Welding Strength Calculation Help

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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) / IShear stress = Force / throat area of weldForce = 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.
Reply:Originally Posted by dnpatterson1A plate (0.25"thick X 1.38"height X 4.74"length) is fillet welded (1/8") all the way around each side.
Reply:Sounds like a question from a college level engineering class for a mechanical engineer or similar.  It's a little early for finals; is this for your homework?Or are you trying to solve a more practical problem?Benson's Mobile Welding - Dayton, OH metro area - AWS Certified Welding Inspector
Reply:This is work related for practice. I've been out of school for a decade and haven't used the math I learned much since then. I've attached a JPEG to help clarify the geometry. The plate is welded to the inside diameter of the branch in a tee. For the sake of this calculation the force is applied at one point in the middle of the plate along the 0.25" thickness surface. The force is equal distance from each end. To get the formulas for stress and strain, the plate is considered fixed at each end.  Last edited by dnpatterson1; 11-22-2013 at 10:33 AM.
Reply:difficult problem, but my gut says the shear will be a larger issue than the plate yielding.  dont have my old statics book to help me confirm it though.
Reply:I'll throw my hat in the ring here.In order to have a torque (force x distance) there has to be a pivot. The drawing has no pivot as it's connected at both sides. Therefore, to determine the failure will occur, you need to see which is less between the rigidity of the material vs the strength of the weld. In this case, it's the weld x2, because each side will pick up 1/2 the force. The force is at 90deg, so it's totally in shear.Your formulas have no units attached, so they are useless at the moment. In other words, how is the thickness in the shear stress measured? It pretty much has to be an area, but is it ft^2, cm^2, etc?Chay
Reply:Here, let me combine both my posts together. Everything is inches and lbs.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?!?):d=1.38", b=0.25", c=0.69", h=0.125", length=4.74" A=0.288"throat area = A = 1.414*h*(b+d)Moment = M =(Force * length)/8  Inertia = I = (d^2/6)(3b+d)Bending Stress = (M * c) / IShear stress = Force / throat area of weldForce = Not sure how to solve for force(F), my best guess is using yield/tensile strength to solve for Force???Stress Combined Equivalent = sq.rt.[ (Bending stress/2)^2 + Shear stress^2]
Reply:A simple fixed end beam calculation will give you a maximum bending failure force in excess of 85000 lbs assuming fixed ends that cannot fail.However, using the formulas at the site below and calculating the multiple welds as one big weld:Lw= 1.38X4 + 0.25*4 = 6.52infw= 0.75 x 0.6 x 60ksi = 27ksi max shear stressa=0.125Max Shear Load=0.75 x 0.707 x a x Lw x fw = 0.75 x 0.707 x 0.125 x 6.52 x 27000 = 11668 lbs max load.  This being so low means bending stress is not a factor worth incorporating into this calculation.  Shear will fail before it even knows it is being bent.  You could also calculate the shear strength of the base metal to confirm it won't fail first, but simple logic says more material at the same shear strength will not be a factor.http://www.egr.msu.edu/~harichan/cla...e405/chap6.pdfDrewLast edited by Drooopy; 11-22-2013 at 02:57 PM.
Reply:I did not check the calculations, but I agree with shear being the likely issue.Dave J.Beware of false knowledge; it is more dangerous than ignorance. ~George Bernard Shaw~ Syncro 350Invertec v250-sThermal Arc 161 and 300MM210DialarcTried being normal once, didn't take....I think it was a Tuesday.
Reply:Effective length of weld = 6.5" (4*[1.375 + .25]).  Effective throat of weld = .088 (.125*.707).  Effective area of weld =.572 (6.5*.088).  Theoretical strength =34.3 KSI (60ksi*.572) before yieldExperience is something you get right after you need it