First slide

Elastic behaviour of solids

Question

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

Moderate

A

$σ_{max} \times \frac{r}{Δr}$
B

$σ_{max} \times \frac{Δr}{r}$
C

$σ_{max}$
D

$σ_{max} \times \frac{Δ 2 r}{r}$

Solution

Consider a small element of the tube.
$2 Tsin θ = Δp \times A$
$where Δp = p_{i} - p_{0} and A is the area of element. As θ is very$
$small, \sin θ \approx θ so, 2 T \times θ = Δp \times l \times (2 rθ)$
$\Rightarrow Δp = \frac{T}{lr} σ (stress developed in tube) = \frac{ΔT}{Δr \times l}$
$where Δr \times l i s the cross-sectional area.$
$σ = \frac{Δp \times lr}{Δr \times l} = Δp \times \frac{r}{Δr}$
$For no rupturing, σ \leq σ_{max}$
$So, Δp \times \frac{r}{Δr} \leq σ_{max}$
$Δp (max, value) = σ_{max} \times \frac{Δr}{r}$

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