Finding Average Shear Strain in a flexible connection of rubber and steel

by dabest1
Tags: average, connection, flexible, rubber, shear, steel, strain
 P: 3 1. The problem statement, all variables and given/known data [See Attachment] Given: Thickness, load, G Find:γ and horizontal displacement. 2. Relevant equations τ=γG Angle change should be the shear strain. 3. The attempt at a solution I tried plugging 800 into tau and the given value of G to get γ, but that isn't working. I also tried to draw a deformation, but I dont know how to deduce how much it will deform from the question. I also dont know where thickness is used in this problem. Attached Thumbnails
 P: 92 Why are you 'plugging' 800 into Tau? Isn't tau the denoted symbol for shear stress? Look carefully at the diagram given to you. Figure (b) is a section view of figure (a), therefore you can deduce the area over which P is applied using that. You can then calculate the shear stress, and since you are given the shear modulus of rubber, you can finally move on to calculate the shear strain of the rubber.
 P: 3 Will the area be 150 x 10? So I calculated the stress = (15000)/(1500 x 10^-3) = 10000 Then did 15000/800000 for y. It is wrong.
P: 92

Finding Average Shear Strain in a flexible connection of rubber and steel

No, it won't. The area of the applied force is 200mm x 150mm right? 10mm is the thickness of the steel plates. Now, try recalulating the shear stress on area where the force P is applied. Remember, the question asks for the average shear stress, so a bit more work is involved after.
 P: 3 Alright, so I did that. stress = force/area = 15000/(200x150x10^-3) = 500 tau = yG 500 = y*800000 y = 6.25 x 10^-4.. Which is still wrong...
 P: 92 Bare with me as I havn't done this in class yet, just read over it whilst looking through a book. It's looking for the average strain, yes? So I'm assuming you have to use the average force that is applied or the average stress you calculate (again, bare with me). Try it with both, and see what answer you get. Oh, and when converting you units of area, do it for both dimensions, not just the one. :)

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