# 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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