# MD 6min #6 - mode of failure



## Firefly (Mar 13, 2010)

Without plotting the design region, can you determine the failure mode?

I did the principal stress equation, and was able to determine that the minimum principal stress has a Safety Factor &lt;2. How am I supposed to know that it fails in Shear and Not Principal stress?

Problem:

a cube with 2" sides, Sy=35,000 psi , Sc= -125,000 psig. Ft=56,000 lbf, Fc=-100,000 lbf, shear=10,000 psi. A Factor of Safety &gt;2 is req'd.

What is the failure mode for this material? Compression, Principal Stress, Tension, Shear.


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## BrianC (Mar 28, 2010)

I have finally gotten around to this problem, and I am struggling with it also.

The thing that confuses me the most is that the problem states the "cube experiences *consecutive forces* of...". To me this says that each load defines a unique stress state, rather than a combined loading scenario. The solution goes on to solve the problem as a combined loading situation. I understand the solution procedure, but I don't understand why it is treated as a combined loading.


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## Bman (Mar 29, 2010)

Just doing this problem now. The solution states that the principle stresses are not a failure mode, but used to determine whether the design will fail due to the shear stress applied. So I guess if it fails due to a principle stress then that indicates that it will fail in shear.

MD is not my strong side, but it looks like the solution analyzes each of the stresses independently by checking the tensile stress and then the compressive stress of which both pass the &gt;2 SF test. They then analyze the shear stress (created by the tensile and compressive stresses) to find that it doesn't meet the criteria. Now I guess if principle stresses aren't a failure mode, you could have answered D) Shear after you determined that the tensile and compressive stresses were within the safety factor since shear is the only acceptable answer left....


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## erniepower (Apr 9, 2010)

I was under the impression that since it is a "Brittle Material" Shear stress limit is hard to calculate. I am not sure why they call the forces consecutive, but if you treat them as all being applied to the block at the same time, you can calculate the Principle stresses. then by drawing the acceptable design area using modified mohr theory (which I have come to find out is used exclusively for "brittle" materials) you can plot the resultant point using the principle stresses as x and y values. you will see that the point falls outside of the acceptable design area. Since the principle stresses do not exceed the ultimate stresses, and the part still fails, I drew the conclusion that it must have failed in shear.

just my $.02


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