 79

Welding Symbols 86

Miscellaneous 90
 LIST OF SYMBOLS

a (mm): Length of long leg for unequal angle. Iy (cm4): Moment of inertia about Y-Y axis.

b (mm): Length of short leg for unequal angle, Iy upper flange (cm4): Moment of inertia for upper flange about Y-Y axis.
or flange width.
J (cm4): Torsion constant.
b1 (mm): Width of upper flange for unsymmetrical welded I-sections.
r (mm, cm): Radius of inner fillet for cold formed section,
b2 (mm): Width of Lower flange for symmetrical welded I-sections. or Radius of gyration for symmetrical section.

c (mm): Height of curved part including flange thickness for channel and I- r1 (mm): Radius of fillet between web and flange.
sections.
ru (cm): Radius of gyration about U-U axis.
D (mm): Outer diameter of pipe.
rv (cm): Radius of gyration about V-V axis.
d (mm): Height of web for built-up I-sections.
rx (cm): Radius of gyration about X-X axis.
dmax. (mm): Maximum diameter of bolt to be used.
ry (cm): Radius of gyration about Y-Y axis.
e (mm): Distance measured from outer surface to neutral axis of section.

ex (mm): Distance from bottom surface to X-X axis.

s (mm): Thickness of web,
or Thickness of angle leg
ey (mm): Distance from outer surface to Y-Y Axis.
S (cm3): Elastic modulus of section.
h (mm): Height of section.
4 Sv (cm3): Elastic modulus of section about V-V axis.
I (cm ): Moment of inertia for symmetrical section.
Sx (cm3): Elastic modulus of section about X-X axis.
Iu (cm4): Moment of inertia about U-U axis.
Sxb (cm3): Elastic modulus of bottom extreme fibers about X-X axis.
Iv (cm4): Moment of inertia about V-V axis.
Sxt (cm3): Elastic modulus of top extreme fibers about X-X axis.
Ix (cm4): Moment of inertia about X-X axis.

i
Sy (cm3): Elastic modulus of section about Y-Y axis.

Sy upper flange (cm3): Elastic modulus of upper flange about Y-Y axis.

t (mm): Thickness of flange,

or Wall thickness.

tG (mm): Thickness of gusset plate.

u1, u2 (cm): Distance between outer fibers of an angle to V-V axis.

Um (m2/m\): Surface area per unit length.

Ut (m2/t): Surface area per unit weight.

v, v1,&v2 (cm): Distance between outer fibers of an angle to U-U axis.

w, w1 (mm): Distance determining bolt hole location relative to the section.

Xm (cm): Distance between centroid of a channel and its shear center.

ii
 Introduction

The aim of presenting these tables is to provide the structural engineers with

wide range of information required in designing steel structures. Previously,

a steel designer had to search in many tables, handbooks, manuals, …etc. to

collect the necessary information needed to complete his/her design.

These tables are divided into nine chapters. Chapter one provides the

geometrical properties of common hot rolled sections which includes the

following:

• Equal and unequal angles: primarily used in trusses for resisting

axial forces.

• Channels (UPN): used as truss members and purlins or side girts.

• I-sections (IPN, IPE, HEA, HEB, HEM):

IPN section is suitable for beam subjected to bending moment about

its major axis.

IPE section used mainly for beams or beam column.

HEA, HEB, and HEM sections are primarily used for members

subjected to bi-moments or for heavy beam-columns.

• T-sections are produced by cutting I-sections into two halves. They

are used as brackets, truss members or light beams.

iii
 • Pipes and hollow sections: primarily used as truss members in

welded trusses.

Hollow square and rectangular sections, sometimes used as roof

purlins or light beam-columns.

• Flat plates used in connections as head plates, gusset plates or it can

be utilized in composing built-up section.

• Rails: used mainly in tracks of moving structures such as crane

bridge.

In unprecedented presentation of section properties in the Egyptian practice,

Chapter two introduces the geometric properties for combined hot rolled

sections and built–up welded sections.

This chapter provides the properties for the following combined sections:

• Two equal or unequal angles back to back:

Used mainly as lower or upper chord of trusses, or bracing members.

• Two channels back to back:

Used mainly as lower or upper chord in heavy trusses.

• Two channels toe to toe:

Used as beam-column especially when the buckling lengths in-plane

and out of plane are nearly equal.

iv
 • IPE and UPN:

This section is used primarily as section for crane girders. The

“UPN” section is provided at the top flange where the lateral shock

of the crane bridge is applied.

• Symmetrical welded I-Section:

This section consists of three parts welded to form symmetrical

I-sections. It is used mainly for beams or beam-columns.

• Unsymmetrical welded I-sections:

This section is formed in away to make the upper flange

approximately twice the lower flange. Thus it is suitable for crane

girders where the lateral shock is applied at the upper flange. The

section might also be used for composite construction when the

upper and lower flanges are switched.

Chapter three introduces the geometrical properties of cold-formed steel

sections, these sections includes:

• Channels (stiffened and unstiffened) and Z sections (with straight

lips and with inclined lips): These sections are used mainly for roof

purlins and side girts. Sometimes, they can be used in light trusses as

web members.

v
Chapter four provides the geometrical properties for combined cold-formed

sections. These sections includes:

• Two channels back to back:

Used as lower or upper chord in light trusses or in bracing members.

• Two channels toe to toe:

Used in light beam columns or as eave struts I bracing system.

Chapter five provides complete information about the size, weight, and grip

length of ordinary and high strength bolts. It also provides the dimension

and weight of washers and nuts used for each type of bolts.

This chapter also includes the dimension and the weight of anchor bolts

with various configurations.

Chapter six provides the geometrical properties of sample of corrugated

sheets whether they are single layer, sandwich panels, or metal decking steel

sheets. In this chapter the allowable spacing between purlins is also

provided based on the strength and serviceability criteria.

Chapter seven includes samples of the wheel loads of crane bridges

covering the following ranges of crane capacities:

• Light cranes with capacity range from 0.5 to 6.3 ton.

• Normal cranes with capacity range from 2.0 to 25 ton.

• Heavy cranes with capacity range from 25 to 63 ton.

vi
Chapter eight provides the common symbols used in welding constructions.

Chapter nine includes various data covering the following areas:

• Conversion tables used to convert engineering units from imperial to

SI or metric systems and vice versa.

• Dimensions of aircraft with various types to assist in choosing the

configuration of aircraft hangar.

• Geometrical properties of common shapes in engineering practice.

vii
 HOT ROLLED SECTIONS

EQUAL ANGLES HALF I.P.E.

UNEQUAL ANGLE HALF H.E.A.

CHANNELS (U.P.N) HALF H.E.B.

I.P.N. PIPES

I.P.E. HOLLOW SQUARE SECTIONS

H.E.A. HOLLOW RECTANGULAR SECTIONS

H.E.B. PLATES
H.E.M. RAILS

1
 Y
U

v
V
s

u2
S u1

d
v
EQUAL ANGLES

a
W2
X X

W1
e
e
a
U a
V
Y
Size Dimensions Axis X-X and Y-Y Axis u-u Axis v-v Details Surface Area
Weight Area
a s e v u1 u2 I S r Iu ru Iv Sv rv w1 w2 dmax Um Ut
mm mm kg/m` cm2 cm cm cm cm cm
4
cm
3
cm cm
4
cm cm
4
cm3 cm mm mm mm -2
x10 m /m
2 \
m2/t
3 1.36 1.74 0.84 1.18 1.04 1.41 0.65 0.90 2.24 1.14 0.57 0.48 0.57 11.60 85.30
30 4 1.78 2.27 0.89 2.12 1.24 1.05 1.81 0.86 0.89 2.85 1.12 0.76 0.61 0.58 N.A. 11.60 65.17
5 2.18 2.78 0.92 1.30 1.07 2.16 1.04 0.88 3.41 1.11 0.91 0.70 0.57 11.60 53.21
3 1.60 2.04 0.96 1.36 1.23 2.29 0.90 1.06 3.63 1.34 0.95 0.70 0.68 20 M10 13.60 85.00
35 4 2.10 2.67 1.00 2.47 1.41 1.24 2.96 1.18 1.05 4.68 1.33 1.24 0.88 0.68 20 N.A. M10 13.60 64.76
5 2.57 3.28 1.04 1.47 1.25 3.56 1.45 1.04 5.63 1.31 1.49 1.10 0.67 20 M10 13.60 52.91
4 2.42 3.08 1.12 1.58 1.40 4.48 1.56 1.21 7.09 1.52 1.86 1.18 0.78 22 M10 15.50 64.00
40 5 2.97 3.79 1.16 2.83 1.64 1.42 5.43 1.91 1.20 8.64 1.51 2.22 1.35 0.77 22 N.A. M10 15.50 52.20
6 3.52 4.48 1.20 1.70 1.43 6.33 2.26 1.19 9.98 1.49 2.67 1.57 0.77 22 M10 15.50 44.03
5 3.38 4.30 1.28 1.81 1.58 7.83 2.43 1.35 12.40 1.70 3.25 1.80 0.87 25 M12 17.40 51.50
45 6 4.00 5.09 1.32 3.18 1.87 1.59 9.16 2.88 1.34 14.50 1.69 3.85 2.05 0.87 25 N.A. M12 17.40 43.50
7 4.60 5.86 1.36 1.92 1.61 10.40 3.31 1.33 16.40 1.67 4.39 2.29 0.87 25 M12 17.40 37.82
5 3.77 4.80 1.40 1.98 1.76 11.00 3.05 1.51 17.40 1.90 4.59 2.32 0.98 30 M12 19.40 51.50
50 6 4.47 5.69 1.45 3.54 2.04 1.77 12.80 3.51 1.50 20.40 1.89 5.24 2.57 0.96 30 N.A. M12 19.40 43.40
7 5.15 6.56 1.49 2.11 1.78 14.60 4.15 1.49 23.10 1.88 6.02 2.85 0.96 30 M12 19.40 37.67
5 4.18 5.32 1.52 2.15 1.93 14.70 3.70 1.66 23.30 2.09 6.11 2.84 1.07 30 M16 21.30 50.96
55 6 4.95 6.31 1.56 3.89 2.21 1.94 17.80 4.40 1.66 27.40 2.08 7.24 3.26 1.07 30 N.A. M16 21.30 33.30
8 6.46 8.23 1.64 2.32 1.97 22.10 5.72 1.64 34.80 2.06 9.35 4.03 1.07 30 M16 21.30 32.97
6 5.42 6.91 1.69 2.39 2.11 22.80 5.29 1.82 36.10 2.29 9.43 3.85 1.17 35 M16 23.30 43.00
60 8 7.09 9.03 1.77 4.24 2.50 2.14 29.10 6.88 1.80 46.10 2.26 12.10 4.84 1.16 35 N.A. M16 23.30 32.90
10 8.69 11.10 1.85 2.62 2.17 34.90 8.41 1.78 55.10 2.23 14.60 5.57 1.15 35 M16 23.30 26.80
7 6.83 8.70 1.85 2.62 2.29 33.40 7.13 1.96 53.00 2.47 13.80 5.27 1.26 35 M16 25.20 36.90
65 8 7.73 9.85 1.89 4.60 2.67 2.31 37.50 8.13 1.95 59.40 2.46 15.60 5.84 1.26 35 N.A. M16 25.20 32.60
9 8.62 11.00 1.93 2.73 2.32 41.30 9.04 1.94 65.40 2.44 17.20 6.30 1.25 35 M16 25.20 29.23
N.A.=not available for this angle size

2
 Y
U

v
V s

u2
S u1

d
v
EQUAL ANGLES

a
W2
X X

W1
e
e
a
U a
V
Y
Size Dimensions Axis X-X and Y-Y Axis u-u Axis v-v Details Surface Area
Weight Area
a s e v u1 u2 I S r Iu ru Iv Sv rv w1 w2 dmax Um Ut
mm mm kg/m` cm2 cm cm cm cm cm
4
cm 3
cm cm
4
cm cm
4
cm3 cm mm mm mm -2
x10 m /m
2 \
m2/t
7 7.38 9.40 1.97 2.79 2.47 42.4 8.43 2.12 67.1 2.67 17.6 6.37 1.37 40 M20 27.20 36.90
70 9 9.34 11.90 2.05 4.95 2.90 2.50 52.6 10.6 2.10 83.1 2.64 22.0 7.59 1.36 40 N.A. M20 27.20 29.10
11 11.20 14.30 2.13 3.01 2.53 61.8 12.7 2.09 97.6 2.61 26.0 8.61 1.35 40 M20 27.20 24.28
7 7.94 10.10 2.03 2.95 2.63 52.4 8.67 2.28 83.6 2.88 21.1 7.15 1.45 40 M20 29.10 36.65
75 8 9.03 11.50 2.13 5.30 3.01 2.65 58.9 11.0 2.26 93.3 2.85 24.4 8.11 1.46 40 N.A. M20 29.10 32.20
10 11.10 14.10 2.21 3.18 2.68 71.4 13.5 2.25 113 2.83 28.8 9.55 1.45 40 M20 29.10 26.21
8 9.66 12.30 2.26 3.20 2.82 72.3 12.6 2.42 115 3.06 29.6 9.25 1.55 45 M20 31.10 32.20
80 10 11.90 15.10 2.34 5.66 3.31 2.85 87.5 15.5 2.41 139 3.03 35.9 10.9 1.54 45 N.A. M20 31.10 26.10
12 14.10 17.90 2.41 3.41 2.89 102 18.2 2.39 161 3.00 43.0 12.6 1.53 45 M20 31.10 22.10
9 12.20 15.50 2.54 3.59 3.18 116 18.0 2.74 184 3.45 47.8 13.3 1.76 50 M20 35.10 28.80
90 11 14.70 18.70 2.62 6.36 3.70 3.21 138 21.6 2.72 218 3.41 57.1 15.4 1.75 50 N.A. M20 35.10 23.88
13 17.10 21.80 2.70 3.81 3.24 158 25.1 2.69 250 3.39 65.9 17.3 1.74 50 M20 35.10 20.53
10 15.10 19.20 2.82 3.99 3.54 177 24.7 3.04 280 3.82 73.3 18.4 1.95 55 M24 39.00 25.80
100 12 17.80 22.70 2.90 7.07 4.10 3.57 207 29.2 3.02 328 3.80 86.2 21.0 1.95 55 N.A. M24 39.00 21.90
14 20.60 26.20 2.98 4.21 3.60 235 33.5 3.00 372 3.77 98.3 23.4 1.94 55 M24 39.00 18.90
10 16.60 21.20 3.07 4.34 3.89 239 30.1 3.36 379 4.23 98.6 22.7 2.16 40 80 M20 43.00 25.90
110 12 19.70 25.10 3.15 7.78 4.45 3.93 280 35.7 3.34 444 4.21 116 25.1 2.15 40 80 M20 43.00 21.83
14 22.80 29.00 3.21 4.54 3.98 319 41.0 3.32 505 4.18 133 29.3 2.14 40 80 M20 43.00 18.86
12 21.60 27.50 3.40 4.80 4.26 368 42.7 3.65 584 4.60 152 31.6 2.35 45 85 M20 46.90 21.70
120 13 23.30 29.90 3.44 8.49 4.86 4.27 394 46.0 3.64 625 4.59 162 33.3 2.34 45 85 M20 46.90 20.13
15 26.60 33.90 3.51 4.96 4.31 446 52.5 3.63 705 4.56 186 37.5 2.34 45 85 M20 46.90 17.60
N.A.=not available for this angle size

3
 Y

v
V s

u2
S u1

d
v
EQUAL ANGLES

a
W2
X X

W1
e
e
a
U a
V
Y
Size Dimensions Axis X-X and Y-Y Axis u-u Axis v-v Details Surface Area
Weight Area
a s e v u1 u2 I S r Iu ru Iv Sv rv w1 w2 dmax Um Ut
mm mm kg/m` cm2 cm cm cm cm cm4 cm3 cm cm
4
cm cm4 cm3 cm mm mm mm x10-2 m2/m\ m2/t
12 23.60 30.00 3.64 5.15 4.60 472 50.4 3.97 750 5.00 194 37.7 2.54 50 90 M20 50.80 21.50
130 14 27.20 34.70 3.72 9.19 5.26 4.63 540 55.2 3.94 857 4.97 223 42.4 2.53 50 90 M20 50.80 18.68
16 30.90 39.30 3.80 5.37 4.66 605 65.8 3.92 959 4.94 251 46.7 2.52 50 90 M20 50.80 16.44
140 13 27.50 35.00 3.92 9.90 5.54 4.96 638 63.3 4.27 1010 5.38 262 47.3 2.74 55 105 M24 54.70 19.90
15 31.40 40.00 4.00 5.66 4.99 723 72.3 4.25 1150 5.36 298 52.7 2.73 55 105 M24 54.70 17.40
14 31.60 40.30 4.21 5.95 5.31 845 78.2 4.58 1340 5.77 347 58.3 2.94 60 110 M24 58.60 18.50
15 33.80 43.00 4.25 6.01 5.33 898 82.5 4.57 1430 5.76 370 61.6 2.93 60 110 M24 58.60 17.33
150 16 35.90 45.70 4.29 10.60 6.07 5.34 949 88.7 4.56 1510 5.74 391 64.4 2.93 60 110 M24 58.60 16.30
18 40.10 51.00 4.36 6.17 5.38 1050 99.3 4.54 1670 5.70 438 71.0 2.93 60 110 M24 58.60 14.61
20 44.20 56.30 4.44 6.28 5.41 1150 109 4.51 1820 5.68 477 76.0 2.91 60 110 M24 58.60 13.25
15 36.20 46.10 4.49 6.35 5.67 1100 95.6 4.88 1750 6.15 453 71.3 3.14 60 120 M27 62.50 17.30
160 17 40.70 51.80 4.57 11.30 6.46 5.70 1230 108 4.86 1950 6.13 506 78.3 3.13 60 120 M27 62.50 15.40
19 45.10 57.50 4.65 6.58 5.73 1350 118 4.84 2140 6.10 558 84.8 3.12 60 120 M27 62.50 13.85
16 43.50 55.40 5.02 7.11 6.39 1680 130 5.51 2690 6.96 679 95.5 3.50 65 135 M27 70.50 16.20
180 18 48.60 61.90 5.10 12.70 7.22 6.41 1870 145 5.49 2970 6.93 757 105 3.49 65 135 M27 70.50 14.50
20 53.70 68.40 5.18 7.33 6.44 2040 160 5.47 3260 6.90 830 113 3.49 65 135 M27 70.50 13.12
22 58.60 74.70 5.26 7.44 6.47 2210 174 5.44 3510 6.86 918 123 3.50 65 135 M27 70.50 12.03
16 48.50 61.80 5.52 7.80 7.09 2340 162 6.15 3740 7.78 943 121 3.91 65 150 M27 78.50 16.20
18 54.30 69.10 5.60 7.92 7.12 2600 181 6.13 4150 7.75 1050 133 3.90 65 150 M27 78.50 14.50
200 20 59.90 76.40 5.68 14.10 8.04 7.15 2850 199 6.11 4540 7.72 1160 144 3.89 65 150 M27 78.50 13.10
24 71.10 90.60 5.84 8.26 7.21 3330 235 6.06 5280 7.64 1380 167 3.90 65 150 M27 78.50 11.04
28 82.00 105.00 5.99 8.47 7.28 3780 270 6.02 5990 7.57 1580 186 3.89 65 150 M27 78.50 9.57

4
 Y V

V1
U

d
A

a
a
UNEQUAL ANGLES X X

W2
V2

W1
s
ex
S
ey
U
b
u2
u1
b

V Y
Size Dimensions Axis X-X Axis Y-Y Axis u-u Axis v-v Details Surface Area
Weight Area
a b s ex ey v1 v2 u1 u2 tan A Ix Sx rx Iy Sy ry Iu ru Iv rv w1 w2 dmax Um Ut
2 4 3 4 3 4 4 -2 2 \ 2
mm mm mm kg/m` cm cm cm cm cm cm cm cm cm cm cm cm cm cm cm cm cm mm mm mm x10 m /m m /t
3 1.11 1.42 0.99 0.50 2.04 1.51 0.86 1.04 0.431 1.25 0.62 0.94 0.44 0.29 0.56 1.43 1.00 0.25 0.42 9.68 87.20
30 20 N.A.
4 1.45 1.85 1.03 0.54 2.02 1.52 0.91 1.03 0.423 1.59 0.81 0.93 0.55 0.38 0.55 1.81 0.99 0.33 0.42 9.68 66.70
3 1.35 1.72 1.43 0.44 2.61 1.77 0.79 1.19 0.259 2.79 1.08 1.27 0.47 0.30 0.52 2.96 1.31 0.30 0.42 22 M10 11.68 86.50
40 20 N.A.
4 1.77 2.25 1.47 0.48 2.57 1.80 0.83 1.18 0.252 3.59 1.42 1.26 0.60 0.39 0.52 3.79 1.30 0.39 0.42 22 M10 11.68 66.00
3 1.72 2.19 1.43 0.70 3.09 2.23 1.21 1.59 0.436 4.47 1.46 1.43 1.60 0.70 0.86 5.15 1.53 0.93 0.65 25 M12 14.64 85.00
45 30 4 2.25 2.87 1.48 0.74 3.07 2.26 1.27 1.58 0.433 5.78 1.91 1.42 2.05 0.91 0.85 6.65 1.52 1.18 0.64 25 N.A. M12 14.64 65.06
5 2.77 3.53 1.52 0.78 3.05 2.27 1.32 1.58 0.430 6.99 2.95 1.41 2.47 1.11 0.84 8.02 1.51 1.44 0.64 25 M12 14.64 52.80
5 3.37 4.29 2.15 0.68 3.90 2.67 1.20 1.77 0.256 15.6 4.04 1.90 2.60 1.12 0.78 16.5 1.96 1.69 0.63 35 M16 17.50 51.90
60 30 N.A.
7 4.59 5.85 2.24 0.76 3.83 2.72 1.28 1.73 0.248 20.7 5.50 1.88 3.41 1.52 0.76 21.8 1.93 2.28 0.62 35 M16 17.50 38.10
5 3.76 4.79 1.96 0.97 4.08 3.01 1.68 2.09 0.437 17.2 4.25 1.89 6.11 2.02 1.13 19.8 2.03 3.50 0.86 35 M16 19.50 51.80
60 40 6 4.46 5.68 2.00 1.01 4.06 3.02 1.72 2.08 0.437 20.1 5.03 1.88 7.12 2.38 1.12 23.1 2.02 4.12 0.85 35 N.A. M16 19.50 37.90
7 5.14 6.55 2.04 1.05 4.04 3.03 1.77 2.07 0.422 23.0 5.79 1.87 8.07 2.74 1.11 26.3 2.00 4.73 0.85 35 M16 19.50 37.93
5 4.74 6.04 2.40 1.17 5.14 3.73 2.03 2.64 0.437 34.4 6.74 2.39 12.3 3.21 1.43 39.6 2.56 7.10 1.08 40 M20 24.44 51.50
75 50 7 6.51 8.30 2.48 1.25 5.10 3.77 2.13 2.63 0.433 46.4 9.24 2.36 16.5 4.39 1.41 53.3 2.53 9.56 1.07 40 N.A. M20 24.44 37.54
9 8.23 10.50 2.56 1.32 5.06 3.80 2.22 2.62 0.427 57.4 11.6 2.34 20.2 5.49 1.39 65.7 2.50 11.9 1.07 40 M20 24.44 29.70
6 5.41 6.89 2.85 0.88 5.21 3.53 1.55 2.42 0.259 44.9 8.7 2.55 7.6 2.44 1.05 47.6 2.63 4.90 0.84 45 M22 23.40 43.20
80 40 N.A.
8 7.07 9.01 2.94 0.95 5.15 3.57 1.65 2.38 0.253 57.6 11.4 2.53 9.7 3.18 1.04 60.9 2.60 6.41 0.84 45 M22 23.40 33.10
6 6.82 8.69 2.89 1.41 6.14 4.50 2.46 3.16 0.442 71.7 11.7 2.87 25.8 5.61 1.72 82.8 3.09 14.6 1.30 50 M24 29.40 43.10
90 60
8 8.96 11.40 2.97 1.49 6.11 4.54 2.56 3.15 0.497 92.5 15.4 2.85 33.0 7.31 1.70 107 3.06 19.0 1.29 50 M24 29.40 32.80
6 6.85 8.73 3.49 1.04 6.50 4.39 1.91 2.98 0.263 89.7 13.8 3.20 15.3 3.86 1.32 95.2 3.30 9.78 1.06 55 M24 29.20 42.60
100 50 8 8.99 11.50 3.59 1.13 6.48 4.44 2.00 2.95 0.258 116 18.0 3.18 19.5 5.04 1.31 123 3.28 12.6 1.05 55 N.A. M24 29.20 32.80
10 11.10 14.10 3.67 1.20 6.43 4.49 2.08 2.91 0.252 141 22.2 3.16 23.4 6.17 1.29 149 3.25 15.5 1.04 55 M24 29.20 26.30
7 8.77 11.20 3.23 1.51 6.83 4.91 2.66 3.48 0.419 113 16.6 3.17 37.6 7.54 1.84 128 3.39 21.6 1.39 55 M24 32.10 36.60
100 65 9 11.10 14.20 3.32 1.59 6.78 4.94 2.76 3.46 0.415 141 21.0 3.15 46.7 9.52 1.82 160 3.36 27.2 1.39 55 N.A. M24 32.10 28.90
11 13.40 17.10 3.40 1.67 6.74 4.97 2.85 3.45 0.410 167 25.3 3.13 55.1 11.4 1.80 190 3.34 32.6 1.38 55 M24 32.10 24.00
N.A.=not available for this angle size

5
 Y V

V1
U

d
A

a
UNEQUAL ANGLES

a
X X

W2
V2

W1
s
ex
S
ey
U
b
u2
u1
b

V Y
Size Dimensions Axis X-X Axis Y-Y Axis u-u Axis v-v Details Surface Area
Weight Area
a b s ex ey v1 v2 u1 u2 tan A Ix Sx rx Iy Sy ry Iu ru Iv rv w1 w2 dmax Um Ut
2 4 3 4 3 4 4 -2 2 \ 2
mm mm mm kg/m` cm cm cm cm cm cm cm cm cm cm cm cm cm cm cm cm cm mm mm mm x10 m /m m /t
8 12.20 15.50 3.83 1.87 8.23 5.99 3.27 4.20 0.441 226 27.6 3.82 80.8 13.2 2.29 261 4.10 45.8 1.72 45 85 M20 39.10 32.00
10 15.00 19.10 3.92 1.95 8.18 6.03 3.37 4.19 0.438 276 34.1 3.80 98.1 16.2 2.27 318 4.07 56.1 1.71 45 85 M20 39.10 26.10
120 80
12 17.80 22.70 4.00 2.03 8.14 6.06 3.46 4.18 0.433 323 40.4 3.77 114 19.1 2.25 371 4.04 66.1 1.71 45 85 M20 39.10 22.00
14 20.50 26.20 4.08 2.10 8.10 6.08 3.55 4.17 0.429 368 46.4 3.75 130 22.0 2.23 421 4.01 75.8 1.70 45 85 M20 39.10 19.07
8 11.90 15.10 4.56 1.37 8.50 5.71 2.49 3.86 0.263 263 31.1 4.17 44.8 8.7 1.72 280 4.31 28.6 1.38 50 90 M20 38.10 32.00
130 65 10 14.60 18.60 4.65 1.45 8.43 5.76 2.58 3.82 0.259 321 38.4 4.15 54.2 10.7 1.71 340 4.27 35.0 1.37 50 90 M20 38.10 26.10
12 17.30 22.10 4.74 1.53 8.37 5.81 2.66 3.80 0.255 376 45.5 4.12 63.0 12.7 1.69 397 4.24 41.2 1.37 50 90 M20 38.10 22.00
10 16.60 21.20 4.15 2.18 8.92 6.69 3.75 4.62 0.472 358 40.5 4.11 141 20.6 2.58 420 4.46 78.5 1.93 50 90 M20 42.97 25.88
130 90
12 19.70 25.10 4.84 2.26 8.88 6.72 3.85 4.60 0.463 420 48.0 4.09 165.0 24.4 2.56 492 4.43 92.6 1.92 50 90 M20 42.97 21.81
9 15.30 19.50 5.28 1.57 9.79 6.62 2.90 4.46 0.265 455 46.8 4.83 78.3 13.2 2.00 484 4.98 50.0 1.60 60 110 M24 44.10 28.80
150 75
11 18.60 23.60 5.37 1.65 9.73 6.66 2.97 4.44 0.261 545 56.6 4.80 93.0 15.9 1.98 578 4.95 59.8 1.59 60 110 M24 44.10 23.00
10 19.00 24.20 4.80 2.34 10.30 7.50 4.10 5.25 0.442 552 54.1 4.78 198 25.8 2.86 637 5.13 112 2.15 60 110 M24 48.90 25.70
150 100 12 22.60 28.70 4.89 2.42 10.20 7.53 4.19 5.24 0.439 650 64.2 4.76 232 30.6 2.84 749 5.10 133 2.15 60 110 M24 48.90 21.60
14 26.10 33.20 4.97 2.50 10.20 7.56 4.28 5.23 0.435 744 74.1 4.73 264 35.2 2.82 856 5.07 152 2.14 60 110 M24 48.90 18.70
10 18.20 23.20 5.63 1.69 10.50 7.06 3.07 4.76 0.263 611 58.9 5.14 104 16.5 2.12 648 5.29 67.0 1.70 60 120 M27 46.90 25.70
160 80 12 21.60 27.50 5.72 1.77 10.40 7.10 3.15 4.75 0.259 720 70.0 5.11 122 19.6 2.10 763 5.26 78.9 1.69 60 120 M27 46.90 21.70
14 25.00 31.80 5.81 1.85 10.30 7.16 3.23 4.72 0.256 823 80.7 5.09 139 22.5 2.09 871 5.23 90.5 1.69 60 120 M27 46.90 18.70
10 20.60 26.20 6.28 1.85 11.80 7.89 3.38 5.42 0.262 880 75.1 5.80 151 21.1 2.40 934 5.97 97.4 1.93 65 135 M27 52.80 25.63
180 90 12 24.50 31.20 6.37 1.93 11.70 7.95 3.48 5.38 0.261 1040 89.3 5.77 177 25.1 2.38 1100 5.94 114 1.92 65 135 M27 52.80 21.55
14 28.30 36.10 6.46 2.01 11.70 8.01 3.57 5.34 0.259 1190 103 5.75 202 28.9 2.37 1260 5.92 131 1.91 65 135 M27 52.80 18.65
10 23.00 29.20 6.93 2.01 13.20 8.76 3.75 5.98 0.266 1220 93.2 6.46 210 26.3 2.68 1300 6.66 133 2.14 65 150 M27 58.70 25.50
12 27.30 34.80 7.03 2.10 13.10 8.82 3.84 5.95 0.264 1440 111 6.43 247 31.3 2.67 1530 6.63 158 2.13 65 150 M27 58.70 21.50
200 100
14 31.60 40.30 7.12 2.18 13.00 8.88 3.93 5.92 0.262 1650 128 6.41 282 36.1 2.65 1760 6.60 181 2.13 65 150 M27 58.70 18.60
16 35.90 45.70 7.20 2.26 12.90 8.93 4.02 5.88 0.259 1860 145 6.38 316 40.8 2.63 1970 6.57 204 2.11 65 150 M27 58.70 16.35

6
 Y W
b
d

c
r1
ey
CHANNEL

h-2c
h

h
X X
Shear Center xm s
(U.P.N.) s
t

c
t b
Y

Dimensions Axis X-X Axis Y-Y Details Surface Area

Sec. Weight Area Aweb
No. h b s t=r1 c h-2c ey Xm Ix Sx rx Iy Sy ry W dmax Um Ut
2 2 4 3 4 3 -2 2 \
kg/m` cm cm mm mm mm mm mm mm cm cm cm cm cm cm cm cm mm mm x10 m /m m 2/t
30x15 1.74 2.21 0.84 30 15 4.0 4.5 9.0 12 0.52 0.74 2.53 1.69 1.07 0.38 0.39 0.42 N.A. N.A. 7.14 41.00
30 4.27 5.44 0.80 30 33 5.0 7.0 14.5 1 1.31 2.22 6.39 4.26 1.08 5.33 2.68 0.99 N.A. N.A. 17.40 40.74
40x20 2.87 3.66 1.45 40 20 5.0 5.5 11.0 18 0.67 1.01 7.58 3.79 1.44 1.14 0.86 0.56 N.A. N.A. 13.71 47.77
40 4.87 6.21 1.30 40 35 5.0 7.0 14.5 11 1.33 2.32 14.1 7.05 1.50 6.68 3.08 1.04 N.A. N.A. 19.47 40.00
50x25 3.86 4.92 1.90 50 25 5.0 6.0 12.5 25 0.81 1.34 16.6 6.73 1.85 2.49 1.48 0.71 N.A. N.A. 17.52 45.40
50 5.59 7.12 1.80 50 38 5.0 7.0 15.0 20 1.37 2.47 26.4 10.6 1.92 9.12 3.75 1.13 N.A. N.A. 22.40 40.06
60 5.07 6.46 2.88 60 30 6.0 6.0 12.5 35 0.91 1.50 31.6 10.5 2.21 4.51 2.16 0.84 N.A. N.A. 21.32 42.06
65 7.09 9.03 2.75 65 42 5.5 7.5 16.0 33 1.42 2.60 57.5 17.7 2.52 14.1 5.07 1.25 22 M10 27.30 38.50
70 6.73 8.57 3.42 70 40 6.0 6.5 16.0 38 1.42 2.20 61.1 17.5 2.67 11.4 4.10 1.15 22 M10 26.67 39.64
80 8.64 11.00 3.84 80 45 6.0 8.0 17.0 47 1.45 2.67 106 26.5 3.10 19.4 6.36 1.33 25 M12 31.20 36.10
100 10.60 13.50 4.98 100 50 6.0 8.5 18.0 64 1.55 2.93 206 41.2 3.91 29.3 8.49 1.47 30 M12 37.20 35.10
120 13.40 17.00 7.14 120 55 7.0 9.0 19.0 82 1.60 3.03 364 60.7 4.62 43.2 11.1 1.59 30 M16 43.40 32.40
140 16.00 20.40 8.40 140 60 7.0 10.0 21.0 97 1.75 3.37 605 86.4 5.45 62.7 14.8 1.75 35 M16 48.90 30.60
160 18.80 24.00 10.43 160 65 7.5 10.5 22.5 116 1.84 3.56 925 116 6.21 85.3 18.3 1.89 35 M20 54.60 29.00
180 22.00 28.00 12.64 180 70 8.0 11.0 23.5 133 1.92 3.75 1350 150 6.95 114 22.4 2.02 40 M20 61.10 37.80
200 25.30 32.20 15.05 200 75 8.5 11.5 24.5 151 2.01 3.94 1910 191 7.70 148 27.0 2.14 40 M20 66.10 26.10
220 29.40 37.40 17.55 220 80 9.0 12.5 26.5 166 2.14 4.20 2690 245 8.48 197 33.6 2.30 45 M20 71.80 24.40
240 33.20 42.30 20.33 240 85 9.5 13.0 28.0 185 2.23 4.39 3600 300 9.22 248 39.6 2.42 45 M24 77.50 23.30
260 37.90 48.30 23.20 260 90 10.0 14.0 30.0 201 2.36 4.66 4820 371 9.99 317 47.7 2.56 50 M24 83.40 22.00
280 41.80 53.30 25.00 280 95 10.0 15.0 32.0 213 2.53 5.02 6280 448 10.90 399 57.2 2.74 50 M24 89.00 21.00
300 46.20 58.80 26.80 300 100 10.0 16.0 34.0 232 2.70 5.41 8030 535 11.70 495 67.8 2.90 55 M27 95.00 20.60
320 59.50 75.80 39.90 320 100 14.0 17.5 37.0 247 2.60 4.82 10870 679 12.10 597 80.6 2.81 58 M27 98.20 16.50
350 60.60 77.30 44.52 350 100 14.0 16.0 34.0 283 2.40 4.45 12840 734 12.90 570 75.0 2.72 58 M27 105.00 17.30
380 63.10 80.40 46.98 380 102 13.5 16.0 33.5 313 2.38 4.58 15760 829 14.00 615 78.7 2.77 60 M27 111.00 17.70
400 71.80 91.50 50.96 400 110 14.0 18.0 38.0 325 2.65 5.11 20350 1020 14.90 846 102 3.04 60 M27 118.00 16.50
N.A.=not available for this channel size

7
 Y
b w

c
r1

h-2c
STANDARD I - BEAMS

h
X X
( I.P.N. ) s
t

c
Y
Dimensions Axis X-X Axis Y-Y Details Surface Area
Sec. Weight Area Aweb
h b s =r1 t c h-2c Ix Sx rx Iy Sy ry w dmax Um Ut
No.
kg/m` cm2 cm2 mm mm mm mm mm mm cm4 cm3 cm cm4 cm3 cm mm mm x10-2 m2/m\ m2/t
80 5.94 7.57 2.66 80 42 3.9 5.9 10.5 59 77.8 19.5 3.20 6.29 3.00 0.91 N.A. 30.40 51.20
100 8.34 10.60 3.89 100 50 4.5 6.8 12.5 75 171 34.2 4.01 12.2 4.88 1.07 N.A. 37.00 44.40
120 11.10 14.20 5.33 120 58 5.1 7.7 14 92 328 54.7 4.81 21.5 7.41 1.23 N.A. 43.90 39.50
140 14.30 18.20 7.00 140 66 5.7 8.6 15.5 109 573 81.9 5.61 35.2 10.7 1.40 34 M10 50.20 35.10
160 17.90 22.80 8.88 160 74 6.3 9.5 17.5 125 935 117 6.40 54.7 14.8 1.55 40 M10 57.50 32.10
180 21.90 27.90 10.98 180 82 6.9 10.4 19 142 1450 161 7.20 81.3 19.8 1.71 44 M12 64.00 29.20
200 26.20 33.40 13.31 200 90 7.5 11.3 20.5 159 2140 214 8.00 117 26.0 1.87 48 M12 70.90 27.00
220 31.10 39.50 15.84 220 98 8.1 12.2 22 176 3060 278 8.80 162 33.1 2.02 52 M12 77.50 24.90
240 36.20 46.10 18.60 240 106 8.7 13.1 24 192 4250 354 9.59 221 41.7 2.20 56 M16 84.40 23.30
260 41.90 53.30 21.79 260 113 9.4 14.1 26 208 5740 442 10.40 288 51.0 2.32 60 M16 90.60 21.60
280 47.90 61.00 25.21 280 119 10.1 15.2 27.5 225 7590 542 11.10 364 61.2 2.45 60 M16 96.60 20.10
300 54.20 69.00 28.90 300 125 10.8 16.2 29.5 241 9800 653 11.90 451 72.2 2.56 64 M20 103.00 19.00
320 61.00 77.70 32.82 320 131 11.5 17.3 31 258 12510 782 12.70 555 84.7 2.67 70 M20 109.00 17.90
340 68.00 86.70 37.01 340 137 12.2 18.3 33 274 15700 923 13.50 674 98.4 2.80 74 M20 115.00 16.90
360 76.10 97.00 41.73 360 143 13 19.5 35 290 19610 1090 14.20 818 114 2.90 76 M20 121.00 15.90
380 84.00 107.00 46.44 380 149 13.7 20.5 37 306 24010 1260 15.00 975 131 3.02 82 M20 127.00 15.10
400 92.40 118.00 51.38 400 s 14.4 21.6 38.5 323 29210 1460 15.70 1160 149 3.13 86 M20 133.00 14.40
425 104.00 132.00 57.99 425 163 15.3 23 41 343 36970 1740 16.70 1440 176 3.30 88 M20 144.40 13.50
450 115.00 147.00 65.03 450 170 16.2 24.3 43.5 363 45850 2040 17.70 1730 203 3.43 94 M24 148.00 12.90
475 128.00 163.00 72.47 475 178 17.1 25.6 45.5 384 56480 2380 18.60 2090 235 3.60 100 M24 155.10 12.11
500 141.00 179.00 80.28 500 185 18 27 48 404 68740 2750 19.60 2480 268 3.72 100 M27 163.00 11.50
550 166.00 212.00 93.10 550 200 19 30 52.5 445 99180 3610 21.60 3490 349 4.02 110 M27 177.30 10.68
600 199.00 254.00 115.60 600 215 21.6 32.4 57.5 485 139000 4630 23.40 4670 434 4.30 120 M27 191.90 9.64
N.A.=not available for this IPN size
8
 Y
b W

c
r1 d

I.P.E.

h-2c
h
X X
s

c
t
Y

Dimensions Axis X-X Axis Y-Y Details Surface Area

Sec. Weight Area Aweb
h b s t r1 c h-2c Ix Sx rx Iy Sy ry w dmax Um Ut
No. 2
kg/m` cm cm2 mm mm mm mm mm mm mm cm
4
cm3 cm cm
4
cm3 cm mm mm -2 2
x10 m /m
\
m2/t
80 6.00 7.64 2.64 80 46 3.8 5.2 5 10.2 59.6 80.1 20.0 3.24 8.49 3.69 1.05 N.A. N.A. 32.80 54.80
100 8.10 10.30 3.63 100 55 4.1 5.7 7 12.7 74.6 171 34.2 4.07 15.9 5.79 1.24 N.A. N.A. 40.00 49.50
120 10.40 13.20 4.73 120 64 4.4 6.3 7 13.3 93.4 318 53.0 4.90 27.7 8.65 1.45 36 M10 47.50 45.60
140 12.90 16.40 5.93 140 73 4.7 6.9 7 13.9 112.2 541 77.3 5.74 44.9 12.3 1.65 38 M10 55.10 42.60
160 15.80 20.10 7.26 160 82 5.0 7.4 9 16.4 127.2 869 109 6.58 68.3 16.7 1.84 44 M12 62.30 39.40
180 18.80 23.90 8.69 180 91 5.3 8.0 9 17.0 146.0 1320 146 7.42 101 22.2 2.05 50 M12 69.80 37.10
200 22.40 28.50 10.25 200 100 5.6 8.5 12 20.5 159.0 1940 194 8.26 142 28.5 2.24 56 M12 76.80 34.30
220 26.20 33.40 11.89 220 110 5.9 9.2 12 21.2 177.6 2770 252 9.11 205 37.3 2.48 60 M16 84.80 32.40
240 30.70 39.10 13.66 240 120 6.2 9.8 15 24.8 190.4 3890 324 9.97 284 47.3 2.69 68 M16 92.20 30.00
270 36.10 45.90 16.47 270 135 6.6 10.2 15 25.2 219.6 5790 429 11.20 420 62.2 3.02 72 M20 104.00 28.80
300 42.20 53.80 19.78 300 150 7.1 10.7 15 25.7 248.6 8360 557 12.50 604 80.5 3.35 80 M20 116.00 27.50
330 49.10 62.60 23.03 330 160 7.5 11.5 18 29.5 271.0 11770 713 13.70 788 98.5 3.55 86 M24 125.00 25.50
360 57.10 72.70 26.77 360 170 8.0 12.7 18 30.7 298.6 16270 904 15.00 1040 123 3.79 90 M24 135.00 23.60
400 66.30 84.50 32.08 400 180 8.6 13.5 21 34.5 331.0 23130 1160 16.50 1320 146 3.95 96 M27 147.00 22.20
450 77.60 98.80 39.56 450 190 9.4 14.6 21 35.6 378.8 33740 1500 18.50 1680 176 4.12 106 M27 161.00 20.70
500 90.70 116.00 47.74 500 200 10.2 16.0 21 37.0 426.0 48200 1930 20.40 2140 214 4.31 110 M27 174.00 19.20
550 106.00 134.00 57.23 550 210 11.1 17.2 24 41.2 467.6 67120 2440 22.30 2670 254 4.45 120 M27 188.00 17.70
600 122.00 156.00 67.44 600 220 12.0 19.0 24 43.0 514.0 92080 3070 24.30 3390 308 4.66 120 M27 202.00 16.60
N.A.=not available for this IPE size

9
 Y
b w1 w w1

c
r1 d

BROAD FLANGE I - BEAMS

h-2c
h
( H.E.A. ) s

c
t
Y

Dimensions Axis X-X Axis Y-Y Details Surface Area

Sec. Weight Area Aweb
h b s t r1 c h-2c Ix Sx rx Iy Sy ry w w1 dmax Um Ut
No. 2 2
kg/m` cm cm mm mm mm mm mm mm mm cm4 cm3 cm cm4 cm3 cm mm mm mm x10-2 m2/m\ m2/t
100 16.7 21.2 4.00 96 100 5.0 8.0 12 20 56 349 73 4.06 134 26.8 2.51 56 N.A. M12 56.10 33.60
120 19.9 25.3 4.90 114 120 5.0 8.0 12 20 74 606 106 4.89 231 38.5 3.02 66 N.A. M16 67.70 34.00
140 24.7 31.4 6.38 133 140 5.5 8.5 12 21 91 1030 155 5.73 389 55.6 3.52 76 N.A. M20 79.40 32.10
160 30.4 38.8 8.04 152 160 6.0 9.0 15 24 104 1670 220 6.57 616 76.9 3.98 86 N.A. M20 90.60 29.80
180 35.5 45.3 9.12 171 180 6.0 9.5 15 25 121 2510 294 7.45 925 103 4.52 100 N.A. M24 102.00 28.70
200 42.3 53.8 11.05 190 200 6.5 10.0 18 28 134 3690 389 8.28 1340 134 4.98 110 N.A. M24 114.00 26.90
220 50.5 64.3 13.16 210 220 7.0 11.0 18 29 152 5410 515 9.17 1950 178 5.51 120 N.A. M24 126.00 24.90
240 60.3 76.8 15.45 230 240 7.5 12.0 21 33 164 7760 675 10.10 2770 231 6.00 90 40 M24 137.00 22.70
260 68.2 86.8 16.88 250 260 7.5 12.5 24 37 176 10450 836 11.00 3670 282 6.50 95 40 M24 148.00 21.70
280 76.4 97.3 19.52 270 280 8.0 13.0 24 37 196 13670 1010 11.90 4760 340 7.00 110 40 M24 160.00 21.00
300 88.3 113 22.27 290 300 8.5 14.0 27 41 208 18260 1260 12.70 6310 421 7.49 120 50 M27 172.00 19.50
320 97.6 124 25.11 310 300 9.0 15.5 27 43 224 22930 1480 13.60 6990 466 7.49 120 50 M27 176.00 18.00
340 105 133 28.22 330 300 9.5 16.5 27 44 242 27690 1680 14.40 7440 496 7.46 120 50 M27 179.00 17.10
360 112 143 31.50 350 300 10.0 17.5 27 45 260 33090 1890 15.20 7890 526 7.43 120 50 M27 183.00 16.40
400 125 159 38.72 390 300 11.0 19.0 27 46 298 45070 2310 16.80 8560 571 7.34 120 50 M27 191.00 15.30
450 140 178 45.77 440 300 11.5 21.0 27 48 344 63720 2900 18.90 9470 631 7.29 120 50 M27 201.00 14.40
500 155 198 53.28 490 300 12.0 23.0 27 50 390 86970 3550 21.00 10370 691 7.24 120 50 M27 211.00 13.60
550 166 212 61.50 540 300 12.5 24.0 27 51 438 111900 4150 23.00 10820 721 7.15 120 50 M27 221.00 13.30
600 178 226 70.20 590 300 13.0 25.0 27 52 486 141200 4790 25.00 11270 751 7.05 120 50 M27 231.00 13.00
650 190 242 79.38 640 300 13.5 26.0 27 53 534 175200 5470 26.90 11720 782 6.97 120 50 M27 241.00 12.70
700 204 260 92.22 690 300 14.5 27.0 27 54 582 215300 6240 28.80 12180 812 6.84 120 50 M27 250.00 12.30
800 224 286 110.10 790 300 15.0 28.0 30 58 674 303400 7680 32.60 12640 843 6.65 120 50 M27 270.00 12.00
900 252 321 132.80 890 300 16.0 30.0 30 60 770 422100 9480 36.30 13550 903 6.50 120 50 M27 290.00 11.50
1000 272 347 153.12 990 300 16.5 31.0 30 61 868 553800 11190 40.00 14000 934 6.35 120 50 M27 310.00 11.40
N.A. =not available for this HEA size

10
 Y
b w1 w w1

c
r1 d

BROAD FLANGE I - BEAMS

h-2c
h
( H.E.B. ) s

c
t
Y
Dimensions Axis X-X Axis Y-Y Details Surface Area
Sec. Weight Area Aweb
h b s t r1 c h-2c Ix Sx rx Iy Sy ry w w1 dmax Um Ut
No.
kg/m` cm2 cm2 mm mm mm mm mm mm mm cm
4
cm
3
cm cm
4
cm
3
cm mm mm mm x10-2 m2/m\ m2/t
100 20.4 26.0 4.80 100 100 6.0 10.0 12 22.0 56 450 89.9 4.16 167 33.5 2.53 56 N.A. M12 56.70 27.80
120 26.7 34.0 6.37 120 120 6.5 11.0 12 23.0 74 864 144 5.04 318 52.9 3.06 66 N.A. M16 68.60 25.70
140 33.7 43.0 8.12 140 140 7.0 12.0 12 24.0 92 1510 216 5.93 550 78.5 3.58 76 N.A. M20 80.50 23.80
160 42.6 54.3 10.72 160 160 8.0 13.0 15 28.0 104 2490 311 6.78 889 111 4.05 86 N.A. M20 91.80 21.50
180 51.2 65.3 12.92 180 180 8.5 14.0 15 29.0 122 3830 426 7.66 1360 151 4.57 100 N.A. M24 104.00 20.30
200 61.3 78.1 15.30 200 200 9.0 15.0 18 33.0 134 5700 570 8.54 2000 200 5.07 110 N.A. M24 115.00 18.80
220 71.5 91.0 17.86 220 220 9.5 16.0 18 34.0 152 8090 736 9.43 2840 258 5.59 120 N.A. M24 127.00 17.80
240 83.2 106 20.60 240 240 10.0 17.0 21 38.0 164 11260 938 10.30 3920 327 6.08 90 40 M20 138.00 16.60
260 93.0 118 22.50 260 260 10.0 17.5 24 41.5 177 14920 1150 11.20 5130 395 6.58 95 40 M20 150.00 16.10
280 103 131 25.62 280 280 10.5 18.0 24 42.0 196 19270 1380 12.10 6590 471 7.09 110 40 M20 162.00 15.70
300 117 149 28.82 300 300 11.0 19.0 27 46.0 208 25170 1680 13.00 8560 571 7.58 120 50 M24 173.00 14.80
320 127 161 32.09 320 300 11.5 20.5 27 47.5 225 30820 1930 13.80 9240 616 7.57 120 50 M24 177.00 13.90
340 134 171 35.64 340 300 12.0 21.5 27 48.5 243 36660 2160 14.60 9690 646 7.53 120 50 M24 181.00 13.50
360 142 181 39.38 360 300 12.5 22.5 27 49.5 261 43190 2400 15.50 10140 676 7.49 120 50 M24 185.00 13.00
400 155 198 47.52 400 300 13.5 24.0 27 51.0 298 57680 2880 17.10 10820 721 7.40 120 50 M24 193.00 12.40
450 171 218 55.72 450 300 14.0 26.0 27 53.0 344 79890 3550 19.10 11720 781 7.33 120 50 M24 203.00 11.90
500 187 239 64.38 500 300 14.5 28.0 27 55.0 390 107200 4290 21.20 12620 842 7.27 120 50 M24 212.00 11.30
550 199 254 73.80 550 300 15.0 29.0 27 56.0 438 136700 4970 23.20 13080 872 7.17 120 50 M24 222.00 12.20
600 212 270 83.70 600 300 15.5 30.0 27 57.0 486 171000 5700 25.20 13530 902 7.08 120 50 M24 232.00 11.00
650 225 286 94.08 650 300 16.0 31.0 27 58.0 534 210600 6480 27.10 13980 932 6.99 120 50 M24 242.00 10.80
700 241 306 108.1 700 300 17.0 32.0 27 59.0 582 256900 7340 29.00 14440 963 6.87 120 50 M24 252.00 10.50
800 262 334 128.5 800 300 17.5 33.0 30 63.0 674 359100 9890 32.80 14900 994 6.68 120 50 M24 271.00 10.40
900 291 371 153.6 900 300 18.5 35.0 30 65.0 770 494100 10980 36.50 15820 1050 6.53 120 50 M24 291.00 10.00
1000 314 400 176.3 1000 300 19.0 36.0 30 66.0 868 644700 12890 40.10 16280 1090 6.38 120 50 M24 311.00 9.90
N.A.=not available for this HEB size

11
 Y
w1 w w1
b

c
r1 d
BROAD FLANGE I - BEAMS

h-2c
h
( H.E.M. ) X
s
X

c
Y
Dimensions Axis X-X Axis Y-Y Details Surface Area
Sec. Weight Area Aweb
h b s t r1 c h-2c Ix Sx rx Iy Sy ry w w1 dmax Um Ut
No. 2 2
kg/m` cm cm mm mm mm mm mm mm mm cm4 cm3 cm cm4 cm3 cm mm mm mm x10-2 m2/m\ m2/t
100 41.8 53.2 9.60 120 106 12 20 12 32 56 1140 190 4.63 400 75.3 2.74 60 N.A. M12 61.90 14.80
120 52.1 66.4 12.25 140 126 12.5 21 12 33 74 2020 288 5.51 700 112 3.25 68 N.A. M16 73.80 14.20
140 63.2 80.6 15.08 160 146 13 22 12 34 92 3290 411 6.39 1140 157 3.77 76 N.A. M20 85.70 13.60
160 76.2 97.1 18.76 180 166 14 23 15 38 104 5100 566 7.25 1760 212 4.26 86 N.A. M20 97.00 12.70
180 88.9 113 22.04 200 186 14.5 24 15 39 122 7480 748 8.13 2580 277 4.77 100 N.A. M24 109.00 12.30
200 103 131 25.50 220 206 15 25 18 43 134 10640 967 9.00 3650 354 5.27 110 N.A. M24 120.00 11.70
220 117 149 29.14 240 226 15.5 26 18 44 152 14600 1220 9.89 5010 444 5.79 120 N.A. M24 132.00 11.30
240 157 200 37.08 270 248 18 32 21 53 164 24290 1800 11.00 8150 657 6.39 100 40 M20 146.00 9.30
260 172 220 40.50 290 268 18 32.5 24 57 176 31310 2160 11.90 10450 780 6.90 110 40 M20 157.00 9.13
280 189 240 22.76 310 288 18.5 33 24 66 196 39550 2550 12.80 13160 914 7.40 116 40 M20 169.00 8.94
300 238 303 55.02 340 310 21 39 27 67 208 59200 3480 14.00 19400 1250 8.00 120 50 M24 183.00 7.70
320 245 312 58.59 359 309 21 40 27 67 225 68130 3800 14.80 19710 1280 7.95 126 50 M24 187.00 7.63
340 248 316 62.37 377 309 21 40 27 67 243 76370 4050 15.60 19710 1280 7.90 126 50 M24 190.00 7.67
360 250 319 66.15 395 308 21 40 27 67 261 84870 4300 16.30 19520 1270 7.83 126 50 M24 193.00 7.77
400 256 326 73.92 432 307 21 40 27 67 298 104100 4820 17.90 19340 1260 7.70 126 50 M24 200.00 7.81
450 263 335 83.58 478 307 21 40 27 67 344 131500 5500 19.80 19340 1260 7.59 126 50 M24 210.00 7.97
500 270 344 93.24 524 306 21 40 27 67 390 161900 6180 21.70 19150 1250 7.46 130 50 M24 218.00 8.07
550 278 354 103.32 572 306 21 40 27 67 438 198000 6920 23.60 19150 1250 7.35 130 50 M24 228.00 8.20
600 285 364 113.40 620 305 21 40 27 67 486 237400 7660 25.60 18980 1240 7.22 130 50 M24 237.00 8.32
650 293 374 123.48 668 305 21 40 27 67 534 281700 8430 27.50 18980 1240 7.13 130 50 M24 247.00 8.42
700 301 383 133.56 716 304 21 40 27 67 582 329300 9200 29.30 18800 1240 7.01 130 50 M24 256.00 8.50
800 317 404 154.14 814 303 21 40 30 70 674 442600 10870 33.10 18630 1230 6.79 132 50 M24 275.00 8.66
900 333 424 174.30 910 302 21 40 30 70 770 570400 12540 36.70 18450 1220 6.60 132 50 M24 293.00 8.80
1000 349 444 194.88 1008 302 21 40 30 70 868 722300 14330 40.30 18450 1220 6.45 132 50 M24 313.00 8.97
N.A.=not available for this HEM size

12
 Y
b

T - SECTION x x

h
Half I.P.E.

ex
Y
Dimensions Axis X-X Axis Y-Y Surface Area
1/2 IPE Weight Area Aweb
h b Ix Sxb Sxt rx ex Iy Sy ry Um Ut
No.
\ 2 2 4 3 3 4 3
kg/m cm cm mm mm cm cm cm cm cm cm cm cm x10-2 m2/m\ m2/m\
80 3.00 3.82 1.32 40 46 4.80 1.58 5.00 1.12 3.04 4.24 1.84 1.05 16.40 54.80
100 4.05 5.15 1.82 50 55 10.3 2.70 8.66 1.41 3.81 7.95 2.89 1.24 20.00 49.50
120 5.20 6.62 2.36 60 64 19.3 4.20 13.79 1.71 4.60 13.80 4.32 1.45 23.70 45.60
140 6.45 8.21 2.97 70 73 33.2 6.14 20.49 2.01 5.38 22.40 6.15 1.65 27.50 42.60
160 7.89 10.10 3.63 80 82 52.9 8.57 28.75 2.29 6.16 34.10 8.34 1.84 31.10 39.40
180 9.40 12.00 4.35 90 91 80.3 11.5 39.17 2.59 6.95 50.40 11.1 2.05 34.90 37.10
200 11.20 14.20 5.12 100 100 117 15.1 52.00 2.87 7.75 71.20 14.2 2.24 38.40 34.30
220 13.10 16.70 5.95 110 110 165 19.3 67.35 3.15 8.55 100.2 18.6 2.48 42.40 32.40
240 15.40 19.60 6.83 120 120 227 24.3 86.31 3.41 9.37 142.0 23.7 2.69 46.00 30.00
270 18.00 23.00 8.24 135 135 346 32.8 116.50 3.88 10.53 210.0 31.1 3.02 52.00 28.80
300 21.10 26.90 9.89 450 150 509 43.6 153.31 4.35 41.68 302.0 40.3 3.35 58.00 27.50
330 24.60 31.30 11.51 165 160 717 55.8 196.44 4.78 12.85 394.0 49.3 3.55 62.50 25.50
360 28.50 36.40 13.38 180 170 992 70.8 255.01 5.22 14.11 521.0 61.3 3.79 67.50 23.60
400 33.20 42.20 16.04 200 180 1450 93.7 320.80 5.86 15.48 659.0 73.4 3.95 73.50 22.20
450 38.80 49.40 19.78 225 190 2220 129 420.45 6.70 17.22 838.0 88.4 4.12 80.50 20.70
500 45.30 57.80 23.87 250 200 3260 172 542.43 7.52 18.99 1070 107.0 4.31 87.00 19.20
550 52.80 67.20 28.62 275 210 4670 225 689.81 8.33 20.73 1330 127.0 4.45 94.00 17.70
600 61.20 78.00 33.72 300 220 6500 288 868.98 9.13 22.52 1690 154.0 4.66 101.00 16.60

13
 Y
b

T - SECTIONS x x
Half H.E.A.

ex
Y
Dimensions Axis X-X Axis Y-Y Surface Area
1/2 HEA Weight Area Aweb
h b Ix Sxb Sxt rx ex Iy Sy ry Um Ut
No.
\ 2 2
kg/m cm cm mm mm cm4 cm3 cm3 cm cm cm4 cm3 cm x10-2 m2/m\ m2/m\
100 8.34 10.60 2.00 48 100 12.4 3.16 13.9 1.08 3.91 67 13.4 2.51 28.00 33.60
120 9.94 12.70 2.45 57 120 21.3 4.52 21.7 1.30 4.72 115 19.2 3.02 33.80 34.00
140 12.30 15.70 3.19 66 140 37.5 6.79 33.2 1.55 5.47 194 27.8 3.52 39.70 32.10
160 15.20 19.40 4.02 76 160 61.5 9.72 48.0 1.78 6.32 308 38.5 3.98 45.30 29.80
180 17.80 22.60 4.56 85 180 89.1 12.4 65.0 1.98 7.13 462 51.4 4.52 51.00 28.70
200 21.10 26.90 5.53 95 200 133 16.6 87.5 2.22 7.98 668 66.8 4.98 57.00 26.90
220 25.30 32.20 6.58 105 220 194 21.9 116.9 2.45 8.84 975 88.8 5.51 63.00 24.90
240 30.20 38.40 7.73 115 240 273 28.8 150.8 2.67 9.69 1380 115 6.00 68.50 22.70
260 34.10 43.40 8.44 125 260 355 33.5 185.9 2.86 10.59 1830 141 6.50 74.00 21.70
280 38.20 48.60 9.76 135 280 477 41.7 231.6 3.13 11.44 2380 170 7.00 80.00 21.00
300 44.20 56.30 11.14 145 300 630 51.2 285.1 3.35 12.29 3150 210 7.49 86.00 19.50
320 48.80 62.20 12.56 155 300 808 61.7 335.3 3.60 13.09 3490 233 7.49 88.00 18.00
340 52.40 66.70 14.11 165 300 1020 73.5 386.4 3.91 13.86 3720 248 7.46 89.50 17.10
360 56.00 71.40 15.75 175 300 1270 86.7 442.5 4.22 14.63 3940 263 7.43 91.50 16.40
400 62.40 79.50 19.36 195 300 1890 118 557.5 4.88 16.11 4280 285 7.34 99.50 15.30
450 69.90 89.00 22.89 220 300 2820 156 715.7 5.62 18.06 4730 315 7.29 100.00 14.40
500 77.50 98.80 26.64 245 300 4020 201 891.4 6.38 19.99 5180 345 7.24 105.00 13.60
550 83.10 106.00 30.75 270 300 5530 253 1070 7.23 21.83 5410 360 7.15 110.00 13.30
600 88.90 113.00 35.10 295 300 7400 313 1261 8.08 23.63 5630 375 7.05 115.00 13.00

14
 Y
b

x x
T - SECTIONS

ex
Half H.E.B.

Y
Dimensions Axis X-X Axis Y-Y Surface Area
1/2 HEB Weight Area Aweb
h b Ix Sxb Sxt rx ex Iy Sy ry Um Ut
No.
\ 2 2 4 3 3 4 3
kg/m cm cm mm mm cm cm cm cm cm cm cm cm x10-2 m2/m\ m2/m\
100 10.20 13.00 2.40 50 100 16.2 4.05 16.2 1.12 4.00 84 16.8 2.53 28.30 27.80
120 13.30 17.00 3.19 60 120 30.9 6.35 27.1 1.35 4.86 159 26.4 3.06 34.30 25.70
140 16.90 21.50 4.06 70 140 53.5 9.36 41.5 1.58 5.71 275 39.2 3.58 40.20 23.90
160 21.30 27.10 5.36 80 160 91.3 14.0 61.7 1.83 6.52 444 55.5 4.05 45.90 21.50
180 25.60 32.60 6.46 90 180 139 18.9 85.8 2.07 7.38 680 75.6 4.57 52.00 20.30
200 30.60 39.00 7.65 100 200 204 24.8 115.3 2.29 8.23 1002 100 5.07 57.00 18.80
220 35.70 45.50 8.93 110 220 289 31.8 150.5 2.52 9.08 1422 129 5.59 63.50 17.80
240 41.60 53.00 10.30 120 240 397 40.0 192.7 2.74 9.94 1961 163 6.08 69.00 16.60
260 46.50 59.20 11.25 130 260 512 47.3 235.9 2.94 10.83 2570 197 6.58 75.00 16.10
280 51.60 65.70 12.81 140 280 673 57.7 290.1 3.20 11.68 3300 235 7.09 81.00 15.70
300 58.50 74.60 14.41 150 300 871 69.5 352.6 3.42 12.53 4280 285 7.58 86.50 14.80
320 63.30 80.70 16.04 160 300 1097 82.3 409.3 3.69 13.32 4620 308 7.57 88.50 13.90
340 67.10 85.40 17.82 170 300 1362 96.7 468.0 3.99 14.09 4840 323 7.53 90.50 13.50
360 70.90 90.30 19.69 180 300 1671 113 530.5 4.30 14.85 5070 338 7.49 92.50 13.00
400 77.60 98.90 23.76 200 300 2437 149 665.8 4.96 16.34 5410 360 7.40 96.50 12.40
450 85.60 109.00 27.86 225 300 3566 195 843.0 5.72 18.27 5860 390 7.33 101.00 11.90
500 93.70 119.00 32.19 250 300 5020 249 1041.5 6.49 20.18 6310 421 7.27 106.00 11.30
550 99.70 127.00 36.90 275 300 6834 311 1244.8 7.33 22.01 6540 436 7.17 111.00 11.20
600 106.00 135.00 41.85 300 300 9060 381 1461.3 8.19 23.80 6760 451 7.08 116.00 11.00

15
 t
PIPES D

Dimensions Surface Area

Pipe Weight Area I S r
D t Um Ut
No.
kg/m` cm2 mm mm cm4 cm3 cm x10-2 m2/m\ m2/t
2.59 3.30 38 3 5.00 2.68 1.24 11.94 46.09
38
3.35 4.27 38 4 6.26 3.29 1.21 11.94 35.64
3.07 3.91 44.5 3 8.46 3.80 1.47 13.98 45.54
44.5 4.00 5.09 44.5 4 10.5 4.74 1.44 13.98 34.95
4.87 6.20 44.5 5 12.3 5.53 1.41 13.98 28.71
4.22 5.37 60 3 21.9 7.29 2.02 18.85 44.67
60 5.52 7.04 60 4 27.7 9.24 1.99 18.85 34.15
6.78 8.64 60 5 32.9 11.0 1.95 18.85 27.80
4.96 6.31 70 3 35.5 10.1 2.37 21.99 44.34
6.51 8.29 70 4 45.3 13.0 2.34 21.99 33.78
70
8.01 10.20 70 5 54.2 15.5 2.31 21.99 27.45
9.47 12.10 70 6 62.3 17.8 2.27 21.99 23.22
5.40 6.88 76 3 45.9 12.1 2.58 23.88 44.21
7.10 9.05 76 4 58.5 15.5 2.55 23.88 33.63
76
8.75 11.20 76 5 70.6 18.6 2.52 23.88 27.29
10.40 13.20 76 6 81.4 21.4 2.48 23.88 22.96
8.38 10.70 89 4 96.7 21.7 3.01 27.96 33.37
10.40 13.20 89 5 117 26.2 2.98 27.96 26.88
89
12.30 15.60 89 6 135 30.4 2.94 27.96 22.73
14.20 18.00 89 7 153 34.3 2.91 27.96 19.69
12.70 16.20 108 5 215 39.8 3.65 33.93 26.72
15.10 19.20 108 6 251 46.5 3.61 33.93 22.47
108
17.40 22.20 108 7 285 52.7 3.58 33.93 19.50
19.70 25.10 108 8 316 58.6 3.55 33.93 17.22
16.80 20.10 133 5 412 62.0 4.53 41.78 24.87
18.80 23.90 133 6 484 72.7 4.5 41.78 22.23
133
21.80 27.70 133 7 552 82.9 4.46 41.78 19.17
24.70 31.40 133 8 616 92.6 4.43 41.78 16.92
19.00 24.20 159 5 718 90.3 5.45 49.95 26.29
22.60 28.80 159 6 845 106 5.41 49.95 22.10
159 26.20 33.40 159 7 967 122 5.38 49.95 19.07
29.80 38.00 159 8 1080 136 5.35 49.95 16.76
36.70 46.80 159 10 1300 164 5.28 49.95 13.61

16
 t
D
PIPES

Dimensions Surface Area

Pipe Weight Area I S r
D t Um Ut
No.
kg/m` cm2 mm mm cm4 cm3 cm x10-2 m2/m\ m2/t
27.8 35.4 194 6 1570 162 6.65 60.95 21.93
32.3 41.1 194 7 1800 186 6.62 60.95 18.89
194
36.7 46.7 194 8 2030 209 6.59 60.95 16.63
45.4 57.8 194 10 2241 253 6.53 60.95 13.43
31.5 40.2 219 6 2280 208 7.53 68.80 21.84
36.6 46.6 219 7 2620 240 7.50 68.80 18.81
219 41.6 53.0 219 8 2960 270 7.46 68.80 16.54
51.5 65.7 219 10 3590 328 7.40 68.80 13.36
61.3 78.0 219 12 4190 383 7.32 68.80 11.22
41.1 52.3 245 7 3710 303 8.42 76.97 18.75
46.8 59.6 245 8 4190 342 8.37 76.97 16.45
245 57.9 73.8 245 10 5110 417 8.32 76.97 13.29
68.9 87.8 245 12 5980 488 8.25 76.97 11.17
79.8 102.0 245 14 6800 555 8.17 76.97 9.65
45.9 58.5 273 7 5180 379 9.40 85.77 18.68
52.5 66.6 273 8 5850 429 9.37 85.77 16.34
273 64.9 82.6 273 10 7150 524 9.30 85.77 13.22
77.2 98.4 273 12 8400 615 9.24 85.77 11.10
89.4 114.0 273 14 9580 702 9.17 85.77 9.59
62.5 79.7 325 8 10010 616 11.20 102.10 16.34
77.7 99.0 325 10 12290 756 11.10 102.10 13.14
325 92.7 118.0 325 12 14470 891 11.10 102.10 11.01
107.0 137.0 325 14 16570 1020 11.00 102.10 9.54
122.0 155.0 325 16 18590 1140 10.90 102.10 8.37
71.0 90.5 368 8 14660 797 12.70 115.61 16.27
88.3 112.0 368 10 18030 980 12.70 115.61 13.09
105.0 134.0 368 12 21290 1160 12.60 115.61 11.01
368
122.0 156.0 368 14 24430 1330 12.50 115.61 9.48
139.0 177.0 368 16 27460 1490 12.50 115.61 8.32
172.0 219.0 368 20 33210 1800 12.30 115.61 6.72
100.5 128.0 419 10 26880 1283 14.50 131.63 13.10
120.1 153.0 419 12 31800 1518 14.40 131.63 10.96
419
159.4 203.0 419 16 41190 1970 14.30 131.63 8.26
197.0 251.0 419 20 50020 2390 14.10 131.63 6.68
115.0 147.0 529 9 49710 1879 18.40 166.19 14.45
529
128.0 163.0 529 10 54920 2077 18.40 166.19 12.99

17
 Y

h
x x
HOLLOW SQUARE SECTION t

h
Y

Dimensions Axis X-X & Y-Y Surface Area

Weight Area J
h t I S r Um Ut
mm mm kg/m\ cm2 cm4 cm3 cm cm4 x10-2 m2/m\ m2/t
40 4.0 4.46 5.68 12.1 6.07 1.46 19.5 15.10 33.87
3.2 4.66 5.94 21.6 8.62 1.91 33.8 19.30 41.39
50 4.0 5.71 7.28 25.5 10.2 1.87 40.4 19.10 33.42
5.0 6.97 8.88 29.6 11.9 1.83 47.6 18.90 27.11
3.2 5.67 7.22 38.7 12.9 2.31 60.1 23.30 41.11
60 4.0 6.97 8.88 46.1 15.4 2.28 72.4 23.10 33.14
5.0 8.56 10.90 54.4 18.1 2.24 86.3 22.90 26.76
3.6 7.46 9.50 69.5 19.9 2.70 108 27.20 36.47
70
5.0 10.13 12.90 90.1 25.7 2.64 142 26.90 26.56
3.6 8.56 10.90 106 26.5 3.11 164 31.20 36.46
80 5.0 11.70 14.90 139 34.7 3.05 217 30.90 26.42
6.3 14.44 18.40 165 41.3 3.00 261 30.60 21.19
3.6 9.73 12.40 154 34.1 3.52 237 35.20 36.16
90 5.0 13.27 16.90 202 45.0 3.46 315 34.90 26.31
6.3 16.41 20.90 242 53.9 3.41 381 34.60 21.09
4.0 12.01 15.30 234 46.8 3.91 361 39.10 32.55
5.0 14.84 18.90 283 56.6 3.87 439 38.90 26.22
100 6.3 18.37 23.40 341 68.2 3.81 533 38.60 21.01
8.0 22.84 29.10 408 81.5 3.74 646 38.30 16.77
10.0 27.87 35.50 474 94.9 3.65 761 37.90 13.60
5.0 17.98 22.90 503 83.8 4.69 775 46.90 26.09
6.3 22.37 28.50 610 102 4.63 949 46.60 20.83
120
8.0 27.87 35.50 738 123 4.56 1159 46.30 16.61
10.0 34.15 43.50 870 145 4.47 1381 45.90 13.44
5.0 21.12 26.90 814 116 5.50 1251 54.90 26.00
6.3 26.30 33.50 994 142 5.45 1538 54.60 20.76
140
8.0 32.89 41.90 1212 173 5.38 1889 54.30 16.51
10.0 40.43 51.50 1441 206 5.29 2269 53.90 13.33

18
 Y

h
HOLLOW SQUARE SECTION x x
t

h
Y

Dimensions Axis X-X & Y-Y Surface Area

Weight Area J
h t I S r Um Ut
mm mm kg/m\ cm2 cm4 cm3 cm cm4 x10-2 m2/m\ m2/t
5.0 22.69 28.90 1009 135 5.91 1548 58.90 25.96
6.3 28.26 36.00 1236 165 5.86 1907 58.60 20.74
8.0 35.40 45.10 1510 201 5.78 2348 58.30 16.47
150
10.0 43.57 55.50 1803 240 5.70 2829 57.90 13.29
12.5 53.38 68.00 2125 283 5.59 3372 57.30 10.73
16.0 66.33 84.50 2500 333 5.44 4029 56.60 8.53
6.3 34.23 43.60 2186 243 7.08 3357 70.60 20.63
8.0 42.94 54.70 2689 299 7.01 4156 70.30 16.37
180 10.0 52.99 67.50 3237 360 6.92 5041 69.90 13.19
12.5 65.16 83.00 3856 428 6.82 6062 69.30 10.64
16.0 81.64 104.00 4607 512 6.66 7339 68.60 8.40
6.3 38.15 48.60 3033 303 7.90 4647 78.60 20.60
8.0 47.96 61.10 3744 374 7.83 5770 78.30 16.32
200 10.0 59.27 75.50 4525 452 7.74 7020 77.90 13.14
12.5 73.01 93.00 5419 542 7.63 8479 77.30 10.59
16.0 91.85 117.00 6524 652 7.48 10330 76.60 8.34
6.3 48.04 61.20 6049 484 9.94 9228 98.60 20.52
8.0 60.52 77.10 7510 601 9.87 11511 98.30 16.24
250 10.0 74.97 95.50 9141 731 9.78 14086 97.90 13.06
12.5 92.63 118.00 11050 884 9.68 17139 97.30 10.50
16.0 116.97 149.00 13480 1078 9.53 21109 96.60 8.26
10.0 91.06 116.00 16150 1077 11.80 24776 118.00 12.96
300 12.5 112.26 143.00 19630 1309 11.70 30290 117.00 10.42
16.0 142.09 181.00 24160 1610 11.60 37566 117.00 8.23
10.0 106.76 136.00 26050 1489 13.90 39840 138.00 12.93
350 12.5 131.88 168.00 31810 1817 13.80 48869 137.00 10.39
16.0 167.21 213.00 39370 2250 13.60 60901 137.00 8.19
10.0 122.46 156.00 39350 1968 15.90 60028 158.00 12.90
400
12.5 151.51 193.00 48190 2409 15.80 73815 157.00 10.36

19
 Y b

HOLLOW RECTANGULAR SECTION

h
x x
t

Dimensions Axis X-X Axis Y-Y Surface Area

Weight Area J
h b t Ix Sx rx Iy Sy ry Um Ut
\ 2 4 3 4 3 4 -2 2 \ 2
mm mm mm kg/m cm cm cm cm cm cm cm cm x10 m /m m /t
50 30 2.6 3.03 3.86 12.4 4.96 1.79 5.5 3.6 1.19 12.1 15.40 50.80
3.2 3.66 4.66 14.5 5.82 1.77 6.3 4.2 1.16 14.2 15.30 41.80
60 40 3.2 4.66 5.94 28.3 9.44 2.18 14.8 7.4 1.58 30.8 19.30 41.40
4.0 5.71 7.28 33.6 11.2 2.15 17.3 8.7 1.54 36.6 19.10 33.40
80 40 3.2 5.67 7.22 58.1 14.5 2.84 19.1 9.6 1.63 46.1 23.30 41.10
4.0 6.97 8.88 69.6 17.4 2.80 22.6 11.3 1.59 55.1 23.10 33.10
90 50 3.6 7.46 9.50 99.6 22.2 3.24 39.1 15.6 2.03 89.3 27.20 36.50
5.0 10.13 12.90 130 28.9 3.18 50.0 20.0 1.97 116 26.90 26.60
3.2 7.17 9.14 117 23.5 3.58 39.1 15.6 2.07 93 29.30 40.80
50 4.0 8.87 11.30 142 28.4 3.55 46.7 18.7 2.03 113 29.10 32.80
5.0 10.91 13.90 170 34.0 3.50 55.1 22.0 1.99 135 28.90 26.50
100 6.3 13.42 17.10 202 40.5 3.44 64.2 25.7 1.94 160 28.60 21.30
3.6 8.56 10.90 147 29.3 3.66 65.4 21.8 2.45 142 31.20 36.30
60 5.0 11.70 14.90 192 38.5 3.60 84.7 28.2 2.39 187 30.90 26.40
6.3 14.44 18.40 230 46.0 3.54 99.9 33.3 2.33 224 30.60 21.30
3.6 9.73 12.40 230 38.3 4.31 76.9 25.6 2.49 183 35.20 36.20
60 5.0 13.27 16.90 304 50.7 4.24 99.9 33.3 2.43 242 34.90 26.20
6.0 16.41 20.90 366 61.0 4.18 118 39.4 2.38 290 34.60 21.10
120 5.0 14.84 18.90 370 61.7 4.43 195 48.8 3.21 401 38.90 26.30
80 6.3 18.37 23.40 447 74.6 4.37 234 58.4 3.16 486 38.60 21.00
8.0 22.84 29.10 537 89.5 4.29 278 69.4 3.09 586 38.30 16.70
10.0 27.87 35.50 628 105 4.20 320 80.0 3.00 688 37.90 13.60
5.0 18.76 23.90 747 99.5 5.59 396 79.1 4.07 806 48.90 26.10
150 100 6.3 23.31 29.70 910 121 5.53 479 95.9 4.02 985 48.60 20.90
8.0 29.12 37.10 1106 147 5.46 577 115 3.94 1202 48.30 16.60
10.0 35.72 45.50 1312 175 5.37 678 136 3.86 1431 47.90 13.40
5.0 17.98 22.90 753 94.1 5.74 251 62.8 3.31 599 46.90 26.10
160 80 6.3 22.30 28.50 917 115 5.68 302 75.6 3.26 729 46.60 20.90
8.0 27.87 35.50 1113 139 5.60 361 90.2 3.19 882 46.30 16.60
10.0 34.15 43.50 1318 165 5.50 419 105 3.10 1041 45.90 13.40
5.0 22.69 28.90 1509 151 7.23 509 102 4.20 1202 58.90 25.90
6.3 28.26 36.00 1851 185 7.17 618 124 4.14 1473 58.60 20.70
200 100 8.0 35.40 45.10 2269 227 7.09 747 149 4.07 1802 58.30 16.50
10.0 43.57 55.50 2718 272 7.00 881 176 3.98 2154 57.90 13.30
12.5 53.38 68.00 3218 322 6.88 1022 204 3.88 2541 57.30 10.70
16.0 66.33 84.50 3808 381 6.71 1175 235 3.73 2988 56.60 8.52
6.3 38.15 48.60 4178 334 9.27 1886 252 6.23 4049 78.60 20.60
8.0 47.96 61.10 5167 416 9.19 2317 309 6.16 5014 78.30 16.30
250 150 10.0 59.27 75.50 6259 501 9.10 2784 371 6.07 6082 77.90 13.10
12.5 73.01 93.00 7518 601 8.99 3310 441 5.97 7317 77.30 10.60
16.0 91.85 117.00 9089 727 8.83 3943 526 5.82 8863 76.60 8.37
6.3 48.04 61.20 7880 525 11.30 4216 422 8.30 8468 98.60 20.50
8.0 60.52 77.10 9898 653 11.30 5219 522 8.23 10550 98.30 16.20
300 200 10.0 74.97 95.50 11940 796 11.20 6331 633 8.14 12890 97.90 13.10
12.5 92.63 118.00 14460 964 11.10 7619 762 8.04 15650 97.30 10.50
16.0 116.97 149.00 17700 1180 10.90 9239 924 7.89 19220 96.60 8.26
10.0 91.06 116.00 24140 1207 14.50 8138 814 8.39 19240 118.00 13.00
400 200 12.5 112.26 143.00 29410 1471 14.30 9820 982 8.29 23410 117.00 10.40
16.0 142.09 181.00 36300 1815 14.20 11950 1195 8.14 28840 117.00 8.24
10.0 106.76 136.00 37180 1653 16.60 14900 1192 10.50 33250 138.00 13.00
450 250 12.5 131.88 168.00 45470 2021 16.50 18100 1448 10.40 40670 137.00 10.40
16.0 167.21 213.00 56420 2508 16.30 22250 1780 10.20 50480 137.00 8.20

20
 MASS (kg/m\) FOR RECTANGULAR STEEL PLATES*

mm
00
10
thickness
width

Width Thickness (mm)

(mm) 5 6 7 8 10 12 14 16 18 20 22 25 28 30 35 40
25 0.98 1.18 1.53 1.57 1.96 2.36 2.75 3.14 3.53 3.93 4.32 4.91 5.50 5.89 6.87 7.85
28 1.10 1.32 1.72 1.76 2.20 2.64 3.08 3.52 3.96 4.40 4.84 5.50 6.15 6.59 7.69 8.79
30 1.18 1.41 1.84 1.88 2.36 2.83 3.30 3.77 4.24 4.71 5.18 5.89 6.59 7.07 8.24 9.42
35 1.37 1.65 2.14 2.20 2.75 3.30 3.85 4.40 4.95 5.50 6.04 6.87 7.69 8.24 9.62 10.99
40 1.57 1.88 2.45 2.51 3.14 3.77 4.40 5.02 5.65 6.28 6.91 7.85 8.79 9.42 10.99 12.56
45 1.77 2.12 2.76 2.83 3.53 4.24 4.95 5.65 6.36 7.07 7.77 8.83 9.89 10.60 12.36 14.13
50 1.96 2.36 3.06 3.14 3.93 4.71 5.50 6.28 7.07 7.85 8.64 9.81 10.99 11.78 13.74 15.70
55 2.16 2.59 3.37 3.45 4.32 5.18 6.04 6.91 7.77 8.64 9.50 10.79 12.09 12.95 15.11 17.27
60 2.36 2.83 3.68 3.77 4.71 5.65 6.59 7.54 8.48 9.42 10.36 11.78 13.19 14.13 16.49 18.84
70 2.75 3.30 4.29 4.40 5.50 6.59 7.69 8.79 9.89 10.99 12.09 13.74 15.39 16.49 19.23 21.98
80 3.14 3.77 4.90 5.02 6.28 7.54 8.79 10.05 11.30 12.56 13.82 15.70 17.58 18.84 21.98 25.12
90 3.53 4.24 5.51 5.65 7.07 8.48 9.89 11.30 12.72 14.13 15.54 17.66 19.78 21.20 24.73 28.26
100 3.93 4.71 6.13 6.28 7.85 9.42 10.99 12.56 14.13 15.70 17.27 19.63 21.98 23.55 27.48 31.40
110 4.32 5.18 6.74 6.91 8.64 10.36 12.09 13.82 15.54 17.27 19.00 21.59 24.18 25.91 30.22 34.54
120 4.71 5.65 7.35 7.54 9.42 11.30 13.19 15.07 16.96 18.84 20.72 23.55 26.38 28.26 32.97 37.68
140 5.50 6.59 8.58 8.79 10.99 13.19 15.39 17.58 19.78 21.98 24.18 27.48 30.77 32.97 38.47 43.96
160 6.28 7.54 9.80 10.05 12.56 15.07 17.58 20.10 22.61 25.12 27.63 31.40 35.17 37.68 43.96 50.24
180 7.07 8.48 11.03 11.30 14.13 16.96 19.78 22.61 25.43 28.26 31.09 35.33 39.56 42.39 49.46 56.52
200 7.85 9.42 12.25 12.56 15.70 18.84 21.98 25.12 28.26 31.40 34.54 39.25 43.96 47.10 54.95 62.80
220 8.64 10.36 13.48 13.82 17.27 20.72 24.18 27.63 31.09 34.54 37.99 43.18 48.36 51.81 60.45 69.08
250 9.81 11.78 15.31 15.70 19.63 23.55 27.48 31.40 35.33 39.25 43.18 49.06 54.95 58.88 68.69 78.50
300 11.78 14.13 18.38 18.84 23.55 28.26 32.97 37.68 42.39 47.10 51.81 58.88 65.94 70.65 82.43 94.20
350 13.74 16.49 21.44 21.98 27.48 32.97 38.47 43.96 49.46 54.95 60.45 68.69 76.93 82.43 96.16 109.90
400 15.70 18.84 24.50 25.12 31.40 37.68 43.96 50.24 56.52 62.80 69.08 78.50 87.92 94.20 109.90 125.60
450 17.66 21.20 27.56 28.26 35.33 42.39 49.46 56.52 63.59 70.65 77.72 88.31 98.91 105.98 123.64 141.30
500 19.63 23.55 30.63 31.40 39.25 47.10 54.95 62.80 70.65 78.50 86.35 98.13 109.90 117.75 137.38 157.00
600 23.55 28.26 36.75 37.68 47.10 56.52 65.94 75.36 84.78 94.20 103.62 117.75 131.88 141.30 164.85 188.40
700 27.48 32.97 42.88 43.96 54.95 65.94 76.93 87.92 98.91 109.90 120.89 137.38 153.86 164.85 192.33 219.80
800 31.40 37.68 49.00 50.24 62.80 75.36 87.92 100.48 113.04 125.60 138.16 157.00 175.84 188.40 219.80 251.20
900 35.33 42.39 55.13 56.52 70.65 84.78 98.91 113.04 127.17 141.30 155.43 176.63 197.82 211.95 247.28 282.60
1000 39.25 47.10 61.25 62.80 78.50 94.20 109.90 125.60 141.30 157.00 172.70 196.25 219.80 235.50 274.75 314.00
*Steel denisty 7850 kg/m3
21
 RAILS

Dimensions
Rail h h1 h2 2ck 2cf d t0 t2 af r1 r3 r4 r5 r6 ar
Type mm mm mm mm mm mm mm mm mm mm mm mm mm mm mm
vst 130 43 33 60 100 12 23 12 67 13 10 80 200 36
I 145 45 37 65 125 14 25 13 6 30 80 300 19
III 159 49 36 70 140 16 30 13 93 13 16 508 80 300 19
IV 161 51 39 67 125 16 30 14 91 13 16 508 80 300 19
VI 172 51 38 72 150 16.5 32 13 100 13 35 120 80 300 21

Y
2Ck
r1 ar

r6
r5

h2

h1
r4

r3
Properties d
X

h
X
Rail Weight Area Ix Sxt Sxb ex Iy Sy
r4
Type r3

ex
kg/m\ cm2 cm4 cm3 cm3 cm cm4 cm3
vst 35.7 45.5 1000 156 153 6.6 157 31

to
t2
I 46.2 58.8 1630 216 235 6.9 298 47.7
af
III 54.4 69.3 2350 279 313 7.5 418 59.6
IV 53.8 68.8 2310 276 298 7.8 341 54.6 2Cf
VI 60.3 76.9 3060 336 377 8.1 513 68.4
Y

22
 RAILS

Dimensions
Rail h 2cf af 2ck d t1 t2 t3 h1 h2 r1 r2 r3 r4 r5
Type mm mm mm mm mm mm mm mm mm mm mm mm mm mm mm
A 45 55 125 54 45 24 14.5 11 8 24 20 4 3 4 5 4
A 55 65 150 66 55 31 17.5 12.5 9 28.5 25 5 5 5 6 5
A 65 75 175 78 65 38 20 14 10 34 30 6 5 5 6 5
A 75 85 200 90 75 45 22 15.4 11 39.5 35 8 6 6 8 6
A 100 95 200 100 100 60 23 16.5 12 45.5 40 10 6 6 8 6
A 120 105 220 120 120 72 30 20 14 55.5 47.5 10 6 10 10 6

Y
2CK

r1
Properties

h2
h1
Rail Weight Area Ix Sxt Sxb ex Iy Sy r2
Type
X
\ 2 4 3 3 4 3
X

h
kg/m cm cm cm cm cm cm cm

r3
A 45 22.2 28.3 91 27.5 41.6 2.2 169 27.0 r4 d

r5

t1
A 55 32.0 40.7 182 46.9 69.5 2.6 337 44.9

ex
A 65 43.5 55.4 327 73.7 106.9 3.1 609 69.6

t2

t3
A 75 56.6 72.1 545 109.0 155.7 3.5 1010 101.0 aF
A 100 75.2 95.6 888 170.0 207.0 4.3 1360 136.0 2CF
A 120 101.3 129.0 1420 249.0 295.8 4.8 2370 215.0

23
 RAILS

Dimensions

Depth Head base Web

Rail k h r R R1 R2
d c c1 b t w g Alpha
Type
mm mm mm mm mm mm mm mm mm mm mm mm mm
USS or BETH 135 146 87 76 132 12 32 63 27 71 356 305 19 19 13
BETH 171 152 109 102 152 16 32 67 32 70 FLAT VERT 19 22 12
USS orBETH175 152 108 102 152 13 38 68 29 79 457 VERT 29 51 12

Y
C
C1

R2

Properties d
h w CL. BOLT HOLE
X N.A. R
X
Rail Weight Area Ix Sxt Sxb ex
Type R1 g
kg/m\ cm2 cm4 cm3 cm3 cm ex
USS & BETH 135 67 85.9 2110 282 295 7.11 .
K
BETH 171 84.9 108 3060 402 400 7.65 t
USS & BETH 175 86.8 110 2920 386 381 7.67
b

24
 RAILS

Dimensions
Depth Head Base Web
Rail k h r rc R
d c b t w g
Type
mm mm mm mm mm mm mm mm mm mm mm
ASCE 40 89 48 89 6 10 40 16 47 305 6.4 305
ASCE 60 108 60 108 7 12 52 19 58 305 6.4 305
ASCE 85 132 65 132 8 14 63 23 70 305 6.4 305
BETH 104 127 64 127 13 25 62 27 62 305 12.7 89
USS 105 132 65 132 10 24 56 25 61 305 6.4 305

Y
C

Properties 13°
r

N.A.
Rail Weight Area Ix Sxt Sxb ex X X
Type 2 4 3 3 d h C.L. of bolt hole
kg/m cm cm cm cm cm W
ASCE 40 19.8 25.4 272 58.8 63.7 4.27 rc ex
g
60 29.8 38.3 606 108 116 5.21 k 13°
85 42.2 53.7 1250 182 199 6.27 b
t
BETH 104 51.6 66.4 1240 175 221 5.61
USS 105 52.1 66.5 1430 203 234 6.12
Y

25
COMBINED HOT ROLLED SECTIONS

TWO EQUAL ANGLES BACK TO BACK

TWO UNEQUAL ANGLES SHORT LEGS BACK TO BACK
TWO UNEQUAL ANGLES LONG LEGS BACK TO BACK
TWO CHANNELS (U.P.N.) BACK TO BACK
TWO CHANNELS (U.P.N.) TOE TO TOE
SYMMETRICAL WELDED I-BEAMS
UNSYMMETRICAL WELDED I-BEAMS
I.P.E +CHANNEL (U.P.N.)

26
 Y

TWO EQUAL ANGLES x

x
Back to Back

ex
tG

size Axis X-X ry (cm) About Axis Y-Y

Weight Area
a s Ix Sx rx ex tG (mm)
\ 2 4 3
mm mm kg/m cm cm cm cm cm 8 10 12 14
3 2.72 3.48 2.82 1.30 0.90 0.84 1.53 1.61 1.70 1.78
30 4 3.56 4.54 3.62 1.72 0.89 0.89 1.57 1.65 1.74 1.82
5 4.36 5.56 4.32 2.08 0.88 0.92 1.59 1.67 1.76 1.84
3 3.20 4.08 4.58 1.80 1.06 0.96 1.72 1.80 1.89 1.97
35 4 4.20 5.34 5.92 2.36 1.05 1.00 1.75 1.83 1.91 2.00
5 5.14 6.56 7.12 2.90 1.04 1.04 1.78 1.86 1.94 2.03
4 4.84 6.16 8.96 3.12 1.21 1.12 1.94 2.02 2.10 2.19
40 5 5.94 7.58 10.86 3.82 1.20 1.16 1.97 2.05 2.13 2.21
6 7.04 8.96 12.66 4.52 1.19 1.20 1.99 2.08 2.16 2.24
5 6.76 8.60 15.66 4.86 1.35 1.28 2.16 2.23 2.31 2.40
45 6 8.00 10.18 18.32 5.76 1.34 1.32 2.18 2.26 2.34 2.42
7 9.20 11.72 20.80 6.62 1.33 1.36 2.21 2.29 2.37 2.45
5 7.54 9.60 22.00 6.10 1.51 1.40 2.35 2.43 2.51 2.59
50 6 8.94 11.38 25.60 7.02 1.50 1.45 2.38 2.46 2.54 2.62
7 10.30 13.12 29.20 8.30 1.49 1.49 2.41 2.49 2.57 2.65
5 8.36 10.64 29.40 7.40 1.66 1.52 2.54 2.61 2.69 2.77
55 6 9.90 12.62 35.60 8.80 1.66 1.56 2.57 2.65 2.72 2.80
8 12.92 16.46 44.20 11.44 1.64 1.64 2.62 2.70 2.78 2.86
6 10.84 13.82 45.60 10.58 1.82 1.69 2.77 2.85 2.93 3.00
60 8 14.18 18.06 58.20 13.76 1.80 1.77 2.82 2.90 2.98 3.06
10 17.38 22.20 69.80 16.82 1.78 1.85 2.87 2.95 3.03 3.11

27
 Y

TWO EQUAL ANGLES

x x
Back to Back

ex
tG

Y
size Axis X-X ry (cm) About Axis Y-Y
Weight Area
a s Ix Sx rx ex tG (mm)
\ 2 4 3
mm mm kg\m cm cm cm cm cm 8 10 12 14
7 13.64 17.40 66.8 14.3 1.96 1.85 2.98 3.06 3.14 3.22
65 8 15.46 19.70 75.0 16.3 1.95 1.89 3.01 3.08 3.16 3.24
9 17.24 22.00 82.6 18.1 1.94 1.93 3.03 3.11 3.19 3.27
7 14.76 18.80 84.8 16.9 2.12 1.97 3.18 3.26 3.33 3.41
70 9 18.68 23.80 105.2 21.2 2.10 2.05 3.23 3.30 3.38 3.46
11 22.40 28.60 123.6 25.4 2.09 2.13 3.28 3.36 3.44 3.52
7 15.88 20.20 104.8 17.3 2.28 2.03 3.33 3.41 3.48 3.56
75 8 18.06 23.00 117.8 22.0 2.26 2.13 3.39 3.47 3.54 3.62
10 22.20 28.20 142.8 27.0 2.25 2.21 3.45 3.52 3.60 3.68
8 19.32 24.60 144.6 25.2 2.42 2.26 3.60 3.67 3.75 3.82
80 10 23.80 30.20 175.0 31.0 2.41 2.34 3.65 3.72 3.80 3.88
12 28.20 35.80 204.0 36.4 2.39 2.41 3.69 3.77 3.84 3.92
9 24.40 31.00 232.0 36.0 2.74 2.54 4.02 4.09 4.17 4.24
90 11 29.70 37.40 276.0 43.2 2.72 2.62 4.06 4.14 4.22 4.29
13 34.20 43.60 316.0 50.2 2.69 2.70 4.10 4.18 4.26 4.34
10 30.20 38.40 354.0 49.4 3.04 2.82 4.43 4.50 4.58 4.65
100 12 35.60 45.40 414.0 58.4 3.02 2.90 4.47 4.55 4.62 4.70
14 41.20 52.40 470.0 67.0 3.00 2.98 4.52 4.59 4.67 4.75
10 33.20 42.40 478.0 60.2 3.36 3.07 4.83 4.90 4.98 5.05
110 12 39.40 50.20 560.0 71.4 3.34 3.15 4.87 4.95 5.02 5.10
14 45.60 58.00 638.0 82.0 3.32 3.21 4.90 4.98 5.05 5.13
12 43.20 55.00 736.0 85.4 3.65 3.40 5.27 5.34 5.42 5.49
120 13 46.60 59.80 788.0 92.0 3.64 3.44 5.29 5.36 5.44 5.51
15 53.20 67.80 892.0 105.0 3.63 3.51 5.34 5.41 5.48 5.56

28
 Y

TWO ANGLES EQUAL

Back to Back x x

ex
tG

size Axis X-X ry (cm) About Axis Y-Y

Weight Area
a s Ix Sx rx ex tG (mm)
mm mm kg\m\ cm2 cm4 cm 3
cm cm 8 10 12 14
12 47.20 60.00 944 100.8 3.97 3.64 5.66 5.74 5.81 5.88
130 14 54.40 69.40 1080 110.4 3.94 3.72 5.70 5.77 5.85 5.92
16 61.80 78.60 1210 131.6 3.92 3.80 5.75 5.82 5.89 5.97
13 55.00 70.00 1276 126.6 4.27 3.92 6.07 6.15 6.22 6.29
140
15 62.80 80.00 1446 144.6 4.25 4.00 6.12 6.19 6.26 6.34
14 63.20 80.60 1690 156.4 4.58 4.21 6.50 6.57 6.64 6.71
15 67.80 86.00 1796 165.0 4.57 4.25 6.52 6.59 6.66 6.74
150 16 71.80 91.40 1898 177.4 4.56 4.29 6.54 6.61 6.69 6.76
18 80.20 102.00 2100 198.6 4.54 4.36 6.58 6.65 6.72 6.80
20 88.40 112.60 2300 218.0 4.51 4.44 6.62 6.69 6.76 6.84
15 72.40 92.20 2200 191.2 4.88 4.49 6.91 6.98 7.05 7.12
160 17 81.40 103.60 2460 216.0 4.86 4.57 6.95 7.02 7.10 7.17
19 90.20 115.00 2700 236.0 4.84 4.65 6.99 7.07 7.14 7.21
16 87.00 110.80 3360 260.0 5.51 5.02 7.73 7.80 7.87 7.94
18 97.20 123.80 3740 290.0 5.49 5.10 7.77 7.84 7.91 7.99
180
20 107.40 136.80 4080 320.0 5.47 5.18 7.81 7.89 7.96 8.03
22 117.20 149.40 4220 348.0 5.44 5.26 7.85 7.92 8.00 8.07
16 97.00 123.60 4680 336.0 6.15 5.52 8.54 8.61 8.68 8.75
18 108.60 138.20 5200 362.0 6.13 5.60 8.58 8.65 8.72 8.79
200 20 119.80 152.80 5700 398.0 6.11 5.68 8.62 8.69 8.76 8.83
24 142.20 181.20 6660 470.0 6.06 5.84 8.70 8.77 8.84 8.92
28 164.00 210.00 7560 540.0 6.02 5.99 8.78 8.85 8.93 9.00

29
 Y

TWO UNEQUAL ANGLES X X

ex
Short Legs Back to Back
tG
``
Y
Size Axis X-X ry (cm) About Axis Y-Y
Weight Area
a b s IX SX rX eX tG (mm)
\ 2 4 3
mm mm mm kg\m cm cm cm cm cm 8 10 12 14
3 2.22 2.84 0.88 0.58 0.56 0.50 1.68 1.76 1.85 1.93
30 20
4 2.90 3.70 1.10 0.76 0.55 0.54 1.71 1.79 1.88 1.96
3 2.70 3.44 0.94 0.60 0.52 0.44 2.23 2.31 2.39 2.48
40 20
4 3.54 4.50 1.20 0.78 0.52 0.48 2.25 2.34 2.42 2.51
3 3.44 4.38 3.20 1.40 0.86 0.70 2.32 2.40 2.48 2.57
45 30 4 4.50 5.74 4.10 1.82 0.85 0.74 2.36 2.44 2.52 2.60
5 5.54 7.06 4.94 2.22 0.84 0.78 2.38 2.46 2.55 2.63
5 6.74 8.58 5.20 2.24 0.78 0.68 3.18 3.26 3.34 3.43
60 30
7 9.18 11.70 6.82 3.04 0.76 0.76 3.24 3.32 3.41 3.49
5 7.52 9.58 12.22 4.04 1.13 0.97 3.02 3.10 3.18 3.26
60 40 6 8.92 11.36 14.24 4.76 1.12 1.01 3.05 3.13 3.21 3.29
7 10.28 13.10 16.14 5.48 1.11 1.05 3.07 3.15 3.24 3.32
5 9.48 12.08 24.60 6.42 1.43 1.17 3.68 3.76 3.84 3.91
75 50 7 13.02 16.60 33.00 8.78 1.41 1.25 3.72 3.80 3.88 3.96
9 16.46 21.00 40.40 10.98 1.39 1.32 3.77 3.85 3.93 4.01
6 10.82 13.78 15.18 4.88 1.05 0.88 4.13 4.21 4.29 4.37
80 40
8 14.14 18.02 19.36 6.36 1.04 0.95 4.19 4.27 4.35 4.43
6 13.64 17.38 51.60 11.22 1.72 1.41 4.37 4.44 4.52 4.60
90 60
8 17.92 22.80 66.00 14.62 1.70 1.49 4.41 4.49 4.57 4.65
6 13.70 17.46 30.60 7.72 1.32 1.04 5.04 5.11 5.19 5.27
100 50 8 17.98 23.00 39.00 10.08 1.31 1.13 5.10 5.18 5.26 5.34
10 22.20 28.20 46.80 12.34 1.29 1.20 5.15 5.23 5.31 5.39
7 17.54 22.40 75.20 15.08 1.84 1.51 4.82 4.90 4.97 5.05
100 65 9 22.20 28.40 93.40 19.04 1.82 1.59 4.87 4.95 5.03 5.11
11 26.80 34.20 110.20 22.80 1.80 1.67 4.92 5.00 5.08 5.16

30
 Y

TWO UNEQUAL ANGLES

X X
Short Legs Back to Back

ex
tG
Y
Size Axis X-X ry (cm) About Axis Y-Y
Weight Area
a b s IX SX rX eX tG (mm)
mm mm mm kg\m\ cm2 cm4 cm3 cm cm 8 10 12 14
8 24.40 31.00 161.6 26.40 2.29 1.87 5.70 5.77 5.85 5.93
10 30.00 38.20 196.2 32.40 2.27 1.95 5.75 5.83 5.91 5.98
120 80
12 35.60 45.40 228.0 38.20 2.25 2.03 5.79 5.87 5.95 6.03
14 41.00 52.40 260.0 44.00 2.23 2.10 5.84 5.92 6.00 6.08
8 23.80 30.20 89.6 17.44 1.72 1.37 6.48 6.56 6.63 6.71
130 65 10 29.20 37.20 108.4 21.40 1.71 1.45 6.54 6.61 6.69 6.77
12 34.60 44.20 126.0 25.40 1.69 1.53 6.59 6.67 6.74 6.82
10 33.20 42.40 282.0 41.20 2.58 2.18 6.13 6.21 6.28 6.36
130 90
12 39.40 50.20 330.0 48.80 2.56 2.26 6.65 6.73 6.81 6.89
9 30.60 39.00 156.6 26.40 2.00 1.57 7.46 7.53 7.61 7.69
150 75
11 37.20 47.20 186.0 31.80 1.98 1.65 7.51 7.58 7.66 7.74
10 38.00 48.40 396.0 51.60 2.86 2.34 7.06 7.14 7.21 7.29
150 100 12 45.20 57.40 464.0 61.20 2.84 2.42 7.12 7.19 7.27 7.34
14 52.20 66.40 528.0 70.40 2.82 2.50 7.16 7.23 7.31 7.38
10 36.40 46.40 208.0 33.00 2.12 1.69 7.92 8.00 8.08 8.15
160 80 12 43.20 55.00 244.0 39.20 2.10 1.77 7.97 8.05 8.13 8.21
14 50.00 63.60 278.0 45.00 2.09 1.85 8.03 8.11 8.19 8.26
10 41.20 52.40 302.0 42.20 2.40 1.85 8.85 8.92 9.00 9.08
180 90 12 49.00 62.40 354.0 50.20 2.38 1.93 8.90 8.97 9.05 9.13
14 56.60 72.20 404.0 57.80 2.37 2.01 8.95 9.03 9.11 9.18
10 46.00 58.40 420.0 52.60 2.68 2.01 9.77 9.85 9.92 10.00
12 54.60 69.60 494.0 62.60 2.67 2.10 9.83 9.90 9.98 10.05
200 100
14 63.20 80.60 564.0 72.20 2.65 2.18 9.88 9.96 10.03 10.11
16 71.80 91.40 632.0 81.60 2.63 2.26 9.92 10.00 10.08 10.15

31
 Y

TWO UNEQUAL ANGLES X X

ex
Long Legs Back to Back
tG
Y

Size Axis X-X ry (cm) About Axis Y-Y

Weight Area
a b s IX SX rX ex tG (mm)
\ 2 4 3
mm mm mm kg\m cm cm cm cm cm 8 10 12 14
3 2.22 2.84 2.50 1.24 0.94 0.99 1.06 1.15 1.23 1.32
30 20
4 2.90 3.70 3.18 1.62 0.93 1.03 1.09 1.18 1.27 1.36
3 2.70 3.44 5.58 2.16 1.27 1.43 0.99 1.07 1.16 1.25
40 20
4 3.54 4.50 7.18 2.84 1.26 1.47 1.02 1.11 1.20 1.29
3 3.44 4.38 8.94 2.92 1.43 1.43 1.40 1.48 1.56 1.64
45 30 4 4.50 5.74 11.56 3.82 1.42 1.48 1.42 1.50 1.59 1.67
5 5.54 7.06 13.98 5.90 1.41 1.52 1.45 1.53 1.62 1.70
5 6.74 8.58 31.20 8.08 1.90 2.15 1.33 1.41 1.50 1.59
60 30
7 9.18 11.70 41.40 11.00 1.88 2.24 1.39 1.47 1.56 1.65
5 7.52 9.58 34.40 8.50 1.89 1.96 1.78 1.85 1.93 2.02
60 40 6 8.92 11.36 40.20 10.06 1.88 2.00 1.80 1.88 1.96 2.04
7 10.28 13.10 46.00 11.58 1.87 2.04 1.83 1.91 1.99 2.07
5 9.48 12.08 68.80 13.48 2.39 2.40 2.12 2.20 2.28 2.35
75 50 7 13.02 16.60 92.80 18.48 2.36 2.48 2.17 2.25 2.33 2.41
9 16.46 21.00 114.8 23.20 2.34 2.56 2.21 2.29 2.37 2.45
6 10.82 13.78 89.80 17.46 2.55 2.85 1.66 1.73 1.81 1.90
80 40
8 14.14 18.02 115.2 22.80 2.53 2.94 1.70 1.78 1.87 1.95
6 13.64 17.38 143.4 23.40 2.87 2.89 2.50 2.57 2.65 2.72
90 60
8 17.92 22.80 185.0 30.80 2.85 2.97 2.54 2.62 2.69 2.77
6 13.70 17.46 179.4 27.60 3.20 3.49 1.95 2.03 2.11 2.18
100 50 8 17.98 23.00 232.0 36.00 3.18 3.59 2.01 2.09 2.17 2.25
10 22.20 28.20 282.0 44.40 3.16 3.67 2.06 2.13 2.21 2.30
7 17.54 22.40 226.0 33.20 3.17 3.23 2.65 2.73 2.80 2.88
100 65 9 22.20 28.40 282.0 42.00 3.15 3.32 2.70 2.77 2.85 2.93
11 26.80 34.20 334.0 50.60 3.13 3.40 2.74 2.82 2.90 2.98
32
 Y

TWO UNEQUAL ANGLES

X X
Long Legs Back to Back

ex
tG
Y

Size Axis X-X ry (cm) About Axis Y-Y

Weight Area
a b s IX SX rX ex tG (mm)
mm mm mm kg\m\ cm2 cm4 cm3 cm cm 8 10 12 14
8 24.40 31.00 452 55.20 3.82 3.83 3.22 3.30 3.37 3.44
10 30.00 38.20 552 68.20 3.80 3.92 3.27 3.34 3.41 3.49
120 80
12 35.60 45.40 646 80.80 3.77 4.00 3.31 3.39 3.46 3.54
14 41.00 52.40 736 92.80 3.75 4.08 3.35 3.43 3.50 3.58
8 23.80 30.20 526 62.20 4.17 4.56 2.47 2.54 2.62 2.69
130 65 10 29.20 37.20 642 76.80 4.15 4.65 2.52 2.59 2.67 2.75
12 34.60 44.20 752 91.00 4.12 4.74 2.57 2.64 2.72 2.80
10 33.20 42.40 716 81.00 4.11 4.15 3.65 3.72 3.79 3.87
130 90
12 39.40 50.20 840 96.00 4.09 4.84 3.69 3.76 3.84 3.91
9 30.60 39.00 910 93.60 4.83 5.28 2.81 2.88 2.95 3.03
150 75
11 37.20 47.20 1090 113.2 4.80 5.37 2.85 2.92 3.00 3.07
10 38.00 48.40 1104 108.2 4.78 4.80 3.96 4.03 4.10 4.17
150 100 12 45.20 57.40 1300 128.4 4.76 4.89 4.00 4.07 4.15 4.22
14 52.20 66.40 1488 148.2 4.73 4.97 4.05 4.12 4.19 4.27
10 36.40 46.40 1222 117.8 5.14 5.63 2.98 3.05 3.12 3.19
160 80 12 43.20 55.00 1440 140.0 5.11 5.72 3.02 3.09 3.17 3.24
14 50.00 63.60 1646 161.4 5.09 5.81 3.07 3.14 3.22 3.30
10 41.20 52.40 1760 150.2 5.80 6.28 3.29 3.36 3.43 3.50
180 90 12 49.00 62.40 2080 178.6 5.77 6.37 3.33 3.40 3.47 3.55
14 56.60 72.20 2380 206.0 5.75 6.46 3.38 3.45 3.53 3.60
10 46.00 58.40 2440 184.4 6.46 6.93 3.60 3.67 3.74 3.81
12 54.60 69.60 2880 222.0 6.43 7.03 3.66 3.73 3.80 3.87
200 100
14 63.20 80.60 3300 256.0 6.41 7.12 3.70 3.77 3.84 3.91
16 71.80 91.40 3720 290.0 6.38 7.20 3.74 3.81 3.89 3.96

33
 Y

TWO CHANNELS (U.P.N.) X X

BACK to BACK
tG

Y
Axis X-X Axis Y-Y
Sec. Weight Area Aweb
Ix Sx rx Iy Sy ry Iy Sy ry Iy Sy ry
No.
kg/m` cm2 cm2 cm4 cm3 cm tG = 8 mm tG = 10 mm tG = 12 mm
30x15 3.48 4.42 1.68 5.06 3.38 1.07 4.50 2.37 1.01 5.36 2.68 1.10 6.30 3.00 1.19
30 8.54 10.88 1.60 12.78 8.52 1.08 42.47 11.48 1.98 46.30 12.19 2.06 50.35 12.91 2.15
40x20 5.74 7.32 2.90 15.16 7.58 1.44 10.66 4.44 1.21 12.30 4.92 1.30 14.09 5.42 1.39
40 9.74 12.42 2.60 28.20 14.10 1.50 50.53 12.96 2.02 54.95 13.74 2.10 59.62 14.54 2.19
50x25 7.72 9.84 3.80 33.20 13.46 1.85 19.39 6.69 1.40 21.87 7.29 1.49 24.54 7.92 1.58
50 11.18 14.24 3.60 52.80 21.20 1.92 62.85 14.96 2.10 68.04 15.82 2.19 73.50 16.71 2.27
60 10.14 12.92 5.76 63.20 21.00 2.21 31.19 9.17 1.55 34.71 9.92 1.64 38.48 10.69 1.73
65 14.18 18.06 5.50 115.0 35.40 2.52 88.02 19.14 2.21 94.78 20.17 2.29 101.89 21.23 2.38
70 13.46 17.14 6.84 122.2 35.00 2.67 79.57 18.09 2.15 85.98 19.11 2.24 92.74 20.16 2.33
80 17.28 22.00 7.68 212.0 53.00 3.10 114.10 23.28 2.28 122.5 24.49 2.36 131.3 25.74 2.44
100 21.20 27.00 9.96 412.0 82.40 3.91 161.27 29.86 2.44 172.1 31.29 2.52 183.4 32.75 2.61
120 26.80 34.00 14.28 728.0 121.4 4.62 222.40 37.69 2.56 236.3 39.39 2.64 251.0 41.14 2.72
140 32.00 40.80 16.80 1210 172.8 5.45 314.00 49.06 2.77 332.0 51.07 2.85 350.7 53.14 2.93
160 37.60 48.00 20.85 1850 232.0 6.21 411.44 59.63 2.93 433.4 61.92 3.00 456.4 64.28 3.08
180 44.00 56.00 25.28 2700 300.0 6.95 529.41 71.54 3.07 556.0 74.13 3.15 583.6 76.79 3.23
200 50.60 64.40 30.09 3820 382.0 7.70 670.04 84.82 3.23 701.7 87.72 3.30 734.7 90.70 3.38
220 58.80 74.80 35.10 5380 490.0 8.48 876.58 104.4 3.42 915.3 107.7 3.50 955.6 111.1 3.57
240 66.40 84.60 40.66 7200 600.0 9.22 1081.2 121.5 3.57 1126.5 125.2 3.65 1173.6 129.0 3.72
260 75.80 96.60 46.40 9640 742.0 9.99 1369.9 145.7 3.77 1424.1 149.9 3.84 1480.4 154.2 3.91
280 83.60 106.60 50.00 12560 896.0 10.90 1713.2 173.0 4.01 1776.7 177.7 4.08 1842.3 182.4 4.16
300 92.40 117.60 53.60 16060 1070 11.70 2120.1 203.9 4.25 2194.2 209.0 4.32 2270.7 214.2 4.39
320 119.00 151.60 79.80 21740 1358 12.10 2558.4 246.0 4.11 2650.9 252.5 4.18 2746.4 259.1 4.26
350 121.20 154.60 89.04 25680 1468 12.90 2352.1 226.2 3.90 2440.2 232.4 3.97 2531.4 238.8 4.05
380 126.20 160.80 93.96 31520 1658 14.00 2472.7 233.3 3.92 2563.7 239.6 3.99 2658.0 246.1 4.07
400 143.60 183.00 101.92 40700 2040 14.90 3394.4 297.8 4.31 3507.8 305.0 4.38 3624.9 312.5 4.45

34
 Y
c

TWO CHANNELS (U.P.N.)

d
X X

TOE to TOE
Y

Axis X-X Axis Y-Y (toe to toe) Axis Y-Y (c=d)

Sec. Weight Area Aweb
Ix Sx rx Iy Sy ry Iy Sy ry
No.
2 2 4 3 4 3 4 3
kg/m` cm cm cm cm cm cm cm cm cm cm cm
30x15 3.48 4.42 1.68 5.06 3.38 1.07 5.00 3.34 1.06 5.00 3.34 1.06
30 8.54 10.88 1.60 12.78 8.52 1.08 53.75 16.29 2.22 N.A.
40x20 5.74 7.32 2.90 15.16 7.58 1.44 15.23 7.61 1.44 15.23 7.61 1.44
40 9.74 12.42 2.60 28.20 14.10 1.50 71.84 20.53 2.41 N.A.
50x25 7.72 9.84 3.80 33.20 13.46 1.85 33.08 13.23 1.83 33.08 13.23 1.83
50 11.18 14.24 3.60 52.80 21.20 1.92 102.3 26.93 2.68 N.A.
60 10.14 12.92 5.76 63.20 21.00 2.21 65.46 21.82 2.25 65.46 21.82 2.25
65 14.18 18.06 5.50 115.0 35.40 2.52 167.8 39.95 3.05 N.A.
70 13.46 17.14 6.84 122.2 35.00 2.67 136.9 34.22 2.83 N.A.
80 17.28 22.00 7.68 212.0 53.00 3.10 243.5 54.10 3.33 N.A.
100 21.20 27.00 9.96 412.0 82.40 3.91 380.0 75.99 3.75 380.0 75.99 3.75
120 26.80 34.00 14.28 728.0 121.4 4.62 603.5 109.7 4.21 744.6 124.1 4.68
140 32.00 40.80 16.80 1210 172.8 5.45 862.4 143.7 4.60 1250 178.6 5.53
160 37.60 48.00 20.85 1850 232.0 6.21 1213 186.6 5.03 1992 249.0 6.44
180 44.00 56.00 25.28 2700 300.0 6.95 1673 239.0 5.47 3035 337.2 7.36
200 50.60 64.40 30.09 3820 382.0 7.70 2237 298.3 5.89 4407 440.7 8.27
220 58.80 74.80 35.10 5380 490.0 8.48 2963 370.3 6.29 6266 569.6 9.15
240 66.40 84.60 40.66 7200 600.0 9.22 3822 449.6 6.72 8571 714.3 10.07
260 75.80 96.60 46.40 9640 742.0 9.99 4893 543.7 7.12 11570 890.0 10.94
280 83.60 106.60 50.00 12560 896.0 10.90 5977 629.1 7.49 14822 1059 11.79
300 92.40 117.60 53.60 16060 1070 11.70 7257 725.7 7.86 18782 1252 12.64
320 119.00 151.60 79.80 21740 1358 12.10 9496 949.6 7.91 28415 1776 13.69
350 121.20 154.60 89.04 25680 1468 12.90 10070 1007 8.07 36390 2079 15.34
380 126.20 160.80 93.96 31520 1658 14.00 11063 1085 8.29 45647 2402 16.85
400 143.60 183.00 101.92 40700 2040 14.90 14451 1314 8.89 56779 2839 17.61
N.A.= Section is not avilable with the specified configuration.
35
 Y
b

d
SYMMETRICAL WELDED I- SECTIONS x x
S

Y
depth Web Flange Axis x-x Axis y-y Surface Area
Weight Area Aweb
h d s b t Ix Sx rx Iy Sy ry Um Ut
2 2 4 3 4 3 -2 2 \
mm mm mm mm mm kg/m` cm cm cm cm cm cm cm cm x10 m /m m 2/t
210 200 5 140 5 18.84 24.0 10.0 1805 172 8.67 228.9 32.7 3.09 97.00 51.49
212 200 5 140 6 21.04 26.8 10.0 2116 200 8.89 274.6 39.2 3.20 97.40 46.30
214 200 5 140 7 23.24 29.6 10.0 2434 227 9.07 320.3 45.8 3.29 97.80 42.09
260 250 5 140 5 20.80 26.5 12.5 2927 225 10.51 228.9 32.7 2.94 107.00 51.44
262 250 5 140 6 23.00 29.3 12.5 3404 260 10.78 274.7 39.2 3.06 107.40 46.69
264 250 5 140 7 25.20 32.1 12.5 3888 295 11.01 320.4 45.8 3.16 107.80 42.78
262 250 5 165 6 25.36 32.3 12.5 3896 297 10.98 449.5 54.5 3.73 117.40 46.30
264 250 5 165 7 27.95 35.6 12.5 4466 338 11.20 524.3 63.6 3.84 117.80 42.15
340 330 5 140 5 23.94 30.5 16.5 5426 319 13.34 229.0 32.7 2.74 123.00 51.37
342 330 5 140 6 26.14 33.3 16.5 6240 365 13.69 274.7 39.2 2.87 123.40 47.21
344 330 5 140 7 28.34 36.1 16.5 7063 411 13.99 320.5 45.8 2.98 123.80 43.69
342 330 5 165 6 28.50 36.3 16.5 7086 414 13.97 449.6 54.5 3.52 133.40 46.81
344 330 5 165 7 31.09 39.6 16.5 8057 468 14.26 524.4 63.6 3.64 133.80 43.04
346 330 5 165 8 33.68 42.9 16.5 9039 522 14.52 599.3 72.6 3.74 134.20 39.85
344 330 5 200 7 34.93 44.5 16.5 9448 549 14.57 933.7 93.4 4.58 147.80 42.31
346 330 5 200 8 38.07 48.5 16.5 10639 615 14.81 1067.0 106.7 4.69 148.20 38.93

36
 Y
b

h
SYMMETRICAL WELDED I- SECTIONS

d
x x
S

Y
depth Web Flange Axis x-x Axis y-y Surface Area
Weight Area Aweb
h d s b t Ix Sx rx Iy Sy ry Um Ut
\ 2 2 4 3 4 3 -2 2 \
mm mm mm mm mm kg/m cm cm cm cm cm cm cm cm x10 m /m m 2/t
414 400 5 140 7 31.09 39.6 20.0 10784 521 16.50 320.6 45.8 2.85 137.80 44.33
412 400 5 165 6 31.24 39.8 20.0 10827 526 16.49 449.7 54.5 3.36 147.40 47.18
414 400 5 165 7 33.83 43.1 20.0 12234 591 16.85 524.6 63.6 3.49 147.80 43.68
416 400 5 165 8 36.42 46.4 20.0 13655 656 17.15 599.4 72.7 3.59 148.20 40.69
414 400 5 200 7 37.68 48.0 20.0 14263 689 17.24 933.8 93.4 4.41 161.80 42.94
416 400 5 200 8 40.82 52.0 20.0 15985 769 17.53 1067.1 106.7 4.53 162.20 39.74
420 400 5 200 10 47.10 60.0 20.0 19480 928 18.02 1333.8 133.4 4.71 163.00 34.61
514 500 6 140 7 38.94 49.6 30.0 18846 733 19.49 321.0 45.9 2.54 157.60 40.48
514 500 6 165 7 41.68 53.1 30.0 21096 821 19.93 525.0 63.6 3.14 167.60 40.21
516 500 6 165 8 44.27 56.4 30.0 23284 902 20.32 599.9 72.7 3.26 168.00 37.95
514 500 6 200 7 45.53 58.0 30.0 24245 943 20.45 934.2 93.4 4.01 181.60 39.89
516 500 6 200 8 48.67 62.0 30.0 26897 1043 20.83 1067.6 106.8 4.15 182.00 37.39
420 500 6 200 10 54.95 70.0 30.0 32263 1241 21.47 1334.2 133.4 4.37 182.80 33.27
520 500 6 250 10 62.80 80.0 30.0 38767 1491 22.01 2605.1 208.4 5.71 202.80 32.29
524 500 6 250 12 70.65 90.0 30.0 45579 1740 22.50 3125.9 250.1 5.89 203.60 28.82

37
 Y
b

d
SYMMETRICAL WELDED I- SECTIONS x x
S

Y
depth Web Flange Axis x-x Axis y-y Surface Area
Weight Area Aweb
h d s b t Ix Sx rx Iy Sy ry Um Ut
2 2 4 3 4 3 -2 2 \
mm mm mm mm mm kg/m` cm cm cm cm cm cm cm cm x10 m /m m 2/t
614 600 7 165 7 51.10 65.1 42.0 33879 1104 22.81 526 64 2.84 187.40 36.67
614 600 7 200 7 54.95 70.0 42.0 38393 1251 23.42 935 94 3.65 201.40 36.65
616 600 7 200 8 58.09 74.0 42.0 42175 1369 23.87 1068 107 3.80 201.80 34.74
620 600 7 200 10 64.37 82.0 42.0 49813 1607 24.65 1335 134 4.03 202.60 31.47
620 600 7 250 10 72.22 92.0 42.0 59117 1907 25.35 2606 208 5.32 222.60 30.82
624 600 7 250 12 80.07 102.0 42.0 68789 2205 25.97 3127 250 5.54 223.40 27.90
624 600 7 330 12 95.14 121.2 42.0 86769 2781 26.76 7189 436 7.70 255.40 26.84
674 660 7 165 7 54.40 69.3 46.2 42464 1260 24.75 526 64 2.75 199.40 36.65
676 660 7 165 8 56.99 72.6 46.2 46223 1368 25.23 601 73 2.88 199.80 35.06
674 660 7 200 7 58.25 74.2 46.2 47914 1422 25.41 935 94 3.55 213.40 36.64
676 660 7 200 8 61.39 78.2 46.2 52470 1552 25.90 1069 107 3.70 213.80 34.83
680 660 7 200 10 67.67 86.2 46.2 61664 1814 26.75 1335 134 3.94 214.60 31.71
680 660 7 250 10 75.52 96.2 46.2 72887 2144 27.53 2606 208 5.20 234.60 31.07
684 660 7 250 12 83.37 106.2 46.2 84515 2471 28.21 3127 250 5.43 235.40 28.24
680 660 7 330 10 88.08 112.2 46.2 90845 2672 28.45 5991 363 7.31 266.60 30.27

38
 Y
b

d
SYMMETRICAL WELDED I- SECTIONS x x
S

Y
depth Web Flange Axis x-x Axis y-y Surface Area
Weight Area Aweb
h d s b t Ix Sx rx Iy Sy ry Um Ut
2 2 4 3 4 3 -2 2 \
mm mm mm mm mm kg/m` cm cm cm cm cm cm cm cm x10 m /m m2/t
764 750 7 200 7 63.19 80.5 52.5 64724 1694 28.36 935.5 93.5 3.41 231.40 36.62
766 750 7 200 8 66.33 84.5 52.5 70576 1843 28.90 1068.8 106.9 3.56 231.80 34.95
770 750 7 200 10 72.61 92.5 52.5 82373 2140 29.84 1335.5 133.5 3.80 232.60 32.03
770 750 7 250 10 80.46 102.5 52.5 96814 2515 30.73 2606.3 208.5 5.04 252.60 31.39
774 750 7 250 12 88.31 112.5 52.5 111713 2887 31.51 3127.1 250.2 5.27 253.40 28.69
770 750 7 330 10 93.02 118.5 52.5 119919 3115 31.81 5991.6 363.1 7.11 284.60 30.59
774 750 7 330 12 103.38 131.7 52.5 139586 3607 32.56 7189.5 435.7 7.39 285.40 27.61
764 750 8 200 7 69.08 88.0 60.0 68240 1786 27.85 936.5 93.7 3.26 231.20 33.47
766 750 8 200 8 72.22 92.0 60.0 74092 1935 28.38 1069.9 107.0 3.41 231.60 32.07
770 750 8 200 10 78.50 100.0 60.0 85888 2231 29.31 1336.5 133.7 3.66 232.40 29.61
770 750 8 250 10 86.35 110.0 60.0 100329 2606 30.20 2607.4 208.6 4.87 252.40 29.23
774 750 8 250 12 94.20 120.0 60.0 115229 2977 30.99 3128.2 250.3 5.11 253.20 26.88
770 750 8 330 10 98.91 126.0 60.0 123435 3206 31.30 5992.7 363.2 6.90 284.40 28.75
774 750 8 330 12 109.27 139.2 60.0 143102 3698 32.06 7190.6 435.8 7.19 285.20 26.10

39
 Y
b

SYMMETRICAL WELDED I- SECTIONS

d
x x
S

Y
depth Web Flange Axis x-x Axis y-y Surface Area
Weight Area Aweb
h d s b t Ix Sx rx Iy Sy ry Um Ut
2 2 4 3 4 3 -2 2 \ 2
mm mm mm mm mm kg/m` cm cm cm cm cm cm cm cm x10 m /m m /t
814 800 7 200 7 65.94 84.0 56.0 75455 1854 29.97 935.6 93.6 3.34 241.40 36.61
816 800 7 200 8 69.08 88.0 56.0 82097 2012 30.54 1069.0 106.9 3.49 241.80 35.00
820 800 7 200 10 75.36 96.0 56.0 95480 2329 31.54 1335.6 133.6 3.73 242.60 32.19
820 800 7 250 10 83.21 106.0 56.0 111883 2729 32.49 2606.5 208.5 4.96 262.60 31.56
824 800 7 250 12 91.06 116.0 56.0 128775 3126 33.32 3127.3 250.2 5.19 263.40 28.93
824 800 7 330 12 106.13 135.2 56.0 160426 3894 34.45 7189.7 435.7 7.29 295.40 27.83
828 800 7 330 14 116.49 148.4 56.0 182941 4419 35.11 8387.6 508.3 7.52 296.20 25.43
814 800 8 200 7 72.22 92.0 64.0 79722 1959 29.44 936.7 93.7 3.19 241.20 33.40
816 800 8 200 8 75.36 96.0 64.0 86364 2117 29.99 1070.1 107.0 3.34 241.60 32.06
820 800 8 200 10 81.64 104.0 64.0 99747 2433 30.97 1336.7 133.7 3.59 242.40 29.69
820 800 8 250 10 89.49 114.0 64.0 116150 2833 31.92 2607.6 208.6 4.78 262.40 29.32
824 800 8 250 12 97.34 124.0 64.0 133042 3229 32.76 3128.4 250.3 5.02 263.20 27.04
824 800 8 330 12 112.41 143.2 64.0 164693 3997 33.91 7190.8 435.8 7.09 295.20 26.26
828 800 8 330 14 122.77 156.4 64.0 187208 4522 34.60 8388.7 508.4 7.32 296.00 24.11

40
 Y
b

h
SYMMETRICAL WELDED I- SECTIONS

d
x x
S

depth Web Flange Axis x-x Axis y-y Surface Area

Weight Area Aweb
h d s b t Ix Sx rx Iy Sy ry Um Ut
4 3 4 3 -2 2 \
mm mm mm mm mm kg/m` cm2
cm2
cm cm cm cm cm cm x10 m /m m2/t
1014 1000 10 200 7 100.48 128.0 100.0 154318 3044 34.72 941.7 94.2 2.71 280.80 27.95
1016 1000 10 200 8 103.62 132.0 100.0 164620 3241 35.31 1075.0 107.5 2.85 281.20 27.14
1020 1000 10 200 10 109.90 140.0 100.0 185347 3634 36.39 1341.7 134.2 3.10 282.00 25.66
1020 1000 10 250 10 117.75 150.0 100.0 210850 4134 37.49 2612.5 209.0 4.17 302.00 25.65
1024 1000 10 250 12 125.60 160.0 100.0 236962 4628 38.48 3133.3 250.7 4.43 302.80 24.11
1020 1000 10 330 10 130.31 166.0 100.0 251655 4934 38.94 5997.8 363.5 6.01 334.00 25.63
1024 1000 10 330 12 140.67 179.2 100.0 286123 5588 39.96 7195.7 436.1 6.34 334.80 23.80
1028 1000 10 330 14 151.03 192.4 100.0 320862 6242 40.84 8393.6 508.7 6.60 335.60 22.22

41
 Y
b1

t
UNSYMMETRICAL WELDED I-SECTIONS x x

d
h
S

ex
t

b2

Y
depth Web Flange Axis x-x Axis y-y Surface Area
Weight Area Aweb
h d s b1 b2 t Ix Sxt Sxb rx ex Iy Sy ry Syupper flange Um Ut
2 2 4 3 3 4 3 3 -2 2 \
mm mm mm mm mm mm kg/m` cm cm cm cm cm cm cm cm cm cm cm x10 m /m m2/t
224 200 5 140 70 12 27.63 35.2 10.0 2943 339 214 9.14 13.73 309 44 2.96 39.20 85.80 31.05
224 200 5 200 100 12 36.11 46.0 10.0 4031 478 289 9.36 13.97 900 90 4.42 80.00 103.80 28.75
274 250 5 140 70 12 29.59 37.7 12.5 4657 432 280 11.11 16.62 309 44 2.86 39.20 95.80 32.37
274 250 5 250 140 12 46.55 59.3 12.5 8184 759 493 11.75 16.62 1837 147 5.57 125.00 131.80 28.31
324 300 5 200 100 12 40.04 51.0 15.0 9203 735 463 13.43 19.87 900 90 4.20 80.00 123.80 30.92
324 300 5 250 140 12 48.51 61.8 15.0 11834 920 606 13.84 19.53 1837 147 5.45 125.00 141.80 29.23
354 330 5 200 100 12 41.21 52.5 16.5 11226 814 520 14.62 21.61 900 90 4.14 80.00 129.80 31.50
354 330 5 250 140 12 49.69 63.3 16.5 14383 1018 676 15.07 21.27 1837 147 5.39 125.00 147.80 29.74
358 330 5 300 140 14 61.31 78.1 16.5 17831 1375 781 15.11 22.83 3470 231 6.67 210.00 158.60 25.87
358 330 5 330 165 14 67.35 85.8 16.5 20171 1520 895 15.33 22.53 4717 286 7.41 254.10 169.60 25.18
428 400 5 250 140 14 58.56 74.6 20.0 24709 1443 962 18.20 25.67 2143 171 5.36 145.83 162.60 27.77
428 400 5 300 165 14 66.80 85.1 20.0 28773 1712 1107 18.39 26.00 3674 245 6.57 210.00 177.60 26.59
432 400 5 330 165 16 77.87 99.2 20.0 33909 2111 1250 18.49 27.14 5391 327 7.37 290.40 184.40 23.68
432 400 5 400 200 16 91.06 116.0 20.0 40401 2547 1478 18.66 27.34 9600 480 9.10 426.67 205.40 22.56
532 500 5 250 140 16 68.61 87.4 25.0 44398 2074 1396 22.54 31.80 2450 196 5.29 166.67 183.40 26.73
532 500 5 300 165 16 78.03 99.4 25.0 51623 2459 1603 22.79 32.21 4199 280 6.50 240.00 198.40 25.43
536 500 5 330 165 18 89.57 114.1 25.0 59816 2982 1783 22.90 33.54 6065 368 7.29 326.70 205.20 22.91
536 500 5 400 200 18 104.41 133.0 25.0 71148 3595 2104 23.13 33.81 10801 540 9.01 480.00 226.20 21.67
540 500 5 500 250 20 137.38 175.0 25.0 97001 4956 2817 23.54 34.43 23438 938 11.57 833.33 257.00 18.71

42
 Y
b1

UNSYMMETRICAL WELDED I-SECTIONS x x

d
h
S

ex
t

b2

depth Web Flange Axis x-x Axis y-y Surface Area

Weight Area Aweb
h d s b1 b2 t Ix Sxt Sxb rx ex Iy Sy ry Sy upper flange Um Ut
2 2 4 3 3 4 3 3 -2 2 \
mm mm mm mm mm mm kg/m` cm cm cm cm cm cm cm cm cm cm cm x10 m /m m2/t
632 600 6 300 165 16 86.66 110.4 36.0 77386 3026 2057 26.48 37.63 4200 280 6.17 240 218.20 25.18
636 600 6 330 165 18 98.20 125.1 36.0 89165 3645 2278 26.70 39.14 6065 368 6.96 327 225.00 22.91
636 600 6 400 200 18 113.04 144.0 36.0 105355 4376 2666 27.05 39.53 10801 540 8.66 480 246.00 21.76
640 600 6 400 250 20 130.31 166.0 36.0 130563 4946 3472 28.05 37.60 13272 664 8.94 533 256.80 19.71
640 600 6 500 250 20 146.01 186.0 36.0 142083 6004 3523 27.64 40.33 23439 938 11.23 833 276.80 18.96
696 660 7 300 165 18 101.97 129.9 46.2 107758 3787 2619 28.80 41.14 4726 315 6.03 270 230.80 22.63
700 660 7 330 165 20 113.98 145.2 46.2 122578 4495 2869 29.06 42.73 6740 408 6.81 363 237.60 20.85
700 660 7 400 200 20 130.47 166.2 46.2 144402 5385 3344 29.48 43.18 12002 600 8.50 533 258.60 19.82
704 660 7 400 250 22 148.52 189.2 46.2 176417 6031 4287 30.54 41.15 14600 730 8.78 587 269.40 18.14
704 660 7 500 250 22 165.79 211.2 46.2 192046 7297 4357 30.15 44.08 25783 1031 11.05 917 289.40 17.46
786 750 7 330 165 18 111.16 141.6 52.5 146831 4699 3101 32.20 47.35 6067 368 6.55 327 254.80 22.92
786 750 7 400 200 18 125.99 160.5 52.5 171984 5604 3590 32.73 47.91 10802 540 8.20 480 275.80 21.89
790 750 7 400 250 20 143.26 182.5 52.5 210035 6332 4583 33.92 45.83 13273 664 8.53 533 286.60 20.01
794 750 7 500 250 22 170.74 217.5 52.5 249797 8344 5050 33.89 49.46 25783 1031 10.89 917 307.40 18.00
1036 1000 10 330 165 18 148.44 189.1 100.0 302113 6897 5053 39.97 59.79 6073 368 5.67 327 304.20 20.49
1040 1000 10 400 200 20 172.70 220.0 100.0 376577 8814 6146 41.37 61.27 12008 600 7.39 533 326.00 18.88
1044 1000 10 400 250 22 190.76 243.0 100.0 445092 9834 7526 42.80 59.14 14606 730 7.75 587 336.80 17.66
1044 1000 10 500 250 22 208.03 265.0 100.0 484442 11647 7713 42.76 62.81 25790 1032 9.87 917 356.80 17.15

43
 Y
CHANNEL (U.P.N.)

BUILT-UP SECTION X X

IPE +Channel (U.P.N.)

ex
IPE

AXIS X-X AXIS Y-Y

IPE CHANNEL Weight Area Aweb
Ix Sxb Sxt rx eX Iy Sy ry Iy upper flange Sy upper fLange
NO. NO.
kg/m\ cm 2
cm 2
cm 4
cm3
cm 3
cm cm cm 4
cm 3
cm cm 4
cm3
160 41.20 52.50 3060 217 458 7.63 14.07 1067 133.38 4.51 996 124.50
200 180 44.40 56.50 10.25 3168 220 495 7.49 14.40 1492 165.78 5.14 1421 157.89
200 47.70 60.70 3269 223 531 7.34 14.69 2052 205.20 5.81 1981 198.10
160 45.00 57.40 4227 279 556 8.58 15.14 1130 141.25 4.44 1028 128.44
220 180 48.20 61.40 11.89 4371 282 599 8.44 15.51 1555 172.78 5.03 1453 161.39
220 55.60 70.80 4648 288 689 8.10 16.16 2895 263.18 6.39 2793 253.86
180 52.70 67.10 5935 359 719 9.41 16.54 1634 181.56 4.93 1492 165.78
240 200 56.00 71.30 13.66 6113 362 768 9.26 16.90 2194 219.40 5.55 2052 205.20
240 63.90 81.40 6473 368 877 8.92 17.57 3884 323.67 6.91 3742 311.83
200 61.40 78.10 8820 474 952 10.63 18.59 2330 233.00 5.46 2120 212.00
270 240 69.30 88.20 16.47 9325 482 1086 10.28 19.36 4020 335.00 6.75 3810 317.50
280 77.90 99.20 9600 484 1173 9.84 19.82 6700 478.57 8.22 6490 463.57
200 67.50 86.00 12366 613 1159 11.99 20.18 2514 251.40 5.41 2212 221.20
300 240 75.40 96.10 19.78 13066 621 1318 11.66 21.04 4204 350.33 6.61 3902 325.17
300 88.40 112.60 13825 630 1527 11.08 21.95 8634 575.60 8.76 8332 555.47
220 78.50 100.00 17419 784 1490 13.20 22.21 3478 316.18 5.90 3084 280.36
330 260 87.00 110.90 23.03 18336 794 1681 12.86 23.09 5608 431.38 7.11 5214 401.08
320 108.60 138.40 20393 820 2142 12.14 24.88 11658 728.63 9.18 11264 704.00

44
 Y
CHANNEL (U.P.N.)

BUILT-UP SECTION X X
IPE +Channel (U.P.N.)

ex
IPE

AXIS X-X AXIS Y-Y

IPE CHANNEL Weight Area Aweb
Ix Sxb Sxt rx eX Iy Sy ry Iy upper flange Sy upper fLange
NO. NO.
kg/m\ cm 2
cm 2
cm 4
cm3
cm3
cm cm cm 4
cm 3
cm cm 4
cm3
240 90.30 115.00 23994 994 1875 14.44 24.15 4640 386.67 6.35 4120 343.33
360 280 98.90 126.00 26.77 24800 997 2046 14.03 24.88 7320 522.86 7.62 6800 485.71
350 117.70 150.00 27667 1034 2600 13.58 26.76 13880 793.14 9.62 13360 763.43
240 99.50 126.80 33256 1267 2262 16.19 26.24 4920 410.00 6.23 4260 355.00
400 320 125.80 160.30 32.08 37849 1310 3025 15.37 28.89 12190 761.88 8.72 11530 720.63
400 138.10 176.00 39420 1325 3383 14.97 29.75 21670 1083.50 11.10 21010 1050.50
260 115.50 147.10 48555 1649 2932 18.17 29.44 6500 500.00 6.65 5660 435.38
450 320 137.10 174.60 39.56 53797 1695 3671 17.55 31.75 12550 784.38 8.48 11710 731.88
400 149.40 190.30 56037 1713 4096 17.16 32.72 22030 1101.50 10.76 21190 1059.50
260 128.60 164.30 67574 2115 3547 20.28 31.95 6960 535.38 6.51 5890 453.08
500 320 150.20 191.80 47.74 74765 2173 4399 19.74 34.41 13010 813.13 8.24 11940 746.25
400 162.50 207.50 77899 2196 4891 19.38 35.47 22490 1124.50 10.41 21420 1071.00
280 147.80 187.30 92823 2665 4385 22.26 34.83 8950 639.29 6.91 7615 543.93
550 320 165.50 209.80 57.23 101204 2735 5217 21.96 37.00 13540 846.25 8.03 12205 762.81
400 177.80 225.50 105432 2764 5778 21.62 38.15 23020 1151.00 10.10 21685 1084.25
300 168.20 214.80 126776 3359 5452 24.29 37.75 11420 761.33 7.29 9725 648.33
600 350 182.60 233.30 67.44 136120 3437 6246 24.15 39.61 16230 927.43 8.34 14535 830.57
400 193.80 247.50 140596 3461 6769 23.83 40.63 23740 1187.00 9.79 22045 1102.25

45
COLD FORMED SECTIONS
UNSTIFFENED CHANNEL

LIPPED CHANNEL

STRAIGHT LIPPED Z-CHANNEL

INCLINED LIPPED Z-CHANNEL

46
 Yb

r
ey
UNSTIFFENED CHANNEL x x

h
(COLD FORMED SECTION) t

Y
Dimensions Axis X-X Axis Y-Y
Weight Area Aweb
h b t r Ix Sx rx Iy Sy ry ey
2 2 4 3 4 3
mm mm mm mm kg/m` cm cm cm cm cm cm cm cm cm
30 25 4 8 1.99 2.54 0.88 2.87 1.91 1.06 1.28 0.89 0.71 1.06
40 20 4 8 1.99 2.54 1.28 4.80 2.40 1.38 0.73 0.58 0.54 0.73
40 24 4 8 2.24 2.86 1.28 5.84 2.92 1.43 1.35 0.90 0.69 0.90
50 25 4 8 2.62 3.34 1.68 10.80 4.32 1.80 1.73 1.05 0.72 0.85
50 35 4 8 3.25 4.20 1.68 15.03 6.01 1.91 4.78 2.14 1.07 1.27
60 30 4 8 3.24 4.14 2.08 20.30 6.77 2.22 3.29 1.62 0.89 0.97
65 38 4 8 3.90 4.98 2.28 30.71 9.45 2.48 6.82 2.68 1.17 1.25
70 25 4 8 3.25 4.14 2.48 25.40 7.26 2.48 2.00 1.13 0.70 0.73
80 45 4 8 4.81 6.14 2.88 58.81 14.70 3.10 12.05 3.91 1.40 1.41
100 50 4 8 5.76 7.34 3.68 109.40 21.88 3.86 17.67 5.00 1.56 1.46
120 60 4 8 7.01 8.94 4.48 195.80 32.63 4.68 31.50 7.34 1.88 1.71
140 65 4 8 7.96 10.14 5.28 300.13 42.88 5.44 41.54 8.78 2.02 1.77
160 65 4 8 8.58 10.94 6.08 411.28 51.41 6.13 43.37 8.95 1.99 1.65
180 80 4 8 10.06 12.94 6.88 637.23 70.80 7.02 80.45 13.60 2.49 2.09
200 80 4 8 10.78 13.74 7.68 816.04 81.60 7.71 83.13 13.80 2.46 1.98
200 100 4 8 12.04 15.34 7.68 969.70 96.97 7.95 154.37 21.17 3.17 2.71
60 40 3.6 8 3.51 4.48 1.90 24.48 8.16 2.34 7.02 2.66 1.25 1.36
60 40 4 8 3.87 4.94 2.08 26.58 8.86 2.32 7.64 2.92 1.24 1.38
80 40 2.25 8 2.61 3.32 1.70 32.33 8.08 3.12 5.24 1.83 1.26 1.14
80 40 2.5 8 2.89 3.68 1.88 35.52 8.88 3.11 5.75 2.02 1.25 1.15
80 40 3 8 3.43 4.37 2.22 41.66 10.42 3.09 6.75 2.39 1.24 1.18
80 40 4 8 4.50 5.74 2.88 53.04 13.26 3.04 8.60 3.09 1.22 1.22
80 50 5 7.5 6.32 8.05 3.50 77.30 19.33 3.10 19.55 5.85 1.56 1.66
100 40 3 8 3.91 4.97 2.82 70.96 14.19 3.78 7.31 2.48 1.21 1.05
100 40 4 8 5.13 6.54 3.68 90.97 18.19 3.73 9.33 3.21 1.19 1.09
100 50 5 8 7.10 9.05 4.50 131.83 26.37 3.82 21.32 6.10 1.53 1.51
120 60 4 8 7.01 8.94 4.48 195.80 32.63 4.68 31.50 7.35 1.88 1.71
140 63 4 8 7.83 9.98 5.28 292.73 41.82 5.42 38.05 8.27 1.95 1.70
160 63 4 8 8.46 10.78 6.08 401.54 50.19 6.10 39.70 8.43 1.92 1.59
180 75 4 8 9.84 12.54 6.88 606.25 67.36 6.95 67.20 12.01 2.32 1.90

47
 Yb

d
ey
LIPPED CHANNEL
x x
(COLD FORMED SECTION)

h
t

Y
Dimensions Axis X-X Axis Y-Y
Weight Area Aweb
h b d t r Ix Sx rx Iy Sy ry ey
2 2 4 3 4 3
mm mm mm mm mm kg/m` cm cm cm cm cm cm cm cm cm
100 40 15 3 8 4.28 5.45 2.82 76.88 15.38 3.76 10.41 3.84 1.38 1.29
140 70 30 3 8 7.34 9.35 4.02 280.23 40.03 5.47 67.30 15.44 2.68 2.64
160 70 30 3 8 7.81 9.95 4.62 384.69 48.09 6.22 70.80 15.70 2.67 2.49
160 80 34 3 8 8.47 10.79 4.62 427.20 53.40 6.29 103.10 20.68 3.09 3.01
160 80 34 3.65 8 10.21 13.01 5.57 510.03 63.75 6.26 121.93 24.46 3.06 3.01
160 80 30 4 8 10.89 13.87 6.08 545.86 68.23 6.27 124.07 24.35 2.99 2.90
160 80 35 4 8 11.20 14.27 6.08 554.90 69.36 6.24 133.39 26.90 3.06 3.04
180 80 25 3.6 8 10.14 12.91 6.22 642.27 71.36 7.05 109.29 20.30 2.91 2.62
180 80 25 4 8 11.20 14.27 6.88 705.60 78.40 7.03 119.15 22.13 2.89 2.62
100 50 20 2 8 3.40 4.33 1.92 65.92 13.18 3.90 15.36 4.85 1.88 1.84
200 60 20 2 8 5.28 6.73 3.92 386.55 38.66 7.58 30.47 6.96 2.13 1.62
250 70 20 2 8 6.38 8.13 4.92 723.51 57.88 9.43 47.56 9.05 2.42 1.74
185 60 25 1.5 8 3.93 5.00 2.73 251.81 27.22 7.09 25.60 6.12 2.26 1.82
185 60 26.4 2 8 5.25 6.69 3.62 333.19 36.02 7.06 34.09 8.22 2.26 1.85
185 60 27.7 2.5 8 6.56 8.36 4.50 412.99 44.65 7.03 42.46 10.32 2.25 1.89
185 60 29.1 3 8 7.89 10.05 5.37 491.57 53.14 7.00 50.82 12.46 2.25 1.92
215 60 25 1.5 8 4.29 5.46 3.18 360.67 33.55 8.13 26.86 6.21 2.22 1.67
215 60 26.4 2 8 5.72 7.29 4.22 477.98 44.46 8.10 35.79 8.34 2.21 1.70
215 60 27.7 2.5 8 7.15 9.11 5.25 593.41 55.20 8.07 44.59 10.47 2.21 1.74
215 60 29.1 3 8 8.59 10.95 6.27 707.49 65.81 8.04 53.41 12.64 2.21 1.78

48
 Y b
r

d
STRAIGHT LIPPED Z-SECTION t
x x
(COLD FORMED SECTION)

h
t

Y
Dimensions Axis X-X Axis Y-Y
Weight Area Aweb
h b d t r Ix Sx rx Iy Sy ry
mm mm mm mm mm kg/m` cm2 cm2 cm4 cm3 cm cm4 cm3 cm
100 50 20 2 8 3.40 4.33 1.92 65.92 13.18 3.90 28.40 5.80 2.56
150 60 20 2 8 4.50 5.73 2.92 195.22 26.03 5.84 46.04 7.80 2.83
150 60 20 2.5 8 5.58 7.10 3.63 239.74 31.97 5.81 55.66 9.47 2.80
200 60 20 2 8 5.28 6.73 3.92 386.55 38.65 7.58 46.04 7.80 2.62
200 60 20 2.5 8 6.56 8.35 4.88 475.94 47.59 7.55 55.66 9.47 2.58
250 70 20 2 8 6.38 8.13 4.92 723.51 57.88 9.43 69.53 10.08 2.92
250 70 20 2.5 8 7.93 10.10 6.13 893.12 71.45 9.40 84.41 12.28 2.89
Y b

d
r

INCLINED LIPPED Z-SECTION t

h
x x
(COLD FORMED SECTION)
12
0°

Dimensions Axis X-X Axis Y-Y

Weight Area Aweb
h b d t r Ix Sx rx Iy Sy ry
2 2 4 3 4 3
mm mm mm mm mm kg/m` cm cm cm cm cm cm cm cm
185 60 20.8 1.5 8 3.83 4.88 2.73 246.10 26.60 7.10 37.66 6.36 2.78
185 60 21.8 2 8 5.10 6.50 3.62 325.13 35.15 7.07 50.12 8.50 2.78
185 60 22.8 2.5 8 6.37 8.12 4.50 402.66 43.53 7.04 62.53 10.64 2.78
185 60 23.8 3 8 7.64 9.73 5.37 478.65 51.75 7.01 74.88 12.80 2.77
215 60 20.8 1.5 8 4.19 5.33 3.18 352.12 32.76 8.13 37.66 6.36 2.66
215 60 21.8 2 8 5.58 7.10 4.22 465.85 43.33 8.10 50.12 8.50 2.66
215 60 22.8 2.5 8 6.96 8.87 5.25 577.73 53.74 8.07 62.53 10.64 2.66
215 60 23.8 3 8 8.34 10.63 6.27 687.74 63.98 8.04 74.88 12.80 2.65

49
COMBINED COLD FORMED SECTIONS

TWO CHANNELS BACK TO BACK

TWO CHANNELS TOE TO TOE

50
 Y
b
r

TWO CHANNELS (Cold Formed)

X X
BACK to BACK

h
t
tG
Y

Axis X-X Axis Y-Y

Dimensions Weight Area Aweb
Ix Sx rx Iy Sy ry Iy Sy ry Iy Sy ry
h b t r kg/m` cm2 cm2 cm4 cm3 cm tG = 8 mm tG = 10 mm tG = 12 mm
30 25 4 8 3.98 5.08 1.76 5.74 3.82 1.06 13.39 4.62 1.62 14.92 4.97 1.71 16.56 5.34 1.81
40 20 4 8 3.98 5.08 2.56 9.60 4.80 1.38 7.95 3.31 1.25 9.15 3.66 1.34 10.45 4.02 1.43
40 24 4 8 4.48 5.72 2.56 11.68 5.84 1.43 12.37 4.42 1.47 13.91 4.80 1.56 15.57 5.19 1.65
50 25 4 8 5.24 6.68 3.36 21.60 8.64 1.80 13.90 4.79 1.44 15.63 5.21 1.53 17.50 5.65 1.62
50 35 4 8 6.50 8.40 3.36 30.06 12.02 1.91 32.99 8.46 1.98 35.88 8.97 2.07 38.93 9.50 2.15
60 30 4 8 6.48 8.28 4.16 40.60 13.54 2.22 22.12 6.51 1.63 24.47 6.99 1.72 26.99 7.50 1.81
65 38 4 8 7.80 9.96 4.56 61.42 18.90 2.48 40.76 9.70 2.02 44.14 10.27 2.11 47.73 10.85 2.19
70 25 4 8 6.50 8.28 4.96 50.80 14.52 2.48 14.57 5.03 1.33 16.53 5.51 1.41 18.65 6.01 1.50
80 45 4 8 9.62 12.28 5.76 117.62 29.40 3.10 64.33 13.13 2.29 68.90 13.78 2.37 73.71 14.45 2.45
100 50 4 8 11.52 14.68 7.36 218.80 43.76 3.86 86.13 15.95 2.42 91.73 16.68 2.50 97.64 17.44 2.58
120 60 4 8 14.02 17.88 8.96 391.60 65.26 4.68 142.60 22.28 2.82 150.33 23.13 2.90 158.41 24.00 2.98
140 65 4 8 15.92 20.28 10.56 600.26 85.76 5.44 178.58 25.88 2.97 187.58 26.80 3.04 196.99 27.75 3.12
160 65 4 8 17.16 21.88 12.16 822.56 102.82 6.13 178.69 25.90 2.86 187.88 26.84 2.93 197.51 27.82 3.00
180 80 4 8 20.12 25.88 13.76 1274.5 141.60 7.02 321.36 38.26 3.52 334.51 39.35 3.60 348.17 40.48 3.67
200 80 4 8 21.56 27.48 15.36 1632.1 163.20 7.71 321.92 38.32 3.42 335.27 39.44 3.49 349.18 40.60 3.56
200 100 4 8 24.08 30.68 15.36 1939.4 193.94 7.95 605.48 58.22 4.44 624.87 59.51 4.51 644.87 60.84 4.58

51
 Y
b
r

TWO CHANNELS BACK to BACK X X

(Cold Formed SECTION)

h
t
tG
Y

Axis X-X Axis Y-Y

Dimensions Weight Area Aweb
Ix Sx rx Iy Sy ry Iy Sy ry Iy Sy ry
h b t r kg/m` cm2 cm2 cm4 cm3 cm tG = 8 mm tG = 10 mm tG = 12 mm
60 40 3.6 8 7.02 8.96 4.22 48.96 16.32 2.34 41.79 9.50 2.16 45.04 10.01 2.24 48.46 10.53 2.33
60 40 4 8 7.74 9.88 4.16 53.16 17.72 2.32 46.58 10.59 2.17 50.20 11.16 2.25 54.01 11.74 2.34
80 40 2.25 8 5.22 6.64 6.04 64.66 16.16 3.12 26.23 5.96 1.99 28.34 6.30 2.07 30.58 6.65 2.15
80 40 2.5 8 5.78 7.36 6.00 71.04 17.76 3.11 29.18 6.63 1.99 31.54 7.01 2.07 34.04 7.40 2.15
80 40 3 8 6.86 8.74 5.92 83.32 20.84 3.09 35.32 8.03 2.01 38.17 8.48 2.09 41.19 8.95 2.17
80 40 4 8 9.00 11.48 5.76 106.08 26.52 3.04 47.33 10.76 2.03 51.16 11.37 2.11 55.23 12.01 2.19
80 50 5 8 12.64 16.10 5.60 154.60 38.66 3.10 107.42 19.89 2.58 114.22 20.77 2.66 121.33 21.67 2.75
100 40 3 8 7.82 9.94 7.52 141.92 28.38 3.78 35.52 8.07 1.89 38.50 8.56 1.97 41.68 9.06 2.05
100 40 4 8 10.26 13.08 7.36 181.94 36.38 3.73 47.70 10.84 1.91 51.73 11.50 1.99 56.02 12.18 2.07
100 50 5 8 14.20 18.10 7.20 263.66 52.74 3.82 108.67 20.12 2.45 115.77 21.05 2.53 123.22 22.00 2.61
120 60 4 8 14.02 17.88 8.96 391.60 65.26 4.68 142.60 22.28 2.82 150.33 23.13 2.90 158.41 24.00 2.98
140 63 4 8 15.66 19.96 10.56 585.46 83.64 5.42 164.12 24.50 2.87 172.71 25.40 2.94 181.69 26.33 3.02
160 63 4 8 16.92 21.56 12.16 803.08 100.38 6.10 164.78 24.59 2.76 173.58 25.53 2.84 182.80 26.49 2.91
180 75 4 8 19.68 25.08 13.76 1212.5 134.72 6.95 267.07 33.81 3.26 278.86 34.86 3.33 291.15 35.94 3.41

52
 Y
c

r
TWO CHANNELS TOE to TOE
X X
(COLD FORMED SECTION)

h
t

b
Y
Axis x-x Axis y-y (toe to toe) Axis y-y (c=h)
Dimensions (mm) Weight Area Aweb
Ix Sx rx Iy Sy ry Iy Sy ry
2 2 4 3 4 3 4 3
h b t r kg/m` cm cm cm cm cm cm cm cm cm cm cm
30 25 4 8 3.99 5.08 1.76 5.74 3.82 1.06 13.09 5.24 1.61 N.A.
40 20 4 8 3.99 5.08 2.56 9.60 4.80 1.38 9.65 4.83 1.38 9.65 4.83 1.38
40 24 4 8 4.49 5.72 2.56 11.68 5.84 1.43 15.57 6.49 1.65 N.A.
50 25 4 8 5.24 6.68 3.36 21.60 8.64 1.80 21.65 8.66 1.80 21.65 8.66 1.80
50 35 4 8 6.59 8.40 3.36 30.06 12.02 1.91 51.33 14.67 2.47 N.A.
60 30 4 8 6.50 8.28 4.16 40.60 13.54 2.22 40.70 13.57 2.22 40.70 13.57 2.22
65 38 4 8 7.82 9.96 4.56 61.42 18.90 2.48 78.40 20.63 2.81 N.A.
70 25 4 8 6.50 8.28 4.96 50.80 14.52 2.48 29.94 11.98 1.90 67.53 19.29 2.86
80 45 4 8 9.64 12.28 5.76 117.62 29.40 3.10 141.35 31.41 3.39 106.48 26.62 2.94
100 50 4 8 11.52 14.68 7.36 218.80 43.76 3.86 219.30 43.86 3.87 219.30 43.86 3.87
120 60 4 8 14.04 17.88 8.96 391.60 65.26 4.68 392.07 65.34 4.68 392.07 65.34 4.68
140 65 4 8 15.92 20.28 10.56 600.26 85.76 5.44 536.80 82.58 5.14 637.80 91.11 5.61
160 65 4 8 17.18 21.88 12.16 822.56 102.82 6.13 601.41 92.52 5.24 969.00 121.12 6.65
180 80 4 8 20.32 25.88 13.76 1274.46 141.60 7.02 1064.84 133.10 6.41 1396.62 155.18 7.35
200 80 4 8 21.57 27.48 15.36 1632.08 163.20 7.71 1162.15 145.27 6.50 1933.78 193.38 8.39
200 100 4 8 24.08 30.68 15.36 1939.40 193.94 7.95 1939.20 193.92 7.95 1939.20 193.92 7.95
N.A.= Section is not available with the specified configuration.

53
 Y
c

TWO CHANNELS (Cold Formed) X X

TOE to TOE

h
t

b
Y
Axis x-x Axis y-y (toe to toe) Axis y-y (c=h)
Dimensions (mm) Weight Area Aweb
Ix Sx rx Iy Sy ry Iy Sy ry
2 2 4 3 4 3 4 3
h b t r kg/m` cm cm cm cm cm cm cm cm cm cm cm
60 40 3.6 8 7.03 8.96 4.22 48.96 16.32 2.34 76.49 19.12 2.92
N.A.
60 40 4 8 7.76 9.88 4.16 53.16 17.72 2.32 83.10 20.78 2.90
80 40 2.25 8 5.21 6.64 6.04 64.66 16.16 3.12 64.79 16.20 3.12 64.79 16.20 3.12
80 40 2.5 8 5.78 7.36 6.00 71.04 17.76 3.11 71.28 17.82 3.11 71.28 17.82 3.11
80 40 3 8 6.86 8.74 5.92 83.32 20.84 3.09 83.00 20.75 3.08 83.00 20.75 3.08
80 40 4 8 9.01 11.48 5.76 106.08 26.52 3.04 105.92 26.48 3.04 105.92 26.48 3.04
80 50 5 8 12.64 16.10 5.60 154.60 38.66 3.10 218.71 43.74 3.69 N.A.
100 40 3 8 7.80 9.94 7.52 141.92 28.38 3.78 101.12 25.28 3.19 169.71 33.94 4.13
100 40 4 8 10.27 13.08 7.36 181.94 36.38 3.73 129.42 32.36 3.15 218.63 43.73 4.09
100 50 5 8 14.21 18.10 7.20 263.66 52.74 3.82 263.10 52.62 3.81 263.10 52.62 3.81
120 60 4 8 14.04 17.88 8.96 391.60 65.26 4.68 392.07 65.34 4.68 392.07 65.34 4.68
140 63 4 8 15.67 19.96 10.56 585.46 83.64 5.42 498.45 79.12 5.00 636.78 90.97 5.65
160 63 4 8 16.92 21.56 12.16 803.08 100.38 6.10 557.69 88.52 5.09 965.26 120.66 6.69
180 75 4 8 19.69 25.08 13.76 1212.50 134.72 6.95 920.91 122.79 6.06 1398.68 155.41 7.47
N.A.= Section is not available with the specified configuration.

54
 BOLTS
ORDINARY BOLTS
HIGH STRENGTH BOLTS
ANCHOR BOLTS

55
 m

Washer

ORDINARY BOLTS

e
r
Bolt Nut n
Head s
l1 b
Nut
k L
Bolt

Nominal Bolt Size M10 M12 M16 M20 M24 M27 M30
k mm 7.0 8.0 10.0 13.0 15.0 17.0 19.0
l1 mm 1.9 2.5 3.0 4.0 4.5 4.5 5.0
b mm 26.0 19.5 23.0 26.0 29.5 32.5 35.0
Bolt
n mm 2.2 2.5 3.0 3.5 4.5 4.5 5.0
Dimensions
r mm 0.5 0.6 0.6 0.8 0.8 1.0 1.0
s mm 17.0 19.0 24.0 30.0 36.0 41.0 46.0
emin. mm 18.7 20.9 26.2 32.9 39.6 45.2 50.8
m mm 8.0 10.0 13.0 16.0 19.0 22.0 24.0
Nut
s mm 17.0 19.0 24.0 30.0 36.0 41.0 46.0
Dimensions
emin. mm 18.7 20.9 26.2 39.5 39.6 45.2 50.8
Hole Diameter d1 mm 12.0 14.0 18.0 22.0 26.0 30.0 33.0

56
 Grip Length

MINIMUM AND MAXIMUM GRIPS

FOR ORDINARY BOLTS, IN MILLIMETERS

Bolt Length = L

Nominal
M10 M12 M16 M20 M24 M27 M30
Bolt Size
L Min.Grip Max. Grip Min.Grip Max. Grip Min.Grip Max. Grip Min.Grip Max. Grip Min.Grip Max. Grip Min.Grip Max. Grip Min.Grip Max. Grip
30 0 9 2 9
35 10 11 6 7
40 10 19 12 19 8 15 5 12 2 9
45 20 21 16 17 13 14 10 11 7 8
50 20 29 22 29 18 25 15 22 12 19 9 16 6 13
55 30 31 26 27 23 24 20 21 17 18 14 15
60 30 39 32 39 28 35 25 32 22 29 19 26 16 23
65 40 41 36 37 33 34 30 31 27 28 24 25
70 40 49 42 49 38 45 35 42 32 39 29 36 26 33
75 50 51 46 47 43 44 40 41 37 38 34 35
80 50 59 52 59 48 55 45 52 42 49 39 46 36 43
85 60 61 56 57 53 54 50 51 47 48 44 45
90 62 69 58 65 55 62 52 59 49 56 46 53

57
 WASHERS FOR ORDINARY BOLTS

S1
Plain Circular Washer d2
D

Nominal
M10 M12 M16 M20 M24 M27 M30
Bolt Size
d2 mm 11 14 18 22 26 30 33

d2
D mm 21 24 30 37 44 50 56
S1 mm 8 8 8 8 8 8 8

Bevelled Square Washers

e
For Channel Sections
c
0. 8
d2
r=

Nominal
M10 M12 M16 M20 M24 M27
Bolt Size
d2 mm 11 14 18 22 26 30 d2
b mm 22 30 36 44 56 56
c mm 2 2.5 3 3.5 4 4
e mm 3.8 4.9 5.9 7 8.5 8.5
f mm 2.9 3.7 4.45 5.25 6.25 6.25

Bevelled Square Washers

c

For IPN Sections

c
0. 8

d2
r=

Nominal
M10 M12 M16 M20 M24 M27
Bolt Size
d2 mm 11 14 18 22 26 30
d2
b mm 22 30 36 44 56 56
c mm 1.5 2 2.5 3 3 3
e mm 4.6 6.2 7.5 9.2 10.8 10.8
f mm 3 4.1 5 6.1 6.9 6.9

58
 s1 m

Washer
D
LARGE DIAMETER
d2
ORDINARY BOLTS

e
r
Bolt Nut n
Head s
l1 b Nut Washer
k L
Bolt
Nominal Bolt Size M30 M33 M36 M39 M42 M45 M48 M52
k mm 19 21 23 25 26 28 30 33
l1 mm 30 33 36 39 42 45 48 52
b mm 4.5 4.5 5 5 5.5 5.5 6.3 6.3
Bolt
n mm 66 72 78 84 90 96 102 116
Dimensions
r mm 5 5 6 6 6.5 6.5 7.5 7.5
s mm 46 50 55 60 65 70 75 80
emin. mm 50.8 55.4 60.8 66.4 72.1 77.7 83.4 89
m mm 24 26 29 31 34 36 38 42
Nut
s mm 46 50 55 60 65 70 75 80
Dimensions
emin. mm 50.8 55.4 60.8 66.4 72.1 77.7 83.4 89
D mm 56 60 66
Washer
d2 mm 33 36 39 N.A.
Dimensions
s1 mm 8 8 8
Hole Diameter d1 mm 33 36 39 42 45 48 51 55
N.A. = Washer plates with appropriate thickness are used with these bolts size

59
 HIGH STRENGTH BOLTS
F Man ufacture's
identification mark Transition Thread
thread length length

D
F
8S
5M
32

Grade marketing H Bolt length

BOLT
W

W
3 3
8S 8S

"S" STRUCTURAL NUT N N

3 DENOTES WEATHERING STEEL

NUT
DIMENSIONS
Heavy Hex Strucural Bolt
Heavy Hex Bolt or Nut Dimension Heavy
Thread Length
Nominal Hex Max. Max.
Bolt Across Flats Across Corners Nut Max. Head Bolt Bolt Transition
Size F or W F/ or W/ Height Height Lengths Lengths Thread
N H <100 >100 Length
Max. Min. Max. Min.
mm mm mm mm mm mm mm mm mm mm
M16 27 26.2 31.18 29.56 17.1 10.75 31 38 6.0
M20 34 33.0 39.26 37.29 20.7 13.40 36 43 7.5
M22 36 35.0 41.57 39.55 23.6 14.90 38 45 7.5
M24 41 40.0 47.34 45.20 24.2 15.90 41 48 9.0
M27 46 45.0 53.12 50.58 27.6 17.90 44 51 9.0
M30 50 49.0 57.74 55.37 30.7 19.75 49 56 10.5
M36 60 58.8 69.28 66.44 36.6 23.55 56 63 12.0

60
 Grip Length

MINIMUM AND MAXIMUM GRIPS

FOR HIGH STRENGTH BOLTS

D
IN MILLIMETERS Bolt Length = L

Nominal
M16 M20 M22 M24 M27 M30 M36
Bolt Size
L
Bolt Min. Max. Min. Max. Min. Max. Min. Max. Min. Max. Min. Max. Min. Max.
Length Grip Grip Grip Grip Grip Grip Grip Grip Grip Grip Grip Grip Grip Grip
(mm)
45 14 26 23 20
50 19 31 14 28 25 24
55 24 36 19 32 17 29 29 25
60 29 41 24 37 22 34 19 34 30 27
65 34 46 29 42 27 39 24 39 21 35 32
70 39 51 34 47 32 44 29 44 26 40 21 37 31
75 44 56 39 52 37 49 34 49 31 45 26 42 36
80 49 61 44 57 42 59 39 54 36 50 31 47 24 41
85 54 66 49 62 47 59 44 59 41 55 36 52 29 46
90 59 71 54 67 52 64 49 64 46 60 41 57 34 51
95 64 76 59 72 57 69 54 69 51 65 46 62 39 56
100 69 81 64 77 62 74 59 74 56 70 51 67 44 61
110 72 91 67 87 65 84 62 84 59 80 54 77 47 71
120 82 101 77 97 75 94 72 94 69 90 64 87 57 81
130 92 110 87 107 85 104 82 103 79 100 74 97 67 91
140 102 120 97 117 95 114 92 113 89 110 84 107 77 101
150 112 130 107 127 105 124 102 123 99 120 94 117 87 111

61
 Grip Length
MINIMUM AND MAXIMUM GRIPS
FOR HIGH STRENGTH BOLTS
IN MILLIMETERS

D
Bolt Length = L

Nominal
M16 M20 M22 M24 M27 M30 M36
Bolt Size
L
Bolt Min. Max. Min. Max. Min. Max. Min. Max. Min. Max. Min. Max. Min. Max.
Length Grip Grip Grip Grip Grip Grip Grip Grip Grip Grip Grip Grip Grip Grip
(mm)
160 122 138 117 135 115 132 112 131 109 128 104 125 97 119
170 132 148 127 145 125 142 122 141 119 138 114 135 107 129
180 142 158 137 155 135 152 132 151 129 148 124 145 117 139
190 152 168 147 165 145 162 142 161 139 158 134 155 127 149
200 162 178 157 175 155 172 152 171 149 168 144 165 137 159
210 172 188 167 185 165 182 162 181 159 178 154 175 147 169
220 182 198 177 195 175 192 172 191 169 188 164 185 157 179
230 192 208 187 205 185 202 182 201 179 198 174 195 167 189
240 202 218 197 215 195 212 192 211 189 208 184 205 177 199
250 212 228 207 225 205 222 202 221 199 218 194 215 187 209
260 222 238 217 235 215 232 212 231 209 228 204 225 197 219
270 232 248 227 245 225 242 222 241 219 238 214 235 207 229
280 242 258 237 255 235 252 232 251 229 248 224 245 217 239
290 252 268 247 265 245 262 242 261 239 258 234 255 227 249
300 262 278 257 275 255 272 252 271 249 268 244 265 237 259

62
 WASHERS FOR HIGH STRENGTH BOLTS

B C S

C
21 NI 2 EPOLS

A
SREHSAW RALUCRIC NIALP U
SREH SA W ERAUQS DELLEVEB T

Metric Outside Diameter Hole Diameter Thickness Metric Width Hole Diameter Thickness
Bolt B (mm) A (mm) T (mm) Bolt C (mm) A (mm) S (mm) U (mm)
T (mm)
Size Max. Min. Max. Min. Max. Min. Size Max. Min. Max. Min. +0.5 +0.5
M16 34 32.4 18.4 18 4.6 3.1 M16 45 43 18.4 18 11.7 8 4.3
M20 41 39.4 22.5 22 4.6 3.1 M20 45 43 22.5 22 11.7 8 4.3
M22 44 42.4 24.5 24 4.6 3.4 M22 45 43 24.5 24 11.7 8 4.3
M24 50 48.4 26.5 26 4.6 3.4 M24 45 43 26.5 26 11.7 8 4.3
M27 56 54.1 30.5 30 4.6 3.4 M27 59 57 30.5 30 12.8 8 3.2
M30 60 58.1 33.5 33 4.6 3.4 M30 59 57 33.5 33 12.8 8 3.2
M36 72 70.1 39.5 39 4.6 3.4 M36 59 57 39.5 39 12.8 8 3.2

63
 ERECTION CLEARENCES
FOR IMPACT WRENCHES

B
EXTENSION BAR

56
A
160 to 380
C

75 E

70
20° for 20
15°for 22,24

F
50

Minimum Clearences

Sockets Min. Clearence

Size C D Bolt Size A B E F
Light Wrenches 16-24 377-356 54 16 80 45 25 28
Heavy Wrenches 24-36 375-438 64 20 85 54 30 34
22 90 57 32 36
24 95 60 34 38
27 100 70 38 42
30 110 75 41 45
36 130 90 48 52

64
 ANCHOR BOLTS

EMBEDMENT LENGTH
A
A

EMBEDEMENT STRAIGHT LENGTH

R R
B

C
C

TYPICAL ANCHOR BOLT PROFILE TYPICAL ANCHOR BOLT PROFILE

Total Embedment Total Embedment

Bolt Bolt
Weight A B C D R Straight Straight Weight A C D R Straight Straight
Nominal Thread Length Length Nominal Thread Length Length
Diameter Pitch kg mm mm mm mm mm mm mm Bolt Size Pitch kg mm mm mm mm mm mm
M16 2.0 0.88 400 110 128 100 48 558 633 M20 2.5 1.48 380 160 90 60 600 330
M20 2.5 1.72 500 140 160 125 60 700 775 M25 3.0 2.90 500 180 125 75 755 400
M24 3.0 3.00 600 170 192 125 72 842 917 M25 3.0 3.30 600 180 125 75 855 540
M30 3.5 7.00 900 240 300 150 120 1265 1340 M32 3.5 8.00 900 240 125 128 1272 840-780
M36 4.0 12.00 1000 400 360 200 144 1510 1635 M32 3.5 8.53 980 240 125 128 1352 890-840

65
SAMPLE OF CORRUGATED SHEETS

SINGLE LAYER CORRUGATED SHEETS

SANDWITCH PANELS
METAL DECK PANELS

66
 SINGLE LAYER CORRUGATED STEEL SHEETS (St.52)

800 mm MODULE

Properties & Purlin Spacing

Panel
Nominal Nominal Live Load (kg/m2)
Nominal Ix Sxt Sxb
Weight Area
Thickness
50 100 150 200
2 2 4 3 3 Allowable Purlin Spacing (cm)
mm kg/m cm cm cm cm
0.5 5.62 7.16 16.19 3.69 13.13 345 245 200 170
0.6 6.74 8.59 22.53 5.34 16.34 415 295 240 200
0.7 7.86 10.02 28.94 7.22 19.57 485 345 280 240

67
 SINGLE LAYER CORRUGATED STEEL SHEETS (St.52)

900 mm MODULE

Properties & Purlin Spacing

Panel
Nominal Nominal Live Load (kg/m2)
Nominal Ix Sxt Sxb
Weight Area
Thickness 50 100 150 200
2 2 4 3 3
mm kg/m cm cm cm cm Allowable Purlin Spacing (cm)
0.5 4.99 6.36 4.05 2.15 4.97 265 185 150 130
0.6 5.99 7.63 5.13 2.96 6.3 310 220 180 155
0.7 6.99 8.91 6.15 3.64 7.61 345 240 200 170

68
 SINGLE LAYER CORRUGATED STEEL SHEETS (St.52)

960 mm MODULE

Properties & Purlin Spacing

Panel
Nominal Nominal Live Load (kg/m2)
Nominal Ix Sxt Sxb
Weight Area
Thickness
50 100 150 200
2 2 4 3 3
mm kg/m cm cm cm cm Allowable Purlin Spacing (cm)
0.5 4.68 5.96 4.47 1.39 7.79 215 150 120 100
0.6 5.62 7.16 6.19 2.07 9.74 260 185 150 130
0.7 6.55 8.35 7.57 2.84 11.76 300 215 175 150

69
 SINGLE LAYER CORRUGATED STEEL SHEETS (St.52)

1000 mm MODULE

Properties & Purlin Spacing

Panel 2
Nominal Nominal Live Load (kg/m )
Nominal Ix Sxt Sxb
Weight Area
Thickness
50 100 150 200
2 2 4 3 3 Allowable Purlin Spacing (cm)
mm kg/m cm cm cm cm
0.5 4.49 5.73 3.8 1.27 6.53 200 145 115 100
0.6 5.39 6.87 5.16 1.91 8.3 245 175 140 120
0.7 6.29 8.02 6.48 2.53 10.13 285 200 165 140

70
 SINGLE LAYER CORRUGATED STEEL SHEETS (St.52)

1065 mm MODULE

Properties & Purlin Spacing

Panel 2
Nominal Nominal Live Load (kg/m )
Nominal Ix Sxt Sxb
Weight Area
Thickness
50 100 150 200
2 2 4 3 3 Allowable Purlin Spacing (cm)
mm kg/m cm cm cm cm
0.5 4.71 6 4.86 1.82 6.91 245 170 140 120
0.6 5.65 7.2 6.45 2.59 8.66 290 205 165 140
0.7 6.59 8.39 7.75 3.38 10.35 330 235 190 160

71
 SINGLE LAYER CORRUGATED STEEL SHEETS

1065 mm MODULE

Properties & Purlin Spacing

Panel
Nominal Live Load (kg/m2)
Nominal Ix
Weight
Thicknes 50 100 150 200
mm kg/m2 cm4 Allowable Purlin Spacing (cm)
0.50 4.75 12.53 240 170 140 125
0.55 5.23 13.79 250 180 150 130
0.60 5.70 15.04 260 190 155 135
0.65 6.18 16.29 270 195 160 140
0.70 6.66 17.51 275 205 170 145

72
 SANDWITCH ROOF PANELS

213 213 213 213 213

30
72 69 72

T
26 167 177.5 177.5 177.5 177.5 188

Properties
Panel
Nominal Weight Ix Sx
Thickness (kg/m2) (cm ) 4
(cm3)
T (mm)
65 9.48 50.4 12
80 10.08 64.4 18.6
105 11.08 191.2 31
130 12.08 289.9 44.2

Allowable Uniform Loads (kg/m2)

Panel
Nominal Span in Meters
Thickness
T (mm)
2.5 3 3.5 4 4.5 5
65 164 137 117 103 91 75
80 224 187 160 140 125 112
105 324 271 232 203 180 162
130 425 354 303 266 236 212

73
 SANDWITCH SIDE WALL PANELS

188 177.5 177.5 177.5 177.5 167 26

T
26 167 177.5 177.5 177.5 177.5 188

1065

Properties
Panel
Nominal Weight Ix Sx
Thickness (kg/m2) (cm4) (cm3)
(mm)
35 8.74 30.5 17.3
50 9.34 64.4 25.3
75 10.34 144.9 37.9
100 11.34 257.9 50.6

Allowable Uniform Loads (kg/m2)

Panel
Nominal Span in Meters
Thickness
(mm)
2.5 3 3.5 4 4.5 5
35 140 117 100 88 76 62
50 200 167 143 125 110 90
75 301 251 215 188 166 135
100 401 334 286 251 222 180

74
 CORRUGATED STEEL DECKING
SHEETS FOR CONCRETE FLOOR SLABS

830
85

55
t 166
45

PROFILE OF STEEL DECKING

EFFECTIVE CHARACTERISTICS OF THE STEEL DECKING SHEET

Thickness t (mm) 0.70 0.75 0.80 0.88 1.00
Weight (kg/m2) 8.57 9.18 9.79 10.77 12.24
4
Moment of inertia (cm ) 47.12 50.49 53.85 59.24 67.32
3
Sxb (cm ) 14.71 15.65 16.81 18.44 21.01
3
Sxt (cm ) 20.52 22.70 23.46 26.14 29.32
9.5 156.6 157 157.6 158.4 159.6
of the roof in (cm)

10 168.8 169.3 169.8 170.6 171.9

Total thickness

11 193.3 193.8 194.3 195.1 196.4

Weight 12 217.8 218.3 218.8 219.6 220.9
2
(kg/m ) 13 242.3 242.8 243.3 244.1 245.4
14 266.8 267.3 267.8 268.6 269.9
15 291.3 291.8 292.3 293.1 294.4
16 315.8 316.3 316.8 317.6 318.8

75
ACCESSORIES

76
 ACCESSORIES*
100

60 40

20
250

150 200
50 150

OUTSIDE GUTTER SECTION INSIDE GUTTER SECTION

25
25 1 58 1 58

29 .2
10
180

154,63

190
100 400 100
170

EAVE GUTTER VALLEY GUTTER

2 76
1: 10
103

25
30

30
70
15
35

50 250 50

15

CAP FLASHING PEAK CAP SHEET

150 30 50
60

60
200

25
25

40

410
50

EXPANSION JOINT TRIM

50
EAVE &PEAK FASCIA

*Indicated dimensions may be altered to fit project requirements

77 M.Korashy
 ACCESSORIES*
`

40
19

50
37

120
89

19
25

89 120 50 40

CORNER TRIM CORNER TRIM

25
145

12
30

15
70
40
FRAMED OPENING TRIM DRIP TRIM

112
30

29
,2
110
150

10 80

13 DOWN SPOUT
OASIS EAVE TRIM

*Indicated dimensions may be altered to fit project requirements

78
CRANES

79
 LIGHT CAPACITY
SINGLE GIRDER

200 mm
Wheel loads in kg
Span S 200 mm
X
g

a1 a2
H

1810
1135
1485
1800 for S<10m
2500 for S<16m
3150 for S<20m

1810
1135
1485
Wheel loads in kg
Capacity Wheel Span in Meters
ton Load 6 8 10 12 14 16 18 20
max. 650 750 800 850 900 950 1050 1150
0.5
min. 350 350 400 450 500 550 700 750
max. 800 850 900 950 1000 1050 1200 1200
0.8
min. 350 400 400 450 500 550 700 750
max. 950 1000 1050 1150 1200 1250 1400 1500
1.0
min. 350 400 450 500 550 600 750 800
max. 1200 1250 1300 1400 1450 1500 1650 1700
1.6
min. 400 400 450 500 550 600 800 850
max. 1500 1600 1650 1750 1850 1900 2050 2150
2.0
min. 450 450 500 550 600 650 850 950
max. 2200 2300 2400 2500 2600 2700 2800 2900
3.2
min. 600 600 650 700 800 900 1000 1100
max. 2450 2550 2650 2800 2850 2900 3100 3200
4.0
min. 600 600 600 650 800 900 1000 1150
max. 3000 3150 3300 3400 3500 3550 3750 3900
5.0
min. 750 750 750 750 850 950 1100 1300
max. 3600 3800 3900 4000 4100 4350 4400 4500
6.3
min. 800 800 850 900 1000 1100 1300 1500

80
 LIGHT CAPACITY
SINGLE GIRDER
Dimensions
Capacity Hmax. S (up to) g a1 a2
ton m m mm mm mm
15.2 850 600 525
17.0 950 600 600
0.5 7.0
18.5 950 600 600
20.0 1050 775 775
14.3 850 600 525
0.8 7.0 17.0 950 600 600
20.0 1050 775 775
13.0 850 600 525
16.0 950 600 600
1.0 7.0
19.0 1050 775 775
20.0 1150 1000 1000
11.8 825 600 525
15.0 925 600 600
1.6 3.5
17.0 1025 775 775
20.0 1125 1000 1000
11.8 1000 650 600
1.6 7.0 17.0 1100 650 600
20.0 1200 800 800
12.0 1300 1000 1000
13.8 825 600 525
2.0 3.5
18.0 925 600 600
20.0 1025 775 775
12.0 1125 1000 1000
13.8 1000 650 600
2.0 7.0
18.0 1100 650 600
20.0 1200 800 800
10.7 1300 1000 1000
13.6 1000 650 600
3.2 3.5
16.4 1100 650 600
20.0 1200 800 800
9.7 1300 1000 1000
12.5 1200 850 700
3.2 7.0
14.4 1300 850 700
20.0 1400 850 800
9.7 1500 1050 1050
12.5 1000 650 600
4.0 3.5
14.4 1100 650 600
20.0 1200 800 800
9.4 1300 1000 1000
12.0 1200 850 700
4.0 7.0 14.8 1300 850 700
18.0 1400 850 800
20.0 1500 1050 1050
8.3 1375 900 800
10.9 1475 900 800
5.0 8.0 13.8 1575 900 850
17.0 1675 1050 1050
20.0 1775 1250 1250
8.3 1150 850 700
10.9 1250 850 700
6.3 3.5 13.8 1350 850 800
17.0 1450 1000 1000
20.0 1550 1200 1200
8.3 1375 900 800
10.9 1475 900 800
6.3 8.0 13.8 1575 900 850
17.0 1675 1050 1050
20.0 1775 1250 1250

81
 NORMAL CAPACITY
DOUBLE GIRDER

b Span S b
gX

a1 a2
H

d1

L
e1
t
Wheel loads in kg
Capacity Wheel Span in Meters
ton Load 6 8 10 12 14 16 18 20 22 24 26 28 30
max. 1450 1550 1600 1750 1900 2000 2150 2250 2950 3050 3350 3600 3900
2.0 min. 450 500 500 600 750 800 950 1050 1700 1850 2100 2400 2300
max. 2050 2200 2300 2400 2550 2650 2850 3100 3600 3900 4150 4300 4600
3.2 min. 550 550 550 650 750 850 1000 1250 1750 2000 2250 2400 2700
max. 2450 2550 2700 2800 3000 3150 3350 3550 4050 4300 4600 4850 5150
4.0 min. 600 600 650 650 800 900 1100 1300 1750 2000 2250 2550 2800
max. 2900 3100 3250 3400 3550 3750 3950 4150 4650 4850 5200 5400 5750
5.0 min. 650 650 700 750 850 1000 1200 1350 1850 2000 2400 2550 2900
max. 3550 3750 3900 4100 4250 4450 4700 4900 5350 5600 5900 6150 6450
6.3 min. 750 700 750 850 950 1100 1300 1500 1900 2150 2400 2650 2900
max. 4600 4850 5050 5250 5450 5700 6050 6300 6550 6800 7100 7500 7800
8.0 min. 950 900 950 1000 1150 1300 1650 1800 2000 2250 2550 2900 3200
max. 5550 5850 6100 6300 6600 6850 7150 7450 7650 7950 8350 8650 8950
10.0 min. 1100 1050 1050 1150 1350 1500 1750 2000 2150 2400 2750 3050 3350
max. 6500 6900 7200 7500 7800 8100 8400 8650 9000 9300 9700 10100 10550
12.5 min. 1500 1400 1350 1450 1600 1750 1950 2150 2450 2700 3000 3400 3850
max. 8350 8800 9200 9700 9900 10350 10600 11000 11300 11700 12150 12600 13000
16.0 min. 1850 1600 1600 1800 1900 2050 2250 2550 2750 3100 3500 3900 4250
max. 10300 10850 11300 11700 12000 12350 12750 13150 13550 14100 14400 15250 1550
20.0 min. 2350 2100 2050 2050 2150 2300 2550 2850 3200 3650 3900 4650 4950
max. 11900 12900 13600 14150 14600 15150 15550 16000 16550 16950
25.0 min. 3950 3350 3050 2950 2950 3100 3250 3500 3900 4150

82
 NORMAL CAPACITY
DOUBLE GIRDER
Dimensions
Capacity Hmax. S (up to) g a1 a2 e1 L d1 t x b
ton m m mm mm mm mm mm mm mm mm mm
12 0 750 750 2000 2470 250 1000 920 200
20 50 900 750 3150 3620 250 1400 920 200
3.5
25 -50 1100 750 4000 4560 400 2240 1040 200
30 -50 1100 750 4560 5210 400 2800 1040 200
2.0 12 150 750 750 2000 2470 250 1000 990 200
7 20 150 900 750 3150 3620 250 1400 990 200
25 50 1100 750 4000 4560 400 2240 1100 200
30 50 1100 750 4560 5210 400 2800 1100 200
12 150 750 750 2000 2470 250 1000 990 200
20 150 900 750 3150 3620 250 1400 990 200
3.5
25 50 1100 750 4000 4560 400 2240 1100 200
30 50 1100 750 4560 5210 400 2800 1100 200
3.2 12 300 750 750 2000 2470 250 1000 1055 200
7 20 300 900 750 3150 3620 250 1400 1055 200
25 200 1100 750 4000 4560 400 2240 1065 200
30 200 1100 750 4560 5210 400 2800 1065 200
12 150 750 750 2000 2470 250 1000 990 200
20 150 900 750 3150 3620 250 1400 990 200
3.5
25 50 1100 750 4000 4560 400 2240 1100 200
30 50 1100 750 4560 5210 400 2800 1100 200
4.0 12 300 750 750 2000 2470 250 1000 1055 200
20 300 900 750 3150 3620 250 1400 1055 200
7
25 200 1100 750 4000 4560 400 2240 1065 200
30 200 1100 750 4560 5210 400 2800 1065 200
12 350 750 750 2000 2470 250 1000 1115 200
16 350 750 750 2500 2970 250 1400 1115 200
5.0 8 20 350 900 750 3150 3620 250 1400 1115 200
25 300 1100 750 4000 4560 400 2240 1225 200
30 300 1100 750 4560 5210 400 2800 1225 200
12 200 750 750 2000 2470 250 1000 1050 200
16 200 750 750 2500 2970 250 1400 1050 200
3.5 20 200 900 750 3150 3620 250 1400 1050 200
25 100 1100 750 4000 4560 400 2240 1160 200
30 100 1100 750 4560 5210 400 2800 1160 200
6.3
12 300 750 750 2000 2470 250 1000 1115 200
16 300 750 750 2500 2970 250 1400 1115 200
8 20 300 900 750 3150 3620 250 1400 1115 200
25 200 1100 750 4000 4560 400 2240 1225 200
30 200 1100 750 4560 5210 400 2800 1225 200
16 100 750 750 2500 3150 400 1400 1165 200
20 100 900 750 3150 3800 400 1400 1165 200
3.5
25 100 1100 750 4000 4650 400 2240 1165 200
30 100 1100 750 4560 5210 400 2800 1165 200
8.0
16 450 750 750 2500 3150 400 1400 1355 200
12 20 450 900 750 3150 3800 400 1400 1355 200
25 450 1100 750 4000 4650 400 2240 1355 200
30 450 1100 750 4560 5210 400 2800 1355 200
16 200 750 750 2500 3150 400 1400 1225 200
20 200 900 750 3150 3800 400 1400 1225 200
4
25 200 1100 750 4000 4650 400 2240 1225 200
30 200 1100 750 4560 5210 400 2800 1225 200
10.0 16 450 750 750 2500 3150 400 1400 1355 200
12 20 450 900 750 3150 3800 400 1400 1355 200
25 450 1100 750 4000 4650 400 2240 1355 200
30 450 1100 750 4560 5210 400 2800 1355 200
14 -150 900 900 2500 3150 400 1400 1385 200
20 -150 900 900 3150 3800 400 1400 1385 200
4
25 -150 1000 900 4000 4650 400 2240 1385 200
30 -250 1100 900 4560 5350 500 2800 1485 200
12.5 14 50 900 900 2500 3150 400 1400 1500 200
6 20 50 900 900 3150 3800 400 1400 1500 200
25 50 1000 900 4000 4650 400 2240 1500 200
30 -50 1100 900 4560 5350 500 2800 1600 250
11 50 900 900 2500 3150 400 1400 1470 200
16.0 6 25 -50 900 900 4000 4790 500 2240 1600 250
30 -50 900 900 4560 5350 500 2800 1600 250
25 -50 900 900 4000 4790 500 2240 1600 250
20.0 6
30 -50 900 900 4560 5350 500 2800 1600 250
20 100 1350 1350 4000 4790 500 2240 1850 250
25.0 6
25 100 1350 1350 4000 4790 500 2240 1850 250
negative (g) dimensions = hook above crane rail

83
 HEAVY CAPACITY
DOUBLE GIRDER

300 Span S 300

X
g

1800
1800

a1 a2
H

1450 1400
1400 1650

d1

e1
L
t

Wheel loads in kg
Capacity Wheel Span in Meters
ton Load 6 8 10 12 14 16 18 20 22 24 26 28 30
max. 13100 14300 15100 15800 16500 17100 17500 18200 1900 19900 21100 21800 22500
25
min. 4900 4200 3800 3600 3600 3700 3500 4000 4300 4900 5800 6300 6800
max. 15700 17100 18100 19000 19700 20400 21000 21700 22600 23700 24600 25300 26300
32
min. 5800 4800 4400 4200 4100 4100 4200 4500 5000 5700 6200 6600 7400
max. 18600 20300 21500 22600 23400 24400 25200 26100 27000 27900 28900 29700 30600
40
min. 7200 5900 5200 4900 4800 4900 5100 5500 5900 6400 6900 7400 8100
max. 23300 25500 27000 28300 29400 30400 31700 32400 33400 34300 35100 35900 37200
50
min. 9000 7200 6400 5900 5800 5800 6300 6600 6800 7200 7500 8000 9000
max. 28100 30900 32800 34300 35500 37000 38000 39000 40000 41000 42000 42900 44000
63
min. 10700 8700 7700 7100 6900 7100 7200 7700 7800 8200 8600 9200 10000

84
 HEAVY CAPACITY
DOUBLE GIRDER
Dimensions
Capacity Hmax. S (up to) g e1 L d1 t X
ton m m mm mm mm mm mm mm
10 25 530 3950 5280 630 1800 2480
15 25 530 4500 5830 630 2800 2480
20 25 530 4850 6180 630 3150 2480
25
10 30 530 4500 5830 630 1800 2480
15 30 530 5500 6830 630 2800 2480
20 30 530 5850 7180 630 3150 2480
10 22 530 3500 4830 630 1800 2480
15 22 530 4500 5830 630 2800 2480
20 22 530 4850 6180 630 3150 2480
10 24 530 3950 5280 630 1800 2480
32 15 24 530 4500 5830 630 2800 2480
20 24 530 4850 6180 630 3150 2480
10 30 530 4500 5830 630 1800 2480
15 30 530 5500 6830 630 2800 2480
20 30 530 5850 7180 630 3150 2480
10 23 530 3500 4830 630 1800 2480
15 23 530 4500 5830 630 2800 2480
20 23 530 4850 6180 630 3150 2480
10 24 530 4500 5830 630 1800 2480
40 15 24 530 5500 6830 630 2800 2480
20 24 530 5850 7180 630 3150 2480
10 30 530 4500 5830 630 1800 2560
15 30 530 5500 6830 630 2800 2560
20 30 530 5850 7180 630 3150 2560
10 13 770 3500 4830 630 1800 2480
15 13 770 4500 5830 630 2800 2480
20 13 770 4850 6180 630 3150 2480
10 19 770 3500 4830 630 1800 2480
50 15 19 770 4500 5830 630 2800 2480
20 19 770 4850 6180 630 3150 2480
10 30 770 4500 5830 630 1800 2480
15 30 770 5500 6830 630 2800 2480
20 30 770 5850 7180 630 3150 2480
10 8 770 3500 4830 630 1800 2480
15 8 770 4500 5830 630 2800 2480
20 8 770 4850 6180 630 3150 2480
10 14 770 3500 4830 630 1800 2480
15 14 770 4500 5830 630 2800 2480
20 14 770 4850 6180 630 3150 2480
63
10 19 770 4500 5830 630 1800 2480
15 19 770 5500 6830 630 2800 2480
20 19 770 5850 7180 630 3150 2480
10 30 690 4700 6150 710 1800 2560
15 30 690 5700 7150 710 2800 2560
20 30 690 6050 7500 710 3150 2560

85
WELDING SYMBOLS

86
M.Korashy
 BASIC wELD SYMBOLS
PLUG GROOVE DR BUTT
BACK FILLET DR FLARE
SLOT SQUARE V BEVEL u J FLARE V
BEVEL

� L CJ I I V V y � ,r If
SUPPLEMENTARY wELD SUMBDLS
\./ELD ALL *FIELD CONTOUR
AROUND \./ELD
FLUSH CONVEX

0 1 DJ
- �

NOTE
Size , welol syMbol, lenghth of welol o.nol spo.cing Must be reo.ol in tho.t oroler froM left to right a.long the refero.nce line.

The perpenoliculo.r leg of�, V , � , Ir.

Neither oriento.tion of refero.nce line nor loco.tion of the o.rrow o.lter this rule.
welol syMbols Must be o.t left.
Size o.nol spo.clng of fillet welols Must be shown on both the Arrow Slole o.nol the other Slole syMbol.
SyMbol o.pplieol between o.brupt cho.nges olirection of weloling unless governeol by the 'o.ll o.rounol' syMbol or otherwise oliMensioneol.

87
 WELDING SYMBOLS
D OUBLE-FILLET SYMBOL I SQUARE-GROVE WELD NG SYMBOL
I

SIZE(LENGTH OF LEG)
SPECI FICATION PROCESS.,}�
OR OTHER REFERENCE
8 7
'e� 306��l�iis°��
OOEN S BETWEEN ABRUPT
D
LD

CHAGES I N DIRECTION
OR f,S DIMENSIONED
OMISSION OF SIZE
N CATES COMPLETE JOINT
I DI
PENETRATION
_/

J}t'
I 8l__ ROOT OPENNI NG

CHAIN-I NTERMITTENT-FILLET WELDING SYMBOL SINGLE-V GROOVE WELDI NG SYMBOL

,-LENGTH OF INCREMENTS
SIZE(LENGTH OF LEG / SIZE (DEPTH OF CHAMFERI NG)
� �-1�
-
( DISTANCE BElWEEN OMISSION I NDI CATES
8 CENTRES OF I NCRMENTS) DEPTH OF CHAMFERI NG" -
--= -- /
EQUAL TO THICKNESS - -- ==:;l'l!'l,f;--
12?' _,.
OF MEMBER GROOVE ANGLE
ROOT OPENI NG/60" -

STAGGERED I NTERMITTENT-FILLET WELDI NG SYMBOL SINGLE-V GROOVE WELDI NG SYMBOL INDICATING ROOT PENETRATION
LENGTH OF I NCREMENTS _;- GROOVE ANGLE
SIZE(LENGTH OF LEG) � DEPTH SIZE OF
------.., 2 80-200-pfTCH (DISTANCE BETWEEN PENTRATION
· 12� 80-200 CENTRES OF I NCRMENTS) �
/
EFFECTIVE THROAT------'
��

BEAD WELD SYMBOL I NDICATING BEAD lYPE BACK WELD D OUBLE-BEVEL GROOVE WELDI NG SYMBOL

OMISSI ON OF SIZE
DI MENSI ON I N DI CATES
A TOTAL DEPTH OF
50"
r ARROW POI NTS TOWARD
MEMBER TO BE CHAMFERED

I
/·
_o
/ � APPLICABLE SINGLE PENTRATION EQUAL L=ROOT OPENI NG
GROOVE WELD SYMBOL TO THICKNESS OF MEMBER 40" ---GROOVE ANGLE

D UAL BEAD WELD SYMBOL INDICATING BUILT-UP SURFACE PLUG WELDI NG SYMBOL
DEPTH OF FIWNG
NCLUDED ANGLE OMISSI ONINDICATES
SIZE (HEIGHT OF DEPOSIT) FIWNG I S COMPLETE
I
ORIENTATION,
/ LOCATIONAN OF COUNTERSINK 45•
IS D ALL
4 OO 12 1 150--PITCH (DISTANCE BETWEEN CENTERS OF WELDS)
DI
�� ;���l�� �� DIMENSI ONS OTHER �
"'

THAN SIZE ARE SHOWN � CEHTERS OF WELD

DESI RE D
ON THE DRAWI NG. SIZE (DIA. OF HOLE ATROOT)

SEAM WELDING SYMBOL LENGTH OF WELDS OR I NCREMENTS SLOT WELDI NG SYMBOL

OMI SSION INDI CATES THAT WELD ORI ENTATION,
/ EXTENDS BETWEEN ABRUPT LOCATION AND ALL
SIZE (WIDTH OF W EPTH OF FIWNG DIMENSI ONS OTHER
CHANGES IN DIRECTION
D

MI N. ACCEPTABLE OMISSION I NDI CATES�THAN SIZE ARE SHOWN

75-225 OR f,S DI MENSIONED
MAY BE USED IN
�6
SHEAR RESI STANCE
X l FIWNG IS COMPLETE ON THE DRAWING

/ PITH (DISTANCE BElWEEN

CENTElls OF INCREMENTS)

88
MISCELLANEOUS

90
 AIRCRAFT HANGERS

Dimensions
L W H
TYPE OF AIRCRAFT
(m) (m) (m)
BOING 727-200 46.680 32.920 8.660
BOING 737-500 29.790 28.890 8.660
BOEING 747-400 COMBI 68.600 64.940 19.580
BOEING 757-200 47.320 38.049 13.564
BOEING 767-300 54.940 47.574 15.849
AIRBUS A340-200 59.422 58.640 16.918
AIRBUS A340-300 63.658 60.304 16.828
McDONNELL-DOUGLAS DC-8-70 57.125 45.237 12.929
McDONNELL-DOUGLAS DC-9-80 45.020 32.850 9.200
McDONNELL-DOUGLAS DC-10 55.499 50.934 17.704
FASLCON 900 19.550 19.330 7.550
LOCKHEAD C-5A GALAXY 75.540 67.882 19.850
LOCKHEAD C-130 HERCULES 29.794 40.411 11.659
LOCKHEAD -141 STRLIFTER 51.178 48.743 11.976
LOCKHEAD L-1011 54.178 47.346 16.866
CONCORDE 62.103 25.552 11.405

91
 Common Conversion Factors

Item Imperial or Metric to SI SI to Imperial or Metric

Length 1 ft. = 0.3048 m 1m = 3.28 ft.
1 in. = 25.4 mm 1 mm = 0.0394 in.
1 in. = 2.54 cm 1 in. = 0.3937 cm
1 yd. = 0.914 m 1m = 1.09 yd.
1 mile = 1.609344 km 1 km = 0.622 mi.

Area 1 acre = 0.404685 6 ha 1 ha = 2.471 acres

1 ft.2 = 0.09290304 m2 1 m2 = 10.764 ft.2
1 in.2 = 645.16 mm2 1 mm2 = 1.55x 10-3 in2
1 in.2 = 6.4516 cm2 1 cm2 = 0.155 in2
1 mi.2 = 2.589 988 km2 1 km2 = 0.3861 mi.2
1 yd.2 = 0.836127 m2 1 m2 = 1.20 yd.2
Volume 1 in.3 = 16387.064mm.3 1 mm 3 = 0.061 x10-3 in.3
1 in.3 = 16.387064 cm3 1 cm3 = 0.061 in.3
1 ft.3 = 28.316 85m.3 1 dm 3 = 0.0353 ft.3
1 yd.3 = 0.764 555m.3 1 m3 = 1.308 yd.3
Mass 1 lb. = 0.45359237 kg 1 kg = 2.20 lb.
1 ton = 0.90718474 Mg 1 Mg = 1.10 ton =2200 lb.

Mass per unit Length 1 lb./in. = 17.858 kg/m 1 kg/m = 0.056 lb./in.
1 lb./ ft. = 1.48816 kg/m 1 kg/m = 0.672 lb./ft.
1 lb./yd. = 0.496055 kg/m 1 kg/m = 2.016 lb./yd.

Mass per unit Area 1 lb./ ft.2 = 4.88243 kg/m2 1 kg/m2 = 0.2051 lb./ft.2
1 lb./ ft.2 = 4.88243 x10-4 kg/cm2 1 kg/cm2 = 2051 lb./ft.2
1 oz./ft.2 = 305.152 g/m2 1 g/m2 = 3.277X10-3 oz./ft.2
1 lb./in.2 = 703.0696 kg/m2 1 kg/m2 = 1.42X 10-3 lb./in.2

Mass per unit volume

1 lb./ft.3 = 16.01846 kg/m3 1 kg/m3 = 62.4X10-3 lb./ft.3
1 lb./in.3 = 27.67990 Mg/m3 1 Mg/m3 = 0.0361 lb./in.3

Force 1 kg force = 9.80665 N 1N = 0.1019716 kg force

1 Kip = 4.448222 kN 1 kN = 0.225 Kip
1 pound force = 4.448222 N 1N = 0.225 pound force

Torque or Moment of Force 1 pound-force foot = 1.355818 N.m 1 N.m = 0.737562 pound-force foot
1 pound-force inch = 0.112985 N.m 1 N.m = 8.850732 pound-force inch

Pressure or Stress 1 psf = 47.88026 pa 1 pa = 0.0209 psf

1 psi = 6.894757 kpa 1 kpa = 0.145 psi
1 ksi = 6.894757 Mpa 1 Mpa = 0.145 ksi
1 kg/cm2 = 0.0980665 Mpa 1 Mpa = 10.19 kg/cm2
1 t/cm2 = 98.0665 Mpa 1 Mpa = 0.01019 t/cm2

Moment of inertia
a) second Moment of 1 in.4 = 416231.4 mm4 1 mm4 = 2.4 x10- 6 in.4
1 in.4 = 41.62314 cm4 1 cm4 = 0.024 in.4
Area 1 in.3 =16387.064 mm3 1 cm4 = 0.024 in.4
b) section Modulus 1 in.3 = 16.387064 cm3 1 mm3 = 0.061 x10-3 in.3
1 cm3 = 0.061 in.3

92
PROPERTIES OF GEOMETRIC SECTIONS
AND STRUCTURAL SHAPES

A = bd
d
c=
2

c
bd 3
I=
12
d
bd 2
S=
6
d
r=
12
b
RECTANGLE

bd
A=
2
c

c = 2d 3
d

bd 3
I=
36
bd 2
S=
24
d
r= b
18
TRIANGLE

93
 d (b + b1 ) b1
A=
2
d (2b + b1 )
c=
3(b + b1 )

c
2
d 3 (b + 4bb1 + b1 )
I=

d
36(b + b1 )
d 2 (b 2 + 4bb1 + b12 )
S=
12( 2b + b1 )
d
r= 2(b 2 + 4bb1 + b12 )
6(b + b1 ) b
TRAPEZOID

π d2
A= = π R2
4
d c
c= =R R
2
π d 4 π R4
d

I= =
64 4
πd 3
π R3
S= =
32 4
d R
r= =
4 2
CIRCLE

94
 π ( d 2 − d12 )
A=
4

c
d R
c=
2

d1
d
π ( d 4 − d14 )
I=
64
π ( d 4 − d14 )
S=
32 d
d 2 + d12
r= HOLLOW CIRCLE
4

A = πDt = 2πRt
c=R
c

I = π D 3t 8 R

S = πD 2 t 8
d

r=D 8
t

THIN WALLED SECTION

95
 h = d + 2t
bo = a + 2b bo
b1

t
A1 = b1t
A = 2( A1 + A2 ) A1
b
[
I xx = 2 I xc + 1 h 3 − d 3
12
]

d
S xx = 2 I xx h
A2

h
rxx = I xx A
A1 2
b1 + 2 A2 ( x + a 2 )
2
I yy = 2 I yc +
6
S yy = 2 I yy bo if b1 < bo
b a x
S yy = 2 I yy b1 if b1 ≥ bo
ryy = I yy A

h = d + 2t
b1
A1 = b1t
a
A = 2( A1 + A2 )

I xx = 2 I xc +
b1 3
12
(
h −d3 )
S xx = 2 I xx h
rxx = I xx A A2 h
A1 2
d

b1 + 2 A2 (a 2 − x )
2
I yy = 2 I yc +
6
S yy = 2 I yy b1 if a < b1
S yy = 2 I yy a if a ≥ b1
x A1
ryy = I yy A
t

Note: Elements of the shape, which are shown in dotted outline, are continuous and if
disconnected the variable defining their size should be set equal to zero.
96
 h = d − 2t
b
A1 = bt a

t
A2 = wh
A = 2( A1 + A2 ) A1
I xx =
1
12
{
b(d 3 − h 3 ) + 2 A2 h 2 } w

d
h
S xx = 2 I xx d
rxx = I xx A A2
c = a − 2w

I yy =
1
12
{
2 A1b 2 + h(a 3 − c 3 ) }
`

t
S yy = 2 I yy b
ryy = I yy A

b
h = d − 2t
t Y t
A1 = bt t
A2 = ht
A = 2 A1 + 3 A2
h

I xx =
1
12
[3 A2 h 2 + b ( d 3 − h 3 ) ] X X
d

S xx = 2 I xx d A2 t
rxx = I xx A

I yy =
1
{ }
2 A1b 2 + A2 t 2 + 2t[b 3 − (b − 2t ) 3 ] A1
12 Y
t

S yy = 2 I yy b
ryy = I yy A

Note: Elements of the shape, which are shown in dotted outline, are continuous and if
disconnected the variable defining their size should be set equal to zero.

97
A1 = bt
A2 = (d − w − 2t ) w 2 d
a b a
A3 = 2 A2 + w 2
A1

t
A = 4 A1 + 2 A2 + A3 t A2 t

a
E = d − 2t

Ix = Iy =
1
{
b(d 3 − E 3 ) + wE 3 + 2tb 3 + Ew 3 − w 4 }

b
d
12 A3

w
S x = S y = 2I x d

a
rx = ry = I x A

t
w

h = d − 2w
A2 = ht b
w

A = A1 + A2
1
x
I xx = 2I yc + A2 h 2 + 2 A1 (d 2 − x) 2
12
A1
S xx = 2I xx d A2
rxx = I xx A t
h

1 A1
I yy = 2I xc + A2 t 2
12
S yy = 2I yy
x

b
w

ryy = I yy A

Note: Elements of the shape, which are shown in dotted outline, are continuous and if
disconnected the variable defining their size should be set equal to zero.

98
A2 = bt bf
Y
A = 2 ( A1 + A2 )

t
A2

I xx = 2 I xI +
1
12
[
b ( d + 2t ) 3 − d 3 ]
A1
S xx = 2 I xx ( d + 2 t )

d
X X
rxx = I xx A
1 1
I yy = 2 I yI + A2 t 2 + A1 x 2
6 2
S yy = 2 I yy (x + b f ) if ( x + b f ) > b Y

t
x
S yy = 2 I yy b if ( x + b f ) ≤ b b

ryy = I yy A

b
d 0 = d + 2t b1
A3 = b1t

t
A = 2( A1 + A3 ) + A2
b1 3
A1 A1
I xx = 2 I x1 + I y 2 + (d 0 − d 3 )
12
S xx = 2 I xx d 0

do
d
A2
rxx = I xx A
A3 2
I yy = I x 2 + 2 I Y 1 + b1 + A1b12 2
6
S yy = 2 I xx b if b ≥ b1 A3
bf
t

S yy = 2 I xx b1 if b < b1

ryy = I yy A

Note: Elements of the shape, which are shown in dotted outline, are continuous and if
disconnected the variable defining their size should be set equal to zero.
99
 1
h=d+ (b1 + w1 )
2
A ( d + w1 / 2) + A2 d / 2
y1 = 1 b
A1 + A2
y 2 = h − y1

w1
A = A1 + A2 A1

b1
I xx = I y1 + I x 2 + A1 ( y 2 − b1 / 2) 2

y2
S x1 = I xx y1
S x 2 = I xx y 2

h
N.A.

d
rxx = I xx / A A2
w
I yy = I x1 + I y 2

y1
S yy = 2 I yy / b
ryy = I yy / A

A = Ac + AI
b

w
do = d + w
AI d / 2 + Ac ( d 0 − x )
x
y1 =
y2

A
y 2 = d 0 − y1 N.A.
I xx = I xI + I yc + AI ( y1 − d / 2) 2 + Ac ( y 2 − x ) 2

do
d
S x1 = I xx y1
t
y1

S x 2 = I xx / y 2
rxx = I xx / A
I yy = I yI + I xc
S yy = 2 I yy / b
ryy = I yy / A

Note: Elements of the shape, which are shown in dotted outline, are continuous and if
disconnected the variable defining their size should be set equal to zero.

100