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MCQ Practice Test & Solutions: Test: Principal Stress & Strain - 2 (30 Questions)

You can prepare effectively for Mechanical Engineering GATE Mechanical (ME) Mock Test Series 2027 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Principal Stress & Strain - 2". These 30 questions have been designed by the experts with the latest curriculum of Mechanical Engineering 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 90 minutes
  • - Number of Questions: 30

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Test: Principal Stress & Strain - 2 - Question 1

When two mutually perpendicular principal stresses are unequal but like, the maximum shear stress is represented by

Test: Principal Stress & Strain - 2 - Question 2

Assertion (A): A plane state of stress does not necessarily result into a plane state of strain as well.

Reason (R): Normal stresses acting along X and Y directions will also result into normal strain along the Z-direction.

Test: Principal Stress & Strain - 2 - Question 3

Principal strains at a point are 100 x 10-6 and -200 x 10-6 . What is the maximum shear strain at the point?

Detailed Solution: Question 3

don' t confusewithMaximumShear stress 

Test: Principal Stress & Strain - 2 - Question 4

The number of strain readings (using strain gauges) needed on a plane surface to determine the principal strains and their directions is:

Detailed Solution: Question 4

Three strain gauges are needed on a plane surface to determine the principal strains and their directions.

Test: Principal Stress & Strain - 2 - Question 5

Assertion (A): Mohr's construction is possible for stresses, strains and area moment of inertia.

Reason (R): Mohr's circle represents the transformation of second-order tensor.

Test: Principal Stress & Strain - 2 - Question 6

On a plane, resultant stress is inclined at an angle of 45o to the plane. If the normal stress is 100 N /mm2, the shear stress on the plane is:

Detailed Solution: Question 6

Test: Principal Stress & Strain - 2 - Question 7

The complementary shear stresses of intensity τ are induced at a point in the material, as shown in the figure. Which one of the following is the correct set of orientations of principal planes with respect to AB?

Detailed Solution: Question 7

It is a case of pure shear so principal planes will be along the diagonal.

Test: Principal Stress & Strain - 2 - Question 8

A body is subjected to a pure tensile stress of 100 units. What is the maximum shear produced in the body at some oblique plane due to the above?

Detailed Solution: Question 8

Test: Principal Stress & Strain - 2 - Question 9

A material element subjected to a plane state of stress such that the maximum shear stress is equal to the maximum tensile stress, would correspond to

Detailed Solution: Question 9

Test: Principal Stress & Strain - 2 - Question 10

A solid circular shaft of diameter 100 mm is subjected to an axial stress of 50 MPa. It is further subjected to a torque of 10 kNm. The maximum principal stress experienced on the shaft is closest to

Detailed Solution: Question 10

Shear Stress 

Maximum principal Stress 

Test: Principal Stress & Strain - 2 - Question 11

The maximum principle stress for the stress state shown in the figure is

Detailed Solution: Question 11

Test: Principal Stress & Strain - 2 - Question 12

 In a Mohr's circle, the radius of the circle is taken as:

Where, σx and σy are normal stresses along x and y directions respectively and τxy is the shear stress.

Detailed Solution: Question 12

Test: Principal Stress & Strain - 2 - Question 13

The Mohr's circle of plane stress for a point in a body is shown. The design is to be done on the basis of the maximum shear stress theory for yielding. Then, yielding will just begin if the designer chooses a ductile material whose yield strength is:

Detailed Solution: Question 13

According to maximum shear stress theory for yielding, for like stresses, we have

σy = 100 MPa

Note: For unlike stresses we have 

Test: Principal Stress & Strain - 2 - Question 14

Determine the directions of maximum and minimum principal stresses at the point “P” from the Mohr's circle

Detailed Solution: Question 14

From the Mohr‟s circle it will give all directions

Test: Principal Stress & Strain - 2 - Question 15

Which one of the following Mohr‟s circles represents the state of pure shear?

Test: Principal Stress & Strain - 2 - Question 16

Assertion (A): If the state at a point is pure shear, then the principal planes through that point making an angle of 45° with plane of shearing stress carries principal stresses whose magnitude is equal to that of shearing stress.

Reason (R): Complementary shear stresses are equal in magnitude, but opposite in direction.

Test: Principal Stress & Strain - 2 - Question 17

When a component is subjected to axial stress the normal stress σn is maximum, if cos θ is _______ . (σnxCos2θ)

1. maximum
2. minimum
3. always one
4. always zero

Test: Principal Stress & Strain - 2 - Question 18

An elastic material of Young‟s modulus E and Poisson‟s ratio ν is subjected to a compressive stress of σ1 in the longitudinal direction. Suitable lateral compressive stress σ 2 are also applied along the other two each of the lateral directions to half of the magnitude that would be under σ1 acting alone. The magnitude of σ2 is

Detailed Solution: Question 18

For an isotropic linear elastic material, the lateral strain under three-dimensional loading is given by ε2 = (1/E)[σ2 - ν(σ1 + σ2)].

Rearranging, ε2 = (1/E)[σ2(1-ν) - νσ1].

When only σ1 acts (no lateral stresses), the lateral strain is ε0 = -νσ1 / E.

The problem states the applied lateral stresses make the lateral strain equal to half the magnitude of that under σ1 alone, so ε2 = -νσ1 / (2E).

Equate the two expressions: (1/E)[σ2(1-ν) - νσ1] = -νσ1 / (2E). Multiplying by E gives σ2(1-ν) - νσ1 = -νσ1 / 2.

So σ2(1-ν) = νσ1 / 2, and therefore σ2 = νσ1 / [2(1-ν)].

Thus the required magnitude is νσ1 / (2(1-ν)), which matches option B.

Test: Principal Stress & Strain - 2 - Question 19

If the principal stresses and maximum shearing stresses are of equal numerical value at a point in a stressed body, the state of stress can be termed as

Test: Principal Stress & Strain - 2 - Question 20

A piece of material is subjected, to two perpendicular tensile stresses of 70 MPa and 10 MPa. The magnitude of the resultant stress on a plane in which the maximum shear stress occurs is

Test: Principal Stress & Strain - 2 - Question 21

Principal stresses at a point in plane stressed element are σx = σ= 500 kg/cm2 . Normal stress on the plane inclined at 45o to x-axis will be:

Detailed Solution: Question 21

When stresses are alike, then normal stress σn on plane inclined at angle 45° is

Test: Principal Stress & Strain - 2 - Question 22

At a point in two-dimensional stress system σx = 100 N/mm2, σy = τxy = 40 N/mm2. What is the radius of the Mohr circle for stress drawn with a scale of: 1 cm = 10 N/mm2?

Detailed Solution: Question 22

Radius of the Mohr circle

Test: Principal Stress & Strain - 2 - Question 23

Assertion (A): Mohr's circle of stress can be related to Mohr's circle of strain by some constant of proportionality

Reason (R): The relationship is a function of yield stress of the material.

Test: Principal Stress & Strain - 2 - Question 24

The principal strains at a point in a body, under biaxial state of stress, are 1000 ×10–6 and –600 × 10–6. What is the maximum shear strain at that point?

Detailed Solution: Question 24

Test: Principal Stress & Strain - 2 - Question 25

Which of the following stresses can be determined using Mohr's circle method?

Test: Principal Stress & Strain - 2 - Question 26

Normal stresses of equal magnitude p, but of opposite signs, act at a point of a strained material in perpendicular direction. What is the magnitude of the resultant normal stress on a plane inclined at 45° to the applied stresses?

Detailed Solution: Question 26

Test: Principal Stress & Strain - 2 - Question 27

A point in two-dimensional stress state, is subjected to biaxial stress as shown in the above figure. The shear stress acting on the plane AB  is

Detailed Solution: Question 27

Test: Principal Stress & Strain - 2 - Question 28

What are the normal and shear stresses on the 45o planes shown?

Test: Principal Stress & Strain - 2 - Question 29

If a body is subjected to stresses in xy plane with stresses of 60N/mm² and 80N/mm² acting along x and y axes respectively. Also the shear stress acting is 20N/mm²Find the maximum amount of shear stress to which the body is subjected.

Detailed Solution: Question 29


  • Given stresses: σx = 60 N/mm², σy = 80 N/mm², τxy = 20 N/mm².

  • The maximum shear stress (τmax) formula is: τmax = 0.5 x √((σx - σy)² + 4τxy²).

  • Calculate: τmax = 0.5 x √((60 - 80)² + 4 × 20²).

  • = 0.5 x √((−20)² + 1600).

  • = 0.5 x √(400 + 1600).

  • = 0.5 x √2000.

  • = 0.5 x 44.72 ≈ 22.36 N/mm².

  • Correct answer: 22.4 N/mm² (Option B).


  •  

Test: Principal Stress & Strain - 2 - Question 30

State of stress in a plane element is shown in figure I. Which one of the following figures -II is the correct sketch of Mohr's circle of the state of stress?

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