Question 1 |
Stresses acting on an infinitesimal soil element are shown in the figure (with \sigma _z \gt \sigma _x). The major and minor principal stresses are \sigma _1
and \sigma _3, respectively. Considering the compressive stresses as positive, which one of the
following expressions correctly represents the angle between the major
principal stress plane and the horizontal plane?
\tan ^{-1}\left ( \frac{\tau _{zx}}{\sigma _1-\sigma _x} \right ) | |
\tan ^{-1}\left ( \frac{\tau _{zx}}{\sigma _3-\sigma _x} \right ) | |
\tan ^{-1}\left ( \frac{\tau _{zx}}{\sigma _1+\sigma _x} \right ) | |
\tan ^{-1}\left ( \frac{\tau _{zx}}{\sigma _1+\sigma _3} \right ) |
Question 1 Explanation:
\begin{aligned} \Sigma F_x &=0 \\ \sigma _x(BC)-\tau _Z \times (AB)\sigma _1 \sin \theta &= 0\\ \sigma _x\left ( \frac{AC \sin \theta }{\cos \theta } \right )+\tau _{zx}\left ( \frac{AC \cos \alpha }{\cos \theta } \right ) &=\sigma _1 \frac{AC \sin \theta }{\cos \theta }\\ \sigma _x \tan \theta +\tau _{zx} &=\sigma _1 \tan \theta \\ \tan \theta(\sigma _1-\sigma _2) &= \tau _{zx}\\ \tan \theta &= \left ( \frac{\tau _{zx}}{\sigma _1-\sigma _x} \right ) \end{aligned}
Question 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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 5 |
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 5 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 6 |
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 6 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 7 |
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 7 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 8 |
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 8 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 9 |
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 9 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 10 |
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 10 Explanation:
Maximum shear stress,
\tau=\frac{\sigma_{1}-\sigma_{2}}{2}=\frac{P-P}{2}=0
There are 10 questions to complete.
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