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Test: Theories of Failure - Civil Engineering (CE) MCQ


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10 Questions MCQ Test - Test: Theories of Failure

Test: Theories of Failure for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Test: Theories of Failure questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Theories of Failure MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Theories of Failure below.
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Test: Theories of Failure - Question 1

A certain steel has proportionality limit of 300 N/mm2 in simple tension. It is subjected to principal stress of 120 N/mm2 (tensile), 60 N/mm2 (tensile) and 30 N/mm2 (compressive). The factor of safety according to maximum shear stress theory is

Detailed Solution for Test: Theories of Failure - Question 1

Proportionality limit shear stress
= 300/2 = 150N/mm2
Maximum shear stress

∴ Factor of safety = 150/75 = 2.0

Test: Theories of Failure - Question 2

All the theories of failure, will give nearly the same result when

Detailed Solution for Test: Theories of Failure - Question 2

When one of the principal stresses at a point is large in comparison to the other, the situation resembles uniaxial tension test. Therefore all theories give nearly the same results.

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Test: Theories of Failure - Question 3

The principal stresses at a point in a critical - section of a machine component are σ1 = 60 MPa, σ2 = 5 MPa and σ3 = – 40 MPa. For the material of the component, the tensile yield strength is σy = 200 MPa. According to the maximum shear stress theory, the factor of safety is

Detailed Solution for Test: Theories of Failure - Question 3

Test: Theories of Failure - Question 4

For ductile material the suitable theory of failure is

Detailed Solution for Test: Theories of Failure - Question 4

For ductile material the most suitable theory is maximum shear stress theory.
Other theories for ductile material, Maximum Strain Energy Theory and Maximum Shear Stress Theory (Most Conservative Theory)
For brittle material the most suitable theory is Maximum Principal Stress Theory.
Other theories for brittle material, Maximum Principal Stress Theory.

Test: Theories of Failure - Question 5

Which of the following theories of failure is most appropriate for a brittle material?

Detailed Solution for Test: Theories of Failure - Question 5

Maximum principal stress theory (Rankine theory) is suitable for brittle materials.

Test: Theories of Failure - Question 6

In a structural member, there are perpendicular tensile stresses of 100 N/mm2 and 50 N/mm2. What is the equivalent stress in simple tension, according to the maximum principal strain theory? (Poisson’s ratio = 0.25)

Detailed Solution for Test: Theories of Failure - Question 6

Equivalent stress
= σ1 - μσ2
= 100 - 0.25 x 50 = 87.5 N/mm2

Test: Theories of Failure - Question 7

Permissible bending moment in a circular shaft under pure bending is M, according to maximum principal stress theory of failure. According to maximum shear theory of failure, the permissible bending moment in the shaft is

Detailed Solution for Test: Theories of Failure - Question 7

According to maximum principal stress theory, 
σ1 = σy
According to maximum shear stress theory,
σ1 - σ2 = σy
Under pure bending, 
Therefore, in both the cases, permissible bending moment is M,

Test: Theories of Failure - Question 8

A shaft subjected to pure torsion is to be designed which of the following theories gives the largest diameter of shaft?

Detailed Solution for Test: Theories of Failure - Question 8

It is clear from the relation T/J = (Shear Stress)/ (Radial Distance) that a shaft subjected to pure torsion is to be designed for maximum shear stress theory.

Test: Theories of Failure - Question 9

If maximum principal stress σ1 = 60 N/mm2, σ2 and σ3 of value zero act on a cube of unit dimensions, then the maximum shear stress energy stored in it would be

Detailed Solution for Test: Theories of Failure - Question 9

Shear strain energy

Test: Theories of Failure - Question 10

In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (tensile) and p2 (compressive). For the materials, yield stress in simple tension and compression is 240 N/mm2 and Poisson’s ratio is 0.25. According to maximum normal strain theory for what value of p2 shall yielding commence?

Detailed Solution for Test: Theories of Failure - Question 10

Yielding may occur either in tension or compression.
For yielding in tension,
According to normal strain theory, 

For yielding in compression, 

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