250+ TOP MCQs on Design Strength of Laterally Supported Beams – II and Answers

Design of Steel Structures Problems on “Design Strength of Laterally Supported Beams – II”.

1. The design bending strength of beams when V ≤ 0.6Vd is given by
a) βb /Zpfy γm0
b) βbZpfy / γm0
c) βbZp /fy γm0
d) βbZpfy γm0
Answer: b
Clarification: The design bending strength of beams when V ≤ 0.6Vd is given by Md = βbZpfy / γm0 , where βb is a constant, Zp = plastic section modulus of cross section, fy is yiled stress of material, γm0 = 1.1,partial safety factor.

2. The value of βb in the equation of design bending strength for plastic section is given by
a) 1.5
b) 2.0
c) 0.5
d) 1.0
Answer: d
Clarification: The value of βb in the equation of design bending strength is 1 for plastic and compact sections. The value of βb in the equation of design bending strength for semi-compact section depends on section modulus.

3. The value of βb in the equation of design bending strength for semi-compact section is given by
a) Ze/Zp
b) ZeZp
c) Zp/ Ze
d) Ze+Zp
Answer: a
Clarification: The value of βb in the equation of design bending strength for semi-compact section is given by βb = Ze/Zp , where Ze, Zp are elastic and plastic moduli of the cross section.

4. The check for design bending strength for simply supported beams is given by
a) Md = 2.4Zpfym0
b) Md < 1.2Zpfym0
c) Md ≤ 1.2Zpfym0
d) Md ≥ 1.2Zpfym0
Answer: c
Clarification: The check for design bending strength for simply supported beams is given by Md ≤ 1.2Zpfym0 to ensure that onset of plasticity under unfactored loads is prevented.

5. The check for design bending strength for cantilever beams is given by
a) Md = 2.4Zpfym0
b) Md ≤ 1.5Zpfym0
c) Md ≤ 1.2Zpfym0
d) Md ≥ 1.5Zpfym0
Answer: b
Clarification: The check for design bending strength for cantilever beams is given by Md ≤ 1.5Zpfym0 to ensure that onset of plasticity under unfactored loads – dead loads, imposed loads and wind load- is prevented.

6. The design bending strength for slender sections is given by
a) Md = Zefy
b) Md = fy
c) Md = Ze /fy
d) Md = Ze +fy
Answer: a
Clarification: The design bending strength for slender sections is given by Md = Zefy‘ , where Ze is elastic section modulus of cross section and fy‘ is reduced design strength for slender sections.

7. IS 800 permits bolt holes in the flanges to be ignored when
a) 0.9fuAnfm1 ≤ 2fyAgf/γm0
b) 0.9fuAnfm1 ≤ fyAgf/γm0
c) 0.9fuAnfm1 ≥ fyAgf/γm0
d) 0.9fuAnfm1 ≤ 0.5fyAgf/γm0
Answer: c
Clarification: IS 800 permits bolt holes in the flanges to be ignored when the tensile fracture strength of flange is at least equal to tensile yield strength i.e. when 0.9fuAnfm1 ≥ fyAgf/γm0 or (Anf/Agf) ≥ (fy/fu)x(γm1m0)x(1/0.9), where Anf/Agf = ratio of net area to gross area of tension flange, fy/fu = ratio of yield stress to ultimate stress of material, γm1m0 = ratio of partial safety factors against ultimate stress to yield stress.

8. The moment capacity of semi-compact section for V > 0.6Vd is given by
a) Md = Zefyγm0
b) Md = Zefy
c) Md = fym0
d) Md = Zefym0
Answer: d
Clarification: Few I-and channel sections are semi-compact because of width-thickness ratio. The moment capacity of semi-compact section for V > 0.6Vd is given by Md = Zefym0, where Ze = elastic section modulus of whole section, fy = yield stress of material, γm0 = partial safety factor.

Design of Steel Structures Problems,

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