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Test: Analysis of Stress & Strain Level - 2 - Mechanical Engineering MCQ


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20 Questions MCQ Test Strength of Materials (SOM) - Test: Analysis of Stress & Strain Level - 2

Test: Analysis of Stress & Strain Level - 2 for Mechanical Engineering 2024 is part of Strength of Materials (SOM) preparation. The Test: Analysis of Stress & Strain Level - 2 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Analysis of Stress & Strain Level - 2 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Analysis of Stress & Strain Level - 2 below.
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Test: Analysis of Stress & Strain Level - 2 - Question 1

The maximum shear due to an axial compression of 100 MPa is

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 1
Maximum shear stress,

τ =

=

τ = 50 MPa

Test: Analysis of Stress & Strain Level - 2 - Question 2

When a rod of cross-sectional area 100 mm2is subjected to an tensile force 6kN, the maximum shear stress in the rod will be

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 2

σ = = 60 MPa

τmax = σ/2 = 30 MPa

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Test: Analysis of Stress & Strain Level - 2 - Question 3

A body is subjected to two normal stresses 20 kN/m2 (tensile) and 10 kN/m2 (compressive) acting mutually perpendicular to each other. The maximum shear stress is

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 3
τmax = = 15 kN/m2
Test: Analysis of Stress & Strain Level - 2 - Question 4

A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. If is also subjected to shear stress of 400 MPa on the same planes. The maximum shear stress will be

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 4

maximum Shear Stress , τmax

τmax =

=

= 500 MPa

Test: Analysis of Stress & Strain Level - 2 - Question 5

A Plane stressed element is subjected to the state of stress given by σx = τxy = 100 kgf/cm2 and σy = 0. Maximum shear stress in the element is equal to

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 5
Maximum shear stress,

τmax =

=

= √12, 500

= 50√5 kgf/cm2

Test: Analysis of Stress & Strain Level - 2 - Question 6

The radius of Mohr’s circle for two unlike principal stresses of magnitude σ is

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 6
Radius of Mohr’s circle, τmax =

τmax = σ

Test: Analysis of Stress & Strain Level - 2 - Question 7

At a point in a stressed material, the stress system is σx = + 800N/mm2 , σy = 0 and q = 0.

The radius of the Mohr’s stress circle for this case is__________ units.

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 7
σx = 800 N/mm2, σy = 0, q = τxy = 0

Radius of Mohr’s circle = τmax

τmax = =400 N/mm2

Test: Analysis of Stress & Strain Level - 2 - Question 8

The principal stresses at a point are 150.0 MPa and − 50.0 MPa; the radius of the corresponding Mohr’s circle will be

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 8
Radius of Mohr’s circle

τmax =

=

= 100 MPa

Test: Analysis of Stress & Strain Level - 2 - Question 9

A spherical ball of volume 106 mm3 is subjected to a hydrostatic pressure of 90 MPa. If the bulk modulus for the material is 180 GPa, the change in the volume of the ball is

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 9
Volume = 106 mm3 Hydrostatic pressure, P = 90 MPa

Bulk modulus, k = 180 kN/m3 + 2

k =

180 × 103 =

δv = 500 mm3

Test: Analysis of Stress & Strain Level - 2 - Question 10

An element is subjected to px = 35 MPa, py = 20 MPa and shear stress q = 7.5 MPa. Then the direction of principal stresses is

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 10

tan 2α =

tan 2α = 1 2α = 45°

α = 45°/2

= 22.5°

Test: Analysis of Stress & Strain Level - 2 - Question 11

For an inclined plane in a rectangular block subjected to two mutually perpendicular normal stresses 1000 MPa and 400 MPa and shear stress 400 MPa, the maximum normal stress will be

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 11

σ1 =

= 700 + √9 × 104 + 16 × 104

= 700 + 500 = 1200MPa

Test: Analysis of Stress & Strain Level - 2 - Question 12

The normal stresses of 55 N/mm2 tensile and 45 N/mm2 compressive on two mutually perpendicular planes are acting at a point in a piece of elastic material. These planes also carry shear stress of 50 N/mm2. The minor principal stress will be

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 12
Minor principal stress, σ2

σ2 =

=

σ2 = 65.71 N/mm2 compressive

σy = 45 N/mm2

Test: Analysis of Stress & Strain Level - 2 - Question 13

A rectangular strain rosette, shown in figure give following reading in a strain measurement task ϵP = 100 × 10−6, ϵQ = 150 × 10−6, ϵR = 200 × 10−6. The major principal strain is

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 13
ϵ1,2 =

ϵx = ϵP = 100 × 10−6

ϵy = ϵR = 200 × 10−6

= 150 × 10−6 − 150 × 10−6

= 0

ϵ1,2 =

ϵ1 =

ϵ2 =

ϵ1 = ϵy = 200 × 10−6

Test: Analysis of Stress & Strain Level - 2 - Question 14

If major and minor principal strains are given as 500 × 10−6 and − 200 × 10−6. Then calculate the maximum shear strain.

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 14

⇒ γmax = (ϵ1 − ϵ2) = [500 − (−200)]10−6

= 700 × 10−6

Test: Analysis of Stress & Strain Level - 2 - Question 15

A rectangular strain rosette, shown in figure gives following reading in a strain measurement task ϵ1 = 1000 × 10−6, ϵ2 = 800 × 10−6 and ϵ3 = 600 × 10−6. The direction of the major principal strain with respect to gauge 1 is

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 15

tan 2θp =

=

= 0

Test: Analysis of Stress & Strain Level - 2 - Question 16

Consider the following failure criteria and match them with their corresponding graphical representations.

P - Maximum principal stress Theory

Q - Maximum distortion energy Theory

R - Maximum principal strain Theory

I.

II.

III.

Common Data for Question No: 17 and 18 If the principal stress at a point in an elastic material are 2σ, σ and − σ/2 the properties are Syt = 200 MPa and μ = 0.3

Test: Analysis of Stress & Strain Level - 2 - Question 17

The value of σ at failure according to maximum principal stress theory

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 17
According to MPST

σ1

[N = 1 at failure]

σ = 100 MPa

Test: Analysis of Stress & Strain Level - 2 - Question 18

The value of σ at failure according to strain energy theory

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 18
According to strain energy theory

σ12 + σ22 + σ32 − 2μ(σ1σ2 + σ2σ3 + σ3σ1) ≤

4.95 σ2 = (200)2

σ = 89.89 MPa ≈ 90 MPa

Test: Analysis of Stress & Strain Level - 2 - Question 19

A spherical pressure shell 3.8 meters diameter and 3 mm thick is subjected to an internal pressure ‘P’. The elastic limit of the shell material in simple tension is 260 MPa. The factor of safety is to 2.8. The internal pressure, if the failure of shell is to be prevented, according to maximum shear stress theory is

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 19

σnoop = σ1 = σ2 =

Syt = 260 MPa

N = 2.8

According to MSST

Abs τmax

Abs τmax =

P = 2.932 bar

Test: Analysis of Stress & Strain Level - 2 - Question 20

The force acting on a bolt consists of two components, an axial pull of 14 kN and a transverse shear load of 7 kN. The bolt material yield strength is 340 MPa and the factor of safety is 3 then, the cross sectional area of bolt required according to maximum shear stress theory is

Detailed Solution for Test: Analysis of Stress & Strain Level - 2 - Question 20

σx =

τxy =

A= 174.69 mm2

A ≈ 175 mm2

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