1. The flange is classified as semi-compact if outstand element of compression flange of rolled section is less than
a) 8.4ε
b) 10.5ε
c) 15.7ε
d) 9.4ε
Explanation: The flange is classified as semi-compact if outstand element of compression flange of rolled section is less than 15.7ε and for a welded section, less than 13.6ε.
2.The flange is classified as plastic if outstand element of compression flange of rolled section is less than
a) 8.4ε
b) 9.4ε
c) 10.5ε
d) 15.7ε
Explanation: The flange is classified as plastic if outstand element of compression flange of rolled section is less than 9.4ε and for a welded section, less than 8.4ε.
3. The outstand element of compression flange of a rolled section is 10.2 (ε=1). The flange will be classified as
a) compact
b) plastic
c) semi-compact
d) slender
Explanation: The flange is classified as compact if outstand element of compression flange of rolled section is less than 10.5ε and for a welded section, less than 9.4ε.
4. The design compressive stress of compression member in IS 800 is given by
a) Rankine Formula
b) Euler Formula
c) Perry-Robertson formula
d) Secant-Rankine formula
Explanation: The design compressive stress of axially loaded compression member in IS 800 is given by Perry-Robertson formula. IS 800:2007 proposes multiple columns curves in nin-dimensional form based on Perry-Robertson approach.
5. 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.
6. The design compressive stress, fcd of column is given by
a) [fy / γm0]/ [φ – (φ2-λ2)2].
b) [fy / γm0] / [φ + (φ2-λ2)].
c) [fy / γm0]/[φ – (φ2-λ2)0.5].
d) [fy / γm0] / [φ + (φ2-λ2)0.5].
Explanation: The design compressive stress, fcd of column is given by fcd = [fy / γm0] / [φ + (φ2-λ2)0.5], where fy is yield stress of material, φ is dependent on imperfection factor, λ is non dimensional effective slenderness ratio.
7. 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.
8. 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.
9. 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.
10. 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.