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 Author: dr funda Time: 01/26/00 09:31 PST This is a reply to message no. 125 by gjacc Reply | Original Message | New Topic | List Topics | List Messages on This Topic
 Current Topic:Bending Stress in Pin It sounds like you're talking about a pin loaded in shear.  The picture that we see is a pin inserted into a joint (basically, two or more parts that can rotate with respect to one another around the axis of the pin). The joint components are pulled apart with a total force F, and the centers of adjoining joint members are a distance t apart. Let's assume that this is the problem you're asking about (if we got the problem interpretation wrong, sorry!!). In this case, the failure mode of the pin/joint assembly will be either shear in the pin cross-section, or shear or tearing in the joint members themselves.  Rarely is bending stress in the pin an issue. This realization may be why your calculated stresses (using a bending stress formula) seem high.  We can see this by taking the ratio of maximum bending stress to shear stress.  We'll do this by dividing their respective formula. As you stated, the maximum bending stress is given by: sbending = (M?i>c)/(I) where c is the radius of the pin, and I for a circle is equal to: I = (p/4)?i>c4 The average shear stress in the pin is given by: sshear = (F/2)/A where we will use the same F/2 factor for the shear force as you used in your calculation for M. (Why the factor of 1/2 multiplying the shear force? Is your connection a butt-joint?) The factor A is the pin cross-sectional area (p?i>c2). If we assume that the effective moment M is given by: M = (F/2)?i>t then we can take the ratio of sbending/sshear to find: sbending/sshear = (2?i>t)/d where d is the pin diameter.  Note that if 2?i>t is significantly larger than the pin diameter d, then the calculated bending stress will be noticeably larger than the shear stress. Hope this helps, or if not, maybe it will at least get you to consider pin shear stress.  Good luck. eFunda Staff
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