Question

A circular rod of diameter d and length 3d is subjected to a compressive force F acting at the top point as shown below. Calculate the stress at the bottom most supports point A.

• A. $\frac{12F}{\pi d_2}$
• B. $\frac{16F}{\pi d_2}$
• C. $-\frac{4F}{\pi d_2}$
• D. $-\frac{12F}{\pi d_2}$​​​​​​

Answer 1

The correct option is A $\frac{12F}{\pi d_2}$

In the given figure,

The given load is equal to a moment and a force at G to the stress

Bending stress,

${\sigma }_b=\frac{M}{Z}$

$WhereM=F\times \frac{d}{2}$

$Z=\frac{I}{y}$

$Where,I=\frac{\pi }{64}{d}^4,y=\frac{d}{2}$

$\therefore Z=\frac{I}{y}=\frac{\pi }{64}\frac{d^4\times 2}{d}=\frac{\pi d^3}{32}$

$\therefore {\sigma }_b=\frac{M}{Z}=\frac{F\times \frac{d}{2}}{\frac{\pi d^3}{32}}$

$=\frac{16F}{\pi d^2}\left(Tensile\right)$

Axial stress,

${\sigma }_a=\frac{Force}{Area}=\frac{F}{\frac{\pi }{4}{d}^2}$

$=\frac{4F}{\pi d^2}\left(Compressive\right)$

Combined stress $={\sigma }_a+{\sigma }_b$

$=-\frac{4F}{\pi d^2}+\frac{16F}{\pi d^2}=\frac{12F}{\pi d^2}$