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# Test: Principal Stresses & Theories Of Elastic Failure

## 10 Questions MCQ Test Machine Design | Test: Principal Stresses & Theories Of Elastic Failure

Description
This mock test of Test: Principal Stresses & Theories Of Elastic Failure for Mechanical Engineering helps you for every Mechanical Engineering entrance exam. This contains 10 Multiple Choice Questions for Mechanical Engineering Test: Principal Stresses & Theories Of Elastic Failure (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Principal Stresses & Theories Of Elastic Failure quiz give you a good mix of easy questions and tough questions. Mechanical Engineering students definitely take this Test: Principal Stresses & Theories Of Elastic Failure exercise for a better result in the exam. You can find other Test: Principal Stresses & Theories Of Elastic Failure extra questions, long questions & short questions for Mechanical Engineering on EduRev as well by searching above.
QUESTION: 1

### The normal stress is perpendicular to the area under considerations, while the shear stress acts over the area.

Solution:

Explanation: This is the convention used.

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.

Solution:

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

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.

Solution:

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

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.

Solution:

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

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.

Solution:

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

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?

Solution:

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.

QUESTION: 7

Maximum Principal Stress Theory is not good for brittle materials.

Solution:

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

QUESTION: 8

The region of safety in maximum shear stress theory contains which of the given shape

Solution:

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).

QUESTION: 9

The total strain energy for a unit cube subjected to three principal stresses is given by?

Solution:

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.

QUESTION: 10

Distortion energy theory is slightly liberal as compared to maximum shear stress theory.

Solution:

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