# Design of Steel Structures Questions and Answers – Design of Compression Members – II

This set of Design of Steel Structures Assessment Questions and Answers focuses on “Design of Compression Members – II”.

1. The design compressive strength of member is given by
a) Aefcd
b) Ae /fcd
c) fcd
d) 0.5Aefcd

Explanation: The design compressive strength of member is given by Pd = Aefcd, where Ae is effective sectional area, fcd is design compressive stress.

2. The design compressive stress, fcd of column is given by
a) [fy / γm0]/ [φ – (φ22)2].
b) [fy / γm0] / [φ + (φ22)].
c) [fy / γm0]/[φ – (φ22)0.5].
d) [fy / γm0] / [φ + (φ22)0.5].

Explanation: The design compressive stress, fcd of column is given by fcd = [fy / γm0] / [φ + (φ22)0.5], where fy is yield stress of material, φ is dependent on imperfection factor, λ is non dimensional effective slenderness ratio.

3. What is the value of imperfection factor for buckling class a?
a) 0.34
b) 0.75
c) 0.21
d) 0.5

Explanation: The value of imperfection factor, α for buckling class a is 0.21. The imperfection factor considers all the relevant defects in real structure when considering buckling, geometric imperfections, eccentricity of applied loads and residual stresses.

4. If imperfection factor α = 0.49, then what is the buckling class?
a) a
b) c
c) b
d) g

Explanation: For buckling class c, the value of imperfection factor is 0.49. The imperfection factor takes into account all the relevant defects in real structure when considering buckling, geometric imperfections, eccentricity of applied loads and residual stresses.

5. The value of φ in the equation of design compressive strength is given by
a) φ = 0.5[1-α(λ-0.2)+λ2].
b) φ = 0.5[1-α(λ-0.2)-+λ2].
c) φ = 0.5[1+α(λ+0.2)-λ2].
d) φ = 0.5[1+α(λ-0.2)+λ2].

Explanation: The value of φ in the equation of design compressive strength is given by φ = 0.5[1+α(λ-0.2)+λ2], where α is imperfection factor(depends on buckling class) and λ is non-dimensional effective slenderness ratio.
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6. Euler buckling stress fcc is given by
a) (π2E)/(KL/r)2
b) (π2E KL/r)2
c) (π2E)/(KL/r)
d) (π2E)/(KLr)2

Explanation: Euler buckling stress fcc is given by fcc = (π2E)/(KL/r)2, where E is modulus of elasticity of material and KL/r is effective slenderness ratio i.e. ratio of effective length, KL to appropriate radius of gyration, r.

7. What is the value of non dimensional slenderness ratio λ in the equation of design compressive strength?
a) (fy /fcc)
b) √(fy fcc)
c) √(fy /fcc)
d) (fy fcc)

Explanation: The value of non dimensional slenderness ratio λ in the equation of design compressive strength is given by λ = √(fy /fcc) , where fy is yield stress of material and fcc = (π2E)/(KL/r)2, where E is modulus of elasticity of material and KL/r is effective slenderness ratio i.e. ratio of effective length.

8. The design compressive strength in terms of stress reduction factor is given by
a) Xfy
b) Xfy / γm0
c) X /fy γm0
d) Xfy γm0

Explanation: The design compressive strength in terms of stress reduction factor is given by fcd = Xfy / γm0 , where X = stress reduction factor for different buckling class, slenderness ratio and yield stress = 1/ [φ + (φ22)0.5], fy is yield stress of material and γm0 is partial safety factor for material strength.

9. The value of design compressive strength is limited to
a) fy + γm0
b) fy
c) fy γm0
d) fy / γm0

Explanation: The value of design compressive strength is given by fcd = [fy / γm0] / [φ + (φ22)0.5] ≤ fy / γm0 i.e. fcd should be less than or equal to fy / γm0.

10. The compressive strength for ISMB 400 used as a column for length 5m with both ends hinged is
a) 275 kN
b) 375.4 kN
c) 453 kN
d) 382 kN

Explanation: K = 1 for both ends hinged, KL = 1×5000 = 5000, r = 28.2mm (from steel table), Ae = 7846 mm2(from steel table)
KL/r = 5000/28.2 = 177.3
h/bf = 400/140 = 2.82, t = 16mm Therefore, buckling class = b
From table in IS code, fcd = 47.85MPa
Pd = Ae fcd = 7846 x 47.85 = 375.43 kN.

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