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) A_{e}f_{cd}

b) A_{e} /f_{cd}

c) f_{cd}

d) 0.5A_{e}f_{cd}

View Answer

Explanation: The design compressive strength of member is given by P

_{d}= A

_{e}f

_{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 – (φ^{2}-λ^{2})^{2}].

b) [f_{y} / γ_{m0}] / [φ + (φ^{2}-λ^{2})].

c) [f_{y} / γ_{m0}]/[φ – (φ^{2}-λ^{2})^{0.5}].

d) [f_{y} / γ_{m0}] / [φ + (φ^{2}-λ^{2})^{0.5}].

View Answer

Explanation: The design compressive stress, f

_{cd}of column is given by f

_{cd}= [f

_{y}/ γ

_{m0}] / [φ + (φ

^{2}-λ

^{2})

^{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

View Answer

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

View Answer

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)+λ^{2}].

b) φ = 0.5[1-α(λ-0.2)-+λ^{2}].

c) φ = 0.5[1+α(λ+0.2)-λ^{2}].

d) φ = 0.5[1+α(λ-0.2)+λ^{2}].

View Answer

Explanation: The value of φ in the equation of design compressive strength is given by φ = 0.5[1+α(λ-0.2)+λ

^{2}], 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) (π^{2}E)/(KL/r)^{2}

b) (π^{2}E KL/r)^{2}

c) (π^{2}E)/(KL/r)

d) (π^{2}E)/(KLr)^{2}

View Answer

Explanation: Euler buckling stress f

_{cc}is given by f

_{cc}= (π

^{2}E)/(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})

View Answer

Explanation: 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}= (π

^{2}E)/(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}

View Answer

Explanation: 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 + (φ

^{2}-λ

^{2})

^{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}

View Answer

Explanation: The value of design compressive strength is given by f

_{cd}= [f

_{y}/ γ

_{m0}] / [φ + (φ

^{2}-λ

^{2})

^{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

View Answer

Explanation: K = 1 for both ends hinged, KL = 1×5000 = 5000, r = 28.2mm (from steel table), A

_{e}= 7846 mm

^{2}(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.

**Sanfoundry Global Education & Learning Series – Design of Steel Structures.**

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