Engineering Questions with Answers - Multiple Choice Questions

# Stress Distribution – Vertical and Horizontal Pressure

1 - Question

In simple radial distribution, the three stress components σr, σθ and τ are given by ___________

a) σr=KQcosθr,σθ=0andτrθ=0
b) σr=KQ, σθ=0 and τ=0
c) σr=Qcosθr,σθ=0andτrθ=0
d) σr=0, σθ=0 and τ= 0

Explanation: At any radial distance r and polar angle θ, Mitchell found that the three stress components σr, σθ and τ are given by,
σr=KQcosθr,σθ=0andτrθ=0.
Where K is a constant to be found by boundary conditions. The above solution is valid only if it satisfies the equilibrium equations and the compatibility equation.

2 - Question

The equilibrium equation in polar coordinates is given by _____________
a) 1rτrθθ+σrσθr=0
b) σrr+τrθθ+σrσθr=0
c) σrr+1rτrθθ+σrσθr=0
d) σrr+1rτrθθ=0

Explanation: The equilibrium equations in polar coordinates are given by,
1. σrr+1rτrθθ+σrσθr=0
2. 1rσθθ+τrθr+2τrθr=0.

3 - Question

The equilibrium equation in polar coordinates is given by _____________
a) 1rτrθθ+σrσθr=0
b) σrr+τrθθ+σrσθr=0
c) σrr+1rτrθθ+σrσθr=0
d) σrr+1rτrθθ=0

Explanation: The equilibrium equations in polar coordinates are given by,
1. σrr+1rτrθθ+σrσθr=0
2. 1rσθθ+τrθr+2τrθr=0.

4 - Question

The compatibility equation in terms of stress components in polar coordinates are given by ____________
a) (2r2+1rr+1r22θ2)(σr+σθ)=0
b) (2r2+1rr+1r22θ2)(σθ)=0
c) (2r2+1rr+1r22θ2)(σr)=0
d) (2r2+1rr+1r22θ2)(σr+σθ)=1

Explanation: The compatibility equation is the additional equation to solve the stress problem. The compatibility equation in terms of stress components in polar coordinates are given by,
(2r2+1rr+1r22θ2)(σr+σθ)=0.

5 - Question

In simple radial distribution, if σr=KQcosθr, then the value of K is ________

a) K=22α+sin2α
b) K=2α+sinα
c) K=2α-sinα
d) K=sinα

Explanation: Considering the equilibrium of the wedge aob,
We have KQ(α+12sin2α)=Q
∴ K=22α+sin2α.

6 - Question

When the ground is horizontal, α=π2 in constant K. What will be the radial stress σr due to vertical line load?

a) σr=Qcosθr
b) σr=2Qcosθπr
c) σr=Qsinθr
d) σr=2Qsinθr

Explanation: At any radial distance r and polar angle θ, Mitchell found that the radial stress component σr is given by,
σr=KQcosθrwhereK=22α+sin2α
When the ground is horizontal =π2,
∴ σr=2Qcosθπr.

7 - Question

The relation between the stress component in x-direction on a horizontal plane in Cartesian coordinates and polar coordinates for vertical line load is ___________
a) σxr tan2⁡θ
b) σxr cosec2⁡θ
c) σxr cos⁡θ
d) σxr sin2⁡θ

Explanation: From the figure,

On a horizontal plane, the relation between the stress component in x-direction in Cartesian coordinates and polar coordinates is,
σxr sin2⁡θ.

8 - Question

The relation between the stress component in z-direction on a horizontal plane in Cartesian coordinates and polar coordinates for vertical line load is ___________
a) σzr cos2⁡θ
b) σzr cosec2⁡θ
c) σzr cos⁡θ
d) σzr sin2⁡θ

Explanation: From the figure,

On a horizontal plane, the relation between the stress component in z-direction in Cartesian coordinates and polar coordinates is,
σzr cos2⁡θ.

9 - Question

The relation between the shear stress component in xz-plane in Cartesian coordinates and polar coordinates for vertical line load is ___________
a) τxzr tan2⁡θ
b) τxzr cosec2⁡θ
c) τxzr sinθcos⁡θ
d) τxzr sin2⁡θ

Explanation: From the figure,

On xz-plane , the relation between the shear stress component in Cartesian coordinates and polar coordinates is,
τxzr sinθcos⁡θ.

10 - Question

The stress component in x-direction on a horizontal plane in Cartesian coordinates for horizontal line load is ___________
a) σx=2Qxzsinθcosθ
b) σx=2Qxz2π(x2+z2)2
c) σx=2Qx3π(x2+z2)2
d) σx=2Qx2zπ(x2+z2)2

Explanation: On a horizontal plane, the relation between the stress component in x-direction in Cartesian coordinates and polar coordinates is,
σx=σrsin2θwhereσr=KQcosθr
And K=22α+sin2α
∴ σx=2Qx3π(x2+z2)2.

11 - Question

The stress component in x-direction on a horizontal plane in Cartesian coordinates for horizontal line load is ___________
a) σx=2Qxzsinθcosθ
b) σx=2Qxz2π(x2+z2)2
c) σx=2Qx3π(x2+z2)2
d) σx=2Qx2zπ(x2+z2)2

Explanation: On a horizontal plane, the relation between the stress component in z-direction in Cartesian coordinates and polar coordinates is,
σz=σrcos2θwhereσr=KQcosθr
And K=22α+sin2α
∴ σx=2Qx3π(x2+z2)2.

12 - Question

The shear stress component in xz-plane in Cartesian coordinates for horizontal line load is ___________
a) τxz=2Qxzsinθcosθ
b) τxz=2Qxz2π(x2+z2)2
c) τxz=2Qx3π(x2+z2)2
d) τxz=2Qx2zπ(x2+z2)2

Explanation: On a xz-plane, the relation between the shear stress component in Cartesian coordinates and polar coordinates is,
τxz=σrsinθcosθwhereσr=KQcosθr
And K=22α+sin2α
τxz=2Qx2zπ(x2+z2)2.

13 - Question

The radial stress component σr due to inclined line load of intensity Q per unit length is given by ___________

a) σr=2Qr(cosβcosθ2α+sin2α)
b) σr=2Qr(cosβcosθ2α+sin2α+sinβsinθ2αsin2α)
c) σr=Qr(cosβcosθ2α+sin2α+sinβsinθ2αsin2α)
d) σr=2Qr(sinβsinθ2αsin2α)

Explanation: The stresses due to inclined load of intensity Q per unit length can be found by resolving the inclined load into horizontal and vertical components, we get,
σr=2Qr(cosβcosθ2α+sin2α+sinβsinθ2αsin2α).

14 - Question

When the ground is horizontal, α=π2 in constant K. What will be the radial stress σr due to inclined line load at the horizontal ground surface?

a) σr=Qcosθr
b) σr=2Qcos(θβ)πr
c) σr=Qsinθr
d) σr=2Qsinθr