Engineering Questions with Answers - Multiple Choice Questions

# Elasticity Elements – Stress Functions

1 - Question

The stress function was introduced by __________
a) G.B Airy
b) Terzaghi
c) Darcy
d) Meyerhof

Explanation: The stress function was introduced by G.B Airy in 1862. The stress function is denoted by Φ and is called Airy stress function. Darcy gave the law of flow of water through soils.

2 - Question

The σx in terms of stress function is given by __________
a) Φz
b) 2Φz2
c) 2Φx2
d) Φx

Explanation: It is convenient to reduce the compatibility and equilibrium equations into a single equation in terms of stress function Φ.
The stress function for stress in the x-direction is given by,
2Φz2.

3 - Question

The σz in terms of stress function is given by __________
a) Φz
b) 2Φz2
c) 2Φx2
d) Φx

Explanation: It is convenient to reduce the compatibility and equilibrium equations into a single equation in terms of stress function Φ.
The stress function for stress in the z-direction is given by,
σz=2Φx2.

4 - Question

The shear stress τxz in terms of stress function is given by __________
a) Φzγx
b) 2Φxzγx
c) 2Φx2γx
d) Φxγx

Explanation: It is convenient to reduce the compatibility and equilibrium equations into a single equation in terms of stress function Φ.
The shear stress function τxz is given by,
τxz=2Φxzγx

5 - Question

For both plane stress as well as plain strain case the equilibrium equation in x-direction is _______
a) σxx+τyxy=0
b) σxx+τyxy+τzxz+X=1
c) σxx+τyxy+X=0
d) σxx+τzxz=0

Explanation: The equilibrium equation in x-direction is σxx+τyxy+τzxz+X=0
In the plain strain case, one dimension (y) is very large in comparison to the other two directions. So, the strain components in this direction are zero. Also in plain stress condition, the stresses in y-direction are considered as zero.
∴ The equation reduces to σxx+τzxz=0

6 - Question

For both plane stress as well as plain strain case the equilibrium equation in z-direction is _______
a) τxzx+σzz+γ=0
b) σxx+τzxz+γ=1
c) σxx+τyxy+γ=0
d) σxx+τzxz=0

Explanation: The equilibrium equation in x-direction is τxzx+τyzy+σzz+γ=0.
In the plain strain case, one dimension (y) is very large in comparison to the other two directions. So, the strain components in this direction are zero. Also in plain stress condition, the stresses in y-direction are considered as zero.
∴ The equation reduces to τxzx+σzz+γ=0

7 - Question

For two dimensional case, for both plane stress as well as plain strain case the compatibility equation is _______
a) 2εxz2+2εzx2=2Γxzxz
b) 2εzz2+2εyx2=2Γzyzy
c) 2εxy2+2εyx2=2Γxyxy
d) 2εzz2+2εyx2=0

Explanation: For two dimensional case, the six compatibility equations are evidently reduced to one single equation;
2εxz2+2εzx2=2Γxzxz.

This is because, in the plain strain case, one dimension (y) is very large in comparison to the other two directions. So, the strain components in this direction are zero. Also in plain stress condition, the stresses in y-direction are considered as zero.

8 - Question

The compatibility equation in terms of plane stress case is given by ________
a) (2x2+2z2)=0
b) (2y2+2z2)(σy+σz)=0
c) (2x2+2y2)(σx+σy)=0
d) (2x2+2z2)(σx+σz)=