Calculating Maximum Distributed Load on Beam Using Failure Theories

[AdamX1980X]

Having a hard time with failure theories. Beam length 20 feet, 6"h x 4"w. sigma yield is 40 ksi. How to calculate the maximum distributed load the beam could carry using von Mises and Tresca failure theories. I have my shear and moment diagrams drawn and know they are right. I found that the max stress will be dead center due to bending. So sigma x= My/I . according to Tresca tmax = sigma yield/2. So how do I make it all relate to find my sigma1 and sigma 2 to plug into the failure theory equations. Any help would be appreciated.

 

[Mapes]

Hi AdamX1980X, welcome to PF. So you're looking for the principal stresses σ₁ and σ₂? If your shear stresses are negligible (a common assumption with long beams), than these are just the largest and second largest normal stresses.

More generally, the principal stresses are the eigenvalues of the stress matrix. Or you can get them graphically from Mohr's circle, if you're familiar with that.

 

[AdamX1980X]

I have not covered the Eigen values of the stress matrix. I have covered Mohr's circle but I think I solved it conceptually. After drawing the shear and moment diagrams and calculating some equations I found the maximum stress to come from bending in the top or bottom of the beam. So sigma yield is sigma x which is sigma 1 I have a 0 value for sigma 2 which is sigma y. After using the stress equations I found that the yield is the same and the load is the same in both cases. Does this sound correct to you. I arrived at this from an example from the book.

 

[Mapes]

Yes, this sounds reasonable.

 

[Studiot]

I am not sure why you are doing this?

Von Mises and Tresca criteria are point failure criteria.
So they are only valid as failure criteria in structural elements for loadings that produce uniform stress distributions eg direct stresses.

The beam that you have described will not fail just because the extreme fibres have reached one of these criteria.

You need to use plastic theory for this.

go well

 

[AdamX1980X]

The reason why I did that is because the moment at the dead center of the bottom or the top is the greatest source of stress. I checked values for other points of the beam but this turned out to have the greatest stress. Indeed this was correct. Thanks for all the help.

 

[Studiot]

The reason why I did that is because the moment at the dead center of the bottom or the top is the greatest source of stress. I checked values for other points of the beam but this turned out to have the greatest stress. Indeed this was correct. Thanks for all the help.

Perhaps you have not yet covered plastic analysis, but at failure of the beam your statement above is not true.

A beam such as you describe has considerable strength above the point where the midspan extreme fibres are stressed beyond their limits and yield.

The failure moment is called the full plastic moment and only occurs when a plastic hinge develops in a simply supported beam.
Other modes of support require more than one plastic hinge (failure section) to develop.