Test: Solid Mechanics- 3 - Civil Engineering (CE) MCQ

Test: Solid Mechanics- 3 - Civil Engineering (CE) MCQ

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10 Questions MCQ Test GATE Civil Engineering (CE) 2025 Mock Test Series - Test: Solid Mechanics- 3

Test: Solid Mechanics- 3 for Civil Engineering (CE) 2024 is part of GATE Civil Engineering (CE) 2025 Mock Test Series preparation. The Test: Solid Mechanics- 3 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Solid Mechanics- 3 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Solid Mechanics- 3 below.
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Test: Solid Mechanics- 3 - Question 1

For state, a stress  mohr circle will have center and radius is

Detailed Solution for Test: Solid Mechanics- 3 - Question 1

σx = 30 : σy = 30 : τxy = 0,

Mohr Circle will be a point with centre (30,0) and radius 0 units

τmax = 0 , radius = 0

Test: Solid Mechanics- 3 - Question 2

Which of the following is the most appropriate theory of failure for mild steel?

Detailed Solution for Test: Solid Mechanics- 3 - Question 2

Maximum principal stress theory (Rankine’s theory)

According to this theory, permanent set takes place under a state of complex stress, when the value of maximum principal stress is equal to that of yield point stress as found in a simple tensile test.

For design criterion, the maximum principal stress (σ1) must not exceed the working stress ‘σy’ for the material.

Note: For no shear failure τ ≤ 0.57 σy

Graphical representation

For brittle material, which do not fail by yielding but fail by brittle fracture, this theory gives satisfactory result.

The graph is always square even for different values of σ1 and σ2.

Maximum principal strain theory (ST. Venant’s theory)

According to this theory, a ductile material begins to yield when the maximum principal strain reaches the strain at which yielding occurs in simple tension.

For no failure in uni – axial loading.

For no failure in tri – axial loading.

For design, Here, ϵ = Principal strain

σ1, σ2 and σ3 = Principal stresses

Graphical Representation

This story over estimate the elastic strength of ductile material.

Maximum shear stress theory

(Guest & Tresca’s Theory)

According to this theory, failure of specimen subjected to any combination of load when the maximum shearing stress at any point reaches the failure value equal to that developed at the yielding in an axial tensile or compressive test of the same material.

Graphical Representation

σ1 and σ2 are maximum and minimum principal stress respectively.

Here, τmax = Maximum shear stress

σy = permissible stress

This theory gives satisfactory result for ductile material.

Maximum strain energy theory (Haigh’s theory)

According to this theory, a body complex stress fails when the total strain energy at elastic limit in simple tension.

Graphical Representation.

This theory does not apply to brittle material for which elastic limit stress in tension and in compression are quite different.

Maximum shear strain energy / Distortion energy theory / Mises – Henky theory.

It states that inelastic action at any point in body, under any combination of stress begging, when the strain energy of distortion per unit volume absorbed at the point is equal to the strain energy of distortion absorbed per unit volume at any point in a bar stressed to the elastic limit under the state of uniaxial stress as occurs in a simple tension / compression test.

It gives very good result in ductile material.

It cannot be applied for material under hydrostatic pressure.

All theories will give same results if loading is uniaxial.

Test: Solid Mechanics- 3 - Question 3

Consider the following statements: 1. On planes having maximum and minimum principal stress, there will be no tangential stress. 2. Shear stresses on mutually perpendicular planes are numerically equal. 3. Maximum shear stress is numerically equal to half the sum of the maximum and minimum principal stress Which of the following statement is/are correct –

Detailed Solution for Test: Solid Mechanics- 3 - Question 3

On plane of principal stress, shear stress is always zero but in the plane of maximum shear stress, normal stress may exists.

Shear stresses on mutually perpendicular planes are equal due to moment equilibrium.

Maximum shear stress is numerically half of the difference between maximum and minimum principal stress.

Test: Solid Mechanics- 3 - Question 4

A cube is subjected to equal tensile stress on the three faces. If the yield stress of the material is ‘σy’ and poison ratio ‘μ’ then based on strain energy theory, the maximum tensile stress will be

Detailed Solution for Test: Solid Mechanics- 3 - Question 4

Let the maximum tensile stress is σ.

According to maximum strain energy theory

Test: Solid Mechanics- 3 - Question 5

At a point in a steel member, major principal stress is 1000 kg/cm2, and minor principal stress is compressive. If uniaxial tensile yield stress is 1500 kg/cm2, then the magnitude of minor principal stress at which yielding will commence, according to maximum shearing stress theory is

Detailed Solution for Test: Solid Mechanics- 3 - Question 5

But σ2 is compressive

Test: Solid Mechanics- 3 - Question 6

The principal stresses developed at a point are +30, -30, 0 MPa. Using shear strain energy theory, a factor of safety obtained is √6. What is yield stress of the material?

Detailed Solution for Test: Solid Mechanics- 3 - Question 6

As per maximum shear strain energy theory,

σy = 90√2 MPa

Test: Solid Mechanics- 3 - Question 7

In a structural member, there are perpendicular tensile stresses of 100 N/mm2. What is the equivalent stress in simple tension, according to the maximum principal strain theory? (Poisson’s ratio = 0. 50)

Detailed Solution for Test: Solid Mechanics- 3 - Question 7

Equivalent stress

= σ1−μσ2

= 100 – 0.50 × 100 = 50 N/mm2

*Answer can only contain numeric values
Test: Solid Mechanics- 3 - Question 8

In metal forming operation when the material has just started yielding, the principal stresses are σ1 = +180 MPa, σ2 = -100 MPa, σ3 = 0. Following the von Mises criterion, the yield stress is ________ MPa.

Detailed Solution for Test: Solid Mechanics- 3 - Question 8

According to von Mises theory, for yielding

σyt = 245.76 MPa

Test: Solid Mechanics- 3 - Question 9

The Mohr’s circle given below corresponds to which one of the following stress condition.

Detailed Solution for Test: Solid Mechanics- 3 - Question 9

Radius of Mohr’s circle = 100 MPa
And centre of Mohr’s circle is at a distance,
from the origin. Here it is in origin.

As, σ+ σ= 0 both the normal stress may be zero which is a pure shear case or opposite in nature which don’t exist in any of the options. So it is the pure shear case, where the radius is

*Answer can only contain numeric values
Test: Solid Mechanics- 3 - Question 10

2D stress matrix at a point is given below:

If the maximum shear stress is 200 Mpa. The value of τxy will be ________ Mpa.

Detailed Solution for Test: Solid Mechanics- 3 - Question 10

Maximum and Minimum Normal stress is given by:

σx = 250

σy = 60

τxy = 175.99 Mpa

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