# Help with bending/shear stress Calculation



## dnpatterson1 (Nov 21, 2013)

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A plate (0.25"thick X 1.38"height X 4.74"length) is fillet welded (1/8") all the way around each side.

The weld material and base material are all Alloy 625, which has a yield strength of 60ksi and a tensile strength of 120ksi.

Both sides are fixed(welded) and the force will be centered in the middle of the bar, equal distance from each side it is welded to.

*What is the maximum force that the weld can take before it breaks?*

_I got stuck after gathering the following info.(if this helps at all?!?):_

_Moment = M =(Force * length)/8 - _

_Inertia = I = (d^2/6)(3b+d)_

_Bending Stress = (M * c) / I_

_Shear stress = Force / throat area of weld_

_Force = No force give, my best guess is using yield/tensile strength to solve for Force???_

_Stress Combined Equivalent = sq.rt.[ (Bending stress/2)^2 + Shear stress^2]_

Please help if you can, I am lost.


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## majormajor (Nov 21, 2013)

The geometry is not clear from your description. If you post an engineering drawing (jpg, dwg, whatever), I can get into detail.

In general, though

1. Look at the beam diagrams in the MERM appendix. From there you can get the moment at the joints.

2. You can calculate the shear in your head, and it looks like you've already done.

3. With the shear and moment in hand, look at the welding section in Shigley. There are two cases, one for welded joints in torsion and one for welded joints in bending. You obviously want bending. I won't walk you through the end of the problem yet, because you need to get intimately familiar with this material.

If you don't have Shigley, buy the international edition today from Amazon. You cannot pass the PE exam without it.

I'm assuming you are doing the machine design depth. If that's not true, ignore everything above and stop working on this problem, because it is never going to be in the morning section. Let me know if you get it.


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## dnpatterson1 (Nov 22, 2013)

I attached a JPEG to clarify the geometry.


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## CRNewsom (Nov 22, 2013)

To follow up on what majormajor said, there is almost this exact same problem in Shigley &amp; Mischke 5th Edition (but if you're buying, I recommend picking up Shigley &amp; Mitchell 4th Edition instead) on page 399.

The key to translating this problem to the example in the text is to take F/2 on a cantilever and split the length in half. Now you are eccentrically loading one weld group instead of two.

EDIT:

I forgot to mention to majormajor that this particular beam example is not in the appendix of the 12th edition of the MERM. I recommend picking up the shear and moment diagrams published by the AF&amp;PA (American Forest and Paper Association) since they are much more comprehensive.


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## dnpatterson1 (Nov 22, 2013)

I have Shigley &amp; Mischke 5th Edition and the example on page 399 is about a bracket with a force acting on the hole in it, which is quite different than this situation. I'm stuck on how to solve without a known force (F)???


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## MA_PE (Nov 22, 2013)

off the top of my head, I say you want to"

Determine the capacity of the weld and then backcalculate the forece needed to create that value in shear.

Based on that force check bending.


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## CRNewsom (Nov 22, 2013)

Similarities:

The bracket has a perimeter fillet weld - Your bar has a perimeter fillet weld

The bracket has a lateral force on the weld - your bar has a lateral force on the weld (*in this case, lateral means parallel to the faying surface)

The bracket has a moment on the weld - your bar has a moment on the weld

Differences:

The bracket has a normal force on the weld - your bar does not have a normal force on the weld.

The problem is the eccentrically loaded weld group. Do not get hung up on irrelevant details (i.e. bracket vs. bar)


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## majormajor (Nov 22, 2013)

All good points. Roarkes formulas for stress and strain also has this case. I just checked and then saw that I had tabbed the page because that case was not in the MERM.


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