Recognitions:
Gold Member
Homework Help

## Beam welded to plate. Plate stress?

 Quote by nvn PhanthomJay: Roark does not seem to contain the given problem. Nice approach. Using your approach, I get the same answer you got, except a different allowable stress (and therefore a different wall plate thickness). I currently do not know how you got your allowable stress. My allowable bending stress is currently, Sta = 197 MPa. Therefore, I currently obtain the following wall plate thickness (t1), using your above approach. t1 = sqrt[3(1 - h2/h1)*L*V/(b1*Sta)] = 12.75 mm,where b1 = wall plate width, h1 = wall plate height, h2 = cantilever beam plate height, L = cantilever beam length, and V = cantilever transverse tip load.
Thanks for the check..the alowable stress of 20,000 pounds per square inch (138 MPa) was later introduced by the OP in post 3.

 Quote by 0xDEADBEEF Just out of interest: How would you use strain gauges with this. Do you glue them on beforehand and then measure the resistance before and after? Do they survive welding temperatures?
Wouldn't you place the strain gauges on the plate next to the beam. Not directly behind the beam attachment face. That attachment face isn't bending.

It'll be interesting to see how the FEA compares to the hand calc. Please let us know.
 Recognitions: Homework Help Science Advisor Engineer_Phil: I found that the following formula currently appears to give a relatively accurate answer to your given problem in post 11, and is more accurate than the equation in post 17. t1 = sqrt[2.522(1 - h2/h1)*L*V/(b1*Sta)]Therefore, if Sta = 197 MPa, then the above equation gives t1 = 11.69 mm. Or if Sta = 138 MPa, the above equation gives t1 = 13.97 mm. Alternately, solving the above equation for stress gives the following wall plate maximum normal stress, sigma1, for your given problem in post 11. sigma1 = C*(1 - h2/h1)*L*V/(b1*t1^2),where C = coefficient = 2.522, currently. E.g., if t1 = 12.7 mm, then the above equation gives a wall plate maximum tensile stress of sigma1 = 166.9 MPa. Ensure sigma1 does not exceed the wall plate allowable tensile stress, Sta. If your current FEA wall plate in-plane normal stress is vastly different from sigma1 above, it might indicate a mistake in your FEM.
 Recognitions: Homework Help Science Advisor Engineer_Phil: OK, I see what you mean, regarding this baffling problem. For your given problem in post 11, if you compute the wall plate stress using analytic methods, you get a wall plate stress of, say, 170 +/-20 MPa. But if you then use FEA, you get a wall plate stress of approximately 260 +/-20 MPa, even if you make an effort to ignore FEA stress concentrations. The two answers are not even close, yet. I have not figured out this discrepancy yet. This is something a person needs to try themselves, to see this odd discrepancy. Interesting question you came up with.
 Recognitions: Homework Help Science Advisor Engineer_Phil: For your given problem in post 11, I assumed a 6.0 mm fillet radius between the cantilever plate and the wall plate, and an integral connection between the cantilever plate and wall plate (for simplicity), because you have not given us any details regarding how you modeled it. I now obtained more accurate stress results, as follows. sigma1 = C*(1 - h2/h1)*L*V/(b1*t1^2)(1) For the maximum von Mises stress on the back of the wall plate, C is currently C = 3.29. I.e., sigma1 = 218 MPa, which does not exceed the wall plate tensile yield strength, Sty = 262 MPa. (Therefore, we see that the approach by PhanthomJay is off by only -8.8 % for stress on the back of the wall plate.) (2) For the maximum von Mises stress on the front of the wall plate, just above the edge of the cantilever plate upper fillet, C is currently C = 4.56. I.e., sigma1 = 302 MPa, which exceeds the wall plate tensile yield strength, Sty. (3) The maximum von Mises stress in the middle of the fillet (not on the wall plate), at the upper left or right corner where the cantilever plate connects to the wall plate (where two fillets intersect, forming a curved, concave, sharp edge), there is a tiny spot where C reaches C = 8.67. I.e., sigma1 = 574 MPa. This tiny spot will reach the tensile yield strength, then redistribute; therefore, it is probably negligible, for static (noncyclic) loading. (4) The deflection of your cantilever tip is y = 1.12 mm, downward. (5) Here are the detailed stresses corresponding to items 1 and 2. For the in-plane, x- and y-direction normal stress on the back of the wall plate, C = -2.84 in the x direction, and C = -3.61 in the y direction. I.e., sigma_x = -188 MPa, and sigma_y = -239 MPa. (The y direction is vertical.) (6) For the in-plane, x- and y-direction normal stress on the front of the wall plate, C = 3.54 in the x direction, and C = 5.15 in the y direction. I.e., sigma_x = 234 MPa, and sigma_y = 341 MPa. The above stress results indicate that your wall plate is currently overstressed, and that you need to add stiffeners. Let us know if you want ideas regarding different methods of adding stiffeners, and then someone might be able to show you. Another benefit of adding stiffeners is, it might make the connection easier to analyze using FEA. Your current design is very difficult to analyze using FEA, and is virtually nonexistent in text books.