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Mechanical Engineering Exam  >  Mechanical Engineering Tests  >  Additional Study Material for Mechanical Engineering  >  Test: Principal Stresses & Theories Of Elastic Failure - Mechanical Engineering MCQ

Test: Principal Stresses & Theories Of Elastic Failure - Mechanical Engineering MCQ


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10 Questions MCQ Test Additional Study Material for Mechanical Engineering - Test: Principal Stresses & Theories Of Elastic Failure

Test: Principal Stresses & Theories Of Elastic Failure for Mechanical Engineering 2024 is part of Additional Study Material for Mechanical Engineering preparation. The Test: Principal Stresses & Theories Of Elastic Failure questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Principal Stresses & Theories Of Elastic Failure MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Principal Stresses & Theories Of Elastic Failure below.
Solutions of Test: Principal Stresses & Theories Of Elastic Failure questions in English are available as part of our Additional Study Material for Mechanical Engineering for Mechanical Engineering & Test: Principal Stresses & Theories Of Elastic Failure solutions in Hindi for Additional Study Material for Mechanical Engineering course. Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free. Attempt Test: Principal Stresses & Theories Of Elastic Failure | 10 questions in 20 minutes | Mock test for Mechanical Engineering preparation | Free important questions MCQ to study Additional Study Material for Mechanical Engineering for Mechanical Engineering Exam | Download free PDF with solutions
Test: Principal Stresses & Theories Of Elastic Failure - Question 1
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The normal stress is perpendicular to the area under considerations, while the shear stress acts over the area.

Detailed Solution for Test: Principal Stresses & Theories Of Elastic Failure - Question 1

Explanation: This is the convention used.

Test: Principal Stresses & Theories Of Elastic Failure - Question 2
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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 for Test: Principal Stresses & Theories Of Elastic Failure - Question 2

Explanation: τ(max)=√( [σ(x)-σ(y) ]²/2² + τ²).

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Test: Principal Stresses & Theories Of Elastic Failure - Question 3
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 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 10N/mm². Find the inclination of the plane in which shear stress is maximal.

Detailed Solution for Test: Principal Stresses & Theories Of Elastic Failure - Question 3

Explanation: tan (2Ǿ)=2τ/[σ(x) – σ(y)].

Test: Principal Stresses & Theories Of Elastic Failure - Question 4
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 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/mm². Find the maximum normal stress.
Detailed Solution for Test: Principal Stresses & Theories Of Elastic Failure - Question 4

Explanation: σ=[σ(x) +σ(y)]/2 + √( [σ(x)-σ(y) ]²/2² + τ²).

Test: Principal Stresses & Theories Of Elastic Failure - Question 5
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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/mm². Find the minimum normal stress.
Detailed Solution for Test: Principal Stresses & Theories Of Elastic Failure - Question 5

Explanation: σ=[σ(x) +σ(y)]/2 – √( [σ(x)-σ(y) ]²/2² + τ²).

Test: Principal Stresses & Theories Of Elastic Failure - Question 6
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If compressive yield stress and tensile yield stress are equivalent, then region of safety from maximum principal stress theory is of which shape?
Detailed Solution for Test: Principal Stresses & Theories Of Elastic Failure - Question 6

Explanation: The equation of four lines is given by σ1=± S(yt), σ2=±S(yc) Now given S(yt)=S(yc), hence the region of safety is of square shape.

Test: Principal Stresses & Theories Of Elastic Failure - Question 7
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Maximum Principal Stress Theory is not good for brittle materials.
Detailed Solution for Test: Principal Stresses & Theories Of Elastic Failure - Question 7

Explanation: Experimental investigations have shown that maximum principle stress theory gives good results for brittle materials.

Test: Principal Stresses & Theories Of Elastic Failure - Question 8
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The region of safety in maximum shear stress theory contains which of the given shape
Detailed Solution for Test: Principal Stresses & Theories Of Elastic Failure - Question 8

Explanation: In maximum shear stress theory we have the following equations: σ1= ±S(yt)
σ2= ±S (yt), σ1 – σ2 =±S (yt) assuming S(yt)=S(yc).

Test: Principal Stresses & Theories Of Elastic Failure - Question 9
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The total strain energy for a unit cube subjected to three principal stresses is given by? 
Detailed Solution for Test: Principal Stresses & Theories Of Elastic Failure - Question 9

Explanation: U= [(σέ) ₁ + (σέ) ₂+ (σέ) ₃]/2 is the basic formula. After substituting values of έ₁, έ₂ and έ₃, we get the expression b.

Test: Principal Stresses & Theories Of Elastic Failure - Question 10
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Distortion energy theory is slightly liberal as compared to maximum shear stress theory.
Detailed Solution for Test: Principal Stresses & Theories Of Elastic Failure - Question 10

Explanation: The hexagon of maximum shear theory falls completely inside the ellipse of distortion energy theorem.

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