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Beam loading question (simple) |
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| Aug1-08, 10:22 AM | #1 |
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Beam loading question (simple)
Hey guys,
Just a basic question. If I have a beam loaded like that in the picture below, I would like to calculate how the beam would break. Now I know the ultimate tensile strenght for the material. I have drawn the bending moment diagram and shear force diagram for the beam. From there I got the maximum bending moment and maximum shear stress I then used the formula (MY)/I and VQ/IB. Now how do I find if the beam would break? Do I just compare these two values to the ultimate tensile strenght? I am a bit lost. Thankyou for any help http://img137.imageshack.us/img137/3720/beammk1.jpg |
| Aug1-08, 10:27 AM | #2 |
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CS |
| Aug1-08, 11:21 AM | #3 |
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Failure theories abound. You have to define what you mean by "break" i.e. what is the failure mode you are most interested? If the beam yielding is enough, then use the yield strength. Do you have a maximum deflection criteria? That could be your limiter. Obviously you are looking at a combined loading scenario. If you go past yield, your equations are going to be invalid and a fair amount of error is introduced. So if you want to keep it simple, use yield as your failure criteria and select a failure criteria like Von Mises-Henke to calculate your actual stresses.
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| Aug1-08, 06:39 PM | #4 |
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Beam loading question (simple)
Thanks guys
edit1- the book has a formula sy=su-30,000psi (or something similar I guess I'll use that) At what point in the beam would I use to calculate the stresses for the mohrs circle? edit2-I should clarify this. I understand that the maximum bending moment and shear force occur at the very end of the beam. But the shear stress and bending moment varies across the cross section of the beam. I.e. maximum bending moment at the neutral axis, zero moment at the top and bottom. Max shear at top and bottom and zero at the neutral axis. It has a square cross section. So wondering where abouts the I calcuate the stresses? |
| Aug1-08, 06:41 PM | #5 |
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Sorry forgot to attach picture
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| Aug2-08, 08:42 AM | #6 |
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| Aug2-08, 09:22 PM | #7 |
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Ahh cheers thanks. So it is alright to calculate the bending stress at the neutral axis and the shear stress at the neutral axis then calculate the principle stresses? I just assumed they would have to be calculated at the same position (e.g. half way between the surface and the neutral axis)
Also, out of curiousity if fracture was to be considered my method of failure how would I approach this? Thanks a lot for your help! |
| Aug3-08, 12:42 PM | #8 |
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Ahh crud. You know, my reading comprehension really stinks sometimes. Even after I quoted you...
The stresses should be calculated at the outermost faces of the beam, not at the neutral axis. That after all is the definition of the neutral axis, i.e. the plane where the stress is equal to zero. The outermost faces of the beam are where the bending stresses are the highest. One other thing...You did not specify the boundary conditions of the beam. From your loading I am assuming you have a cantilevered beam |
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