# 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
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
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
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) MpL(wL2/8)[1-k+(1+k)0.5].
b) MpL(wL2/8)[1+k+(1-k)0.5].
c) MpL(wL2/8)[1+k-(1+k)0.5].
d) MpL(wL2/8)[1+k+(1+k)0.5].
Clarification: When gravity load governs the design, a good estimate of required section may be obtained by using following formulae :
For pin based frames : Mp =γL(wL2/8)[1+k+(1+k)0.5]
For fixed=-base frame : Mp =γL(wL2/8)[1/[1+0.5k+(1+k)0.5]], where γL = 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
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
Clarification: The presence of axial equation implies that sum of tension and compression forces is not zero and hence following equation is used : ∫A fydA – 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) (Mpr/Mp) + (N/Np)2 = 1
b) (Mpr/Mp) – (N/Np)2 = 1
c) (Mpr/Mp) + (N/Np) = 1
d) (Mpr/Mp) – (N/Np) = 1
Clarification: For a rectangular section of width b and depth d subjected to axial force N together with moment M, (Mpr/Mp) + (N/Np)2 = 1, where Mpr is moment with axial force, Mp is moment with axial force, Np 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/Np) – (1/1.18)(M/Mp) ≤ 1, when N/Np > 0.15
b) (N/Np) – (1/1.18)(M/Mp) ≤ 1, when N/Np < 0.15
c) (N/Np) + (1/1.18)(M/Mp) ≤ 1, when N/Np < 0.15
d) (N/Np) + (1/1.18)(M/Mp) ≤ 1, when N/Np > 0.15
Clarification: For I- section subjected to axial force N together with moment M,
(N/Np) + (1/1.18)(M/Mp) ≤ 1, when N/Np > 0.15
M = Mp, when N/Np < 0.15 .

9. When a member is subjected to uniaxial tensile or compressive stress in presence of shear stress τ , yield occurs when ___
a) fy2 = f2 – 3 τ
b) fy2 = f2 + 3 τ2
c) fy2 = f2 – 3 τ2
d) fy2 = f2 + 3 τ
Clarification: When a member is subjected to uniaxial tensile or compressive stress in presence of shear stress τ, yield occurs when fy2 = f2 + 3 τ2.

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

11. At full plasticity, the stress in web is given by
a) fw = fy√[1+(τwy)2 ].
b) fw = fy√[(τwy)2 ].
c) fw = fy√[1-(τwy)2 ].
d) fw = fy√[1+2(τwy)2 ].