Mohr’s Circle

Question 1
A linear elastic structure under plane stress condition is subjected to two sets of loading, I and II. The resulting states of stress at a point corresponding to these two loadings are as shown in the figure below. If these two sets of loading are applied simultaneously, then the net normal component of stress \sigma _{xx} is ________.

A
\frac{3\sigma }{2}
B
\sigma \left (1+\frac{1}{\sqrt{2}} \right )
C
\frac{\sigma }{2}
D
\sigma \left (1-\frac{1}{\sqrt{2}} \right )
GATE ME 2022 SET-2   Strength of Materials
Question 1 Explanation: 




\sigma _x=0,\sigma _y=0,\tau _{xy}=0,\theta =-45^{\circ}
\begin{aligned} \sigma _{xx}&=\sigma +\sigma _\theta \\ \sigma _\theta&=\left [ \frac{\sigma _x +\sigma _y}{2} \right ] +\left [ \frac{\sigma _x -\sigma _y}{2} \right ] \cos 2\theta +\tau _{xy} \sin 2\theta \\ \sigma _\theta &= \frac{0+\sigma }{2} +\left [\frac{0-\sigma }{2} \right ] \cos 2(-45)+0\\ \sigma _\theta &= \frac{\sigma }{2}\\ \sigma _{xx} &=\sigma +\sigma _\theta =\sigma +\frac{\sigma }{2}=\frac{3\sigma }{2} \end{aligned}
Question 2
The stress state at a point in a material under plane stress condition is equi-biaxial tension with a magnitude of 10 MPa. If one unit on the \sigma -\tau plane is 1 MPa, the Mohr's circle representation of the state-of-stress is given by
A
a circle with a radius equal to principal stress and its center at the origin of the \sigma -\tau plane
B
a point on the \sigma axis at a distance of 10 units from the origin
C
a circle with a radius of 10 units on the \sigma -\tau plane
D
a point on the \tau axis at a distance of 10 units from the origin
GATE ME 2020 SET-1   Strength of Materials
Question 2 Explanation: 


The given state of stress is represented by a point on \sigma -\tau graph which is located on \sigma-axis at a distance of 10 units from origin.
Question 3
The state of stress at a point in a component isrepresented by a Mohr's circle of radius 100MPa centered at 200 MPa on the normal stress axis. On a plane passing through the same point, the normal stress is 260 MPa. The magnitude of the shear stress on the same plane at the same point is ______ MPa.
A
48
B
63
C
96
D
80
GATE ME 2019 SET-2   Strength of Materials
Question 3 Explanation: 


In triangle CEF
\begin{array}{l} \mathrm{CF}^{2}=\mathrm{CE}^{2}+\mathrm{EF}^{2} \\ 100^{2}=60^{2}=\mathrm{EF}^{2}\\ \mathrm{EF}^{2}=100^{2}-60^{2}=6400 \\ \mathrm{EF}=80 \mathrm{MPa} \end{array}
\mathrm{EF} \rightarrow Represents shear stress at the same point =\mathrm{EF}=\tau=80 \mathrm{MPa}
Question 4
The state of stress at a point, for a body in plane stress, is shown in the figure below. If the minimum principal stress is 10 kPa, then the normal stress \sigma_{y} (in kPa) is
A
9.45
B
18.88
C
37.78
D
75.5
GATE ME 2018 SET-1   Strength of Materials
Question 4 Explanation: 
\begin{aligned} \sigma_{x} &=100 \mathrm{kPa}, \tau_{x y}=50 \mathrm{kPa} \\ \text { Minimum principal stress } &=\frac{\sigma_{x}+\sigma_{y}}{2}-\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau_{x y}^{2}} \\ 10 &=\frac{100+\sigma_{y}}{2}-\sqrt{\left(\frac{100-\sigma_{y}}{2}\right)^{2}+50^{2}} \\ \therefore \quad \sqrt{\left(50-\frac{\sigma_{y}}{2}\right)^{2}+50^{2}} &=50+\frac{\sigma_{y}}{2}-10=40+\frac{\sigma_{y}}{2} \end{aligned}
By squaring
2500+\frac{\sigma_{y}^{2}}{4}-50 \sigma_{y}+2500=1600+\frac{\sigma_{y}^{2}}{4}+40 \sigma_{y}
\begin{aligned} \therefore \quad 90 \sigma_{y} &=3400 \\ \sigma_{y} &=37.78 \mathrm{MPa} \end{aligned}
Question 5
If \sigma _{1} and \sigma _{3} are the algebraically largest and smallest principal stresses respectively, the value of the maximum shear stress is
A
\frac{\sigma _{1} + \sigma _{3}}{2}
B
\frac{\sigma _{1} - \sigma _{3}}{2}
C
\sqrt{\frac{\sigma _{1} + \sigma _{3}}{2}}
D
\sqrt{\frac{\sigma _{1} - \sigma _{3}}{2}}
GATE ME 2018 SET-1   Strength of Materials
Question 5 Explanation: 
Maximum shear stress =\frac{\sigma_{1}-\sigma_{3}}{2}
Question 6
The state of stress at a point is \sigma _{x}=\sigma _{y}=\sigma _{z}=t_{xz}=t_{zx}=t_{yz}=t_{zy}=0 and t_{xy}=t_{yx}=50MPa . The maximum normal stress (in MPa) at that point is_____.
A
49
B
50
C
55
D
60
GATE ME 2017 SET-2   Strength of Materials
Question 6 Explanation: 
Given state of stress condition indicates pure shear state of stress.
For pure shear state of stress,
Max. tensile stress = Max. comp. stress = Max. Shear stress
=\tau_{X Y}=50 \mathrm{MPa} Hence, Max. normal stress =50 \mathrm{MPa}
Question 7
In a plane stress condition, the components of stress at a point are \sigma_{x}=20 MPa ,\sigma_{y}=80 MPa and \tau _{xy}=40 MPa . The maximum shear stress (in MPa) at the point is
A
20
B
25
C
50
D
100
GATE ME 2015 SET-2   Strength of Materials
Question 7 Explanation: 
\begin{array}{c} \sigma_{1,2}=\frac{1}{2}\left[\left(\sigma_{x}+\sigma_{y}\right) \pm \sqrt{\left(\sigma_{x}-\sigma_{y}\right)^{2}+4 \tau_{x y}^{2}}\right] \\ =\frac{1}{2}[100 \pm \sqrt{(60)^{2}+4 \times 40^{2}}] \\ \sigma_{1}=100 \\ \sigma_{2}=0 \\ \tau_{\max }=\sigma_{1} / 2=50 \mathrm{MPa} \end{array}
Question 8
The state of stress at a point under plane stress condition is \sigma _{xx} = 40MPa, \sigma _{yy} = 100MPa and \tau _{xy} = 40MPa. The radius of the Mohr's circle representing the given state of stress in MPa is
A
40
B
50
C
60
D
100
GATE ME 2012   Strength of Materials
Question 8 Explanation: 
Mohr's circle

R=\sqrt{(40)^{2}+(30)^{2}}=50 \mathrm{MPa}
Question 9
A two dimensional fluid element rotates like a rigid body. At a point within the element, the pressure is 1 unit. Radius of the Mohr's circle, characterizing the state of stress at the point, is
A
0.5 unit
B
0 unit
C
1 unit
D
2 unit
GATE ME 2008   Strength of Materials
Question 9 Explanation: 
since the fluid element will be subjected to hydrostatic loading therefore Mohr circle will
reduce into a point on \sigma\text{-axis}.
\therefore Radius of mohr circle =0 unit
Question 10
The Mohr's circle of plane stress for a point in a body is shown. The design is to be done on the basis of the maximum shear stress theory for yielding. Then, yielding will just begin if the designer chooses a ductile material whose yield strength is
A
45 Mpa
B
50 Mpa
C
90 Mpa
D
100 Mpa
GATE ME 2005   Strength of Materials
Question 10 Explanation: 
As per maximum shear stress theory,
\left(\tau_{\max }\right)_{\text {absolute }} \leq\left(\frac{S_{y t}}{2}\right)_{\pi}
and when \sigma_{1} and \sigma_{2} are like in nature
\begin{aligned} \sigma_{1} & \leq S_{y t} \\ S_{y t} &=100 \mathrm{MPa} \end{aligned}
There are 10 questions to complete.

11 thoughts on “Mohr’s Circle”

  2. Question 12, Explanation given is right but Answer given is wrong. Answer would be option B (175 MPa, 175 MPa) …However Answer Provided is Opion D (0,0) which is wrong.

    Reply

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