Recent content by Rybose

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"That description doesn't make sense to me. But that might be because I am applying that definition to the entire cross section instead of just a differential element." Stress states are defined for differential elements so that equilibrium is met. You tried to define a single stress...

T = F*(r-h) To derive the formula that you found, simply take the sum of the moments about the tip of the obstacle. sum of Moments: F*(r-h) - mg*sqrt[r^2 - (r-h)^2] = 0

I don't have any practical experience with this, but your first step is to calculate the mass moment of inertia of the arm + the mass rotating about the pivot point. Second, use your motion requirements calculate the required angular acceleration of the arm. T = I * alpha Check out...

Attached is a picture of what I attempted to explain in my previous post. Yes, on the surface and underneath the load, the lateral stress is simply the load divided by the area. On the top surface with no load, the lateral stress is zero. On the bottom surface the lateral stress is zero...

I think it's important to clarify that the existence of \sigma_y depends on the type of load applied, not the geometry of the part. A beam in direct tension will only have \sigma_x stress. A beam in torsion will only have \tau_{xy} stress. A beam in direct shear will only have \tau_{xy}...

This problem difficult because you can't determine the reaction moments through equilibrium alone (statically indeterminate). Therefore you need to use information about the deflection to solve the problem. With the symmetry of the problem you know: y(0) = y(L) = 0 \frac{dy}{dx}_{x =...

I used Rao as an undergrad. I wasn't very helpful. Try "Mechanical Vibrations" - Hartog (1985) It's a very old book, but the explanations about the fundamentals are the best I've read. It's available for around $12 on amazon, and it's free here...