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
2007-2008
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
2007-2008
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
2007-2008
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
2007-2008
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
2007-2008
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
2007-2008
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
2007-2008
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
2007-2008
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
2007-2008
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
2007-2008
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
2007-2008
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
2007-2008
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
2007-2008
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
2007-2008
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
2007-2008
