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 τrθ are given by ___________
a) σr=KQcosθr,σθ=0andτrθ=0
b) σr=KQ, σθ=0 and τrθ=0
c) σr=Qcosθr,σθ=0andτrθ=0
d) σr=0, σθ=0 and τrθ= 0
View Answer
Answer: a
Explanation: At any radial distance r and polar angle θ, Mitchell found that the three stress components σr, σθ and τrθ 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) 1r∂τrθ∂θ+σr−σθr=0
b) ∂σr∂r+∂τrθ∂θ+σr−σθr=0
c) ∂σr∂r+1r∂τrθ∂θ+σr−σθr=0
d) ∂σr∂r+1r∂τrθ∂θ=0
View Answer
Answer: c
Explanation: The equilibrium equations in polar coordinates are given by,
1. ∂σr∂r+1r∂τrθ∂θ+σr−σθr=0
2. 1r∂σθ∂θ+∂τrθ∂r+2τrθr=0.
3 - Question
The equilibrium equation in polar coordinates is given by _____________
a) 1r∂τrθ∂θ+σr−σθr=0
b) ∂σr∂r+∂τrθ∂θ+σr−σθr=0
c) ∂σr∂r+1r∂τrθ∂θ+σr−σθr=0
d) ∂σr∂r+1r∂τrθ∂θ=0
View Answer
Answer: c
Explanation: The equilibrium equations in polar coordinates are given by,
1. ∂σr∂r+1r∂τrθ∂θ+σr−σθr=0
2. 1r∂σθ∂θ+∂τrθ∂r+2τrθr=0.
4 - Question
The compatibility equation in terms of stress components in polar coordinates are given by ____________
a) (∂2∂r2+1r∂∂r+1r2∂2∂θ2)(σr+σθ)=0
b) (∂2∂r2+1r∂∂r+1r2∂2∂θ2)(σθ)=0
c) (∂2∂r2+1r∂∂r+1r2∂2∂θ2)(σr)=0
d) (∂2∂r2+1r∂∂r+1r2∂2∂θ2)(σr+σθ)=1
View Answer
Answer: a
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,
(∂2∂r2+1r∂∂r+1r2∂2∂θ2)(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
View Answer
Answer: a
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
View Answer
Answer: b
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) σx=σr tan2θ
b) σx=σr cosec2θ
c) σx=σr cosθ
d) σx=σr sin2θ
View Answer
Answer: d
Explanation: From the figure,
On a horizontal plane, the relation between the stress component in x-direction in Cartesian coordinates and polar coordinates is,
σx=σr sin2θ.
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) σz=σr cos2θ
b) σz=σr cosec2θ
c) σz=σr cosθ
d) σz=σr sin2θ
View Answer
Answer: a
Explanation: From the figure,
On a horizontal plane, the relation between the stress component in z-direction in Cartesian coordinates and polar coordinates is,
σz=σr cos2θ.
9 - Question
The relation between the shear stress component in xz-plane in Cartesian coordinates and polar coordinates for vertical line load is ___________
a) τxz=σr tan2θ
b) τxz=σr cosec2θ
c) τxz=σr sinθcosθ
d) τxz=σr sin2θ
View Answer
Answer: c
Explanation: From the figure,
On xz-plane , the relation between the shear stress component in Cartesian coordinates and polar coordinates is,
τxz=σr sinθcosθ.
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
View Answer
Answer: c
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
View Answer
Answer: b
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
View Answer
Answer: d
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+sinβsinθ2α−sin2α)
c) σr=Qr(cosβcosθ2α+sin2α+sinβsinθ2α−sin2α)
d) σr=2Qr(sinβsinθ2α−sin2α)
View Answer
Answer: b
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+sinβsinθ2α−sin2α).
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(θ−β)πr
c) σr=Qsinθr
d) σr=2Qsinθr
View Answer
Answer: b
Explanation: At any radial distance r and polar angle θ, the radial stress component σr due to inclined line load is given by,
σr=2Qr(cosβcosθ2α+sin2α+sinβsinθ2α−sin2α).
When the ground is horizontal =π2,
∴ σr=2Qcos(θ−β)πr.