cylindrical pressure vessel with inner pressure P.(adsbygoogle = window.adsbygoogle || []).push({});

Take a infinitesimal cube on the inner-wall of the vessel. When calculating the normal stress along the radial direction it is assumed that that it is equal to P on the both the negative [itex]\widehat{r}[/itex] and positive [itex]\widehat{r}[/itex] directions.

For an outer wall infinitesimal cube it is assumed that the normal stresses in radial directions are zero.

I was told in class that the normal stress in the radial direction will decrease linearly from P to 0 along the thin wall.

My question is this: If the pressure is slowly decreasing in radial direction shouldn't there be a shear force to compensate for the fact that that normal stress is decreasing. So for the infinitesimal cube on the inner wall I picture the face in negative [itex]\widehat{r}[/itex] direction to have Pressure P and the face on the positive [itex]\widehat{r}[/itex] direction to have pressure P-dP. This change in pressure should cause a shearing force on the 4 faces normal to the [itex]\widehat{r}[/itex] direction in the negative [itex]\widehat{r}[/itex] direction.

Why is this shear force neglected. If it is because it is so small compared to the pressure stresses so it is neglected then can someone explain to me why it is small. It can't reason why this force would be of such small magnitude when the shear forces must compensate for this Pressure over the small distance of the thin wall.

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# Thin wall Pressure Vessel(cylinder)

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