Design of Steel Structures Questions and Answers – Plastic Theory

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This set of Design of Steel Structures Multiple Choice Questions & Answers (MCQs) focuses on “Plastic Theory”.

1. Structures designed using elastic analysis may be ______ than those designed using plastic analysis
a) lighter
b) heavier
c) of same weight
d) almost half times the weight
View Answer

Answer: b
Explanation: In elastic design structures are designed for allowable stress level well below the elastic limit, whereas in plastic design structures are designed using ultimate load rather than yield stress, Hence, structures designed using elastic analysis may be heavier than those designed using plastic analysis.
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2. Both elastic and plastic methods neglect ________
a) live load acting on structure
b) dead load acting on structure
c) deformations due to load
d) influence of stability
View Answer

Answer: d
Explanation: Both elastic and plastic methods neglect the influence of stability, which may significantly affect the load carrying capacity of structures or elements which are slender and subjected to compressive stresses.

3. What is buckling?
a) Structural behaviour in which a deformation develops in direction of plane perpendicular to that of load which produced it
b) Structural behaviour in which a deformation does not develop in direction of plane perpendicular to that of load which produced it
c) Structural behaviour in which a deformation develop in direction of plane parallel to that of load which produced it
d) Structural behaviour in which a deformation develops in direction of plane along that of load which produced it
View Answer

Answer: a
Explanation: Buckling may be defined as structural behaviour in which a deformation develops in direction of plane perpendicular to that of load which produced it. This deformation changes rapidly with variations in the applied load.

4. Which of the following relation about plastic moment is correct?
a) Mp = Zp /fy
b) Mp = Zp + fy
c) Mp = Zpfy
d) Mp = Zp – fy
View Answer

Answer: c
Explanation: When every fibre of the section has a strain equal to or greater than εy=fy/Es , nominal moment strength is referred as plastic moment and is given by Mp = Zpfy , where Zp = ∫ydA is plastic section modulus and fy = yield stress.

5. What is plastic moment of resistance?
a) maximum moment in stress strain curve, the point where the curvature can increase indefinitely
b) maximum moment in stress strain curve, the point where the curvature can decrease indefinitely
c) minimum moment in stress strain curve, the point where the curvature can increase indefinitely
d) minimum moment in stress strain curve, the point where the curvature can decrease indefinitely
View Answer

Answer: a
Explanation: In stress strain curve, the maximum moment is reached at a point where the curvature can increase indefinitely, neglecting the strain hardening benefits. This maximum moment is called plastic moment of resistance and the portion where this moment occurs is called as plastic hinge.
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6. In elastic stage, equilibrium condition is achieved when neutral axis ___________ and in fully plastic stage, it is achieved when neutral axis ___________
a) is above centroid of the section, divides the section into two parts of one-third area and two-third area
b) is below centroid of the section, divides the section into two parts of one-third area and two-third area
c) is above centroid of the section, divides the section into two equal areas
d) passes through centroid of the section, divides the section into two equal areas
View Answer

Answer: d
Explanation: In elastic stage, when bending stress varies from zero at neutral axis to maximum at extreme fibres, equilibrium condition is achieved when neutral axis passes through centroid of the section. In fully plastic stage, because the stress is uniformly equal to yield stress, equilibrium condition is achieved when neutral axis divides the section into two equal areas.

7. Which of the following relation is correct for plastic section modulus, Zo ?
a) Zp = 2A(y1+y2)
b) Zp = A(y1+y2)/2
c) Zp = A(y1+y2)/4
d) Zp = 4A(y1+y2)
View Answer

Answer: b
Explanation: Zp = A(y1+y2)/2, where A= area of cross section, y1 and y2 are centroids of portion above and below neutral axis respectively. Plastic modulus is defined as combined statical moment of cross sectional area above and below the equal-area axis.

8. Which of the following relation is correct about shape factor, v?
a) v = Zp+Ze
b) v = ZpZe
c) v = Zp/Ze
d) v = Ze/Zp
View Answer

Answer: c
Explanation: Shape factor, v = Zp/Ze = Mp/My , where Zp and Ze are plastic and elastic section modulus respectively, Mp and My are plastic and elastic moments respectively.

9. The shape factor does not depend on ___
a) material properties
b) cross sectional shape
c) moment of resistance
d) section modulus
View Answer

Answer: a
Explanation: Shape factor, v = Zp/Ze = Mp/My, where Zp and Ze are plastic and elastic section modulus respectively, Mp and My are plastic and elastic moments respectively. This ratio is the property of cross-sectional shape and is independent of material properties.
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10. Match the pairs with correct shape factor

		Cross section				Shape factor (average or maximum)
	A) Circular					(i) 1.8
	B) Rectangular					(ii) 1.14
	C) wide flange I-section (about major axis)	(iii) 1.7
	D) Channels (about minor axis) 			(iv) 1.5

a) A-i, B-ii, C-iii, D-iv
b) A-iv, B-iii, C-ii, D-i
c) A-iii, B-iv, C-ii, D-i
d) A-iii, B-ii, C-iv, D-i
View Answer

Answer: c
Explanation: Shape factor, v = Zp/Ze = Mp/My. The shape factor for various cross section are (i) for circular = 1.7, (ii) for rectangular = 1.5, (iii) wide flange I-section (about major axis) = 1.09-1.18, average is 1.14, (iv) wide flange I-section (about minor axis) = 1.67, (v) channels (about major axis) = 1.16-1.22, average is 1.18, (vi) channels (about minor axis) = 1.8 .

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

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To practice all areas of Design of Steel Structures, here is complete set of 1000+ Multiple Choice Questions and Answers.

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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