Design of Steel Structures Questions and Answers – Plastic Design of Portal Frames, Effect of Axial and Shear force on Plastic Moment Capacity

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This set of Design of Steel Structures online test focuses 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
View Answer

Answer: b
Explanation: Single bay portal frames with fixed bases have three redundancies and require four hinges to produce a mechanism.
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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
View Answer

Answer: c
Explanation: 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
View Answer

Answer: a
Explanation: 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].
View Answer

Answer: d
Explanation: 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
View Answer

Answer: a
Explanation: 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.
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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
View Answer

Answer: c
Explanation: 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
View Answer

Answer: a
Explanation: 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
View Answer

Answer: d
Explanation: 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
View Answer

Answer: b
Explanation: When a member is subjected to uniaxial tensile or compressive stress in presence of shear stress τ, yield occurs when fy2 = f2 + 3 τ2.
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10. Yield in pure shear occurs when ______
a) 0.58 fy
b) 1.58 fy
c) 2.8 fy
d) 3.5 fy
View Answer

Answer: a
Explanation: 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 ].
View Answer

Answer: c
Explanation: When full plasticity is produced, the stress in web will be equal to fw, fw =√(fy2 -3τw2 ) = fy√[1-(wy)2 ] where τy is shear strengths when yield is in pure shear.

12. Loss of web capacity is given by ___
a) Zpw / (fy – fw)
b) Zpw (fy – fw)
c) Zpw (fy + fw)
d) Zpw /(fy + fw)
View Answer

Answer: b
Explanation: Loss of web capacity is given by Zpw (fy – fw) = Zpw fy { 1-√[1-(τwy)2] }.

Sanfoundry Global Education & Learning Series – Design of Steel Structures.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn