GATE ME 2014 SET-3

3\times 3 real symmetric matrix such that two of its
eigen value are a \neq 0 b \neq 0 with respective eigen
vectors \left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]\left[\begin{array}{l}y_{1} \\ y_{2} \\ y_{3}\end{array}\right] \text{ if } a \neq b then
x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}=0 because they are orthogonal.
\therefore \qquad x^{T} y =0 \quad(\text { since } a \neq b)
\left[x_{1} x_{2} x_{3}\right]\left[\begin{array}{c}y_{1} \\y_{2} \\y_{3}\end{array}\right]=0
x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}=0

f(x) is continuous at any point
\text{if} \quad \lim_{x \rightarrow a^{-}} f(x)=\lim_{x \rightarrow a^{+}} f(x)=f(a)

\begin{aligned} \vec{F} &=x^{2} z \hat{i}+x y \hat{j}-y z^{2} \hat{k} \\ \nabla \cdot \vec{F} &=\frac{\partial}{\partial x}\left(x^{2} z\right)+\frac{\partial}{\partial y}(x y)-\frac{\partial}{\partial z}\left(y z^{2}\right) \\ \nabla \cdot \vec{F} &=2 x z+x-2 y z \\ \left.\therefore \nabla \cdot \vec{F}\right|_{1,-1,1} &=2 \times 1 \times 1+1-2 \times-1 \times 1 \\ &=2+1+2=5 \end{aligned}

Total \% of employed person
\begin{aligned} &=100-\frac{20+50}{2} \% \\ &=100-35 \%=65 \% \\ \text { Required prob } &=\frac{65}{100}=0.65 \end{aligned}

\begin{aligned} I &=\int_{1}^{3} \frac{1}{x} \mathrm{d} x=\frac{h}{2}\left[\left(y_{0}+y_{2}\right)+2 y_{1}\right] \\ &=\frac{1}{2}[1+0.33+2 \times 0.5]=\frac{2.33}{2}=1.165 \end{aligned}

Same fiber under point load at the center of a rotating steel shaft supported at ends is subjected to reversal of stresses. Here only maximum stress is given, its sign (nature) is not given
\begin{aligned} \therefore \qquad \sigma_{\min }&=-100 \mathrm{MPa}\\ \sigma_{\max } &=100 \mathrm{MPa} \\ \sigma_{m} &=\frac{\sigma_{\max }+\sigma_{\min }}{2}=0 \\ \sigma_{a} &=\frac{\sigma_{\max }-\sigma_{\min }}{2} \\ &=\frac{100+100}{2}=100 \mathrm{MPa} \end{aligned}
As the steel is ductile material, Soderberg criteria will be applied.
\therefore \quad \frac{\sigma_{m}}{\sigma_{y}}+\frac{\sigma_{a}}{\sigma_{e}}=\frac{1}{N}
\text{or }\quad 0+\frac{100}{200}=\frac{1}{N} \Rightarrow N=2

\begin{aligned} \tau _{max}&=\frac{2T}{\pi R^3}\\ \frac{\tau _2}{\tau _1}&=\left ( \frac{R_1}{R_2} \right )^3\\ &=2^3=8 \;\;(\because \; T_1=T_2=T) \end{aligned}

X: Displacement
F_{0}: Excitation force
c \omega X: Damping force
k X: Spring force
m \omega^{2} X: Inertia force

Coefficient of restitution,
C_{R}=\frac{\text { Relative speed after collision }}{\text { Relative speed before collision }}
After collision, both the masses move with same velocity. So, there is zero relative velocity between them,
\therefore \quad C_{R}=\frac{0}{5}=0

The maximum shear stress theory predicts that shear yield value \tau_{y} is 0.5 times the tensile yield value. This is about 15 % less than the value predicted by the distortion energy (or the octahedral shear) theory. The maximum shear stress theory gives values for design on the safe side. Also, because of its simplicity, this theory is widely used in machine design dealing with ductile materials.