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_{e}f_{cd}

b) A_{e} /f_{cd}

c) f_{cd}

d) 0.5A_{e}f_{cd}

Answer: a

Clarification: 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}].

Answer: d

Clarification: 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

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)+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}].

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)+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}

Answer: a

Clarification: 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})

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} = (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}

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 + (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}

Answer: d

Clarification: 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

Answer: b

Clarification: 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.

Design of Steel Structures Assessment Questions,