Q1GATE 2024MCQ

The three-dimensional state of stress at a point is given by

\sigma=\left(\begin{array}{ccc}10 & 0 & 0 \\ 0 & 40 & 0 \\ 0 & 0 & 0\end{array}\right) \mathrm{MPa}

. The maximum shear stress at the point is

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📖 Explanation

In the given matrix, shear stresses are zero. So, the elements on diagonal of this matrix are principal stress.So,

\begin{aligned} \tau_{\max } & =\left|\left(\frac{\sigma_{p 1}-\sigma_{p 2}}{2}\right),\left(\frac{\sigma_{p 2}-\sigma_{p 3}}{2}\right),\left(\frac{\sigma_{p 3}-\sigma_{p 1}}{2}\right)\right|_{\max } \\ & =\left|\left(\frac{10-40}{2}\right),\left(\frac{40-0}{2}\right),\left(\frac{0-10}{2}\right)\right|_{\max } \\ & =|(-15), 20,(-5)| \\ & =20 \mathrm{MPa} \end{aligned}

Q2GATE 2023NAT

A $\quad 2 \mathrm{D}$ thin plate with modulus of elasticity, $E=1.0$ $\mathrm{N} / \mathrm{m}^{2}$ , and Poisson's ratio, $\mu=0.5$ , is in plane stress condition. The displacement field in the plate is given by $\mu=C^{2} y$ and $v=0$ , where us and $v$ are displacements (in $m$ ) along the $X$ and $Y$ directions, respectively, and $C$ is constant (in $\mathrm{m}^{-2}$ ). The distance $x$ and $y$ along $X$ an $Y$ , respectively, are in $\mathrm{m}$ . The stress in the $X$ direction is $\sigma_{x x}=40 x y \mathrm{~N} / \mathrm{m}^{2}$ , and the shear stress is $\tau_{x y}=a x^{2} \mathrm{~N} / \mathrm{m}^{2}$ . What is the value of $\alpha$ (in $\mathrm{N} / \mathrm{m}^{4}$ , in integer) ?

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📖 Explanation

\begin{aligned} & \epsilon_{x}=\frac{\sigma_{x}}{E}-\frac{\mu \sigma_{y}}{E} \\ & \epsilon_{y}=\frac{\sigma_{y}}{E}-\frac{\mu \sigma_{x}}{E} \\ & \tau_{x y}=\frac{\gamma_{x y}}{G_{1}} \\ & G=\frac{E}{2(1+\mu)} \\ & \epsilon_{x}+\mu \epsilon_{y}=\frac{\sigma_{x}}{E}-\frac{\mu \sigma_{y}}{E}+\frac{\mu \sigma_{y}}{E}-\frac{\mu^{2} \sigma_{x}}{E} \\ & \epsilon_{x}+\mu \epsilon_{y}=\frac{\sigma_{x}}{E}\left(1-\mu^{2}\right) \\ & \sigma_{x}=\frac{E}{1-\mu^{2}}\left[\epsilon_{x}+\mu \epsilon_{y}\right] \\ & \sigma_{x}=\frac{E}{1-\mu^{2}}\left[\frac{\partial u}{\partial x}+\mu \frac{\partial v}{\partial y}\right] \\ & \sigma_{x}=\frac{E}{1-\mu^{2}}[2(x y+0)] \\ & \sigma_{x}=\frac{2 \mathrm{ECxy}}{1-\mu^{2}} \\ & \sigma_{x}=40 x y \text { (Given) } \\ & \Rightarrow \frac{2 \mathrm{ECxy}}{1-\mu^{2}}=40 x y \\ & \Rightarrow \quad \frac{2 \mathrm{EC}}{1-\mu^{2}}=40 \\ & Y_{x y}=\frac{\tau_{x}}{G} \\ & G \gamma_{x y}=\tau_{x y} \\ & \frac{E}{2(1+\mu)}\left[\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\right]=\tau_{x y} \\ & \frac{E}{2(1+\mu)}\left[C^{2}+0\right]\$=\alpha x^{2} \\ & \alpha=\frac{C E}{2(1+\mu)} \\ & \alpha=\frac{40\left(1-\mu^{2}\right)}{2 E} \frac{E}{2(1+\mu)} \\ & \alpha=10(1-\mu)=10(1-0.5)=5 \end{aligned}

Q3GATE 2022MCQ

The hoop stress at a point on the surface of a thin cylindrical pressure vessel is computed to be 30.0 MPa. The value of maximum shear stress at this point is

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📖 Explanation

Given, Hoop stress $(\sigma _h)=\frac{pd}{2t}=30MPa$ Maximum shear stress in plane $(\tau _{max})_{\text{\in plane}}=\frac{\frac{pd}{2t}-\frac{pd}{4t}}{2}=7.5MPa$

Q4GATE 2022NAT

The components of pure shear strain in a sheared material are given in the matrix form:

\varepsilon =\begin{bmatrix} 1 &1 \\ 1& -1 \end{bmatrix}

Here, $Trace(\varepsilon)=0$ . Given, $P=Trace(\varepsilon ^8)$ and $Q=Trace(\varepsilon ^{11})$ . The numerical value of $(P + Q)$ is ________. (in integer)

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📖 Explanation

\begin{aligned} |A-\lambda I|&=0 \\ \begin{vmatrix} 1-\lambda &1 \\ 1 & 1-\lambda \end{vmatrix}&=0 \\ (1-\lambda)(-1-\lambda)-1&=0 \\ (\lambda^{2-}1)-1&=0 \\ \lambda&=\pm \sqrt{2} \end{aligned}

Eigen values of $\varepsilon$ are $\sqrt{2}$ and $-\sqrt{2}$ Eigen values of $\varepsilon ^8$ are $(\sqrt{2})^8$ and $( -\sqrt{2})^8$ Eigen values of $\varepsilon ^{11}$ are $(\sqrt{2})^{11}$ and $(-\sqrt{2})^{11}$ $P=Trace(\varepsilon ^8) =$ sum of Eigen values $= ( \sqrt{2})^{8+}( -\sqrt{2})^8=32$ $Q=Trace(\varepsilon ^{11}) =$ sum of Eigen values $= ( \sqrt{2})^{11}+( -\sqrt{2})^{11}=0$ $P+Q=32+0=32$

Q5GATE 2019MCQ

The cross-section of a built-up wooden beam as shown in the figure is subjected to a vertical shear force of 8 kN. The beam is symmetrical about the neutral axis (N.A.) shown, and the moment of inertia about N.A. is $1.5 \times 10^9 mm^4$ . Considering that the nails at the location P are spaced longitudinally (along the length of the beam) at 60 mm, each of the nails at P will be subjected to the shear force of

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📖 Explanation

Shear stress acting at the joint at which the nail 'P' is resisting the shear.

\begin{aligned} \tau &=\frac{FA\bar{y}}{Ib}\\ &=\frac{(8 \times 10^3)(50 \times 100)(150)}{1.5 \times 10^9 \times 50}\\ &=0.08N/mm^2 \end{aligned}

The shear force acting on nail at 'P'. $F=\tau (60\times 50)=240N$

Q6GATE 2018MCQ

A cantilever beam of length 2 m with a square section of side length 0.1 m is loaded vertically at the free end. The vertical displacement at the free end is 5 mm. The beam is made of steel with Young's modulus of $2.0 \times 10^{11} N/m^{2}$ . The maximum bending stress at the fixed end of the cantilever is

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📖 Explanation

\begin{aligned} I&=\frac{(0.1)^4}{12} \\ \text{Deflection }\delta &=\frac{Pl^2}{3eI} \\ 5 \times 10^{-3}&=\frac{P(2)^3}{3 \times 2 \times 10^11 \times \frac{(0.1)^4}{12}} \\ P&=3125N \\ \text{Now, } M&= Pl=3125 \times 2\\ &= 6250Nm\\ \text{As, }\frac{M}{I} &=\frac{\sigma }{y}=\frac{E}{R} \\ \sigma _{max}&=\frac{M}{Z}=\frac{6250}{\frac{(0.1)^3}{6}}\\ &=37.5 \times 10^6 N/m^2=37.5MPa \end{aligned}

Q7GATE 2016NAT

A 450 mm long plain concrete prism is subjected to the concentrated vertical loads as shown in the figure. Cross section of the prism is given as 150 mm x 150 mm. Considering linear stress distribution across the cross-section, the modulus of rupture (expressed in MPa) is _______ .

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📖 Explanation

\begin{aligned} \mathrm{BM}_{\mathrm{Q}} &=11.25 \times 150 \\ &=1.6875 \times 10^{6} \mathrm{N}-\mathrm{mm} \\ \Rightarrow\quad \frac{\sigma}{y} &=\frac{M_{Q}}{I} \\ \text{where}\quad y &=\frac{150}{2}=75 \mathrm{mm} \text { and } 1=\frac{(150)^{4}}{12} \\ \Rightarrow\quad \sigma &=\frac{1.6875 \times 10^{6} \times 75}{(150)^{4} / 12}=3 \mathrm{MPa} \end{aligned}

Q8GATE 2013MCQ

A symmetric I-section (with width of each flange = 50 mm, thickness of each flange = 10 mm, depth of web = 100 mm, and thickness of web = 10 mm) of steel is subjected to a shear force of 100 kN. Find the magnitude of the shear stress(in $N/mm^{2}$ ) in the web at its junction with the top flange. __________

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📖 Explanation

\begin{aligned} 1 &=\frac{50 \times 120^{3}}{12}-\frac{40 \times 100^{3}}{12} \\ &=3.866 \times 10^{6} \mathrm{mm}^{4} \\ q &=\frac{S A \bar{y}}{16}=\frac{100 \times 10^{3} \times 50 \times 10 \times 55}{3.866 \times 10^{6} \times 10} \\ &=71.12 \mathrm{N} / \mathrm{mm}^{2} \end{aligned}

Q9GATE 2013MCQ

The 'plane section remains plane' assumption in bending theory implies:

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Q10GATE 2008MCQ

The maximum tensile stress at the section X-X shown in the figure below is

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📖 Explanation

The section at X-X may be shown as in the figure below: The maximum tensile stress at the section X-X is

\begin{aligned} \sigma &=\frac{P}{A}+\frac{M}{Z} \\ &=\frac{P}{b \times (d / 2)}+\frac{P \times (d / 4) \times 6}{b \times \left(d^{2} / 4\right)} \\ &=\frac{2 P}{b d}+\frac{6 P}{b d}=\frac{8 P}{b d} \end{aligned}

Q11GATE 2007MCQ

The shear stress at the neutral axis in a beam of triangular section with a base of 40 mm and height 20 mm, subjected to a shear force of 3 kN is

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📖 Explanation

$\text { Shear stress, } \tau=\frac{S A \bar{y}}{1 b}$ Width at a distance of $\frac{40}{3} \mathrm{mm}$ from the top

\begin{aligned} &=\frac{40}{20} \times \frac{40}{3}=\frac{80}{3} \mathrm{mm}\\ \therefore \tau &=\frac{3 \times 10^{3} \times \left(\frac{1}{2} \times \frac{80}{3} \times \frac{40}{3}\right) \times \left(\frac{1}{3} \times \frac{40}{3}\right)}{\left(\frac{40 \times 20^{3}}{36}\right) \times \frac{80}{3}} \\ &=\frac{3 \times 10^{3} \times 3200 \times 40 \times 36 \times 3}{162 \times 3200 \times 20^{3}} \\ &=10 \mathrm{MPa} \end{aligned}

Q12GATE 2006MCQ

If a beam of rectangular cross-section is subjected to a vertical shear force V , the shear force carried by the upper one-third of the cross-section is

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📖 Explanation

Shear stress at distance y from neutral axis,

\begin{aligned} \tau&=\frac{S A \bar{y}}{1 b} =\frac{V \times \left(\frac{d}{2}-y\right) \times b \times \left(\frac{(d / 2)+y}{2}\right)}{16} \\ \Rightarrow \quad \tau&= \frac{V \times \left(\frac{d^{2}}{4}-y^{2}\right)}{21}\\ \therefore \quad d F&=\tau \times b d y=\frac{V \times \left(\frac{d^{2}}{4}-y^{2}\right)}{21} \times b d y\\ \end{aligned}

Integrating both sides, we get

\begin{aligned} \Rightarrow \quad F &=\frac{V b^{d / 2}}{21} \int_{d / 6}^{d}\left(\frac{d^{2}}{4}-y^{2}\right) d y \\ &=\frac{V b}{21}\left[\frac{d^{2}}{4} y-\frac{y^{3}}{3}\right]_{d / 6}^{d / 2} \\ &=\frac{V b}{21}\left[\frac{d^{3}}{8}-\frac{d^{3}}{24}-\frac{d^{3}}{24}+\frac{d^{3}}{648}\right]\$ \\ &=\frac{V b}{21} \times \frac{d^{3}}{8} \times \frac{28}{81} \\ &=\frac{V b}{2 b d^{3}} \times \frac{d^{3}}{8} \times \frac{28}{81} \times 12=\frac{7 V}{27} \end{aligned}

Q13GATE 2006MCQ

A beam with the cross-section given below is subjected to a positive bending moment (causing compression at the top) of 16 kN-m acting around the horizontal axis. The tensile force acting on the hatched area of the cross-section is

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📖 Explanation

\begin{aligned} f &=\frac{M}{1} \times y \\ &=\frac{16 \times 10^{6} \times 12}{100 \times 150^{3}} \times 25=14.22 \mathrm{N} / \mathrm{mm}^{2} \end{aligned}

From similar triangles, we have

\begin{aligned} \therefore \text { Tensile force } &=\frac{1}{2} \times 25 \times 14.22 \times 50 \times 10^{3} \\ &=8.88=8.9 \mathrm{kN} \end{aligned}

Q14GATE 2006MCQ

I-section of a beam is formed by gluing wooden planks as shown in the figure below. If this beam transmits a constant vertical shear force of 3000 N, the glue at any of the four joint will be subjected to a shear force (in kN per meter length) of

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📖 Explanation

Considering shaded area 'abcd' of plank. $\text { Shear flow, } \quad a=\frac{V Q}{1}$ Shear flow is measured as force per unit length along the longitudinal axis of a beam and over the cross-section is as shown above.

\begin{aligned} Q &=A \bar{y} \\ Q &=(75 \times 50) \times 125 \\ &=468750 \mathrm{mm}^{3} \end{aligned}

Moment of inertia of I-section about NA.

\begin{aligned} 1&=350 \times 10^{6} \mathrm{mm}^{4}\\ V &=3000 N \\ q &=\frac{468750 \times 3000}{350 \times 10^{6}} \\ &=4.018 N / m m \end{aligned}

So, resistance offered by glue to keep I-section intact. $\mathrm{q}=4.018 \mathrm{N} / \mathrm{mm} \simeq 4 \mathrm{kN} / \mathrm{m}$

Q15GATE 2004MCQ

A homogeneous, simply supported prismatic beam of width B, depth D and span L is subjected to a concentrated load of magnitude P. The load can be placed anywhere along the span of the beam. The maximum flexural stress developed in beam is

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📖 Explanation

When the concentrated load is placed at the midspan, maximum bending moment will develop at the mid span

\begin{aligned} \text{Now}\quad f&=\frac{M}{l} y \qquad\left[\because M=\frac{P L}{4}\right]\\ &=\frac{\frac{P L}{4} \times \frac{D}{2}}{\frac{B D^{3}}{12}}=\frac{3 P L}{2 B D^{2}} \end{aligned}

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