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Mechanical Engineering
Single shear element in stress tensor: Finding Von Mises
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[QUOTE="jasc15, post: 5454516, member: 49304"] Thanks for the reply. How am I to understand this from the point of view of a simple strength of materials problem, where I am asked to find the shear stress in a shaft with an applied torque, and to compare it to the shear allowable for that material. For example, a torque of 150 in*lb and a shaft radius of 0.15 in results in an applied shear stress of τ = T*c/J = 150*0.15/(Π*.15[SUP]2[/SUP]) = 28,294 psi. The shear allowable of the material I am working with is 31000 psi, and the uniaxial yield allowable is 36000 psi. Using the simple shear stress calculated, I get a margin of safety of 31,000/28,294 -1 = 0.10 Using Von Mises, I get √3*28,294 psi = 49,007 psi, with a resulting margin of 36,000/49,007 -1 = -0.27 Is this simply a matter of choosing how conservative you want to be in design? I wouldn't be comfortable using only the first method knowing the second method results in negative margin. [/QUOTE]
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Single shear element in stress tensor: Finding Von Mises
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