The one thing that comes to my mind is that you have to be able to calculate ultimate tensile strength. For example, you have a .75" x .75" coupon that broke at a load of 50,000 lbs, what is the ultimate tensile strength? Quite common question.
.75 x .75 = .5625 in2. (Area)
50,000 lb/.5625 in2 = 88,888 lb/in2.
Also, be able to back calculate that question, if it were asked like, calculate the load the test coupon broke at if the ultimate tensile strength is 88,888 lb/in2 and the area of the coupon is .5625 in2.
x/.5625 in2 = 88,888 lb/in2
x = 88,888 lb/in2 * .5625 in2
x = 50,000 lb
What is the area of the coupon if the load is 50,000 lb and the ultimate tensile strength is 88,888 lb/in2?
50,000 lb/x = 88,888 lb/in2
x = 50,000 lb/88,888 lb/in2
x = .5625 in2