Principal Stress and Principal Strain

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 ________________
A
96.2
B
54.6
C
48.2
D
28.7
GATE CE 2021 SET-1   Solid Mechanics
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
A
\sigma _{xx}\sigma _{yy} -\tau^2 _{xy}=0
B
\tau _{xy}=0
C
\sigma _{xx}+\sigma _{yy}=0
D
(\sigma _{xx}-\sigma _{yy})^2 +4\tau^2 _{xy}=0
GATE CE 2020 SET-1   Solid Mechanics
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?
A
13.3
B
26.6
C
31.7
D
48.3
GATE CE 2019 SET-2   Solid Mechanics
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 ______
A
0
B
1
C
2
D
4
GATE CE 2018 SET-2   Solid Mechanics
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
A
10.0 MPa
B
5.0 MPa
C
2.5 MPa
D
0.0 MPa
GATE CE 2016 SET-2   Solid Mechanics
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:
A
-50 MPa, on a plane 45^{\circ} clockwise w.r.t. x-axis
B
-50 MPa, on a plane 45^{\circ} anti-clockwise w.r.t. x-axis
C
50 MPa, at all orientations
D
Zero, at all orientations
GATE CE 2015 SET-2   Solid Mechanics
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.
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.
A
0
B
1
C
2
D
3
GATE CE 2015 SET-1   Solid Mechanics
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.
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?
A
50
B
75
C
100
D
110
GATE CE 2013   Solid Mechanics
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
A
0
B
p/2
C
p
D
2p
GATE CE 2012   Solid Mechanics
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?
A
A
B
B
C
C
D
D
GATE CE 2011   Solid Mechanics
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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