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
View Answer
Answer: a
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) ∂Φ∂z
b) ∂2Φ∂z2
c) ∂2Φ∂x2
d) ∂Φ∂x
View Answer
Answer: b
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,
∂2Φ∂z2.
3 - Question
The σz in terms of stress function is given by __________
a) ∂Φ∂z
b) ∂2Φ∂z2
c) ∂2Φ∂x2
d) ∂Φ∂x
View Answer
Answer: c
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=∂2Φ∂x2.
4 - Question
The shear stress τxz in terms of stress function is given by __________
a) ∂Φ∂z−γx
b) −∂2Φ∂x∂z−γx
c) ∂2Φ∂x2−γx
d) ∂Φ∂x−γx
View Answer
Answer: b
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=−∂2Φ∂x∂z−γx
5 - Question
For both plane stress as well as plain strain case the equilibrium equation in x-direction is _______
a) ∂σx∂x+∂τyx∂y=0
b) ∂σx∂x+∂τyx∂y+∂τzx∂z+X=1
c) ∂σx∂x+∂τyx∂y+X=0
d) ∂σx∂x+∂τzx∂z=0
View Answer
Answer: d
Explanation: The equilibrium equation in x-direction is ∂σx∂x+∂τyx∂y+∂τzx∂z+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 ∂σx∂x+∂τzx∂z=0
6 - Question
For both plane stress as well as plain strain case the equilibrium equation in z-direction is _______
a) ∂τxz∂x+∂σz∂z+γ=0
b) ∂σx∂x+∂τzx∂z+γ=1
c) ∂σx∂x+∂τyx∂y+γ=0
d) ∂σx∂x+∂τzx∂z=0
View Answer
Answer: a
Explanation: The equilibrium equation in x-direction is ∂τxz∂x+∂τyz∂y+∂σz∂z+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 ∂τxz∂x+∂σz∂z+γ=0
7 - Question
For two dimensional case, for both plane stress as well as plain strain case the compatibility equation is _______
a) ∂2εx∂z2+∂2εz∂x2=∂2Γxz∂x∂z
b) ∂2εz∂z2+∂2εy∂x2=∂2Γzy∂z∂y
c) ∂2εx∂y2+∂2εy∂x2=∂2Γxy∂x∂y
d) ∂2εz∂z2+∂2εy∂x2=0
View Answer
Answer: a
Explanation: For two dimensional case, the six compatibility equations are evidently reduced to one single equation;
∂2εx∂z2+∂2εz∂x2=∂2Γxz∂x∂z.
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) (∂2∂x2+∂2∂z2)=0
b) (∂2∂y2+∂2∂z2)(σy+σz)=0
c) (∂2∂x2+∂2∂y2)(σx+σy)=0
d) (∂2∂x2+∂2∂z2)(σx+σz)=
View Answer
Answer: d
Explanation: For two dimensional case, the six compatibility equations are evidently reduced to one single equation;
∂2εx∂z2+∂2εz∂x2=∂2Γxz∂x∂z———————(1)
From the Hooke’s law equation,
εx=1E(σx−μσz),
εz=1E(σz−μσx) and
Γxz=2(1+μ)Eτzx
Substituting these values in (1) and simplifying further we get,
(∂2∂x2+∂2∂z2)(σx+σz)=0.