|May22-06, 10:27 AM||#1|
Mechanical Principles. Engineering Components
Hi i am need of some help with a Mechanical Principle question.
The maximum load in a tensile test on a mild steel specimen is 94kN. If the test piece is 12mm diameter calculate the ultimate tensile stress.
(u.t.s) What is the working stress and greatest allowable load on a 36mm diameter bar? Use a factor of safety of 4.
I have n idea how to answer this question.
|May22-06, 11:23 AM||#2|
Stress is just Force/Area
Then you know the stress the steel can take before it breaks (it should be yield strength instead of ultimate however, since I think the maximum force occurs right before the sample begins to yield), so you can then use that to find the force it would take to duplicate that force in the sample.
Then apply the safety factor... Done.
|May23-06, 03:12 PM||#3|
The ultimate tensile stress is independent of the force and dimensions of a particular material. It is relatively a constant determined bythe material. For example, a piece of Alumnim - T6061 with a diameter of 12 cm has the same ultimate tensile stress as a similar piece with a diameter of 1264213 km. Granted that stress is Force/Area, the larger piece will be able sustain a much larger force. You can just look that number up online for the particular alloy of steel you are using as steel is quite common.
For a factor of safety of 4, this means the maximum allowable stress on the specimen must be 1/4 the ultimate stress. So if the ultimate stress was 40 Mpa, then the maximum allowable stress would be 10Mpa. Then we just set 10Mpa = Force (MegaNewtons)/ Area (meters^2). you know the diamater in mm so you'll need to do some basic converting. You can then solve this equation for the force (which will come out in MegaNewtons) and then convert that to whatever you need.
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