Derivation of the expression for hoop and longitudinal stresses in cylindrical vessels PDF

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Hoop Stress...

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Derivation of the expression for hoop and longitudinal stress in cylindrical vessels. (Supporting material for lecture 6). Let us first consider the force that exerted on each half of a thin walled cylindrical section as a consequence of the internal pressure. In order to determine this force we need to integrate the xcomponent (i.e. in an arbitrarily chosen radial direction) of the pressure force (i.e 𝑃𝛿𝐴) acting over the side of the cylindrical vessel as shown b elow….

𝛿𝜙

However, we can write that 𝛿𝐴 = 𝐿 𝜋𝐷 2𝜋 where 𝛿𝜙 is a small increment of angle, hence, 𝜋

𝐿𝐷 2 𝐹= ∫ 𝑃 cos 𝜙 𝑑𝜙 2 −𝜋 2

Hence, 𝐹=

𝜋

𝑃𝐿𝐷 2 ∫ cos 𝜙 𝑑𝜙 2 −𝜋

𝐹=

2

𝑃𝐿𝐷 [sin 𝜙]𝜋/2 −𝜋/2 2 𝐹 = 𝑃𝐿𝐷

i.e. the projected area of the surface in the plane perpendicular to the x axis multiplied by the pressure. Now, if the vessel is to maintain its shape this force must be balances by an equal and opposite force in the wall of the vessel as shown below….

But this force will be distributed over an area equal to the product of the cylindrical section length L and the wall thickness t so we have …. 𝜎𝐻𝑜𝑜𝑝 =

𝑃𝐿𝐷 𝑃𝐷 = 2𝐿𝑡 2𝑡

There will also be a force acting along the length of the vessel which will again be equal to the product of pressure and projected area perpendicular to the axis of interest, i.e.. 𝐹=

𝑃𝜋𝐷2 4

Again, this must be balanced by a force in the shell of the vessel which will be distributed over the wall cross section. The cross section perpendicular to the axis of interest is.. 𝐴= 𝐴=

𝜋(𝐷 + 2𝑡)2 𝜋(𝐷)2 − 4 4

𝜋 2 𝜋 [𝐷 + 4𝐷𝑡 + 4𝑡 2 − 𝐷2 ] = [4𝐷𝑡 + 4𝑡 2 ] = 𝜋𝑡[𝐷 + 𝑡 ] 4 4

Hence the longitudinal stress will be

𝜎𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙

𝑃𝜋𝐷2 4 = 𝜋𝑡[𝐷 + 𝑡]

𝜎𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 =

𝑃𝐷 2 4𝑡 [𝐷 + 𝑡 ]

But since, for a thin walled vessel 𝐷 ≈ 𝐷 + 𝑡 𝜎𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 = These stresses are shown on the diagram below….

𝑃𝐷2 𝑃𝐷 = 4𝑡𝐷 4𝑡...