Design of Steel Structures Questions and Answers – Limit State Method

This set of Design of Steel Structures Multiple Choice Questions & Answers (MCQs) focuses on “Limit State Method”.

1. Limit State Method is based on _____________
a) calculations on service load conditions alone
b) calculations on ultimate load conditions alone
c) calculations at working loads and ultimate loads
d) calculations on earthquake loads
View Answer

Answer: c
Explanation: Working stress method is based on calculations on service load conditions alone. Ultimate Strength method is based on calculations on ultimate load conditions alone. In Limit State method, safety at ultimate loads and serviceability at working loads are considered.

2. What is limit state?
a) Acceptable limits for safety and serviceability requirements before failure occurs
b) Acceptable limits for safety and serviceability requirements after failure occurs
c) Acceptable limits for safety after failure occurs
d) Acceptable limits for serviceability after failure occurs
View Answer

Answer: a
Explanation: Acceptable limits for safety and serviceability requirements before failure occurs is called limit state. In Limit State design, structures are designed on the basis of safety against failure and are checked for serviceability requirements.

3. Which of the following format is used in limit state method?
a) Single safety factor
b) Multiple safety factor
c) Load factor
d) Wind factor
View Answer

Answer: b
Explanation: Limit state method uses multiple safety factor format that helps to provide adequate safety at ultimate loads and adequate serviceability at service loads, by considering all possible limit states. Multiple safety factor format is also called partial safety factor format.
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4. Which of the following factors is included in the limit state of strength?
a) Fire
b) Failure by excessive deformation
c) Corrosion
d) Repairable damage or crack due to fatigue
View Answer

Answer: b
Explanation: Limit state of strength are prescribed to avoid collapse of structure which may endanger safety of life and property. It includes (i) loss of equilibrium of whole or part of structure, (ii) loss of stability of structure as a whole or part of structure, (iii) failure by excessive deformation, (iv) fracture due to fatigue , (v) brittle fracture.

5. Which of the following factors is included in the limit state of serviceability?
a) Brittle facture
b) Fracture due to fatigue
c) Failure by excessive deformation
d) Deformation and deflection adversely affecting appearance or effective use of structure
View Answer

Answer: d
Explanation: Limit state of serviceability includes (i) deformation and deflection adversely affecting appearance or effective use of structure, (ii) vibrations in structure or any part of its compound limiting its functional effectiveness, (iii) repairable repair or crack due to fatigue, (iv) corrosion, (v) fire.
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6. What is permanent action according to classification of actions by IS code?
a) due to self weight
b) due to construction and service stage loads
c) due to accidents
d) due to earthquake loads
View Answer

Answer: a
Explanation: Permanent actions are actions due to self weight of structural and non structural components, fittings, ancillaries, fixed equipments etc.

7. What is variable action according to classification of actions by IS code?
a) due to self weight
b) due to accidents
c) due to construction and service stage loads
d) due to earthquake loads
View Answer

Answer: c
Explanation: Variable actions are actions due to construction and service stage loads such as imposed loads, wind loads, earthquake loads, etc.
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8. Which of the following relation is correct?
a) Design Load = Characteristic Load
b) Design Load = Characteristic Load + Partial factor of safety
c) Design Load = Characteristic Load / Partial factor of safety
d) Design Load = Characteristic Load x Partial factor of safety
View Answer

Answer: d
Explanation: Design Load = Partial factor of safety x Characteristic Load.
This partial safety factor accounts for possibility of unfavourable deviation of load from characteristic value, inaccurate assessment of load, uncertainty in assessment of effects of load and in assessment of limit state being considered.

9. Which of the following relation is correct?
a) Design Strength = Ultimate strength + Partial factor of safety
b) Design Strength = Ultimate strength – Partial factor of safety
c) Design Strength = Ultimate strength /Partial factor of safety
d) Design Strength = Ultimate strength x Partial factor of safety
View Answer

Answer: c
Explanation: Design Strength = Ultimate strength /Partial factor of safety.
This partial safety factor accounts for possibility of unfavourable deviation of material strength from characteristic value, variation of member sizes, reduction in member strength due to fabrication and tolerances and uncertainty in calculation of strength of members.
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10. Which of the following criteria is to be satisfied in selection of member in limit state method?
a) Factored Load > Factored Strength
b) Factored Load ≤ Factored Strength
c) Factored Load ≥ Factored Strength
d) Sometimes Factored Load < Factored Strength (or) Factored Load > Factored Strength
View Answer

Answer: b
Explanation: Limit Sate method is also known as load and resistance factor design. Load factors are applied to service loads and then theoretical strength of member is reduced by application of resistance factor. The criteria is to be satisfied in selection of member in limit state method is factored load ≤ factored strength.

11. The partial factor of safety for resistance governed by yielding is :
a) 1.10
b) 1.5
c) 2.0
d) 1.25
View Answer

Answer: a
Explanation: Partial factor of safety for resistance governed by yielding and resistance of member to buckling is 1.10. The loads are multiplied or resistances are divided by this factor to get design values.

12. The partial factor of safety for resistance governed by ultimate strength is :
a) 1.10
b) 1.5
c) 2.0
d) 1.25
View Answer

Answer: d
Explanation: Partial factor of safety for resistance governed by ultimate strength is 1.25. Factors affecting ultimate strength are stability, fatigue and plastic collapse. The loads are multiplied or resistances are divided by this factor to get design values.

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

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 lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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