I have a steel plate 3' long x 2 wide x 3/16 thick. I have a 1/2 diameter hole with the center 3/4 from the end. Tensile strength is 38,000psi. A chain goes through the hole and is connected to a load. how does one determine how much tension the chain can have before it rips through the hole?
oo that's a tough one. dont remember... just finished that class...sorry
Assuming that the chain can take the breaking load of the plate, the plate will fail at the area between the hole and the end, in double shear. The shear strength of the material should therefore be known for us to be able to calculate the breaking force. Approximately, the shear strength of steel is about 60% of the tensile strength. For ordinary mild steel the tensile strength is about 60,000 psi, not 38,000 psi. 38,000 psi is just about the yield point of mild steel. Using 60,000 psi tensile strength, the approximate shear strength of steel therefore is 36.000 psi. The net area between the hole and the end of the plate is: A = 3/16(3/4 - 1/4) = 0.09375 in^2 The force necessary to rip through the area which is also the tension on the chain is: T = 2 x 0.09375 x 36,000 = 6750 lbs
Your title asks for the working load, yet your problem asks for the failure load (ripping the plate). These are two different things, since there must be a factor of safety on the failure load to get to the working load. The factor of safety varies from code to code and depending on what the plate/chain is being used for. The failure load would be the net cross section of the plate (after subtracting out the hole) x the strength of the plate. In this case it would be (2 - .5)*3/16*38000psi=10,687.5 lbs You would also need to check the strength of the chain to make sure that it doesn't break before the plate.
