Week 11 Assignment 11 Solution
Q.1
Full Marks 20
Determine the plastic section modulus of ISHB 400 @ 82.2 kg/m about
the strong axis and weak axis. (Neglect the fillets)
4
1601.5 cm3, 407.4 cm3
(d)
𝑡𝑤 (𝐷−2𝑡 )2
𝑓
Sol: 𝑍𝑝𝑧 = 𝑏𝑓 𝑡𝑓 (𝐷 − 𝑡𝑓 ) +
4
Zpz = 250×12.7×(400 – 12.7) + 10.6×(400-2×12.7)2/4 = 1229.67×103 + 371.86×103
= 1601.53×103 mm3 =1601.5 cm3
𝑍𝑝𝑦 =
=
𝑡𝑓 𝑏𝑓2
+
2
(ℎ−2𝑡𝑓 )𝑡𝑤
2
12.7×2502
2
4
(400−2×12.7)×10.62
+
4
= 407.4 cm3
Q.2
When electric operated crane carry up to 50t then deflection of the
gantry girder should not exceed
1
(c)
L/750
Sol: Lec 56, Slide 6.
Q.3
Gantry girders are usually designed
1
(d)
both (b) and (c)
Sol: Lec 56.
Q.4
Which of the following loads are considered in designing a gantry girder
in an industrial building?
1
(1)
Gravity loads
(2)
Lateral loads
(3)
Longitudinal loads
(4)
Wind loads
Select the correct answer using the codes given below:
(b)
1,2 and 3
Q.5
What is the bending stress reduction factor (XLT) to account for lateral
torsion buckling for a rolled steel section if the slenderness ratio, λLT
=1.2?
2
(a)
0.53
Sol: ∅𝐿𝑇 = 0.5[1 + 𝛼𝐿𝑇 (𝜆𝐿𝑇 − 0.2) + 𝜆2𝐿𝑇 ]
∅𝐿𝑇 = 0.5[1 + 0.21(1.2 − 0.2) + 1.22 ] = 1.325
1
Χ𝐿𝑇 =
2 −𝜆2
∅𝐿𝑇 +√𝜑𝐿𝑇
𝐿𝑇
𝑋𝐿𝑇 =
1
1.325+√1.3252 −1.22
= 0.53
Q.6
A proposed simply supported beam is built into a concrete wall. It
supports a dead load of 20 kN/m and a live load 15 kN/m. The length of
the beam is 6 m. But the beam depth is restricted to 300 mm, so ISHB 250
@ 54.7 kg/m has been chosen as beam section. Calculate the minimum
width of the plate that has to be connected to both flanges if the thickness
of the plates is 10 mm. (Neglect Deflection check)
5
(c)
130 mm
Sol: Dead load = 1.5 × 20 = 30 kN/m
Live load = 1.5 × 15 = 22.5 kN/m
Total load = (30 + 22.5) = 52.5 kN/m
𝑤𝑙 2
52.5×62
BM =
=
= 236.25 kN-m
8
8
SF = w×l/2 = 52.5×6/2 = 157.5 kN
𝑍𝑝,𝑟𝑒𝑞𝑑 =
𝑀×𝛾𝑚0
𝑓𝑦
=
236.25×106 ×1.1
250
= 1039.5×103 mm3
Properties of ISHB 250 @ 54.7 kg/m
Zpz = 708.43 cm3
Additional plastic section modulus to be provided by the plate = (1039.5×10 3 –
708.43×103) = 331 × 103 mm3
Thickness of the plate is 10 mm
Thus, the total depth of the beam = 250+20 = 270 mm.
Distance between the c/c of the plates = 260 mm.
So, required area of plate = 331 × 103 /260 = 1273 mm2
Thus the width of plate = 1273/10 = 127.3 mm  130 mm
Q.7
Purlins are provided in industrial buildings, over roof trusses to carry
dead loads, live loads and wind loads. As per IS code, what are they
assumed to be?
1
(c)
Continuous
Q.8
What is the maximum bending moment in purlin (span of purlin = L, Total
load = W)?
1
(a)
WL/10
Q.9
An industrial building situated in the outskirt of a city, is to support a
galvanized iron sheet roof for the following data:
Slope of truss = 30°
Spacing of truss c/c = 6.0 m
spacing of purlins c/c = 2 m
wind speed = 45 m/s
Weight of galvanized sheets = 140 N/m2
Dead load of purlin = 100 N/m
Calculate the magnitude of the factored load normal to the roof. 4
(a)
4.14 kN/m
Sol: For steel of grade Fe 410: fy = 250 MPa
Weight of galvanized corrugated iron sheets = 140 × 2 = 280 N/m
Dead load of purlin = 100 N/m
Total dead load = 280 + 100 = 380 N/m.
The dead load acts vertically downwards.
The component of dead load normal to roof = 380cos30° = 329.1 N/m
Wind pressure, 𝑝𝑧=0.6𝑉𝑧2 = 0.6×452=1215 N/m2
Wind load is assumed to act normal to the roof.
Wind load = 1215×2×1 = 2430 N/m
Total load on purlin normal to roof = 2430+329.1 = 2759.1 N/m
Factored load normal to roof, P = 1.5×2759.1 = 4138.65 N/m  4.14 kN/m