Question

Maximum excess pressure inside a thin-walled steel tube of radius $r$ and thickness $\Delta r(<<r)$, so that tube would not rupture would be (breaking stress of steel is ${\sigma }_{max}$)

• A. ${\sigma }_{max}\times \frac{r}{△r}$
• B. ${\sigma }_{max}\times \frac{△r}{r}$
• C. ${\sigma }_{max}$
• D. ${\sigma }_{max}\times \frac{2△r}{r}$

Answer 1

Answer: (B)

${\sigma }_{max}\times \frac{△r}{r}$

Consider a small elements of the tube.
$2T\sin \theta =△p\times A$
where $△p=p_i-p_0$ and $A$ is the area of element. As $\theta$ is very small, $\sin \theta \approx \theta$ so, $2T\times \theta =△p\times l\times \left(2r\theta \right)$
$\Rightarrow △p=\frac{T}{lr}\sigma$ (stress developed in tube) $=\frac{△T}{△r\times l}$
where $△r\times i$ is the cross-sectional area.
$\sigma =\frac{△p\times lr}{△r\times l}=△p\times \frac{r}{△r}$
For no rupturing, $\sigma \le {\sigma }_{max}$
So, $△p\times \frac{r}{△r}\le {\sigma }_{max}$
$△p(max,value)={\sigma }_{max}\times \frac{△r}{r}$.