# Bending and Shear Stresses

 Question 1
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) ?
 A 30 B 40 C 45 D 55
GATE CE 2023 SET-2   Solid Mechanics
Question 1 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}
 Question 2
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)
 A 12 B 28 C 32 D 46
GATE CE 2022 SET-2   Solid Mechanics
Question 2 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 $= ( \sqrt{2})^{11}+( -\sqrt{2})^{11}=0$
$P+Q=32+0=32$

 Question 3
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
 A 7.5 MPa B 15.0 MPa C 30.0 MPa D 22.5 MPa
GATE CE 2022 SET-1   Solid Mechanics
Question 3 Explanation:
Given,
Hoop stress $(\sigma _h)=\frac{pd}{2t}=30MPa$
Maximum shear stress in plane $(\tau _{max})_{\text{in plane}}=\frac{\frac{pd}{2t}-\frac{pd}{4t}}{2}=7.5MPa$
 Question 4
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 A 60N B 120N C 240N D 480N
GATE CE 2019 SET-1   Solid Mechanics
Question 4 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'.
$F=\tau (60\times 50)=240N$
 Question 5
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
 A 20.0 MPa B 37.5 MPa C 60.0 MPa D 75.0 MPa
GATE CE 2018 SET-1   Solid Mechanics
Question 5 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}

There are 5 questions to complete.