#brittle materials

[engineeringmath]

Theory of Failure List of Questions with Answers

Theory of Failure List of Questions with Answers

σ_1 is the only stress to use to find the factor of safety by the maximum normal stress theory of failure

True

False

A load is changing on a part from 1,000 lb to 5,000 lb, the average and range loads are

6,000 lb and 4,000 lb

5,000 lb and 1,000 lb

3,000 lb and 2,000 lb

1,000 lb and 5,000 lb

A part is loaded under service stress for 25,000 reversals, then under a higher stress for 35,000…

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[opusgoodinvestorsuv]

Crusher principles and others

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[engineeringtldr]

We’ve done this problem before for ductile materials, but let’s try it for brittle materials. The forces and dimensions are the same (The  length of the rod l = 5 in, the length of the arm a = 6 in, the rod diameter d = 1.25 in, and the force F = 1000 lb.) The only thing we’ll change is the material. We’ll say that the rod is cast iron with S_ut = 52500 psi and S_uc = -164000 psi. (Problem and image adapted from Machine Design: An Integrated Approach, 4th Ed., by Robert L. Norton.)

We already found the stresses for this situation. They won’t change because we changed materials, we’ll just be comparing them to something different. Here are the principal stresses we got:

If we plot these, here’s what we get:

The load line intersects the portion of the failure envelope controlled by the tensile strength, so we can calculate our safety factor based on that.

We could also calculate this out without graphing it.

Since this is a short rod, we should do transverse shear too. We found this earlier as:

This is a pure shear loading, so our principal stresses here are just +16709 psi and -16709 psi. Here’s what we get if we graph it:

And here’s what we get calculating it out:

[engineeringtldr]

More Than You Ever Wanted to Know About Mechanical Engineering, Part 22: Failure of Uneven Brittle Materials Under Static Loading

We’ve talked about the differences between even and uneven materials. Even materials have equal responses to tension and compression, while uneven materials are stronger in one than the other. Most ductile materials are even materials. Brittle materials may be even or uneven. If they are even, their point of failure can be found easily by comparing the normal stress they experience to their tensile strengths.

Things get a little more complicated when we talk about uneven brittle materials. These are typically stronger in compression and shear than in tension. When they’re subjected to a purely tensile stress, we can ascribe their failure purely to normal stresses like we would for even brittle materials. However, when they’re in compression, a failure is considered to be due to a combination of normal and shear stresses.

Let’s think about this for a second. For even brittle materials, we can say that their failure envelope is defined purely by their normal stress relative to their ultimate tensile strength. If we plot that, for a 2D stress situation, we get something that looks like this:

Combinations of σ1 and σ3 which fall inside the square (defined by the ultimate tensile strength of the material, S_ut) are safe. Failures occur on its borders.

If we do a similar plot for the max shear theory for failure of ductile materials, where we said that failure was due to shear only, we get a failure envelope that looks like this:

If we’re saying that failure of uneven brittle materials is due purely to normal stresses in tension and to a combination of shear and normal stresses in compression, we should expect the first quadrant of its failure envelope (where stresses are tensile) to look like the normal stress failure envelope, and the other quadrants, where some compression occurs, to look more like the hexagonal failure envelope we got for max shear theory. Moreover, we should expect our failure envelope to be asymmetric, since we know these materials are stronger in compression than in tension.

That is, in fact, exactly what we get.

Note that the corners of this failure envelope are defined by separate tensile and compressive strengths.

So, how do we use this to actually figure out safety factors for uneven brittle materials?

We’ll consider a 2D case and say our principal stresses are ordered from highest to lowest as σ3, σ2, and σ1, with σ2 being equal to zero, as the situation is 2D. This makes life easy for us, since we only have to consider the first and fourth quadrants of the failure envelope.

Let’s pick a random point inside this envelope that represents some particular state of stress. Call it A. If you draw a line from the origin through A, it will intersect the failure envelope at some point. This is called a load line.

Note that there are basically two regions to this graph that A can fall in:

It can fall in a region where its load line intersects a line of the failure envelope defined by the tensile strength alone.

Or it can call in an area where the load line intersects a line of the failure envelope defined by both tensile strength and compressive strength.

If state of stress A falls within the first region, then its safety factor can be calculated as purely the comparison of σ1 to the ultimate tensile stress.

However, if A falls in the second region, then its safety factor depends on both σ1 and σ3 and both S_ut and S_uc. (Basically, you’re solving for the intersection of the load line with the diagonal of the failure envelope.)

If you don’t want to go to the trouble of graphing this out, there’s another way to do this. If you recall, back when we were doing ductile safety factors, we used von Mises stresses as a way of getting a one-to-one comparison between a complex state of stress and simple tensile yield strength. We can do something similar here.

I won’t walk through the derivation, but you can calculate three separate effective stresses based on principal stresses. You compare all three of these and all three of your principal stresses, and whichever is largest is counted as the effective stress, which can be compared to ultimate tensile strength. Note that, if your largest stress is less than zero, the effective stress is deemed to be zero.

Note that this approach lets you deal with 3D stresses pretty easily.

[engineeringtldr]

More Than You Ever Wanted to Know About Mechanical Engineering, Part 21: Failure of Brittle Materials Under Static Loading - Even and Uneven Materials

We’ve talked about the mechanisms involved in the failure of ductile materials, which typically show significant yielding before their ultimate failure. Brittle materials may behave very differently, fracturing suddenly with very little yielding beforehand.

We’ll separate brittle materials into two classes: even and uneven brittle materials. Even brittle materials generally have compressive strengths equal to their tensile strengths. (Most ductile materials, which we’ve discussed earlier, are also even materials.) This can happen with wrought materials, like some tool steels, or other metals which have undergone hardening. (Many ductile metals can be worked to increase their hardness - this frequently leads to them becoming less ductile and more brittle.) Even brittle materials can be evaluated for failure by comparing their maximum normal stress to their tensile strengths.

Uneven materials have compressive strengths greater than their tensile strengths. This is typical of many materials which are cast. These materials normally contain microscopic voids due to the casting process. When they’re subjected to tension, these voids can produce stress concentrations and lead to rapid fracture. However, when they’re subjected to compression, the flaws of the casting are pushed together and their faces can support each other, allowing the material to survive much more force before failing.

These materials may also be much stronger in shear than in tension (although not as strong in shear as in compression). Moreover, if the material is subjected to compression, its shear strength will increase. You can imagine how this occurs - compression on the voids inside the material makes it harder for their faces to slip past each other.

You can see just how weirdly uneven brittle materials behave if you look at Mohr’s circle.

(Image from Machine Design: An Integrated Approach, 4th Ed., by Robert L. Norton.)

The Mohr’s circle on the left hand side shows the stresses seen in tension and compression tests for an even material. The one on the right shows the same two tests for an uneven material. The important takeaway from this is that, for uneven materials, their normal and shear strengths are interdependent.

Experimentally, this holds true in the compressive regime. Where the primary stresses are compressive, both shear and compressive stresses are responsible for failure. However, where the primary stresses are tensile, tensile stress alone causes failure in uneven brittle materials - shear does not play a part.

Clearly, we can’t handle uneven brittle materials like we did ductile materials or even brittle materials. We’ll need a theory which can account for these differences in behavior in tension and compression.

[charterpestcontrolrightj]

Crusher principles and others

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