Help with finding cross-sectional area when calculating stress, Please explain h

Question

Help with finding cross-sectional area when calculating stress, Please explain how to get these areas, I thought cross-sec area of a cylinder is just pi*r^2 and I dont understand how A2 for the plate was found these area formulas don't make sence please someone explain

A load P is applied to a steel rod supported as shown by an aluminum plate into which a 0.6-in.-diameter hole has been drilled. Knowing that the shearing stress must not exceed 18 ksi in the steel rod and 10 ksi in the aluminum plate, determine the largest load P which may be applied to the rod. For steel A1 = pi dt = pi(0, 6)(0, 4) = 0.7540 in2 tau1 = p / A1 P = A1 tau1 = (0.7540)(18) = 13.57 kips For aluminum A2 = pi dt = pi(1.6)(0.25) = 1.2566 in2 tau2 = p / A2 P = A2 tau2 = (1.2566)(10) = 12.57 kips Limiting value of P is the smaller value P = 12.57 kips

Explanation / Answer

Well, A1 and A2 are "Shearing" areas.

Imagine that when you apply the load, the steel rod fails in shear (instead of tension). Had it failed in tension, you are right that we should use pi*r^2 for calculating stress. Its all about which cross-section fails.

(pi*d) is the cicumference of rod. And when you multiply it by thickness t, it'll give you the shearing surface area.

Still got doubts?? Feel free to ask.

Cheers!!

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