Thin-walled Pressure Vessels

A D V E R T I S E M E N T


A tank or pipe carrying a fluid or gas under a pressure is subjected to tensile forces, which resist bursting, developed across longitudinal and transverse sections.

TANGENTIAL STRESS, σt (Circumferential Stress)
Consider the tank shown being subjected to an internal pressure $ p $. The length of the tank is $ L $ and the wall thickness is $ t $. Isolating the right half of the tank:

Free body diagram of cylindrical tank under pressure

The forces acting are the total pressures caused by the internal pressure $ p $ and the total tension in the walls $ T $.
$ F = pA = pDL $
$ T = \sigma_t A_{wall} = \sigma_t \, tL $

$ \Sigma F_H = 0 $
$ F = 2T $
$ pDL = 2(\sigma_t \, tL) $

$ \sigma_t = \dfrac{pD}{2t} $

If there exist an external pressure $ p_o $ and an internal pressure $ p_i $, the formula may be expressed as:

$ \sigma_t = \dfrac{(p_i - p_o) D}{2t} $

LONGITUDINAL STRESS, $ \sigma_L $
Consider the free body diagram in the transverse section of the tank:

Longitudinal stress of cylindrical tank

The total force acting at the rear of the tank $ F $ must equal to the total longitudinal stress on the wall $ P_T = \sigma_L \, A_{wall} $. Since $ t $ is so small compared to $ D $, the area of the wall is close to $ \pi Dt $

$ F = pA = p\dfrac{\pi}{4} D^2 $
$ P_T = \sigma_L \pi Dt $

$ \Sigma F_H = 0 $
$ P_T = F $
$ \sigma_L \, \pi Dt = p\dfrac{\pi}{4} D^2 $

$ \sigma_t = \dfrac{pD}{4t} $

If there exist an external pressure $ p_o $ and an internal pressure $ p_i $, the formula may be expressed as:

$ \sigma_t = \dfrac{(p_i - p_o) D}{4t} $

It can be observed that the tangential stress is twice that of the longitudinal stress.

$ \sigma_t = 2 \sigma_L $

SPHERICAL SHELL

Spherical tank under pressure

If a spherical tank of diameter $ D $ and thickness $ t $ contains gas under a pressure of $ p $, the stress at the wall can be expressed as:

$ \sigma_t = \dfrac{(p_i - p_o) D}{4t} $

A D V E R T I S E M E N T

 

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