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

Advanced Design of Steel Structures Questions and Answers 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
Answer: b
Clarification: 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
Answer: d
Clarification: 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) λ_LT = √(β_bZ_pf_y/M_cr)
b) λ_LT = √(β_bZ_pf_yM_cr)
c) λ_LT = √(β_bZ_p/M_cr)
d) λ_LT = √(β_bZ_pf_y)
Answer: a
Clarification: Non-dimensional slenderness ratio is given by λ_LT = √(β_bZ_pf_y/M_cr), where β_b = 1 for plastic and compact sections, β_b = Ze/Z_p for semi-compact sections, Ze = elastic section modulus, Z_p = plastic section modulus, M_cr is elastic critical moment.

4. The check for non- dimensional slenderness ratio is given by
a) λ_LT = 2.4 √(Zef_y/M_cr)
b) λ_LT > 2 .4 √(Zef_y/M_cr)
c) λ_LT ≤ 1.2 √(Zef_y/M_cr)
d) λ_LT ≥ 1.2 √(Zef_y/M_cr)
Answer: c
Clarification: The non- dimensional slenderness ratio is given by λ_LT = √(β_bZ_pf_y/M_cr). The check for it is given by λ_LT ≤ 1.2 √(Zef_y/M_cr), where Ze = elastic section modulus, M_cr is elastic critical moment.

5. Which of the following relation is correct?
a) λ_LT = √(f_y/f_cr,b)
b) λ_LT = f_y/f_cr,b
c) λ_LT = (f_y/f_cr,b)²
d) λ_LT = √(f_y f_cr,b)
Answer: a
Clarification: λ_LT = √(β_bZ_pf_y/M_cr) = √(f_y/f_cr,b), where β_b = 1 for plastic and compact sections, β_b = Ze/Z_p for semi-compact sections, Ze = elastic section modulus, Z_p = plastic section modulus, M_cr is elastic critical moment, f_cr,b is extreme compressive elastic buckling stress.

6. The elastic critical moment is given by
a) M_cr = β_b f_cr,b
b) M_cr = β_bZ_p / f_cr,b
c) M_cr = β_bZ_p
d) M_cr = β_bZ_p f_cr,b
Answer: d
Clarification: The elastic critical moment is given by M_cr = √{[π²EI_y/ L²M_LT][ GIt + (π²EIw/L²_LT)]} = β_bZ_p f_cr,b , I_y = 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+β_f)β_f I_y h²_f
b) (1-β_f)β_f I_y h²_f
c) β_f I_y h²_f
d) (1-β_f)/β_f I_y h²_f
Answer: b
Clarification: Warping constant in elastic critical moment is given by Iw = (1-β_f)β_f I_y h²_f , where β_f is ratio of moment of inertia of compression flange to sum of moments of inertia of compression and tension flanges, I_y = moment of inertia about minor axis, h_f = centre-to-centre distance between flanges.

8. St. Venant’s constant is given by
a) ∑b_it_i²/3
b) ∑b_it_i²
c) ∑b_it_i³/3
d) ∑b_it_i
Answer: c
Clarification: St. Venant’s constant is given by It = ∑b_iti3/3. For open section (e.g. I -section) : It = 2bft3f/3 + bft3w/3.

9. The value of f_cr,b is given by
a) f_cr,b = [1.1π²E/(L_LT/r_y)²]{1+1/20[(L_LT/r_y)/(h_f/t_f)]²}
b) f_cr,b = [1.1π²E/(L_LT/r_y)]{1-1/20[(L_LT/r_y)/(h_f/t_f)]}
c) f_cr,b = [1.1π²E/(L_LT/r_y)²]{1+1/20[(L_LT/r_y)/(h_f/t_f)]²}^0.5
d) f_cr,b = [1.1π²E/(L_LT/r_y)²]{1-1/20[(L_LT/r_y)/(h_f/t_f)]²}^0.5
Answer: c
Clarification: The value of f_cr,b is given by f_cr,b = [1.1π²E/(L_LT/r_y)²]{1+1/20[(L_LT/r_y)/(h_f/t_f)]²}^0.5, where r_y = radius of gyration about weaker axis, L_LT = effective length for lateral-torsional buckling, t_f = thickness of flange, h_f = centre-to-centre distance between flanges.