GATE CE · Solid Mechanics

Principal Stress And Principal Strain – GATE CE Solid MechanicsPractice Questions & PYQs

Generate GATE-level questions on Principal Stress And Principal Strain in Solid Mechanics. Focus on core concepts, previous year patterns, and numerical problem-solving techniques.

23 questions · 20 PYQs · 0 AI practice · GATE CE 2027

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Q1GATE 2025MCQ

Let the state of stress at a point in a body be the difference of two plane states of stress shown in the figure. Consider all the possible planes perpendicular to the x-y plane and passing through that point. The magnitude of the maximum compressive stress on any such plane is $k \sigma_0$ , where $k$ is equal to _____ (round off to one decimal place).

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📖 Explanation

The total stress is the difference between the two stress states shown. Left figure (normal stress): $\sigma_x = 3\sigma_0$ $\sigma_y = 3\sigma_0$ $\tau_{xy} = 0$ Right figure is a rotated element by $45^\circ$ showing pure shear: It represents: $\sigma_x = 0$ $\sigma_y = 0$ $\tau_{xy} = 3\sigma_0$ Net stress state (subtracting second from first): $\sigma_x = 3\sigma_0$ $\sigma_y = 3\sigma_0$ $\tau_{xy} = -3\sigma_0$ Now, calculate principal stresses: $\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 }$ $= \frac{6\sigma_0}{2} \pm \sqrt{ 0 + 9\sigma_0^2 }$ $= 3\sigma_0 \pm 3\sigma_0$ $\Rightarrow \sigma_1 = 6\sigma_0,\quad \sigma_2 = 0$ Now, compute maximum compressive stress: This is the maximum normal stress on any inclined plane: $\sigma_{\text{max}} = \sqrt{ \sigma_{\text{avg}}^2 + \tau_{xy}^2 }$ Where: $\sigma_{\text{avg}} = 3\sigma_0,\quad \tau_{xy} = -3\sigma_0$ $\Rightarrow \sigma_{\text{max}} = \sqrt{ (3\sigma_0)^2 + (-3\sigma_0)^2 } = \sqrt{18}\sigma_0$ $\Rightarrow \sigma_{\text{max}} \approx 4.24\sigma_0$ So, $k = \frac{4.24\sigma_0}{2\sigma_0} = \boxed{2.1}$

Q2GATE 2024MCQ

Find the correct match between the plane stress states and the Mohr's circles.

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Q3GATE 2023MCQ

In a two-dimensional stress analysis, the state of stress at a point is shown in the figure. The values of length of PQ, QR, and RP are 4, 3, and 5 units, respectively. The principal stresses are (round off to one decimal place)

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📖 Explanation

$\Rightarrow \quad \tan \theta=\frac{\mathrm{QR}}{\mathrm{PQ}}=\frac{3}{4}$ $\Rightarrow \quad \theta=\tan ^{-1}\left(\frac{3}{4}\right)=36.87^{\circ}$ Using transformation equations: $\sigma_{x^{\prime}}=\frac{\sigma_{x}+\sigma_{y}}{2}+\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \cos 2 \theta$ $120=\frac{\sigma_{x}+\sigma_{y}}{2}+\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \cos \left(2 * 36.87^{\circ}\right)$ $\tau_{x^{\prime} y^{\prime}}=-\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \sin \left(2 * 36.87^{\circ}\right)$ From above equations, we get

\begin{aligned} & \sigma_{x}=67.5 \mathrm{MPa} \\ & \sigma_{y}=213.3 \mathrm{MPa} \end{aligned}

Explanation Figure 2

Q4GATE 2023MCQ

A hanger is made of two bars of different sizes. Each bar has a square cross-section. The hanger is loaded by three-point loads in the mid vertical plane as shown in the figure. Ignore the self-weight of the hanger. What is the maximum tensile stress in $\mathrm{N} / \mathrm{mm}^{2}$ anywhere in the hanger without considering stress concentration effects?

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📖 Explanation

$\sigma_{A B}=\frac{P_{A B}}{A_{A B}}=\frac{250 \times 10^{3}}{100 \times 100}=25 \mathrm{~N} / \mathrm{mm}^{2}$ $\sigma_{B C}=\frac{P_{B C}}{A_{B C}}=\frac{50 \times 10^3}{50 \times 50}=20 \mathrm{N} / \mathrm{mm}^{2}$ $\sigma_{\max }=\sigma_{\mathrm{AB}}=25 \mathrm{~N} / \mathrm{mm}^{2}$

Explanation Figure 2

Q5GATE 2023NAT

The infinitesimal element shown in the figure (not to scale) represents the state of stress at a point in a body. What is the magnitude of the maximum principal stress (in $\mathrm{N} / \mathrm{mm}^{2}$ , in integer) at the point?

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📖 Explanation

On plane $x^{\prime} \rightarrow \quad \sigma_{x^{\prime}}=5 \mathrm{~N} / \mathrm{mm}^{2}$ $\tau_{x^{\prime} y^{\prime}}=4 \mathrm{~N} / \mathrm{mm}^{2}$ Using transformation equations:

\begin{aligned} & \sigma_{x^{\prime}}=\left(\frac{\sigma_{x}+\sigma_{y}}{2}\right)+\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \cos 2 \theta+\tau_{x y} \sin 2 \theta \\ & 5=\left(\frac{\sigma_{x}+6}{2}\right)+\left(\frac{\sigma_{x}-6}{2}\right) \cos \left(2 \times 45^{\circ}\right)+3 \sin \left(2 \times 45^{\circ}\right) \\ & \Rightarrow \sigma_{x}=-2 \mathrm{~N} / \mathrm{mm}^{2} \\ & \sigma_{\text {major } / \text { minor }}=\frac{\sigma_{x}+\sigma_{y}}{2} \pm \sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau_{x y}^{2}} \end{aligned}

$\sigma_{\text {major } / \text { minor }}=\frac{-2+6}{2} \pm \sqrt{\left(\frac{-2-6}{2}\right)^{2}+3^{2}}$ $\sigma_{\text {major } / \text { minor }}=2 \pm 5$ $\Rightarrow \quad \sigma_{\text {major }}=7 \mathrm{~N} / \mathrm{mm}^{2}, \sigma_{\text {minor }}=-3 \mathrm{MPa}$ Hence, magnitude of maximum principal stress is $7 \mathrm{~N} / \mathrm{mm}^{2}$ .

Explanation Figure 2

Q6GATE 2022MCQ

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?

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📖 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}

Explanation Figure 2

Q7GATE 2021NAT

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 ________________

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📖 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 $\tau_{x y}=96.186 \mathrm{MPa}$

Explanation Figure 2

Explanation Figure 3

Q8GATE 2020MCQ

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

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📖 Explanation

In pure shear condition $\sigma _x=0, \sigma _y=0, \tau _{xy}=\tau$ For this condition $\sigma _{xx}+ \sigma _{yy}=0$ is true.

Explanation Figure 1

Explanation Figure 2

Q9GATE 2019MCQ

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?

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📖 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}

Explanation Figure 3

Q10GATE 2018NAT

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 ______

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📖 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 $=90 N/mm^2=90MPa$

Explanation Figure 1

Explanation Figure 2

Explanation Figure 3

Q11GATE 2016MCQ

For the stress state (in MPa) shown in the figure, the major principal stress is 10 MPa. The shear stress $\tau$ is

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📖 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}

Explanation Figure 2

Q12GATE 2015MCQ

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

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📖 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.

Q13GATE 2015NAT

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

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📖 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.

Explanation Figure 2

Q14GATE 2013MCQ

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?

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📖 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}

Q15GATE 2012MCQ

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

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📖 Explanation

Maximum shear stress, $\tau=\frac{\sigma_{1}-\sigma_{2}}{2}=\frac{P-P}{2}=0$

Explanation Figure 1

Q16GATE 2011MCQ

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?

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📖 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

Explanation Figure 3

Explanation Figure 4

Explanation Figure 5

Q17GATE 2010MCQ

The major and minor principal stresses at a point are 3MPa and -3MPa respectively. The maximum shear stress at the point is

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📖 Explanation

Maximum shear stress at the point is given by $\tau_{\max }=\frac{\sigma_{1}-\sigma_{2}}{2}=\frac{3-(-3)}{2}=3 \mathrm{MPa}$

Q18GATE 2009MCQ

Consider the following statements : I. On a principal plane, only normal stress acts. II. On a principal plane, both normal and shear stresses act. III. On a principal plane, only shear stress acts IV. Isotropic state of stress is independent of frame of reference. Which of the above statements is/are correct ?

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Q19GATE 2007MCQ

An axially loaded bar is subjected to a normal stress of 173 MPa. The stress in the bar is

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📖 Explanation

\begin{aligned} \text { Shear stress }&=\frac{\sigma_{1}-\sigma_{2}}{2} \\ \therefore \quad&=\frac{173-0}{2}=86.5 \mathrm{MPa} \end{aligned}

Q20GATE 2006MCQ

Mohr's circle for the state of stress defined by

\begin{bmatrix} 30 &0 \\ 0&30 \end{bmatrix}

MPa is a circle with

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