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A thin-walled long cylindrical tank of inside radius r is subjected simultaneously to internal gas pressure p and axial compressive force F at its ends. In order to produce ‘pure shear’ state of stress in the wall of the cylinder, F should be equal to (a) 2prπ (b) 22prπ [CE: GATE-2006] (c) 23prπ (?
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A thin-walled long cylindrical tank of inside radius r is subjected si...
Solution:

The stress state in the wall of a thin-walled cylindrical tank is a combination of hoop stress and longitudinal stress. To produce pure shear state of stress in the wall of the cylinder, the longitudinal stress should be zero. This can be achieved by applying an axial compressive force F at the ends of the cylinder.

Calculation:

The hoop stress in the wall of the cylinder is given by:

σh = pd/2t

where p is the internal gas pressure, d is the diameter of the cylinder, and t is the thickness of the wall.

The longitudinal stress in the wall of the cylinder is given by:

σL = F/A

where A is the cross-sectional area of the cylinder.

To produce pure shear state of stress in the wall of the cylinder, the longitudinal stress should be zero. This can be achieved by making F equal to 2prπt.

Therefore, the answer is (a) 2prπ.
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A thin-walled long cylindrical tank of inside radius r is subjected simultaneously to internal gas pressure p and axial compressive force F at its ends. In order to produce ‘pure shear’ state of stress in the wall of the cylinder, F should be equal to (a) 2prπ (b) 22prπ [CE: GATE-2006] (c) 23prπ (?
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