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+σyydy2)
b) decreased to (σyσyydy2)
c) equal to σY
d) equal to σz

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+σyydy2) .

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+σyydy2)
b) decreased to (σyσyydy2)
c) equal to σY
d) equal to σz

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 (σyσyydy2).

3 - Question

The equilibrium equation obtained by summing all forces on x-direction is ________
a) σxx+τyxy+τzxz+X=0
b) τxyx+σyy+τzyz+Y=0
c) τxzx+τyzy+σzz+Z=0
d) σxx+τyxy+τzxz=0

Explanation: The sum of forces acting on the element in the x-direction is given by,
{(σx+σxxdx2)dy.dz(σxσxxdx2)dy.dz}+{(τyx+τyxydy2)dx.dz(τyxτyxydy2)dx.dz}+
{(τzx+τzxzdz2)dx.dy(τzxτzxzdz2)dx.dy}+Xdx.dy.dz=0
Now dividing all the terms by dx.dy.dz we get,
σxx+τyxy+τzxz+X=0.

4 - Question

The equilibrium equation obtained by summing all forces on y-direction is ________
a) σxx+τyxy+τzxz+X=0
b) τxyx+σyy+τzyz+Y=0
c) τxzx+τyzy+σzz+Z=0
d) σxx+τyxy+τzxz=0

Explanation: The sum of forces acting on the element in the y-direction is given by,
{(σy+σyydy2)dx.dz(σyσyydy2)dx.dz}+{(τzy+τzyzdz2)dy.dx(τzyτzyzdz2)dy.dx}+
{(τxy+τxyxdx2)dy.dz(τxyτxyxdx2)dy.dz}+Ydx.dy.dz=0
Now dividing all the terms by dx.dy.dz we get,
τxyx+σyy+τzyz+Y=0.

5 - Question

The equilibrium equation obtained by summing all forces on z-direction is ________
a) σxx+τyxy+τzxz+X=0
b) τxyx+σyy+τzyz+Y=0
c) τxzx+τyzy+σzz+Z=0
d) σxx+τyxy+τzxz=0

Explanation: The sum of forces acting on the element in the z-direction is given by,
{(σz+σzzdz2)dx.dy(σzσzzdz2)dx.dy}+{(τzx+τzxxdx2)dz.dy(τzxτzxxdx2)dz.dy}+
{(τyz+τyzydy2)dz.dx(τyzτyzydy2)dz.dx}+Zdx.dy.dz=0
Now dividing all the terms by dx.dy.dz we get,
τxzx+τyzy+σzz+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

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

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) σxx+τyxy+τzxz+X=0
b) τxyx+σyy+τzyz+Y=0
c) τxzx+τyzy+σzz+Z=0
d) σxx+τyxy+τzxz=0

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,
σxx+τyxy+τzxz+X=0
Therefore, the equilibrium equations in terms of total stresses formed by summing all forces on x-direction is,
σxx+τyxy+τzxz=0.

9 - Question

The equilibrium equations in terms of total stresses formed by summing all forces on y-direction is ________
a) σxx+τyxy+τzxz+X=0
b) τxyx+σyy+τzyz=0
c) τxzx+τyzy+σzz+Z=0
d) σxx+τyxy+τzxz=0

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,
τxyx+σyy+τzyz+Y=0
Therefore, the equilibrium equations in terms of total stresses formed by summing all forces on x-direction is,
τxyx+σyy+τzyz=0.

10 - Question

The equilibrium equations in terms of total stresses formed by summing all forces on z-direction is ________
a) σxx+τyxy+τzxz+Z=0
b) τxyx+σyy+τzyz=0
c) τxzx+τyzy+σzz+γ=0
d) σxx+τyxy+τzxz=0

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,
τxzx+τyzy+σzz+Z=0
Therefore, the equilibrium equations in terms of total stresses formed by summing all forces on x-direction is,
τxzx+τyzy+σzz+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

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
∴ 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)

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
∴ 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)

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
∴ 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) σxx
b) σxxγwhx
c) σxx+γwhx
d) σxxγwhx

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,
σxx=σxx+γw(hhe)x
But since hex=0,
∴ σxx=σxx+γwhx.

15 - Question

The partial differential of normal stress in y-direction in terms of effective stress is given by __________
a) σyy
b) σyyγwhy
c) σyy+γwhy
d) σyyγwhy