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


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10 Questions MCQ Test - Test: Principal Stresses & Theories Of Elastic Failure

Test: Principal Stresses & Theories Of Elastic Failure for Mechanical Engineering 2024 is part of 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.
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Test: Principal Stresses & Theories Of Elastic Failure - Question 1

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

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

 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

 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

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

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

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

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

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

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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