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 ________.

\frac{3\sigma }{2} | |

\sigma \left (1+\frac{1}{\sqrt{2}} \right ) | |

\frac{\sigma }{2} | |

\sigma \left (1-\frac{1}{\sqrt{2}} \right ) |

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 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 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.

48 | |

63 | |

96 | |

80 |

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

9.45 | |

18.88 | |

37.78 | |

75.5 |

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}

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

\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 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_____.

49 | |

50 | |

55 | |

60 |

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}

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

20 | |

25 | |

50 | |

100 |

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

40 | |

50 | |

60 | |

100 |

Question 8 Explanation:

Mohr's circle

R=\sqrt{(40)^{2}+(30)^{2}}=50 \mathrm{MPa}

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

0.5 unit | |

0 unit | |

1 unit | |

2 unit |

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

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

45 Mpa | |

50 Mpa | |

90 Mpa | |

100 Mpa |

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}

\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.

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,

Yes.. the correct answer should be 90.. if I am wrong, please reply with explanation..

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

Q.10 answer is wrong .

Correct answer is 120 .

-100-(-10)=90