250+ TOP MCQs on Plastic Design of Portal Frames, Effect of Axial and Shear force on Plastic Moment Capacity and Answers

Design of Steel Structures online test on “Plastic Design of Portal Frames, Effect of Axial and Shear force on Plastic Moment Capacity”.

1. Single bay portal frames with fixed bases have _______
a) two redundancies
b) three redundancies
c) four redundancies
d) zero redundancies
Answer: b
Clarification: Single bay portal frames with fixed bases have three redundancies and require four hinges to produce a mechanism.

2. If order of indeterminacy is r, then minimum number of plastic hinges required for total collapse is _______
a) r-1
b) r
c) r+1
d) r+2
Answer: c
Clarification: If order of indeterminacy is r, then minimum number of plastic hinges required for total collapse is r+1.

3. Which method is used when mechanism is applied to structures with sloping members?
a) method of instantaneous centre
b) method of centre
c) method of seismic centre
d) method of metacentre
Answer: a
Clarification: When mechanism is applied to structures with sloping members, the determination of displacements in the direction of applied forces is required. It is done using method of instantaneous centre or centre-of-rotation technique.

4. Which of the following relation is correct for pin based frames?
a) M_p =γ_L(wL²/8)[1-k+(1+k)^0.5].
b) M_p =γ_L(wL²/8)[1+k+(1-k)^0.5].
c) M_p =γ_L(wL²/8)[1+k-(1+k)^0.5].
d) M_p =γ_L(wL²/8)[1+k+(1+k)^0.5].
Answer: d
Clarification: When gravity load governs the design, a good estimate of required section may be obtained by using following formulae :
For pin based frames : M_p =γL(wL²/8)[1+k+(1+k)^0.5]
For fixed=-base frame : M_p =γL(wL²/8)[1/[1+0.5k+(1+k)^0.5]], where γ_L = 1.7 global load factor, k = h₂/h₁.

5. Which of the following statement is true?
a) combined mechanism is combination of elementary mechanism
b) elementary mechanism is combination of combined mechanism
c) combined mechanism is not combination of elementary mechanism
d) elementary mechanism is combination of elementary and combined mechanism
Answer: a
Clarification: The possible mechanisms can be classified into two types : elementary and combined mechanism. Elementary mechanism is independent of each other, combined mechanism is linear combination of elementary mechanisms.

6. The presence of axial equation implies that _________
a) sum of tension forces is always zero
b) sum of compression forces is always zero
c) sum of tension and compression forces is not zero
d) sum of tension and compression forces is zero
Answer: c
Clarification: The presence of axial equation implies that sum of tension and compression forces is not zero and hence following equation is used : ∫A f_ydA – P = 0, where P is the axial force.

7. Which of the following relation is correct for rectangular section of width b and depth d subjected to axial force N together with moment M?
a) (M_pr/M_p) + (N/N_p)² = 1
b) (M_pr/M_p) – (N/N_p)² = 1
c) (M_pr/M_p) + (N/N_p) = 1
d) (M_pr/M_p) – (N/N_p) = 1
Answer: a
Clarification: For a rectangular section of width b and depth d subjected to axial force N together with moment M, (M_pr/M_p) + (N/N_p)2 = 1, where M_pr is moment with axial force, M_p is moment with axial force, N_p is axial force without any moment.

8. Which of the following relation is correct for I- section of width b and depth d subjected to axial force N together with moment M?
a) (N/N_p) – (1/1.18)(M/M_p) ≤ 1, when N/N_p > 0.15
b) (N/N_p) – (1/1.18)(M/M_p) ≤ 1, when N/N_p < 0.15
c) (N/N_p) + (1/1.18)(M/M_p) ≤ 1, when N/N_p < 0.15
d) (N/N_p) + (1/1.18)(M/M_p) ≤ 1, when N/N_p > 0.15
Answer: d
Clarification: For I- section subjected to axial force N together with moment M,
(N/N_p) + (1/1.18)(M/M_p) ≤ 1, when N/N_p > 0.15
M = M_p, when N/N_p < 0.15 .

9. When a member is subjected to uniaxial tensile or compressive stress in presence of shear stress τ , yield occurs when ___
a) f_y² = f² – 3 τ
b) f_y² = f² + 3 τ²
c) f_y² = f² – 3 τ²
d) f_y² = f² + 3 τ
Answer: b
Clarification: When a member is subjected to uniaxial tensile or compressive stress in presence of shear stress τ, yield occurs when f_y² = f² + 3 τ².

10. Yield in pure shear occurs when ______
a) 0.58 f_y
b) 1.58 f_y
c) 2.8 f_y
d) 3.5 f_y
Answer: a
Clarification: Yield in pure shear occurs when τ_y = f_y/√3 = 0.58 f_y.

11. At full plasticity, the stress in web is given by
a) f_w = f_y√[1+(τ_w/τ_y)² ].
b) f_w = f_y√[(τ_w/τ_y)² ].
c) f_w = f_y√[1-(τ_w/τ_y)² ].
d) f_w = f_y√[1+2(τ_w/τ_y)² ].
Answer: c
Clarification: When full plasticity is produced, the stress in web will be equal to f_w, f_w =√(f_y² -3τ_w² ) = f_y√[1-(_w/τ_y)² ] where τ_y is shear strengths when yield is in pure shear.

12. Loss of web capacity is given by ___
a) Z_pw / (f_y – f_w)
b) Z_pw (f_y – f_w)
c) Z_pw (f_y + f_w)
d) Z_pw /(f_y + f_w)
Answer: b
Clarification: Loss of web capacity is given by Z_pw (f_y – f_w) = Z_pw f_y { 1-√[1-(τ_w/τ_y)2] }.

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