MCQs on Principal Stresses & Theories of Elastic Failure
1 - Question
The normal stress is perpendicular to the area under considerations, while the shear stress acts over the area.
a) True
b) False
View Answer
Answer: a
Explanation: This is the convention used.
2 - Question
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.
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.
View Answer
Answer: a
Explanation: τ(max)=√( [σ(x)-σ(y) ]²/2² + τ²).
3 - Question
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.
a) 45’
b) 30’
c) 60’
d) 15’
View Answer
Answer: a
Explanation: tan (2Ǿ)=2τ/[σ(x) – σ(y)].
4 - Question
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.
a) 90
b) 92.4
c) 94.2
d) 96
View Answer
Answer: b
Explanation: σ=[σ(x) +σ(y)]/2 + √( [σ(x)-σ(y) ]²/2² + τ²).
5 - Question
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.
a) 45.4
b) 47.6
c) 48.2
d) 50.6
View Answer
Answer: b
Explanation: σ=[σ(x) +σ(y)]/2 – √( [σ(x)-σ(y) ]²/2² + τ²).
6 - Question
If compressive yield stress and tensile yield stress are equivalent, then region of safety from maximum principal stress theory is of which shape?
a) Rectangle
b) Square
c) Circle
d) Ellipse
View Answer
Answer: b
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.
7 - Question
Maximum Principal Stress Theory is not good for brittle materials.
a) True
b) False
View Answer
Answer: b
Explanation: Experimental investigations have shown that maximum principle stress theory gives good results for brittle materials.
8 - Question
The region of safety in maximum shear stress theory contains which of the given shape
a) Hexagon
b) Rectangle
c) Square
d) None of the mentioned
View Answer
Answer: a
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).
9 - Question
The total strain energy for a unit cube subjected to three principal stresses is given by?
a) U= [(σέ) ₁ + (σέ) ₂+ (σέ) ₃]/3
b) U= [(σ₁²+σ₂²+σ₃²)/2E] – (σ₁σ₂+σ₂σ₃+σ₃σ₁)2μ
c) U= [(σέ) ₁ + (σέ) ₂+ (σέ) ₃]/4
d) None of the mentioned
View Answer
Answer: b
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.
10 - Question
Distortion energy theory is slightly liberal as compared to maximum shear stress theory.
a) True
b) False
View Answer
Answer: a
Explanation: The hexagon of maximum shear theory falls completely inside the ellipse of distortion energy theorem.