Question

. A closed cylindrical vessel made of steel plates 4 mm thick with plane ends, carries fluid under pressure

of 3 N/mm2 The diameter of the cylinder is 25cm and length is 75 cm. Calculate the longitudinal and hoop

stresses in the cylinder wall and determine the change in diameter, length and Volume of the cylinder. Take

E =2.1x105 N/mm2 and 1/m = 0.286​

Answer 1

The longitudinal stress in the cylinder wall is caused by the internal pressure, and can be calculated using the formula:

σ_longitudinal = P/A

Where P is the internal pressure, and A is the cross-sectional area of the cylinder.

The hoop stress in the cylinder wall is caused by the internal pressure acting circumferentially, and can be calculated using the formula:

σ_hoop = P/2t

Where t is the thickness of the cylinder wall.

The diameter of the cylinder is 25 cm, or 0.25 m. The length of the cylinder is 75 cm, or 0.75 m. The thickness of the cylinder wall is 4 mm, or 0.004 m.

The internal pressure is 3 N/mm² .

The cross-sectional area of the cylinder is:

A = π * $(d/2)^2$ = π * (0.25/2)² = 0.196 m²

So the longitudinal stress is:

σ_longitudinal = $P/A = 3 / 0.196 = 15.306$N/mm²

The hoop stress is:

σ_hoop = P/2t = $3 / (2 * 0.004) = 750$ N/mm²

Regarding the change in diameter, length and volume, we can use the formula

ΔL/L = (1/E) * (σ_longitudinal - σ_hoop)

ΔD/D = (1/E) * (σ_hoop)

ΔV/V = (1/E) * (σ_longitudinal + 2 * σ_hoop)

By substituting the value we have:

ΔL/L =$(1/2.1x105) * (15.306 - 750) = -0.0037$

ΔD/D = $(1/2.1x105) * (750) = 0.000357$

ΔV/V =$(1/2.1x105) * (15.306 + 2 * 750) = 0.0014$

With the longitudinal stress of 15.306 N/mm², hoop stress of 750 N/mm², change in length -0.0037, change in diameter 0.000357 and change in volume of 0.0014.

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