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While learning about normal and tangential stress, we had a task, where I had to calculate the cross sectional area of a beam that could withstand the force applied to it, if the beam was sliced in the middle on an angle (Like two 90 deg trapezoids glued together).

I hope this makes sense.. having a language barrier here.

I was given the maximum allowed normal and tangential stresses under which it would not break.

As we know, Maximum tangential stress = 0.5 maximum normal stress.

I was given two values. For example, the max normal stress was 7.6 MPa and the tangential stress was 4.1 MPa and according to the rule, if tangential stress is 4.1 then the max normal stress can be 8.2 MPa. So the calculations are done based on the lowest value, since it can't hold any higher...

But if I'm only basing on one of the values and they're both connected to the equation, why does the task give me both? Is it just for the task, or in real life there are there materials that somehow have specifications for the tangential stress and normal stress boarder values which are different when using the equation?

I hope any of this makes sense.

The question basically is- How can the same material have two different values of stress it can withstand? Is it just for the task or..?

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  • $\begingroup$ Could you define what you say is "tangential stress"? Are you meaning "shear stress"? Also, in general, if you are having trouble with something that is "as we know ..." that is a good place to start defining more precisely. $\endgroup$ – hazzey Mar 13 '17 at 3:48
  • $\begingroup$ in some composites the maximum normal strength (in some orientation) is 100x the shear strength. $\endgroup$ – agentp Mar 15 '17 at 1:53
  • $\begingroup$ Editing the question so that people can't see what it asked doesn't really help. $\endgroup$ – JMac Mar 15 '17 at 10:55
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I'm not quite sure what you mean by tangential stress. If you mean shear stress, then shear strength does not necessarily have to be $0.5\cdot yield strength$.

The relationship between ultimate tensile strength and ultimate shear strength depend on what kind of material is used. A rule of thumb is, that ductile materials fail due to shear and brittle materials due to tension.

On Wikipedia you can find a table with some values for shear/yield relationships. E.g. for cast iron the ultimate shear strength is about 1.3 times the ultimate yield strength, for steels only about 0.75.

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There are two separate considerations: applied stress and allowable stress.

Allowable normal stress and allowable shear stress are a function of the material. They are not functions of each other. This is why the problem would need to give both values.

Now for applied stress. I assume you're looking at something like a bar under axial load and considering the state of stress for an inclined plane.

bar under axial load

stress on inclined plane

The shear stress and normal stress for the inclined plane are both a function of the pure axial stress (P/A) and the angle of the inclined plane.

normal and shear stress

So, by virtue of statics, maximum shear stress occurs on a plane inclined at 45 degrees and is half the pure axial stress.

The maximum normal stress occurs on a plane inclined at 0 degrees and is equal to the pure axial stress.

Back to your numbers:

Allowable normal stress = 7.6 MPa

Allowable shear stress = 4.1 MPa

Axial stress associated with allowable shear stress = 8.2 MPa

(Which is greater than the allowable normal stress, so the 7.6MPa controls and should be used to calculate the required cross sectional area.)

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Think of it like this. When you place a load on a bar, the bar will want to bend. The bending puts the top in compression and the bottom in tension, these are the tangential stresses.

You can bend a bar and it will spring back, but if you bend it too far it will stay bent.

The reason the bend will become permanent is from when you exceed the strength of the tangential stresses.

Once you understand the concept, the rest is just math!

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    $\begingroup$ Compression and tension are longitudinal stress. $\endgroup$ – Wasabi Mar 17 '17 at 19:02

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