Question 1 |

The state of stress in a deformable body is shown in the figure. Consider transformation of the stress from the x-y coordinate system to the X-Y coordinate system. The angle \theta, locating the X-axis, is assumed to be positive when measured from the x-axis in counter-clockwise direction.

The absolute magnitude of the shear stress component \sigma_{\mathrm{xy}} (in MPa,round off to one decimal place) in x-y coordinate system is ________________

The absolute magnitude of the shear stress component \sigma_{\mathrm{xy}} (in MPa,round off to one decimal place) in x-y coordinate system is ________________

96.2 | |

54.6 | |

48.2 | |

28.7 |

Question 1 Explanation:

\begin{aligned} \sigma_{x}^{\prime}&=\frac{\sigma_{x}+\sigma_{y}}{2}+\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \cos 2 \theta+\tau_{x y} \sin 2 \theta \\ \text { Here } \theta=60^{\circ} \\ \sigma_{x}&=40 \mathrm{MPa}, \sigma_{y}=35.6, \sigma_{x}^{\prime}=120, \tau_{x^{\prime} y^{\prime}}=-50 \end{aligned}

Substituting the values in above equation, we get

\tau_{x y}=96.186 \mathrm{MPa}

Question 2 |

In a two-dimensional stress analysis, the state of stress at a point P is

[\sigma ]=\begin{bmatrix} \sigma _{xx} &\tau _{xy} \\ \tau _{xy}& \sigma _{yy} \end{bmatrix}

The necessary and sufficient condition for existence of the state of pure shear at the point P, is

[\sigma ]=\begin{bmatrix} \sigma _{xx} &\tau _{xy} \\ \tau _{xy}& \sigma _{yy} \end{bmatrix}

The necessary and sufficient condition for existence of the state of pure shear at the point P, is

\sigma _{xx}\sigma _{yy} -\tau^2 _{xy}=0 | |

\tau _{xy}=0 | |

\sigma _{xx}+\sigma _{yy}=0 | |

(\sigma _{xx}-\sigma _{yy})^2 +4\tau^2 _{xy}=0 |

Question 2 Explanation:

In pure shear condition \sigma _x=0, \sigma _y=0, \tau _{xy}=\tau

For this condition \sigma _{xx}+ \sigma _{yy}=0 is true.

Question 3 |

For a plane stress problem, the state of stress at a point P is represented by the stress element as shown in figure.

By how much angle (\theta) in degrees the stress element should be rotated in order to get the planes of maximum shear stress?

By how much angle (\theta) in degrees the stress element should be rotated in order to get the planes of maximum shear stress?

13.3 | |

26.6 | |

31.7 | |

48.3 |

Question 3 Explanation:

\begin{aligned} \sigma _x&=80\\ \sigma _y&=-20\\ \sigma _z&=-25 \end{aligned}

Angle of plane of max shear

\begin{aligned} \theta &=\theta _p+45^{\circ}\\ \tan 2\theta _p&=\frac{2\tau _{xy}}{\sigma _x-\sigma _y}\\ &=\frac{-50}{100}\\ \theta _p&=-13.28^{\circ}\\ \therefore \;\;\theta &=31.71^{\circ} \end{aligned}

Question 4 |

An 8 m long simply-supported elastic beam of rectangular cross-section (100 mm x 200mm) is subjected to a uniformly distributed load of 10 kN/m over its entire span. The maximum principal stress (in MPa, up to two decimal places) at a point located at the extreme compression edge of a cross-section and at 2 m from the support is ______

0 | |

1 | |

2 | |

4 |

Question 4 Explanation:

\begin{aligned} V_A+V_B&=80\\ V_A&=V_B=40\\ &\text{[Due to symmetry]}\\ M_A&=(-10\times 2 \times 1)+40 \times 2\\ &=60 kNm \end{aligned}

At extreme compression edge,

Bending stress,

\begin{aligned} \sigma &=\frac{My}{I}\\ &=\frac{(60 \times 10^6) \times \frac{200}{2}}{100 \times \frac{200^3}{12}}\\ &=90MPa \end{aligned}

Direct share stress = 0

Principal stress,

\sigma_{p1}/\sigma_{p2}=\frac{90+0}{2}\pm \frac{1}{2}\sqrt{(0-90)^2+y(0)^2}

\sigma_{p1}=90 MPa

So, pricipal stress =90 N/mm^2=90MPa

Question 5 |

For the stress state (in MPa) shown in the figure, the major principal stress is 10 MPa.

The shear stress \tau is

The shear stress \tau is

10.0 MPa | |

5.0 MPa | |

2.5 MPa | |

0.0 MPa |

Question 5 Explanation:

\begin{aligned} \sigma_{x}+\sigma_{y}&=\sigma_{1}+\sigma_{2} \\ \Rightarrow \quad 5+5&=10+\sigma_{2} \\ \Rightarrow \quad \sigma_{2}&=0\\ \text { Now, } \quad \sigma_{1 / 2}&=\frac{\sigma_{x}+\sigma_{y}}{2} \pm \sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau_{x y}^{2}} \\ \therefore \quad \sigma_{1}&=\frac{5+5}{2}+\sqrt{\left(\frac{5-5}{2}\right)^{2}+\tau_{x y}^{2}} \\ \Rightarrow \quad 10&=5+\tau_{x y} \\ \therefore \quad \tau_{x y}&=5 \mathrm{MPa} \end{aligned}

Question 6 |

For the plane stress situation shown in the figure, the maximum shear stress and the plane on which it acts are:

-50 MPa, on a plane 45^{\circ} clockwise w.r.t. x-axis | |

-50 MPa, on a plane 45^{\circ} anti-clockwise w.r.t. x-axis | |

50 MPa, at all orientations | |

Zero, at all orientations |

Question 6 Explanation:

Under hydrostatic loading condition, stresses at a point in all directions are equal and hence no shear stress.

Alternatively,

\tau=\frac{\sigma_{1}-\sigma_{2}}{2}=\frac{50-50}{2}=0

Thus, Mohr's circle reduces to a point.

Hence shear stress at all orientations is zero.

Alternatively,

\tau=\frac{\sigma_{1}-\sigma_{2}}{2}=\frac{50-50}{2}=0

Thus, Mohr's circle reduces to a point.

Hence shear stress at all orientations is zero.

Question 7 |

Two triangular wedges are glued together as shown in the following figure. The stress acting normal to the interface, \sigma _{n} is _____ MPa.

0 | |

1 | |

2 | |

3 |

Question 7 Explanation:

As plane AB and BC are principle planes, therefore
Mohr's circle for the given condition is

Here, normal stress is zero at 45^{\circ} to the principle plane.

Here, normal stress is zero at 45^{\circ} to the principle plane.

Question 8 |

The state of 2D-stress at a point is given by the following matrix of stresses:

\begin{bmatrix} \sigma _{xx} & \sigma _{xy}\\ \sigma _{xy} & \sigma _{yy} \end{bmatrix}=\begin{bmatrix} 100 &30 \\ 30& 20 \end{bmatrix}MPa

What is the magnitude of maximum shear stressin MPa?

\begin{bmatrix} \sigma _{xx} & \sigma _{xy}\\ \sigma _{xy} & \sigma _{yy} \end{bmatrix}=\begin{bmatrix} 100 &30 \\ 30& 20 \end{bmatrix}MPa

What is the magnitude of maximum shear stressin MPa?

50 | |

75 | |

100 | |

110 |

Question 8 Explanation:

\begin{aligned} \sigma_{1 / 2} &=\frac{\sigma_{x}+\sigma_{y}}{2} \pm \sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau_{x y}^{2}} \\ &=\frac{100+20}{2} \pm \sqrt{\left(\frac{100-20}{2}\right)^{2}+(30)^{2}} \\ &=60 \pm 50 \\ \sigma_{1} &=110 ; \sigma_{2}=10 \\ \therefore \tau_{\max } &=\frac{\sigma_{1}-\sigma_{2}}{2}=\frac{110-10}{2}=50 \mathrm{MPa} \end{aligned}

Question 9 |

If a small concrete cube is submerged deep in still water in such a way that the pressure exerted on
all faces of the cube is p, then the maximum shear stress developed inside the cube is

0 | |

p/2 | |

p | |

2p |

Question 9 Explanation:

Maximum shear stress,

\tau=\frac{\sigma_{1}-\sigma_{2}}{2}=\frac{P-P}{2}=0

Question 10 |

Consider a simply supported beam with a uniformly distributed load having
a neutral axis (NA) as shown.

For points P (on the neutral axis) and Q (at the bottom of the beam) the state of stress is best represented by which of the following pairs?

For points P (on the neutral axis) and Q (at the bottom of the beam) the state of stress is best represented by which of the following pairs?

A | |

B | |

C | |

D |

Question 10 Explanation:

Point P:

Point P lies on NA, hence bending stress is zero at point P.

Point P also lies at mid span, so shear force, V=0

\Rightarrow Shear stress, \tau=0

\therefore State of stress of point P will be

Point Q :

At point Q flexural stress is maximum and nature of which is tensile due to downward loading. Point Q lies at the extreme of beam, therefore, shear stress at point Q is zero.

\therefore State of stress of point Q will be

There are 10 questions to complete.

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