STRUCTURAL DESIGN AND DETAILED DRAWING OF A STEEL INDUSTRIAL BUILDING
NAME: ANUJ KUMAR MONDAL
ROLL NO.: 34901323049
DesiGn oF PURLin
STRUCTURAL DESIGN AND DETAILED DRAWING OF A STEEL INDUSTRIAL BUILDING
NAME: ANUJ KUMAR MONDAL
ROLL NO.: 34901323049
➢ DESIGN OF PURLIN :
Data :
Spacing of the Truss (S) = 4.5 m
Roof pitch angle (θ) = 20°
Spacing of Purlin = 2.26 m
Weight of 1mm thick Galvalume sheets = 140 N/m2 = 0.14 KN/m2
❖ Load Calculation :
Dead Load (WD): Weight of sheeting + Self-weight of Purlin
=(0.14×2.26 + 0.12)
= 0.436 KN/m
Live Load intensity on roof = 0.55 KN/m2
Live Load (WL) = 0.55×2.26×(cos 20°) =1.168 KN/m
Total Factored Load = {1.5×(0.436+1.168)} = 2.406 KN/m
Component of load parallel to Sheeting = Wsin𝜃
= 2.406×𝑠𝑖𝑛20°
= 0.82 KN/m
Component of load perpendicular to Sheeting = Wcos𝜃
= 2.406×𝑐𝑜𝑠20°
= 2.26 KN/m
Maximum Nodal Suction Load 17.95
Wind Load (Suction) (WS) = = = 3.98 KN/m
L 4.5
Factored Wind Load = 1.5×3.98 = 5.97 KN/m
Total load perpendicular to Sheeting = (2.26+5.97)=8.23 KN
Total load parallel to Sheeting = 0.82 KN
❖ Determine Moment
𝑊𝐿2 5.97×4.52
Mzz = = = 15.11 KN
8 8
𝑊𝐿2 0.82×4.52
Myy = = = 2.07 KN
8 8
STRUCTURAL DESIGN AND DETAILED DRAWING OF A STEEL INDUSTRIAL BUILDING
NAME: ANUJ KUMAR MONDAL
ROLL NO.: 34901323049
❖ Calculate Zpz required :
Adopt , b = 75 mm ; d = 150 mm
𝑀𝑧𝑧 ×ϒ𝑚0 𝑑 𝑀𝑧𝑧
Zpz required = + 2.5× × = 107.884 cm3
𝑓𝑦 𝑏 ϒ𝑚0
❖ Selection of Section :
Considering ISMC 150 @16.4 Kg/m
Here, h= D = 150 mm
B = bf = 75 mm
tf = 9 mm, tw = 5.4 mm [From IS 800:2007, Pg- 140, Table – 46]
Shape Factor = 1.153
Zyy = 19.4 cm3 , Zzz = 103.9 cm3 [Sp -6, Pg- 6-7, Table- II]
3
Ixx = 779.4 cm
R = r1 = 10 mm, h1 = 106.7 mm
❖ Classify the Section :
250 250
ε =√ =√ =1
𝑓𝑦 250
𝑑 126.7
= = 23.46 (<84ε) [Here, d =h1+r1 = 106.7+(2×10) = 126.7]
𝑡𝑤 5.4
𝑏 75
= = 8.33 (<9.4ε)
𝑡𝑓 9
[Hence, the section is Non-Slender (Plastic) [Table 2, Page 18, IS 800:2007]
Find Mdz & Mdy
𝑍𝑝𝑧 ×𝑓𝑦
Mdz = = 27.23 KN.m Mzz < Mdz [Hence, OK]
ϒ𝑚0
Here, Zpy = Zzz × Shape Factor = (103.9 × 103) × 1.153 = 119.79×103
𝑍𝑝𝑦 ×𝑓𝑦
Mdy = = 5.08 KN.m Myy < Mdy [Hence, OK]
ϒ𝑚0
Here, Zpy = Zyy × Shape Factor = ( 19.4 × 103) × 1.153 = 22.368 × 103
STRUCTURAL DESIGN AND DETAILED DRAWING OF A STEEL INDUSTRIAL BUILDING
NAME: ANUJ KUMAR MONDAL
ROLL NO.: 34901323049
Apply Cheaks
1. Biaxial Bending Cheak :
𝑀 𝑀𝑦𝑦
(𝑀𝑧𝑧 ) + (𝑀 ) ≤ 1
𝑑𝑧 𝑑𝑦
15.11 2.07
+ 5.08 = 0.96 (Which<1) [As per Cl. 9.3.1.1 of IS 800:2007]
27.23
……..Hence OK
2. Cheak For defection :
𝑆𝑝𝑎𝑛 4500
Max Permissible Deflection = = = 25 mm
180 180
[IS 800:2007, Pg- 31, Table- 6, For purlin considering supporting as Brittle cladding]
5𝑊𝐿4
Deflection due to load = 384𝐸𝐼 = 7.91 mm (<25 mm)
……..Hence OK
3. Shear Capacity Check :
𝑓𝑦 2𝑏𝑓 𝑡𝑓 250 2×75×9
Shear Capacity in the Z direction (Vdz) = × = × = 177.14 KN
√3 ϒ𝑚0 √3 1.10
𝑓𝑦 ℎ𝑡𝑤 250 150×5.4
Shear Capacity in the Y direction (Vdy) = × = × = 106.28 KN
√3 ϒ𝑚0 √3 1.10
𝑤𝑢𝑧 ×L 0.82×4.5
Vuz = = = 1.84 KN <0.6Vdz
2 2
𝑤𝑢𝑦 ×L 8.23×4.5
Vuy = = = 18.51 KN <0.6Vdy
2 2
……….Hence OK, It is safe & in low in shear.
4. Web Buckling Cheak :
ℎ 150
b1 = (b-tw) = (75 – 5.4) = 69.6 mm ; n1 = 2 = = 75 mm
2
∴ Ab = (𝑏1+ n1) × tw = (69.6 + 75) × 5.4 = 780.84 mm2
3
𝑏1 ×𝑡𝑤
I= = 913.29 mm4
12
𝐼 913.29
rmin = √𝐴 = √780.84 = 1.08 mm
𝑏
0.7×L 0.7×4500
Web slenderness, 𝜆 = = = 51.85
r 61.1
{From table 9(c) of IS 800:2007, For fy = 250 Mpa; fcd = 180.22}
𝑓𝑐𝑑 ×A 𝑏
∴ buckling resistance = = 140.72 KN (<Vdy)
103
……..Hence OK
5. Web Bearing Cheak :
b1 = b-tw = 75 – 5.4 = 69.6 mm ; n2 = 2.5( R+tf ) = 2.5 (10+9) = 47.5 mm
1
Fw = (b1 +n2) × tw × fy × = 158.08 KN (<Vdy)
ϒ𝑚0
……..Hence OK
