Explanation Given information: A thin-walled spherical pressure vessel having an inner radius r and thickness t . The spherical pressure vessel subjected to an internal pressure p . Calculation: The thin wall analysis is used to find the normal stress. For the plane stress σ 3 = 0 . Find the normal stress as shown below. σ 1 = σ 2 = p r 2 t Here, p is the gage pressure, r is the inner radius of the pressure vessel, and t is the thickness of the pressure vessel. Apply Hooke’s law as shown below. Find the normal strain ε 1 . ε 1 = 1 E ( σ 1 − ν ( σ 2 + σ 3 ) ) Here, E is the modulus of elasticity and ν is the Poisson’s ratio. Substitute p r 2 t for σ 1 , p r 2 t for σ 2 , and 0 for σ 3 . ε 1 = 1 E ( p r 2 t − ν ( p r 2 t + 0 ) ) = p r 2 t E ( 1 − ν ) Find the normal strain ε 2 . ε 2 = 1 E ( σ 2 − ν ( σ 1 + σ 3 ) ) Substitute p r 2 t for σ 1 , p r 2 t for σ 2 , and 0 for σ 3 . ε 2 = 1 E ( p r 2 t − ν ( p r 2 t + 0 ) ) = p r 2 t E ( 1 − ν ) Find the normal strain ε 3 . ε 3 = 1 E ( σ 3 − ν ( σ 1 + σ 2 ) ) Substitute p r 2 t for σ 1 , p r 2 t for σ 2 , and 0 for σ 3 . ε 3 = 1 E ( 0 − ν ( p r 2 t + p r 2 t ) ) = − ν p r t E Volume of the spherical pressure vessel. V = 4 3 π r 3 Change in volume of the spherical pressure vessel. V + Δ V = 4 3 π ( r + Δ r ) 3 = 4 3 π r 3 ( 1 + Δ r r ) 3 Here, Δ V < < V and Δ r < < r

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Chapter 10.6, Problem 10.55P

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To show that: The increase in the volume within the vessel is ΔV=2pπr4Et(1−ν) using small-strain analysis.

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Chapter 10.6, Problem 10.54P

Chapter 10.6, Problem 10.56P

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To show that: The increase in the volume within the vessel is Δ V = 2 p π r 4 E t ( 1 − ν ) using small-strain analysis.

Solution Summary: The author explains how a thin-walled spherical pressure vessel has an inner radius r and thickness t.

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To show that: The increase in the volume within the vessel is ΔV=2pπr4Et(1−ν) using small-strain analysis.

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Chapter 10 Solutions