Engineering Questions with Answers - Multiple Choice Questions
Elasticity – Equilibrium Equations
1 - Question
The normal stress component acting at the centre, in the given diagram, will be _________ to the face (B D D1 B1).
a) increased to (σy+∂σy∂ydy2)
b) decreased to (σy−∂σy∂ydy2)
c) equal to σY
d) equal to σz
View Answer
Answer: a
Explanation: The stress on each face will be equal to the stress at the centre, increased, if the face is along the direction of stress, by the distance to the centre to the face times the spatial derivative of the stress. Therefore on face (B D D1 B1), the stresses is (σy+∂σy∂ydy2) .
2 - Question
The normal stress component acting at the centre, in the given diagram, will be _________ to the face (A C C1 A1).
a) increased to (σy+∂σy∂ydy2)
b) decreased to (σy−∂σy∂ydy2)
c) equal to σY
d) equal to σz
View Answer
Answer: b
Explanation: The stress on each face will be equal to the stress at the centre, decreased, if the face is opposite to the direction of stress, by the distance to the centre to the face times the spatial derivative of the stress. Therefore on face (A C C1 A1), the stresses is (σy−∂σy∂ydy2).
3 - Question
The equilibrium equation obtained by summing all forces on x-direction is ________
a) ∂σx∂x+∂τyx∂y+∂τzx∂z+X=0
b) ∂τxy∂x+∂σy∂y+∂τzy∂z+Y=0
c) ∂τxz∂x+∂τyz∂y+∂σz∂z+Z=0
d) ∂σx∂x+∂τyx∂y+∂τzx∂z=0
View Answer
Answer: a
Explanation: The sum of forces acting on the element in the x-direction is given by,
{(σx+∂σx∂xdx2)dy.dz−(σx−∂σx∂xdx2)dy.dz}+{(τyx+∂τyx∂ydy2)dx.dz−(τyx−∂τyx∂ydy2)dx.dz}+
{(τzx+∂τzx∂zdz2)dx.dy−(τzx−∂τzx∂zdz2)dx.dy}+Xdx.dy.dz=0
Now dividing all the terms by dx.dy.dz we get,
∂σx∂x+∂τyx∂y+∂τzx∂z+X=0.
4 - Question
The equilibrium equation obtained by summing all forces on y-direction is ________
a) ∂σx∂x+∂τyx∂y+∂τzx∂z+X=0
b) ∂τxy∂x+∂σy∂y+∂τzy∂z+Y=0
c) ∂τxz∂x+∂τyz∂y+∂σz∂z+Z=0
d) ∂σx∂x+∂τyx∂y+∂τzx∂z=0
View Answer
Answer: b
Explanation: The sum of forces acting on the element in the y-direction is given by,
{(σy+∂σy∂ydy2)dx.dz−(σy−∂σy∂ydy2)dx.dz}+{(τzy+∂τzy∂zdz2)dy.dx−(τzy−∂τzy∂zdz2)dy.dx}+
{(τxy+∂τxy∂xdx2)dy.dz−(τxy−∂τxy∂xdx2)dy.dz}+Ydx.dy.dz=0
Now dividing all the terms by dx.dy.dz we get,
∂τxy∂x+∂σy∂y+∂τzy∂z+Y=0.
5 - Question
The equilibrium equation obtained by summing all forces on z-direction is ________
a) ∂σx∂x+∂τyx∂y+∂τzx∂z+X=0
b) ∂τxy∂x+∂σy∂y+∂τzy∂z+Y=0
c) ∂τxz∂x+∂τyz∂y+∂σz∂z+Z=0
d) ∂σx∂x+∂τyx∂y+∂τzx∂z=0
View Answer
Answer: c
Explanation: The sum of forces acting on the element in the z-direction is given by,
{(σz+∂σz∂zdz2)dx.dy−(σz−∂σz∂zdz2)dx.dy}+{(τzx+∂τzx∂xdx2)dz.dy−(τzx−∂τzx∂xdx2)dz.dy}+
{(τyz+∂τyz∂ydy2)dz.dx−(τyz−∂τyz∂ydy2)dz.dx}+Zdx.dy.dz=0
Now dividing all the terms by dx.dy.dz we get,
∂τxz∂x+∂τyz∂y+∂σz∂z+Z=0.
6 - Question
The problem of elasticity is _________
a) strictly determinate
b) strictly indeterminate
c) in some cases indeterminate
d) cannot be classified as determinate or indeterminate
View Answer
Answer: b
Explanation: There are six independent stresses components acting at a point and the complete solution of the problem requires the determination of these six unknown stress components. Since there are only three equilibrium equations available, the problem of elasticity is strictly of indeterminate nature.
7 - Question
The three equations of static equilibrium of the problem of elasticity are not sufficient to solve the six unknown stress components.
a) True
b) False
View Answer
Answer: a
Explanation: The problem of elasticity is strictly of indeterminate nature. To solve for unknowns, it should be supplemented with equations of compatibility of deformations.
8 - Question
The equilibrium equations in terms of total stresses formed by summing all forces on x-direction is ________
a) ∂σx∂x+∂τyx∂y+∂τzx∂z+X=0
b) ∂τxy∂x+∂σy∂y+∂τzy∂z+Y=0
c) ∂τxz∂x+∂τyz∂y+∂σz∂z+Z=0
d) ∂σx∂x+∂τyx∂y+∂τzx∂z=0
View Answer
Answer: d
Explanation: The equilibrium equations in terms of total stresses formed by summing all forces on x-direction will include seepage force; the body forces will be equal to those due to gravity in respective directions.
∴X=0,
The equilibrium equation obtained by summing all forces on x-direction is,
∂σx∂x+∂τyx∂y+∂τzx∂z+X=0
Therefore, the equilibrium equations in terms of total stresses formed by summing all forces on x-direction is,
∂σx∂x+∂τyx∂y+∂τzx∂z=0.
9 - Question
The equilibrium equations in terms of total stresses formed by summing all forces on y-direction is ________
a) ∂σx∂x+∂τyx∂y+∂τzx∂z+X=0
b) ∂τxy∂x+∂σy∂y+∂τzy∂z=0
c) ∂τxz∂x+∂τyz∂y+∂σz∂z+Z=0
d) ∂σx∂x+∂τyx∂y+∂τzx∂z=0
View Answer
Answer: b
Explanation: The equilibrium equations in terms of total stresses formed by summing all forces on y-direction will include seepage force; the body forces will be equal to those due to gravity in respective directions.
∴ Y=0,
The equilibrium equation obtained by summing all forces on y-direction is,
∂τxy∂x+∂σy∂y+∂τzy∂z+Y=0
Therefore, the equilibrium equations in terms of total stresses formed by summing all forces on x-direction is,
∂τxy∂x+∂σy∂y+∂τzy∂z=0.
10 - Question
The equilibrium equations in terms of total stresses formed by summing all forces on z-direction is ________
a) ∂σx∂x+∂τyx∂y+∂τzx∂z+Z=0
b) ∂τxy∂x+∂σy∂y+∂τzy∂z=0
c) ∂τxz∂x+∂τyz∂y+∂σz∂z+γ=0
d) ∂σx∂x+∂τyx∂y+∂τzx∂z=0
View Answer
Answer: c
Explanation: The equilibrium equations in terms of total stresses formed by summing all forces on z-direction will include seepage force; the body forces will be equal to those due to gravity in respective directions.
∴Z=γ,
The equilibrium equation obtained by summing all forces on z-direction is,
∂τxz∂x+∂τyz∂y+∂σz∂z+Z=0
Therefore, the equilibrium equations in terms of total stresses formed by summing all forces on x-direction is,
∂τxz∂x+∂τyz∂y+∂σz∂z+γ=0.
11 - Question
The normal stress in x-direction in terms of effective stress is given by __________
a) σx= σx’+γw(h-he)
b) σx= σx’*γw(h-he)
c) σx= σx’-γw(h-he)
d) σx= σx’
View Answer
Answer: a
Explanation: The total stress is equal to effective stress plus pore pressure. In x- direction, the normal stress is given by,
σx= σx’+u
u= γw(h-he)
where, h=total head at the point
he=elevation head
∴ σx= σx’+γw(h-he).
12 - Question
The normal stress in y-direction in terms of effective stress is given by __________
a) σy= σy’-γw(h-he)
b) σy= σy’+γw(h-he)
c) σy= σy’/γw(h-he)
d) σy= σy’*γw(h-he)
View Answer
Answer: b
Explanation: The total stress is equal to effective stress plus pore pressure. In y- direction, the normal stress is given by,
σy= σy’+u
u= γw(h-he)
where, h=total head at the point
he=elevation head
∴ σy= σy’+γw(h-he).
13 - Question
The normal stress in z-direction in terms of effective stress is given by __________
a) σz= γw(h-he)
b) σz= σz’
c) σz= σz’+γw(h-he)
d) σz= σz’-γw(h-he)
View Answer
Answer: c
Explanation: The total stress is equal to effective stress plus pore pressure. In x- direction, the normal stress is given by,
σz= σz’+u
u= γw(h-he)
where, h=total head at the point
he=elevation head
∴ σz= σz’+γw(h-he).
14 - Question
The partial differential of normal stress in x-direction in terms of effective stress is given by __________
a) ∂σx‘∂x
b) ∂σx‘∂x−γw∂h∂x
c) ∂σx‘∂x+γw∂h∂x
d) ∂σx‘∂x∗γw∂h∂x
View Answer
Answer: c
Explanation: Since, the normal stress in x-direction in terms of effective stress is given by,
σx= σx’+γw(h-he)
differentiating partially with respect to x,
∂σx∂x=∂σx‘∂x+γw∂(h−he)∂x
But since ∂he∂x=0,
∴ ∂σx∂x=∂σx‘∂x+γw∂h∂x.
15 - Question
The partial differential of normal stress in y-direction in terms of effective stress is given by __________
a) ∂σy‘∂y
b) ∂σy‘∂y−γw∂h∂y
c) ∂σy‘∂y+γw∂h∂y
d) ∂σy‘∂y∗γw∂h∂y
View Answer
Answer: c
Explanation: Since, the normal stress in y-direction in terms of effective stress is given by,
σy= σy’+γw(h-he)
differentiating partially with respect to y,
∂σy∂y=∂σy‘∂y+γw∂(h−he)∂y
But since ∂he∂y=0,
∴ ∂σy∂y=∂σy‘∂y+γw∂h∂y.