Question 1 |
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 circle with a radius equal to principal stress and its center at the origin of the
\sigma -\tau plane | |
a point on the \sigma axis at a distance of 10 units from the origin | |
a circle with a radius of 10 units on the \sigma -\tau plane | |
a point on the \tau axis at a distance of 10 units from the origin |
Question 1 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 2 |
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.
48 | |
63 | |
96 | |
80 |
Question 2 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 3 |
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
9.45 | |
18.88 | |
37.78 | |
75.5 |
Question 3 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}
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 4 |
If \sigma _{1} and \sigma _{3} are the algebraically largest and smallest principal stresses respectively, the value of the maximum shear stress is
\frac{\sigma _{1} + \sigma _{3}}{2} | |
\frac{\sigma _{1} - \sigma _{3}}{2} | |
\sqrt{\frac{\sigma _{1} + \sigma _{3}}{2}} | |
\sqrt{\frac{\sigma _{1} - \sigma _{3}}{2}} |
Question 4 Explanation:
Maximum shear stress =\frac{\sigma_{1}-\sigma_{3}}{2}
Question 5 |
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_____.
49 | |
50 | |
55 | |
60 |
Question 5 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}
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 6 |
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
20 | |
25 | |
50 | |
100 |
Question 6 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 7 |
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
40 | |
50 | |
60 | |
100 |
Question 7 Explanation:
Mohr's circle
R=\sqrt{(40)^{2}+(30)^{2}}=50 \mathrm{MPa}
R=\sqrt{(40)^{2}+(30)^{2}}=50 \mathrm{MPa}
Question 8 |
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
0.5 unit | |
0 unit | |
1 unit | |
2 unit |
Question 8 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
reduce into a point on \sigma\text{-axis}.
\therefore Radius of mohr circle =0 unit
Question 9 |
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
45 Mpa | |
50 Mpa | |
90 Mpa | |
100 Mpa |
Question 9 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}
\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}
Question 10 |
The figure shows the state of stress at a certain point in a stressed body. The magnitudes of normal stresses in the x and y direction are 100 MPa and 20 MPA respectively. The radius of Mohr's stress circle representing this state of stress is
120 | |
80 | |
60 | |
40 |
Question 10 Explanation:
\begin{aligned} \text { Radius } &=\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau_{x y}} \\ \text{Given,} \quad \sigma_{x} &=100 \mathrm{MPa} \\ \sigma_{y}&=-20 \mathrm{MPa} \\ \tau_{x y} &=0 \\ \therefore \text { Radius } &=\sqrt{\left(\frac{100+20}{2}\right)^{2}+0} \\ &=60 \mathrm{MPa} \end{aligned}
There are 10 questions to complete.
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.
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,
We have updated the answer.
Question number 9 answer is not accurate
No bro, it’s correct,
Q.10 answer is wrong .
Correct answer is 120 .