51

CHAPTER 4
ANALYSIS AND DESIGN RESULTS OF 8 STOREYED STEEL BUILDING

4.1. General
The proposed building is 8-storeyed steel building and located in seismic zone
4. The applied loads are gravity loads (dead load, live load, superimposed dead load)
and lateral loads (earthquake load, wind load). Earthquake and wind loading data are
referenced by using UBC-97. Wind exposure type B and wind speed 80 mph are
considered for proposed building. The design concepts and load combinations are
considered by using Load and Resistance Factor Design AISC-LRFD 1999. The
design results for proposed building are described in this chapter.

4.2. Design Results for Superstructure

Design results for superstructure are presented in the followings.

4.2.1. Design Results for Beam Sections

In this study, Wide Flange W section are used for beams. Beams are classified
as type B1and B2. Typical beams sections are mentioned in Table 4.1. and typical
beam layout plan is shown in Fig 4.1.

4.2.2. Design Results for Column Sections

Column are classified as type C1 and C2. Wide Flange W sections are used.
Column sections are changed in storey 1-4 and storey 5-8. Typical column sections
are mentioned in Table 4.1. and column layout plan is shown in Fig 4.2.

Table 4.1. Typical Beam and Column Sections

Beam Type Storey 1-7 Storey 8 Column Type Storey 1-4 Storey 5-8

B1 W10x26 W8x21 C1 W12x152 W12x136

B2 W8x10 W8x10 C2 W12x136 W12x120

 49

Figure 4.1. Beam Layout Plan of Proposed Building

Figure 4.2. Column Layout Plan of Proposed Building

 50

4.3. Checking for Superstructure

In design of structure, stability checking is required according to UBC-97. In

this study, the stability of the building with static analysis is checked with the
following cases.

(a) P-D effect

(b) Overturning moment
(c) Resistance to Sliding
(d) Storey drift and
(e) Torsional irregularity.

4.3.1. Checking for P- Effect

For SMRF buildings, P- effects need to be considered whenever the storey
drifts do not satisfy the following criterion.
s/h £ 0.02/R
0.24647/114 £ 0.02/8.5
0.002162 < 0.0024 \Satisfied
Checking of P- effects for relative storey is shown in table 4.2.
Table 4.2. Checking of P-Δ effects for Relative Storey

Storey Height(in) Drift X Drift Y ∆Limit (in)

8 114 0.000944 0.000652 0.002352941
7 114 0.001171 0.000868 0.002352941
6 114 0.001502 0.001102 0.002352941
5 114 0.001793 0.001307 0.002352941
4 114 0.001999 0.001438 0.002352941
3 114 0.00214 0.001558 0.002352941
2 114 0.002162 0.001692 0.002352941
1 180 0.001558 0.002062 0.002352941
GL 96 0.000438 0.000799 0.002352941

Since s/h < 0.02/R, P-delta effect does not need to consider for the proposed
building.

4.3.2. Checking for Overturning Moment

The distribution of earthquake forces over the height of a structure causes to

experience overturning effect.The summation of moment due to the distributed lateral
 51

forces is the overturning moment. The UBC-97 requires that every designed structure
be capable of resisting overturning effect induced by earthquake forces.
For X-direction,

Overturning moment = (Base shear ×story height) + Mt

= (128.093 × 96) + 82966.376
= 95263.304 kip-in
Total dead weight = 1573.76 kips
Center of mass in X-direction, XCCM = 300.38 in
Resisting moment = 0.9 × Total dead weight × XCCM
= 0.9 × 1573.76 × 300.38
= 425453.4259 kip-in
Resisting moment
Factor of safety =
Overturning moment

425453.4259
=
95263.304

= 4.47 > 1.5 \Satisfied

For Y-direction,
Overturning moment = (Base shear ×story height) + Mt
= (135.104 × 96) + 87316.401
= 100286.385 kip-in
Center of mass in Y-direction, YCCM = 383.949in
Resisting moment = 0.9 × Total dead weight × Y CCM
= 0.9 × 1573.76 × 383.949
= 543819.2204 kip-in

Resisting moment
Factor of safety =
Overturning moment

543819.2204
=
100286.385

= 5.42 > 1.5 \Satisfied

The structure is capable of resisting overturning effect.
 52

4.3.3. Checking for Resistance to Sliding

The ratio of resistance force to driving force must be greater than 1.5 for both
directions. In this building, soil profile type S D is used so the friction coefficient  is
taken as 0.25.According to UBC-97, the resistance for sliding is calculated as follows:
For X-direction,
Frictional coefficient  = 0.25
Total dead weight = 1573.76 kips
Base shear (Sliding force) = 128.093 kips
Resistance due to friction = 0.25 × 0.9 × Total dead weight
= 0.25 × 0.9 ×1573.76
= 354.095 kips
Resistance due to friction
Factor of safety =
Sliding force
354.095
=
128.093

= 2.76 > 1.5 \Satisfied

For Y-direction,
Base shear = 135.104 kips
Total dead weight = 1573.76 kips
Resistance due to friction = 354.095 kips
Resistance due to friction
Factor of safety =
Sliding force
354.095
=
135.104

= 2.62 > 1.5 \Satisfied

There is no sliding occurrence in both directions.

4.3.4. Checking for Storey Drift

Storey drift shall not exceed 0.025 times the storey height for structures
having the fundamental period of less than 0.7 second. For structures having a
fundamental period of 0.7 second or greater, the storey drift shall not exceed 0.02
times the storey height.
In this structure, fundamental period t is greater than 0.7 second. Therefore,
storey drift shall not exceed 0.02 times the storey height.
 53

ΔM = 0.7R Δs
Checking of story drift for relative storey is shown in Table 4.3.
Table 4.3. Calculation for Storey Drift

Story Height(in) Drift X Drift Y ∆Sx(in) ∆Sy(in) ∆Mx(in) ∆My(in) ∆Limit(in)

8 114 0.000944 0.000652 0.107616 0.074328 0.64 0.44 2.28

7 114 0.001171 0.000868 0.133494 0.098952 0.79 0.59 2.28

6 114 0.001502 0.001102 0.171228 0.125628 1.02 0.75 2.28

5 114 0.001793 0.001307 0.204402 0.148998 1.22 0.89 2.28

4 114 0.001999 0.001438 0.227886 0.163932 1.36 0.98 2.28

3 114 0.00214 0.001558 0.24396 0.177612 1.45 1.06 2.28

2 114 0.002162 0.001692 0.246468 0.192888 1.47 1.15 2.28

1 180 0.001558 0.002062 0.28044 0.37116 1.67 2.21 3.6

GL 96 0.000438 0.000799 0.042048 0.076704 0.25 0.46 1.92

All storey drifts are within the storey drift limitation.

4.3.5. Check for Torsional Irregularity

Diagonal points of node point 4 and node point 20 for x direction and
diagonal points of node point 1 and 4 for y direction are checked. Points for torsional
irregularity check are shown in Fig. 4.3.
For X-direction,
For point 4 Ux = 1.614868 in
For point 20 Ux = 1.254202 in
Maximum displacement, ΔMax = 1.614868 in
1.614868  1.254202
Average displacement of two points, Δavg =
2

= 1.434535 in

Δ max 1.614868
=
Δ avg 1.434535

= 1.126 ˂ 1.2
 54

For Y-direction,
For point 1 Uy = 1.369336 in
For point 4 Uy = 1.367859 in
Maximum displacement, ΔMax = 1.369336 in
1.369336  1.367859
Average displacement of two points, ∆avg =
2

= 1.3685975 in
Δ max 1.369336
=  1.0 ˂ 1.2
Δ avg 1.3685975

Therefore, torsional irregularity can be neglected in this structure.

Figure 4.3. Points for Torsional Irregularity Check

4.4. Design Calculation of Slabs

All floor slabs are designed as metal steel deck supporting concrete slab
system. In the design of composite slab, common 3 types of slab have been considered in
accordance with span length. Typical slabs plan is shown in Fig. 4.4. Normal weight
concrete (145 pcf) concrete slab over 2.5 inches metal deck and overall slab depth of 5
inches in composite floor construction is used.
 55

Figure 4.4. Slab Layout Plan of Proposed Building

Yield strength of structural steel, Fy = 50 ksi

Compressive strength of concrete, f c' = 3.0 ksi

Modulus of elasticity of concrete, Ec = 33W1.5 f c'

= 33(145)1.5 3000

= 3155.92 ksi
Es 29000
Modular ratio, n = = = 9
Ec 3155.92

Typical design calculation for type 1 slab is shown below.

5.0625'
f c' = 3000 psi
10' Fy = 50000 psi
ts = 5in
L= 10ft, bo =5.0625ft
 56

The section properties of the W 10 × 26 steel beam are,

d = 10.3 in, A = 7.61 in2, bf = 5.77 in, tw = 0.26 in, tf = 0.44 in, Ix = 144 in4,
Sx = 27.96 in3, Zx = 31.3 in3
Beam weight = 26 lb/ft
Weight of concrete + deck = 39 psf
Construction Live Load = 20 psf
Superimposed Dead Load = 25 psf
Superimposed Live Load = 40 psf

(a) Check Mn³ Mu (for without temporary shoring)

When temporary shores are not used during construction, the steel section
must have adequate strength to support all loads applied prior to the concrete attaining

75% of its specified f c' . Therefore, a controlling limit state should be checked.
u = 1.4 (concrete slab) + 1.2 (steel beam) + 1.6 (construction L.L)
= (1.4 × 39) + (1.2 × 26) + (1.6 × 20)
= 117.8 lb/ft2
u l2 117.8x5.06 25x10 2
Mu = =
8 8x1000

= 7.455 kip-ft
Design strength,
Mn = bFyZx
1
= 0.9 × 50 × 31.3 ×
12
= 117.375 kip-ft > Mu = 7.455 kip-ft
\Mn > Mu \Satisfied
(b) Check Mn³ Mu (after concrete have cured)
L 10  12
bE = = = 30 in (control)
4 4
bE = bo = 5.0625 × 12 = 60.75 in
u = 1.2 D + 1.6 L
= 1.2 (26 + 39 + 25) + 1.6 (40) = 172 lb/ft
u l2 2
Mu = = 172  5.0625  10 = 10.88 kip-ft
8 8  1000
 57

Stress block depth,

Asf y
a =
0.85f c' b E
7.61  50
=
0.85  3  30
= 4.97 in < ts = 5 in
a < ts (plastic neutral axis within the slab)
d a
Mn = Asfy  + ts – 
2 2
10.3 4.97  1
= 0.9 × 7.61 × 50  5
 2 2  12

= 218.74 kip-ft > Mu = 10.88 kip-ft \Satisfied

(c) Check dead load deflection (during construction)

The calculated deflection based on the weight of concrete plus deck is limited
to the smaller value L/180 or 3/4 in, in which L is the clear span of the deck [3].
D = (26 + 39) 5.0625
= 329.0625 lb/ft
5l4
D =
384EI
5  329.0625  10 4  123
=
384  29000  144  1000
L 10  12
= 0.0177 in < = = 0.667 in \Satisfied
180 180

(d) Check live load deflection

For calculating deflections, it is not necessary to consider the construction
loads. The calculated deflection should be less than L/360.

ts = 5" 2.5"
2.5" yt
C.G. of composite
section
y
W10 x 26 yb
C.G. of W section

Figure 4.5. Floor Section for Type 1 Slab

 58

width of equivalent steel,

bE 30
b = = = 3.33 in
n 9
Table 4.4. Dimensions and Properties of Type 1 Slab
Transformed Moment Arm
Element Area from centroid Ay Ay2 Io
A y
Slab 16.65 7.65 127.373 974.40 34.69

W10 x 26 7.61 0 0 0 144

Total 24.26 7.65 127.373 974.40 178.69


y = C.G of composite section measured from C.G of W section
 Ay
y =
A
127.373
=
24.26

= 5.25 in
Ix = Io + Ay2

= 178.69+974.40
= 1153.09 in4

Itr = Ix – Ay 2

= 1153.09 – (16.65× 5.252)

= 694.17 in4

L = 40 × 5.0625

= 202.5 lb/ft

5L4
DL =
384 EI

5  202.5  10 4  12 3
=
384  29000  1153.09  1000
L 10  12
= 0.00136 in < = = 0.33 in \Satisfied
360 360
 59

(e) Check stress in concrete

d
c = yt =  y  slab thickness
2
10.3
=  5.25  5
2
= 4.9 in
I tr
St =
c
694.17
=  141.67in 3
4.9

Service load,
D = (26 + 39+ 25+40) 5.0625
= 658.125 lb/ft
l 2
M =
8
658.125  10 2  12
=
8  1000
= 98.72 kip-in

M
Compressive stress in concrete =
ns t
98.72
=
9  141.67

= 0.0774 ksi < 0.45 f c' = 1.35 ksi 

Satisfied
(f) Design of shear connector
The compressive force that must be carried by the shear connector, Cc.
3
Use in for diameter of stud connector, ds = 0.75 in, Fub = 60 ksi
4
'
Cc = 0.85 f c b Ea
= 0.85 × 3.0 × 30 × 4.97

= 380.205kips

Strength of one stud, Qn

Ec = 1746 f c'

=
 60

1746 3 = 3024.16 ksi

 
Qn = 0.5  d s2  f c' E c
4 

 2 
= 0.5  0.75  3 3024.16
 4 

= 21.04 kip/stud (control)

 2 b
Qn =  d s  Fu
4 

 2 
=   0.75   60
 4 

= 26.51 kip/stud
Number of shear connector,
Cc
N =
Qn

380.205
=
21.04

= 19 Nos (for half span)

For entire span = 2 N = 2 × 19 = 38 Nos
Using uniform spacing with two studs at each location,
L
Required spacing, P =
N
10  12
=
19
= 6.32 in ( use 7 in )

Maximum spacing = 8 ts = 8×4 = 32 in > 7 in

Minimum spacing = 6 ds = 6 ×0.75 = 4.5 in < 7 in
Stud height = 4ds = 4 × 3/4 = 3 in < 4 in
= 4 in (Use)

Use 3/4 in  stud connectors @ 7" c/c spacing with two studs at each location.
 Table 4.5. Design Results for Floor Slabs

4.5.

Allow- Allow-
able Compre- Number Deflection able
Design strength after Deflection deflect- ssive stress of shear Centre to under super- deflect- Stud Stud
(k-ft) (in) eter (in)
concrete have cured based on ion of concrete stud along centre imposed live ion height diam-
(L/180) (ksi) the beam load (in) (L/360)
(in) (in)

218.74 0.0177 0.67 0.0774 38 7 0.001 0.33 4 3/4

61

239.00 0.0681 0.93 0.1205 38 9 0.01 0.47 4 3/4

219.60 0.0147 0.68 0.0716 38 7 0.003 0.34 4 3/4

of slab are designed as similar to the above slabs and the results are described in Table
There are three types of slab in this structure according to span length. The other types
 62

Design strength for

without temporary
shoring

117.375

117.375

117.375
Type (k-ft)

3
slab of

4.5. Design of Connection

Design of connection consists of beam to column connections, beam to girder

connection and column splices. In this study, joint types are classified according to
their location, namely corner joint (Type 1), exterior joint (Type 2) and interior joint
(Type 3) as shown in Fig. 4.6. In this study, beam to column connections that have
maximum moment are designed according to storey levels. In beam to column flange
connections and in beam to column web connections, both methods of bolted-
connection and welded-connection are used. There are 4 corner joints, 12 exterior
joints and 8 interior joints at each storey. Bearing type connection with A 325-N high
strength bolts and E 70 electrodes with SMAW process are used.
 63

Figure 4.6. Location of Joint Types

4.5.1. Calculation of Beam to Column Flange Corner Connection (Type 1)

The corner connection that has maximum moment of C 2 (W12×136) and

B1 (W10×26) at storey 4(grid line 2D) is designed according to LRFD. Beam to
column flange corner connection at storey 4 is shown in Figure 4.7.

Figure 4.7. Beam to Column Flange Corner Connection

 64

Fy = 50 ksi, Fu = 65 ksi, Fub = 120 ksi, FEXX= 70 ksi

The section properties of W 10 × 26 beam are,
bf = 5.77 in, d = 10.3 in, tf = 0.44 in , tw = 0.26 in, A = 7.61 in2, k = 0.74 in
From analysis data,
Mu = 648.899 kips-in
Vu = 13.22 kips
The section properties of W 12 × 136 column are,
bf = 12.4 in, d = 13.4 in, tf = 1.25 in, tw = 0.79 in, k = 1.85 in

Compute the magnitude of the internal tension and compression force,

Mu 648.899
Tu = = = 65.81 kips
d  t fb 10.3 - 0.44

In this study, ¾ in diameter of high strength bolts are used in connection design.
Ab =  x ( 3 )2 = 0.4418 in2
4 4

Nominal strength of bolt in tension,

 Rnt = (0.75A )Fub

b
= 0.75 x (0.75 x 0.4418) x 120
= 29.82 kips/bolt
Tu 65.81
Required number of bolt = = = 2.207
R nt 29.82

Try 6 Nos (for both sides)

Vu 13.22
Shear stress per bolt, fuv = = = 4.98 ksi
nA b 6 x 0.4418
Check combined shear and tension
For A 325 - N
' = 85 -1.8f  68 ksi
Allowable tensile stress, Fut uv

F'ut = 85 - 1.8 x 4.98 = 76.036 ksi > 68 ksi (control)

Tu 65.81
Tensile stress, fut = =
A 3 x 0.4418

= 49.63 ksi < Allowable = 68 ksi Satisfied

 65

Use c/c spacing = 3.5 in > 3d = 3 × 0.75 = 2.25 in

Used end distance = 1.5 in >1.5d = 1.5 × 0.75 = 1.125in
Weld thickness,
Nominal strength of weld per inch of length,
Tu 65.81
Required,  Rnw = = = 5.83 kip/in
2bf  t w (2 x 5.77 ) - 0.26
Design strength,
 Rn =  (0.707 a) (0.6 FEXX)

5.83 = 0.75(0.707a) (0.6×70)

a = 0.262 in
 Use 5/16 in weld thickness.

The required length Lw of fillet weld,

T 65.81
Lw = = = 11.29 in
R nw 5.83

Available weld length,

Lw = 4 bf + 2d –2tw
= (4x5.77) + (2x10.3) – (2x0.26)
= 43.16 in

Total design strength =11.29×5.83

= 65.82 kips > 13.22 kips
Determination of plate thickness
S = 0.74 +5/16 =1.0525 in
db
b' = S  weld size
4
0.75 5
= 1.0525 - - = 0.55
4 16
Fy = 50 ksi ,
A 325  ca = 1.09
b
f 5.77
cb = b

6.77
s

 0.92
1 1
 A 3  b ' 4
m = ca c b  f   
 Aw   db 
 66

1/ 3 1/ 4
 0.44  5.77   0.55 
= 1.09 × 0.92 ×  
 (10.3  2  0.44)0.26   0.75 


= 0.94
 m Tu b '
Me =
4
0.94  65.81  0.55
=
4

= 8.51 kip-in

4.44 M e
tp =
wf y (1  )

4.44  8.51
=
6.77  50(1  0)

= 0.3341 in
Use 0.5 in.

Factored compression force from beam flange

Pbf = Fyc t wc (t fb  6k  2t p  2a)

= 50  0.79 0.44  6  1.85  2  0.5  2  0.3125

= 520.02 kips > Compressive force, C = 65.81 kips

 Stiffener is not required.

4.5.2. Calculation of Beam to Column Web Corner Connection (Type 1)

The corner connection that has maximum moment of C 2 (W12x136) and B 1
(W10×26) at storey 4 (grid line 2D) is designed. Beam to column web corner
connection at storey 4 is shown in Figure 4.8.
 67

Figure 4.8. Beam to Column Web Corner Connection

Fy = 50 ksi, Fu = 65 ksi, Fub = 120 ksi, FEXX= 70 ksi

The section properties of W 10 × 26 beam are,
bf = 5.77 in, d = 10.3 in, tf = 0.44 in , tw = 0.26 in, A = 7.61 in2, k = 0.74 in
From analysis data,
Mu = 253.102 kips-in
Vu = 6.576 kips
The section properties of W 12 × 136 column are,
bf = 12.4 in, d = 13.4 in, tf = 1.25 in, tw = 0.79 in, k = 1.85 in
Compute the magnitude of the internal tension and compression force,
Mu 253.102
Tu = = = 25.67 kips
d  t fb 10.3 - 0.44

In this study, ¾ in diameter of high strength bolts are used in connection design.
Ab =  x ( 3 )2 = 0.4418 in2
4 4

Nominal strength of bolt in tension,

 Rnt = (0.75A )Fub

b
= 0.75 x (0.75 x 0.4418) x 120
= 29.82 kips/bolt

Tu 25.67
Required number of bolt = = = 0.861
R nt 29.82

Try 6 Nos (for both sides)

Vu 6.576
Shear stress per bolt, fuv = = = 2.48 ksi
nA b 6 x 0.4418

Check combined shear and tension

For A 325 - N
' = 85 -1.8f  68 ksi
Allowable tensile stress, Fut uv

F'ut = 85 - 1.8 x 2.48 = 80.536 ksi > 68 ksi (control)

Tu 25.67
Tensile stress, fut = =
A 3 x 0.4418
 68

= 19.37 ksi < Allowable = 68 ksi Satisfied

Use c/c spacing = 3.5 in > 3d = 3 × 0.75 = 2.25 in
Used end distance = 1.5 in >1.5d = 1.5 × 0.75 = 1.125in
Weld thickness,
Nominal strength of weld per inch of length,
Tu 25.67
Required,  Rnw = = = 2.28 kip/in
2bf  t w (2 x 5.77 ) - 0.26

Design strength,
 Rn =  (0.707 a) (0.6 FEXX)

2.28 = 0.75(0.707a) (0.6×70)

a = 0.102 in
 Use 1/4 in weld thickness.
The required length Lw of fillet weld,
T 25.67
Lw = = = 11.26 in
R nw 2.28

Available weld length,

Lw = 4 bf + 2d –2tw
= (4x5.77) + (2x10.3) – (2x0.26)
= 43.16 in

Total design strength =11.26×2.28

= 25.67 kips > 6.576 kips

Determination of plate thickness

S = 0.74 +1/4 =0.99 in
db
b' = S  weld size
4
0.75 1
= 0.99 - - = 0.55
4 4
Fy = 50 ksi ,
A 325  ca = 1.09
b
f 5.77
cb = b

6.77
s

 0.92
 69

1 1
 A   b' 
3 4
m = ca c b  f   
 Aw   db 
1/ 3 1/ 4
 0.44  5.77   0.55 
= 1.09 × 0.92 ×   
 (10.3  2  0. 44) 0. 26   0.75 


= 0.94
 m Tu b '
Me =
4
0.94  25.67  0.55
=
4

= 3.34 kip-in

4.44 M e
tp =
wf y (1  )

4.44  3.34
=
6.77  50(1  0)

= 0.209 in
Use 0.5 in.
Factored compression force from beam flange

Pbf = Fyc t wc (t fb  6k  2t p  2a)

= 50  0.79 0.44  6  1.85  2  0.5  2  0.25

= 515.08 kips > Compressive force, C = 25.67 kips

 Stiffener is not required.

4.5.3. Calculation of Beam to Column Flange Exterior Connection (Type 2)

The corner connection that has maximum moment of C 2 (W12×136) and
B1 (W10×26) at storey 2 (grid line 3D) is designed. Beam to column flange exterior
connection at storey 2 is shown in Figure 4.9.
 70

Figure 4.9. Beam to Column Flange Exterior Connection

Fy = 50 ksi, Fu = 65 ksi, Fub =120 ksi, FEXX = 70 ksi

The section properties of W 10 x 26 beam are,
bf = 5.77 in, d = 10.3 in, tf = 0.44 in, tw = 0.26 in, A = 7.61 in2, k = 0.74 in
From analysis data,
Mu = 891.314 kips-in
Vu = 20.28 kips
The section properties of W 12 x 136 column are,
bf = 12.4 in, d = 13.4 in, tf = 1.25 in, tw = 0.79 in, k = 1.85 in
Compute the magnitude of the internal tension and compression force,
Mu 891.314
Tu = = = 90.4 kips
d  t fb 10.3 - 0.44

In this study, ¾ in diameter of high strength bolts are used in connection design.
Ab =  x ( 3 )2 = 0.4418 in2
4 4

Nominal strength of bolt in tension,

 Rnt = (0.75A )Fub
b

= 0.75 x (0.75 x 0.4418) x 120

= 29.82 kips/bolt
Tu 90.4
Required number of bolt = = = 3.031
R nt 29.82

Try 8 Nos (for both sides)

 71

Vu 20.28
Shear stress per bolt, fuv = = = 5.74 ksi
nA b 6 x 4418

Check combined shear and tension

For A 325 - N
' = 85 -1.8f  68 ksi
Allowable tensile stress, Fut uv

F'ut = 85 - 1.8 x 5.74 = 74.668 ksi > 68 ksi (control)

Tu 90.4
Tensile stress, fut = =
A 3 x 0.4418

= 51.15 ksi < Allowable = 68 ksi Satisfied

Use c/c spacing = 3.5 in > 3d = 3 × 0.75 = 2.25 in
Used end distance = 1.5 in >1.5d = 1.5 × 0.75 = 1.125in
Weld thickness,
Nominal strength of weld per inch of length,
Tu 90.4
Required,  Rnw = = = 8.01 kip/in
2bf  t w (2 x 5.77 ) - 0.26

Design strength,
 Rn =  (0.707 a) (0.6 FEXX)

8.01 = 0.75(0.707a) (0.6×70)

a = 0.36 in
 Use 3/8 in weld thickness.
The required length Lw of fillet weld,
T 90.4
Lw = = = 11.29 in
R nw 8.01

Available weld length,

Lw = 4 bf + 2d –2tw
= (4x5.77) + (2x10.3) – (2x0.26)
= 43.16 in

Total design strength =11.29×8.01

= 90.43 kips > 20.28 kips

Determination of plate thickness

S = 0.74 +3/8 =1.115 in
 72

db
b' = S  weld size
4
0.75 1
= 1.115 - - = 0.55
4 4
Fy = 50 ksi ,
A 325  ca = 1.09
b
f 5.77
cb = b

6.77
s

 0.92
1 1
 A 3  b ' 4
m = ca c b  f   
 Aw   db 
1/ 3 1/ 4
 0.44  5.77   0.55 
= 1.09 × 0.92 ×   0.75 
 (10.3  2  0. 44) 0. 26 

= 0.94
 m Tu b '
Me =
4
0.94  90.4  0.55
=
4
= 11.75 kip-in

4.44 M e
tp =
wf y (1  )

4.44  11.75
=
6.77  50(1  0)

= 0.3926 in

Use 0.5 in.

Factored compression force from beam flange

Pbf = Fyc t wc (t fb  6k  2t p  2a)

= 50  0.79 0.44  6  1.85  2  0.5  2  0.375

= 524.96 kips > Compressive force, C = 90.4 kips

 Stiffener is not required.

 73

4.5.4. Calculation of Beam to Column Web Exterior Connection (Type 2)

The corner connection that has maximum moment of C 2 (W12×136) and B 1
(W10×26) at storey 2 (grid line 3D) is designed. Beam to column web exterior
connection at storey 2 is shown in Figure 4.10.

Figure 4.10. Beam to Column Web Exterior Connection

Fy = 50 ksi, Fu = 65 ksi, Fub = 120 ksi, FEXX= 70 ksi

The section properties of W 10 × 26 beam are,
bf = 5.77 in, d = 10.3 in, tf = 0.44 in, tw = 0.26 in, A = 7.61 in2, k = 0.74 in
From analysis data,
Mu = 204.348 kips-in
Vu = 6.064 kips
The section properties of W 12 × 136 column are,
bf = 12.4 in, d = 13.4 in, tf = 1.25 in, tw = 0.79 in, k = 1.85 in
Compute the magnitude of the internal tension and compression force,
Mu 204.348
Tu = = = 20.72 kips
d  t fb 10.3 - 0.44

In this study, ¾ in diameter of high strength bolts are used in connection design.
Ab =  x ( 3 )2 = 0.4418 in2
4 4

Nominal strength of bolt in tension,

 Rnt = (0.75A )Fub

b
= 0.75 x (0.75 x 0.4418) x 120
= 29.82 kips/bolt
 74

Tu 20.72
Required number of bolt = = = 0.695
R nt 29.82

Try 6 Nos (for both sides)

Vu 6.064
Shear stress per bolt, fuv = = = 2.29 ksi
nA b 6 x 0.4418

Check combined shear and tension

For A 325 - N
' = 85 -1.8f  68 ksi
Allowable tensile stress, Fut uv

F'ut = 85 - 1.8 x 2.29 = 80.878 ksi > 68 ksi (control)

Tu 20.72
Tensile stress, fut = =
A 3 x 0.4418

= 15.64 ksi < Allowable = 68 ksi Satisfied

Use c/c spacing = 3.5 in > 3d = 3 × 0.75 = 2.25 in
Used end distance = 1.5 in >1.5d = 1.5 × 0.75 = 1.125in
Weld thickness,
Nominal strength of weld per inch of length,
Tu 20.72
Required,  Rnw = = = 1.84kip/in
2bf  t w (2 x 5.77 ) - 0.26
Design strength,
 Rn =  (0.707 a) (0.6 FEXX)

1.84 = 0.75(0.707a) (0.6×70)

a = 0.083 in
 Use 1/4 in weld thickness.

The required length Lw of fillet weld,

T 20.72
Lw = = = 11.26 in
R nw 1.84

Available weld length,

Lw = 4 bf + 2d –2tw
= (4x5.77) + (2x10.3) – (2x0.26)
= 43.16 in

Total design strength =11.26×1.84

= 20.72 kips > 6.064 kips
 75

Determination of plate thickness

S = 0.74 +1/4 =0.99 in
db
b' = S  weld size
4
0.75 1
= 0.99 - - = 0.55
4 4
Fy = 50 ksi ,
A 325  ca = 1.09
b
f 5.77
cb = b

6.77
s

 0.92
1 1
 A 3  b ' 4
m = ca c b  f   
 Aw   db 
1/ 3 1/ 4
 0.44  5.77   0.55 
= 1.09 × 0.92 ×  
 (10.3  2  0. 44) 0. 26   0.75 


= 0.94
 m Tu b '
Me =
4
0.94  20.72  0.55
=
4
= 2.69 kip-in

4.44 M e
tp =
wf y (1  )

4.44  2.69
=
6.77  50(1  0)

= 0.1878 in

Use 0.5 in.

Factored compression force from beam flange

Pbf = Fyc t wc (t fb  6k  2t p  2a)

= 50  0.79 0.44  6  1.85  2  0.5  2  0.25

= 515.08 kips > Compressive force, C = 20.72 kips

 Stiffener is not required.

 76

4.5.5. Calculation of Beam to Column Flange Interior Connection (Type 3)

The corner connection that has maximum moment of C 1 (W12×152) and
B1 (W10×26) at storey 3 (grid line 3B) is designed. Beam to column flange interior
connection at storey 1 is shown in Figure 4.11.

Figure 4.11. Beam to Column Flange Interior Connection

Fy = 50 ksi, Fu = 65 ksi, Fub = 120 ksi, FEXX= 70 ksi

The section properties of W 10 × 26 beam are,

bf = 5.77 in, d = 10.3 in, tf = 0.44 in , tw = 0.26 in, A = 7.61 in2, k = 0.74 in
From analysis data,
Mu = 891.670 kips-in
Vu = 19.778 kips
The section properties of W 12 × 152 column are,
bf = 12.5 in, d = 13.7 in, tf = 1.4 in, tw = 0.87 in, k = 2 in
Compute the magnitude of the internal tension and compression force,
Mu 891.670
Tu = = = 90.43 kips
d  t fb 10.3 - 0.44

In this study, ¾ in diameter of high strength bolts are used in connection design.
Ab =  x ( 3 )2 = 0.4418 in2
4 4

Nominal strength of bolt in tension,

 Rnt = (0.75A )Fub
b
= 0.75 x (0.75 x 0.4418) x 120
= 29.82 kips/bolt
 77

Tu 90.43
Required number of bolt = = = 3.033
R nt 29.82

Try 8 Nos (for both sides)

Vu 19.778
Shear stress per bolt, fuv = = = 5.6 ksi
nA b 6 x 0.4418
Check combined shear and tension
For A 325 - N
' = 85 -1.8f  68 ksi
Allowable tensile stress, Fut uv

F'ut = 85 - 1.8 x 5.6 = 74.92 ksi > 68 ksi (control)

Tu 90.43
Tensile stress, fut = =
A 3 x 0.4418

= 51.17 ksi < Allowable = 68 ksi Satisfied

Use c/c spacing = 3.5 in > 3d = 3 × 0.75 = 2.25 in
Used end distance = 1.5 in >1.5d = 1.5 × 0.75 = 1.125in
Weld thickness,
Nominal strength of weld per inch of length,
Tu 90.43
Required,  Rnw = = = 8.02 kip/in
2bf  t w (2 x 5.77 ) - 0.26

Design strength,
 Rn =  (0.707 a) (0.6 FEXX)

8.02 = 0.75(0.707a) (0.6×70)

a = 0.35 in
 Use 3/8 in weld thickness.
The required length Lw of fillet weld,
T 90.43
Lw = = = 11.28 in
R nw 8.02

Available weld length,

Lw = 4 bf + 2d –2tw
= (4x5.77) + (2x10.3) – (2x0.26)
= 43.16 in

Total design strength =11.28×8.02

 78

= 90.47 kips > 19.778 kips

Determination of plate thickness
S = 0.74 +3/8 =1.115 in
db
b' = S  weld size
4
0.75 1
= 1.115 - - = 0.55
4 4
Fy = 50 ksi ,
A 325  ca = 1.09
b
f 5.77
cb = b

6.77
s

 0.92
1 1
 A 3  b ' 4
m = ca c b  f   
 Aw   db 
1/ 3 1/ 4
 0.44  5.77   0.55 
m = 1.09 × 0.92 ×    0.75 
 (10.3  2  0.44)0.26 

= 0.94
 m Tu b '
Me =
4
0.94  53.65  0.55
=
4
= 11.75 kip-in

4.44 M e
tp =
wf y (1  )

4.44  11.75
=
6.77  50(1  0)

= 0.3926 in
Use 0.5 in.

Factored compression force from beam flange

Pbf = Fyc t wc (t fb  6k  2t p  2a)

= 50  0.87 0.44  6  2  2  0.5  2  0.375

= 617.27 kips > Compressive force, C = 90.43 kips

 79

 Stiffener is not required.

4.5.6. Calculation of Beam to Column Web Interior Connection (Type 3)

The corner connection that has maximum moment of C 1 (W12×152) and B 1
(W10×26) at storey 3 (grid line 3B) is designed. Beam to column web interior
connection at storey 1 is shown in Figure 4.12.

Figure 4.12. Beam to Column Web Interior Connection

Fy = 50 ksi, Fu = 65 ksi, Fub = 120 ksi, FEXX= 70 ksi

The section properties of W 10 × 26 beam are,
bf = 5.77 in, d = 10.3 in, tf = 0.44 in , tw = 0.26 in, A = 7.61 in2, k = 0.74 in
From analysis data,
Mu = 738.874 kips-in
Vu = 32.763 kips
The section properties of W 12 × 152 column are,
bf = 12.5 in, d = 13.7 in, tf = 1.4 in, tw = 0.87 in, k = 2 in
Compute the magnitude of the internal tension and compression force,
Mu 738.874
Tu = = = 74.94 kips
d  t fb 10.3 - 0.44
In this study, ¾ in diameter of high strength bolts are used in connection design.
Ab =  x ( 3 )2 = 0.4418 in2
4 4

Nominal strength of bolt in tension,

 Rnt = (0.75A )Fub

b
 80

= 0.75 x (0.75 x 0.4418) x 120

= 29.82 kips/bolt

Tu 74.94
Required number of bolt = = = 2.513
R nt 29.82

Try 6 Nos (for both sides)

Vu 32.763
Shear stress per bolt, fuv = = = 12.36 ksi
nA b 6 x 0.4418
Check combined shear and tension
For A 325 - N
' = 85 -1.8f  68 ksi
Allowable tensile stress, Fut uv

F'ut = 85 - 1.8 x 12.36 = 62.752 (control)

Tu 74.94
Tensile stress, fut = =
A 3 x 0.4418

= 56.52 ksi < Allowable = 62.752 ksi Satisfied

Use c/c spacing = 3.5 in > 3d = 3 × 0.75 = 2.25 in
Used end distance = 1.5 in >1.5d = 1.5 × 0.75 = 1.125in
Weld thickness,
Nominal strength of weld per inch of length,
Tu 74.94
Required,  Rnw = = = 6.64 kip/in
2bf  t w (2 x 5.77 ) - 0.26
Design strength,
 Rn =  (0.707 a) (0.6 FEXX)

6.64 = 0.75(0.707a) (0.6×70)

a = 0.298 in
 Use 5/16 in weld thickness.
The required length Lw of fillet weld,
T 74.94
Lw = = = 11.29 in
R nw 6.64

Available weld length,

Lw = 4 bf + 2d –2tw
= (4x5.77) + (2x10.3) – (2x0.26)
= 43.16 in
 81

Total design strength =11.29×6.64

= 74.97 kips > 32.763 kips

Determination of plate thickness

S = 0.74 +5/16 = 1.0525 in
db
b' = S  weld size
4
0.75 1
= 1.0525 - - = 0.55
4 4
Fy = 50 ksi ,
A 325  ca = 1.09
b
f 5.77
cb = b

6.77
s

 0.92
1 1
 A 3  b ' 4
m = ca c b  f   
 Aw   db 
1/ 3 1/ 4
 0.44  5.77   0.55 
= 1.09 × 0.92 ×  
 (10.3  2  0. 44) 0. 26   0.75 


= 0.94
 m Tu b '
Me =
4
0.94  74.94  0.55
=
4
= 9.69 kip-in

4.44 M e
tp =
wf y (1  )

4.44  9.69
=
6.77  50(1  0)

= 0.3565 in
Use 0.5 in.

Factored compression force from beam flange

Pbf = Fyc t wc (t fb  6k  2t p  2a)

= 50  0.87 0.36  6  2.125  2  0.5  2  0.3125

 Table 4.6. Design Results of Beam to Column Flange Connection

Table 4.8. Design results of beam to column web connections

Moment Distance End Design

Sections Required
(kips) Shear (in) and Diameter Center to Welded
Strength
Length (in)
Table 4.8. Design results of beam to column web connections
Column Beam

W12x136 W10x26
 Stiffener is not required.

13.22 648.899 3/4 6 3.5 1.5 11.29 65.82

W12x136 W10x26 20.28 891.314 3/4 8 3.5 1.5 11.29 90.43

W12x152 W10x26
mentioned in Table 4.6. and Table 4.7.

19.778 891.670 3/4 8 3.5 1.5 11.28 90.47

W12x120
82

Table 4.8. Design results W10x26

of beam to column connections
13.182web645.645 3/4 6 3.5 1.5 11.27 65.48
W12x120 W10x26 20.179 878.918 3/4 6 3.5 1.5 11.28 89.11
W12x136 W10x26 19.599 877.667 3/4 6 3.5 1.5 11.28 52.34
= 611.83 kips > Compressive force, C = 74.94 kips

beam to column web connection for storey level 1-4 and storey level 5-8 are
Design results of beam to column flange connection and Design results of
 Table 4.7. Design Results of Beam to Column Web Connection

Table 4.8. Design results of beam to column web connections

Table 4.7. Design results of beam to column flange connection
Table 4.8. Design results of beam to column web connections

Moment Distance End Design

Sections Required
(kips) Shear (in) and Diameter Center to Storey
Welded
Strength Type Joint Location
Level
Length (in)
Column Beam

W12x136 W10x26 6.576 253.102 3/4 6 3.5 1.5 11.29 25.67 Corner 2D

W12x136 W10x26 1- 4
6.064 204.348 3/4 6 3.5 1.5 11.26 20.72 Exterior 3D
W12x152 W10x26 32.763 738.874 3/4 6 3.5 1.5 11.29 74.97 Interior 3B
W12x120 W10x26
83

Table 4.8. Design results of beam to6.384 web connections

column 238.474 3/4 6 3.5 1.5 11.3 24.18 Corner 2D
W12x120 W10x26 5.138 154.778 3/4 6 3.5 1.5 11.29 15.69 5-8 Exterior 3D
W12x136 W10x26 21.829 483.254 3/4 6 3.5 1.5 11.42 49.02 Interior 3B
 Table 4.7. Design results of beam to column flange connection
84

Location

2D

3D

2D
3D
3B

3B
Type Joint

Exterior

Exterior
Interior

Interior
Corner

Corner
Storey
Level

1- 4

5-8
4.5.7. Calculation of Beam to Girder Connection
When beams frame transversely to other beams or girders, they may be
Table 4.8. Design results of beam to column web connections

attached to other or both sides of the girder web using framed beam connections. For
beam to girder connections, the principle objective is to provide a mean of allowing
the tensile forces developed in one beam flange to be carried across to the adjacent
framing opposite the girder web. In this study, beam to girder connection of B 1
(between grid line 3A and 3B) and B2 (between grid line A and B and between grid
line 2and 3) at story 7 is designed.

W 8 × 10 W 8 × 10

W10×26

Figure 4.13. Beam to girder Connection

The section properties of W 8 × 10 beam are,

tw = 0.17 in, tf = 0.205 in, bf = 3.94 in, d = 7.89 in,
From analysis data,
Vu = 4.844 kips
Use 3/4in  bolts as a bearing type connection (A 325-N) having threads in the
shear plane.
Rn (bearing) = 2.4 Fu d tw
= 0.75 × 2.4× 65 × 0.75 × 0.17 = 14.92 kips/bolt
b
Rn(double shear) =  (0.45 Fu ) m Ab
 85

= 0.65 × 0.45× 120 ×2× 0.4418 = 31.01 kips/bolt

4.844
Number of bolts =
14.92
= 0.325 Take 2 bolts.
Minimum c/c spacing = 3d = 3 × 0.75 = 2.25 in (use 2.5 in)
Minimum end distance = 1.5d = 1.5 × 0.75 = 1.125in (use 1.25in)
Angle thickness;
P 4.844
t ≥ = = 0.02
 Fu l 4x0.75x65x1.25

Try 2-L 4" × 3" × 1/4".

When the material being attached by bolts is ¼ in or less, a tearing failure
limit state, knowns as block shear should be checked.
Check Block Shear
Nominal reaction, Pn = 0.6 Fy Avg + Fu Ant (Shear yielding, Tension fracture)
Nominal reaction, Pn = 0.6 Fu Ans + Fy Atg (Shear fracture, Tension yielding)

L- 4 × 3 × 1/4
3.75 in
a 1.25 in
b
c
Figure 4.14. Block Shear Rapture for Beam to Girder Connection
Avg = 3.75× 0.25 = 0.9375 in2
 3 1  1
Ant = 1.25× 0.25 – 0.5    = 0.211 in2
4 16  4

 3 1  1
Ans = 3.75×0.25-1.5    = 0.633 in2
4 16  4

Atg = 1.25×0.25 = 0.3125 in2

Pn = 0.6 × 50 × 0.9375 + 65 × 0.211 = 41.855 kips

Pn = 0.6 × 65 × 0.633 + 50 × 0.3125 = 40.312 kips

Pn = 0.75 × 40.312 = 30.23 kips > Pu = 4.844 kips  Satisfied

Use 2-L 4 × 3 × 1/4 in for beam to girder connection.

4.5.8. Design of Column Splices

Column splices are classified as two types according to column type C 1, C2.
Column splices (Type 1) and (Type 2), are shown in Fig. 4.15. and Fig. 4.16. Both
 86

methods of bolted connection and welded connection are used in column splice
connection. For bolted connection, A325-N bolts is used for bearing type connection.
For welded connection, E 70 electrode are used with SMAW (Shield Metal Arc
Welding) process.

4.5.8.1. Design calculation for column splice (Type 1)

Column type C1 that have maximum load at storey 5 (grid line 2B) is designed.
5/16"
Vu = 9.553 kips E70
W12´136

4-¾ "A 325 bolts 1.5"

A A 7.9"

W12´152 1.5"

Section A-A

Figure 4.15. Column splices

Shear force = 9.553 kips (From analysis data)

Use 3/4 in  A-325 bolts as a bearing type connection (A 325-N) having
threads in shear plane.
The design strength in single shear (m=1) is

= (0.45Fu )mA b
b
Rn

= 0.65×0.45×120×1×0.442 = 15.51 kips/bolt

9.553
The number of bolts = = 0.31
15.51x2
Use 2 bolts on each side of the connection.
Use c/c spacing = 3.5 in > 3d = 2.25 in
Use end distance = 1.5 in > 1.5 d = 1.125 in
Total depth,d = 13.4 in
Width of flange, bf = 12.4 in
Thickness of web, tw = 0.79 in
Try plate thickness = 1 in
 87

Use 5/16 in fillet weld produced by SMAW and E 70 electrodes having

minimum tensile strength FEXX of 70 ksi.
Effective throat dimension, te = 0.707 a = 0.707×5/16 = 0.22 in
The design strength per unit length of a fillet weld,
Rnw = te (0.6FEXX)
= 0.75×0.22×0.6×70 = 6.93 kip/in
The total length Lw of fillet weld,
Lw = 4 bf + 2d –2tw
= 4 × 12.4 + 2 × 13.4– 2 × 0.79 = 74.82 in
Total design strength = 6.93 ×74.82
= 518.5 kips > 9.553 kips  Satisfied

4.5.8.2. Design calculation for column splices (Type 2)

Column type C2 that have maximum load at storey 5 (grid line 3A) is designed.
5/16"
Vu = 8.808 kips E70
W12´120

4-¾ "A 325 bolts 1.5"

A A 7.9"

W12´136 1.5"

Section A-A

Figure 4.16. Column splices

Shear force = 8.808 kips (From analysis data)

1.5"
Use 3/4"  A-325 bolts as a bearing type connection (A 325-N) having threads
3.5"
in shear plane.
The design strength in single shear (m=1) is 1.5"

Rn = (0.45Fu )mA b

b

= 0.65×0.45×120×1×0.442 = 15.51 kips/bolt

8.808
The number of bolts = = 0.28
15.51  2
 88

Use 2 bolts on each side of the connection.

Use c/c spacing = 3.5 in > 3d = 2.25 in
Use edge distance = 1.5 in > 1.5 d = 1.125 in
Total depth,d = 13.1 in
Wiidth of flange, bf = 12.3 in
Thickness of web, tw = 0.71 in
Try plate thickness = 1 in
Use 5/16 in fillet weld produced by SMAW and E 70 electrodes having
minimum tensile strength FEXX of 70 ksi.
Effective throat dimension,
te = 0.707 a = 0.707×5/16 = 0.22 in
The design strength per unit length of a fillet weld,
Rnw = te (0.6FEXX)
= 0.75×0.22×0.6×70
= 6.93 kip/in
The total length Lw of fillet weld,
Lw = 4 bf + 2d –2tw
= 4 × 12.3+ 2 × 13.1 – 2 × 0.71 = 73.98 in
Total design strength = 6.93×73.98
= 512.68 kips > 8.808 kips  Satisfied

4.6. Design Calculation for Base Plates

Column base plates distribute the concentrated loads acting in the elements of
columns to the supporting medium. The heavy loads must be distributed to prevent
crushing of the concrete support. Base plate performs a complete analysis and design
of centered loaded rectangular base plates with vertical compression load and bi-axial
bending moments. For control, thickness of base plate is chosen greater of t 1 and t2
based on greater of n and m. In this study, base plates are classified as two types
according to column types C1 (Type 1) and C2 (Type 2).

4.6.1. Design of Base Plate for Type 1 Column

Base plates are designed by using maximum load acting on the each of column
types.The maximum load for type 1 column is 510.547 kips.
W 12×152column of A572 Grade 50 steel
 89

bf = 12.5 in
d = 13.7 in
f c' = 3 ksi, Fy = 50 ksi

From analysis data, compressive load, Pu = 510.547 kips

(a) Determine required plate area
Assume concrete pedestal dimension 2 in larger than the base plate in each direction.
Required plate area,
Pu
A1 =
(0.85f c' )

510.547
A1 = = 333.69 in2
0.6  0.85  3

0.8 bf = 0.8 × 12.5 = 10 in

0.95 d = 0.95 × 13.7 = 13.02 in
Try B = 18 in, N = 19 in  A1 = 342 in2

Concrete pedestal dimension  B+2 = 20 in

N+2 = 21 in
A2 = 20 × 21 = 420 in2

A2 420
= = 1.11 < 2
A1 342

preliminary required A1
Required A1 =
1.11
333.69
= = 300.62 in2
1.11

Try B = 17 in, N = 18 in  A1 = 306 in2

Concrete pedestal dimension  B+2 = 19 in
N+2 = 20 in
A2 = 19 × 20 = 380 in2

A2 380
= = 1.11
A1 306

Check bearing strength

 90

A2
Pn = (0.85) f c' A1
A1

= 0.6 × 0.85 × 3 × 306 × 1.11

= 519.68 kips > Pu = 510.547 kips \Satisfied
Use B = 17 in, N = 18 in
(b) Design of plate thickness by using cantilever method
determine the plate thickness
m = 0.5 (N – 0.95d)
= 0.5 (18 – 0.95 × 13.7)
= 2.49 in
n = 0.5 (B – 0.8bf)
= 0.5(17 – 0.8 × 12.5)
= 3.5 in
Pu
Required t = 1.5  n
BNFy

510.547
= 1.5 × 3.5 x
17  18  50
= 0.95 in Use 1 in.
Use 17 in × 18 in × 1 in base plate.

4.6.2. Design of Base Plate for Type 2 Column

Base plates are designed by using maximum load acting on the each of column
types.The maximum load for type 1 column is 236.313 kips.
W 12×136 column of A572 Grade 50 steel
bf = 12.4 in
d = 13.4 in
f c' = 3 ksi, Fy = 50 ksi

From analysis data, compressive load, Pu = 236.313 kips

(a) Determine required plate area
Assume concrete pedestal dimension 2 in larger than the base plate in each direction.
Required plate area,
Pu
A1 =
(0.85f c' )
 91

236.313
=
0.6  0.85  3

= 154.45 in2

0.8 bf = 0.8 × 12.4 = 9.92 in

0.95 d = 0.95 × 13.4 = 12.73 in
Try B = 13 in, N = 14 in  A1 = 182 in2
Concrete pedestal dimension  B+2 = 15 in
N+2 = 16 in
A2 = 15 × 16 = 240 in2

A2 240
= = 1.15 < 2
A1 182

preliminary required A1
Required A1 =
1.15
154.45
= = 134.3 in2
1.15
Try B = 15 in, N = 16 in  A1 = 240 in2
Concrete pedestal dimension  B+2 = 17 in
N+2 = 18 in
A2 = 17 × 18 = 306 in2

A2 306
= = 1.13
A1 240

Check bearing strength

A2
Pn '
= (0.85) f c A1
A1

= 0.6 × 0.85 × 3 × 240 × 1.13

= 414.936 kips > Pu = 236.313 kips \Satisfied
Use B = 15 in, N = 16 in
(b) Design of plate thickness by using cantilever method
determine the plate thickness
m = 0.5 (N – 0.95 d)
= 0.5 (16 – 0.95 × 13.4) = 1.64 in
n = 0.5 (B – 0.8bf)
 92

= 0.5 (15– 0.8 × 12.4) = 2.54 in

Pu
Required t = 1.5  n
BNFy

236.313
= 1.5 x 2.54x
15  16  50
= 0.53 in Use 0.75 in.
Use 15 in × 16 in × 0.75 in base plate.

4.7. Design Calculation for Short Column

Short columns, for which the strength is governed by the strength of the
materials and the geometry of the cross section. Although slender columns are more
common now because of the wider use of high-strength materials and improved
methods of dimensioning members, it is still true that most columns in ordinary
practice can be considered short columns. ACI Code limits the maximum design
strength to 0.8  Po for tied columns (with  =0.7) and to 0.85  Po for spirally
reinforced columns (with  =0.75) where Po is the nominal strength of axially loaded
column with zero eccentrically. In this study, there are two types of short columns
according to base plate sizes.
(a) Type 1 column
17 in × 18 in base plate, fy = 40 ksi, f c = 3 ksi
Size of column based on base plate = 20 in × 20 in
Use 20 in × 20 in column.
The axial load of 510.547 kips and bending moment of 762.015 kip-in are
obtained from analysis results.

20"

20"

2.5" 15" 2.5"

Figure 4.17. Short Column Detail for Type 1

 93

The area of steel in the column is provided 2% of the area of column according to
ACI code.
As = 0.02 × 20 × 20 = 8 in2
Use 12 # 8®As = 9.42 in2
15
Centre to centre spacing of steel = = 5 in
3
4 # 8®As = 3.14 in2

2 # 8®As = 1.57 in2

Pn = 0.7 × (0.85 f c Ac + Asfy)

= 0.7 × (0.85 × 3 × 20 × 20 + 9.42 × 40)

= 977.76 kips> Pu = 510.547 kips Satisfied

The bending moment capacity about origin of the column,

M n = 0.9 Asfyl
= 0.9 × 40 (3.14 × 7.5 + 1.57 × 2.5)
= 989.1 kip-in > Mu = 762.015 kip-in Satisfied
(b) Type 2 column
15 in × 16 in base plate, fy = 40 ksi, f c = 3 ksi
Size of column based on base plate = 18 in × 18 in
Use 18 in × 18 in column.
The axial load of 236.313 kips and bending moment of 577.612 kip-in are
based on analysis results.
18"

18"

2.5" 13" 2.5"

Figure 4.18. Short Column Detail for Type 2

The area of steel in the column is provided 2% of the area of column according to
ACI code.
As = 0.02 × 18 × 18 = 6.48 in2
 94

Use 12 # 7®As = 7.22 in2

13
Centre to centre spacing of steel = =4.33 in
3
4 # 7®As = 2.41 in2
2 # 7®As = 1.2 in2

Pn = 0.7 × (0.85 f c Ac + Asfy)

= 0.7 × (0.85 × 3 × 18 × 18 + 7.22 × 40)

= 780.5 kips> Pu = 236.313 kips Satisfied
The bending moment capacity about origin of the column,
M n = 0.9 Asfyl
= 0.9 × 40 (2.41 × 6.5 + 1.2 × 2.167)
= 657.54 kip-in > Mu = 577.615 kip-in Satisfied