ProblemSet2_solutions

# Note that we can write the area da as θ rd da the

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Note that we can write the area dA as θ rd dA = . The force component along the 2 direction is therefore

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θ θ θ d pr dA p dA p dF cos cos 2 2 = = = . We integrate over θ from -90° to +90°: = = 2 / 2 / 2 2 cos π π θ θ pr d pr F . Since we only looked at a cross section, we have to multiply this with the length of the cylinder, l. The force acting on the wall is therefore 2prl and the hoop stress is therefore t pr = 2 σ . a.3 Radial stress The stress in the through thickness direction (radially outward) is simply p. If you look at the stresses in the 1 and 2 direction, we have ratios of r/t times p. Since r/t is in our case of the order of 1000, the stress in the through thickness direction is about 1000 times smaller than the stresses in the 1 and 2 direction and we neglect the stress in the through thickness or 3 direction in the following. ===================================================================== Step 2: Use stresses for the generalized Hooke’s law: We now use the three stresses radial 3 hoop 2 axial 1 , , σ = σ σ = σ σ = σ for the generalized Hooke’s law and obtain: ( ) [ ] + ν = σ + σ ν σ = ε = ε p t pr t

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• Spring '11
• Cardona
• mechanics, Ratio, axial strain, Cylinder stresses

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