# 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”

1. Question 6 and 10, questions are wrong, in 6 th question Txy=40, not 20
And in 10 , data incomplete

• Thank you for your suggestions. We have updated the correction suggested by You.

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.

• Thank You DURGA SINGH,

3. Question number 9 answer is not accurate

• No bro, it’s correct,

• • Maximum shear is radius of mohr circle.
So (-100-(-10))/2=-90
But we will just take absolute value of 90MPa

• {Maximum–minimumStress}=(yield lodaing /2)
{-100-(-10)}/2=(yeild/2)
90ans

4. 5. 