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Date : 2016-06-23 12:24:21

Introduction:

For then walled pressure vessel the third stress is much smaller than the other two stresses and for this reason in can be neglected.

 

Description:

In order to analyse the thin walled cylinders, let us make the following assumptions :


• There are no shear stresses acting in the wall.
• The longitudinal and hoop stresses do not vary through the wall.
• Radial stresses σr which acts normal to the curved plane of the isolated element are neglibly small as compared to other two stresses especially when

The state of tress for an element of a thin walled pressure vessel is considered to be biaxial, although the internal pressure acting normal to the wall causes a local compressive stress equal to the internal pressure, Actually a state of tri-axial stress exists on the inside of the vessel. However, for then walled pressure vessel the third stress is much smaller than the other two stresses and for this reason in can be neglected.

 

Thin Cylinders Subjected to Internal Pressure:


When a thin – walled cylinder is subjected to internal pressure, three mutually perpendicular principal stresses will be set up in the cylinder materials, namely


• Circumferential or hoop stress


• The radial stress


• Longitudinal stress


now let us define these stresses and determine the expressions for them


Hoop or circumferential stress:

 

This is the stress which is set up in resisting the bursting effect of the applied pressure and can be most conveniently treated by considering the equilibrium of the cylinder.

 

In the figure we have shown a one half of the cylinder. This cylinder is subjected to an internal pressure p.

i.e. p = internal pressure
d = inside diametre
L = Length of the cylinder
t = thickness of the wall

Total force on one half of the cylinder owing to the internal pressure 'p'
= p x Projected Area
= p x d x L
= p .d. L ------- (1)

The total resisting force owing to hoop stresses σH set up in the cylinder walls
= 2 .σH .L.t ---------(2)

Because σ H.L.t. is the force in the one wall of the half cylinder. the equations (1) & (2) we get
2 . σH . L . t = p . d . L
σH = (p . d) / 2t

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