Here
l=1.5m ,\delta = 10 cm , x = 0.5 m ,y =?
depression at the free end
\delta =\frac{Wl^3}{3YI}\\ \\ \frac{W}{YI}=\frac{3\delta}{l^3}Depression at a distance x from the fixed end
y=\frac{Wx^2}{2YI} \\ \\ =\;\frac{Wx^2(3l-x)}{YI \;\;\; 2\times3}=\left (\frac{3l-x}{2l^3} \right )x^2\delta \\ \\\;\;y=\left ( \frac{(3\times1.5)-0.5}{2\times(1.5)^3} \right )\times(0.5)^2\times0.1 \\ \\ =0.0148m.A steel wire of 1 mm radius is bent to form a circle of 10 cm radius. what is the bending moment and maximum stress, if Y= 2 x 10^11 N-m^2.
Solution: bending moment =YI/R
Here
I=\frac{\pi r^4}{4}= \frac{3.14 \times(10^{-3})^4}{4}bending\;moment\;= \frac{2 \times10^{11} \times3.14 \times(10^{-3})^4}{4 \times0.1}=1.57\;Nmstress= \frac {Yz}{R}For maximum stress the value of z should be maximum, i.e, z=r
maximum stress
\frac{Yr }{R}= \frac{2 \times10^ {11} \times(10^{-3})}{0.1}=2 \times 10^9 \;N m^{-2}