Design of Steel Structures Questions and Answers – Design Strength of Laterally Unsupported Beams – II

This set of Advanced Design of Steel Structures Questions and Answers focuses on “Design Strength of Laterally Unsupported Beams – II”.

1. Imperfection factor for rolled section is
a) 0.1
b) 0.21
c) 2.1
d) 4.9
View Answer

Answer: b
Explanation: Imperfection factor for rolled section is 0.21. 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. It depends on the buckling curve.

2. Imperfection factor for welded section is
a) 4.9
b) 0.21
c) 2.1
d) 0.49
View Answer

Answer: d
Explanation: Imperfection factor for welded section is 0.49. The imperfection factor depends on the buckling curve and takes into account all the relevant defects in real structure when considering buckling, geometric imperfections, eccentricity of applied loads and residual stresses.

3. Non-dimensional slenderness ratio is given by
a) λLT = √(βbZpfy/Mcr)
b) λLT = √(βbZpfyMcr)
c) λLT = √(βbZp/Mcr)
d) λLT = √(βbZpfy)
View Answer

Answer: a
Explanation: Non-dimensional slenderness ratio is given by λLT = √(βbZpfy/Mcr), where βb = 1 for plastic and compact sections, βb = Ze/Zp for semi-compact sections, Ze = elastic section modulus, Zp = plastic section modulus, Mcr is elastic critical moment.
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4. The check for non- dimensional slenderness ratio is given by
a) λLT = 2.4 √(Zefy/Mcr)
b) λLT > 2 .4 √(Zefy/Mcr)
c) λLT ≤ 1.2 √(Zefy/Mcr)
d) λLT ≥ 1.2 √(Zefy/Mcr)
View Answer

Answer: c
Explanation: The non- dimensional slenderness ratio is given by λLT = √(βbZpfy/Mcr). The check for it is given by λLT ≤ 1.2 √(Zefy/Mcr), where Ze = elastic section modulus, Mcr is elastic critical moment.

5. Which of the following relation is correct?
a) λLT = √(fy/fcr,b)
b) λLT = fy/fcr,b
c) λLT = (fy/fcr,b)2
d) λLT = √(fy fcr,b)
View Answer

Answer: a
Explanation: λLT = √(βbZpfy/Mcr) = √(fy/fcr,b), where βb = 1 for plastic and compact sections, βb = Ze/Zp for semi-compact sections, Ze = elastic section modulus, Zp = plastic section modulus, Mcr is elastic critical moment, fcr,b is extreme compressive elastic buckling stress.
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6. The elastic critical moment is given by
a) Mcr = βb fcr,b
b) Mcr = βbZp / fcr,b
c) Mcr = βbZp
d) Mcr = βbZp fcr,b
View Answer

Answer: d
Explanation: The elastic critical moment is given by Mcr = √{[π2EIy/ L2MLT][ GIt + (π2EIw/L2LT)]} = βbZp fcr,b , Iy = moment of inertia about minor axis, Iw = warping constant, It = St. Venant’s constant, G = Shear modulus.

7. Warping constant in elastic critical moment is given by
a) (1+βff Iy h2f
b) (1-βff Iy h2f
c) βf Iy h2f
d) (1-βf)/βf Iy h2f
View Answer

Answer: b
Explanation: Warping constant in elastic critical moment is given by Iw = (1-βff Iy h2f , where βf is ratio of moment of inertia of compression flange to sum of moments of inertia of compression and tension flanges, Iy = moment of inertia about minor axis, hf = centre-to-centre distance between flanges.
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8. St. Venant’s constant is given by
a) ∑biti2/3
b) ∑biti2
c) ∑biti3/3
d) ∑biti
View Answer

Answer: c
Explanation: St. Venant’s constant is given by It = ∑biti3/3. For open section (e.g. I -section) : It = 2bft3f/3 + bft3w/3.

9. The value of fcr,b is given by
a) fcr,b = [1.1π2E/(LLT/ry)2]{1+1/20[(LLT/ry)/(hf/tf)]2}
b) fcr,b = [1.1π2E/(LLT/ry)]{1-1/20[(LLT/ry)/(hf/tf)]}
c) fcr,b = [1.1π2E/(LLT/ry)2]{1+1/20[(LLT/ry)/(hf/tf)]2}0.5
d) fcr,b = [1.1π2E/(LLT/ry)2]{1-1/20[(LLT/ry)/(hf/tf)]2}0.5
View Answer

Answer: c
Explanation: The value of fcr,b is given by fcr,b = [1.1π2E/(LLT/ry)2]{1+1/20[(LLT/ry)/(hf/tf)]2}0.5, where ry = radius of gyration about weaker axis, LLT = effective length for lateral-torsional buckling, tf = thickness of flange, hf = centre-to-centre distance between flanges.
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