Problem: A vertical rod is attached at point A to the cast iron hanger shown below.  The allowable normal stresses in the hanger are = + 5 ksi and - 12 ksi.  Determine the largest downward force and the largest upward force which may be exerted by the rod.

Solution: First, I find the centroid and the moment of inertia:

A table is a convenient way to get the centroid:
 

Now for the moment of inertia...note that the base of the rectangles is parallel to the N.A.:

When the force is downward, we have the free body diagram:

where the moment M is equal to P(1.5+1.7) = 3.2P kip-in. We need to use the allowable stresses at C and D to compute the allowable force P. For the force downward as shown, D is in compression; the stress found here will be compared to the -12 ksi. Whatever stress is found at C will be compared with the 5 ksi. (Note: cast iron is brittle; it can handle much more in compression than in tension.)

The correct answer is the smaller value, 7.86 kip. Now we do the same thing for the upwardforce. The free body above changes - the forces and moment get reversed.

Again, we take the smaller value - the 9.15 kip.

³This is Problem 4.128 in Beer & Johnston’s Mechanics of Materials, 2nd Edition, McGraw-Hill.