250+ TOP MCQs on Design of Compression Members – II and Answers

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

1. The design compressive strength of member is given by
a) A_ef_cd
b) A_e /f_cd
c) f_cd
d) 0.5A_ef_cd
Answer: a
Clarification: The design compressive strength of member is given by P_d = A_ef_cd, where A_e is effective sectional area, f_cd is design compressive stress.

2. The design compressive stress, f_cd of column is given by
a) [f_y / γ_m0]/ [φ – (φ²-λ²)²].
b) [f_y / γ_m0] / [φ + (φ²-λ²)].
c) [f_y / γ_m0]/[φ – (φ²-λ²)^0.5].
d) [f_y / γ_m0] / [φ + (φ²-λ²)^0.5].
Answer: d
Clarification: The design compressive stress, f_cd of column is given by f_cd = [f_y / γ_m0] / [φ + (φ²-λ²)^0.5], where f_y 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
Answer: c
Clarification: 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
Answer: b
Clarification: 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].
Answer: d
Clarification: 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.

6. Euler buckling stress f_cc is given by
a) (π²E)/(KL/r)²
b) (π²E KL/r)²
c) (π²E)/(KL/r)
d) (π²E)/(KLr)²
Answer: a
Clarification: Euler buckling stress f_cc is given by f_cc = (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) (f_y /f_cc)
b) √(f_y f_cc)
c) √(f_y /f_cc)
d) (f_y f_cc)
Answer: c
Clarification: The value of non dimensional slenderness ratio λ in the equation of design compressive strength is given by λ = √(f_y /f_cc) , where f_y is yield stress of material and f_cc = (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) Xf_y
b) Xf_y / γ_m0
c) X /f_y γ_m0
d) Xf_y γ_m0
Answer: b
Clarification: The design compressive strength in terms of stress reduction factor is given by f_cd = Xf_y / γ_m0 , where X = stress reduction factor for different buckling class, slenderness ratio and yield stress = 1/ [φ + (φ²-λ²)^0.5], f_y is yield stress of material and γ_m0 is partial safety factor for material strength.

9. The value of design compressive strength is limited to
a) f_y + γ_m0
b) f_y
c) f_y γ_m0
d) f_y / γ_m0
Answer: d
Clarification: The value of design compressive strength is given by f_cd = [f_y / γ_m0] / [φ + (φ²-λ²)^0.5] ≤ f_y / γ_m0 i.e. f_cd should be less than or equal to f_y / γ_m0.

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
Answer: b
Clarification: K = 1 for both ends hinged, KL = 1×5000 = 5000, r = 28.2mm (from steel table), A_e = 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, f_cd = 47.85MPa
P_d = A_e f_cd = 7846 x 47.85 = 375.43 kN.

Design of Steel Structures Assessment Questions,