Question

Maximum excess pressure inside a thin-walled steel tube of radius r and thickness `/_ (lt lt r)`, so that the tube would not rupture would be (breaking stress of steel is `sigma_("max")`
A. `sigma_("max")xxr/(/_ )`
B. `sigma_("max")xx(/_ )/r`
C. `sigma_("max")`
D. `sigma_("max")xx(/_2r)/r`

Answer 1

Correct Answer - B
Consider a small element of the tube.
`2Tsintheta=/_pxxA`
Where `/_p=p_(i)-p_(0)` and `A` is the area of element. As `theta` is very small `sintheta~~theta` so, `2T xxtheta=/_pxxlxx(2rtheta)`

`implies/_p=T(lr)sigma` (stress developed in tube) `=(/_ )/(/_ xxl)`
where `/_ xxi` is the cross sectional area
`sigma=(/_pxxlr)/(/_ xxl)=/-pxxr/(/_ )`
so, `/_pxxr/(/_ )lesigma_("max")`
`/_p("max,value")=sigma_("max")xx(/_ )/r`