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[SOUND] Okay.

Â We've come to the last topic in the course for mechanics and materials part four.

Â And that's to go over again the failures, or theories of failure.

Â And so the learning outcome is to review the normal outcome stress failure,

Â which we've talked about in my before, the maximum shear stress theory,

Â which we've also talked about before,

Â in fact we talked about these in my mechanics and materials part three course.

Â The maximum normal stress we talked about all the way back to my mechanics and

Â materials part one course.

Â And then finally,

Â we're going to explain something called the maximum distortion energy theory.

Â 0:40

and so the failure theory, we went all the way back to my mechanical

Â materials part one course today, simple tension test, and

Â it was easy to perform and provided very good results for a variety of materials.

Â So we put a tensal force, we looked at the stress strain curve,

Â we predicted yield, we predicted fracture, we can find Young's Modulus,

Â a very nice test, very useful test.

Â But what about when we have more complex loading conditions,

Â when it's not just the simple axial load?

Â By have bi-axial loading or tri-axial loading.

Â Well, in those cases, the cause of failure may be unknown and

Â there are several theories that can be used for predicting failure for

Â these various types of loading.

Â And I'm going to focus on just three.

Â The first one we will review is the Maximum Normal Stress Theory, that says,

Â that when the normal stress in our actual engineering element is greater

Â than what's defined as our failure stress, then we're going to experience failure.

Â And it assumes that the material is subject to a combination of loads.

Â And it fails, can be a combination loads and

Â it fails when the maximum normal stress at any point exceeds the actual

Â failure stress as determined by that simple tension test, and

Â we said this by experiences was generally good for just brittle materials.

Â It was not good for ductile materials like steel, aluminum, plastics, etc.

Â For those ductile materials we also looked at the maximum shear stress theory

Â which is also called Tresca's Yield Crieterion and

Â this said that failure occurs when the shear stress, the actual shear stress,

Â is greater than what we define as our failure shear stress.

Â And this is good for ductile material and

Â that's because in yield, ductile materials.

Â Usually the yield is caused by

Â a slippage of crystal planes along the maximum sheer stress surfaces.

Â And so here's our simple tensional test.

Â Here's a stress block with that tension on it.

Â If we draw Mohr's Circle we see that tau failure is one half of

Â the normal failure based on the simple tension test.

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So failure will occur in this case when the maximum shear

Â stress at any point for a complex load reaches the failure shear stress,

Â which is equal to one half the normal yield stress,

Â or failure stress, as determined by the simple tension test for the same material.

Â So for the actual conditions the Mohr's Circle may look different but

Â this is the condition for failure.

Â 3:24

Finally, another very common and useful theory of failure is

Â the maximum distortion energy theorem or what's called von Mises Yield Criterion.

Â Now I'm not going to go into the details of the development of the theory.

Â You can do that on your own or take a more advanced class.

Â But this is generally regarded now as the best yield criteria for ductile materials.

Â Better than the the Tresca criterion in most cases.

Â And this states that failure occurs when the strain energy of distortion or

Â the change of shape of the element reaches a critical value.

Â Now when I talk about strain energy that's a concept that again I haven't

Â covered in this course.

Â You would have to cover it in a more advanced course.

Â But alternatively, you can also think of this theory, and

Â it gives the same results as yielding occurring when the sheer

Â stress on what we call an octahedral plane, which is a critical value.

Â And the octahedral.

Â Octahedral plains or

Â any plain whose normal makes equal angles with the three principle axis.

Â And so if you're interested in this, I'd recommend that you look at more

Â advanced discussion of the max distortion energy theorem or

Â the Von Mices Yield Criterion on your own.

Â 4:48

And if you do that you'll find

Â we're just dealing in this course with 2D plane stress.

Â We're going to say that the outer plane stress is equal to zero and so

Â by the maximum distortion energy theory this is equation we'll look at.

Â The failure stress squared is equal to the principle stresses squared-

Â the first principle stress times the second principle stress.

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Okay, and so, now let's wrap up by applying these failure

Â theories to the combined loading example that we did in the last module.

Â This was the loading condition, this was the stress

Â at point a and then we looked at more circle to find and orientation and

Â a prediction of the maximum principles or the principles stresses and

Â the maximum in plain share stress shown here, and so there are those values.

Â If we were using the various failure theories, let's see what we would say.

Â For the maximum normal stress theory, our maximum normal stress in nine point one

Â two mega pascals in tension and that would be, need to be,

Â less than or equal to what we defined as the normal stress in failure.

Â 6:05

For the maximum sheer stress theory, in this case we found the maximum

Â sheer stress for this complex loading was 4.79 mega Pascals, and

Â that would need to be less than the failure sheer stress as defined earlier.

Â Or, not earlier, but as defined for whatever material you're using and

Â then finally for maximum energy, distortion energy

Â theory I'm going to use that formula that I just showed on the last slide.

Â I'll put the values in and I find out that it ends up being nine,

Â there should be an equal sign here, and this is sigma failure

Â 6:48

and it's equal to the square root of these values, and I've put those in.

Â You find that the calculated stress for this theory must be less than,

Â which is 9.36 must be less than or equal to whatever the failure.

Â Our yield stress is, if yield is the definition of failure, for

Â the material that you're using.

Â And so, that's a good over view of the theories of failure.

Â >> [NOISE]

Â