Beam welded to plate. Plate stress?

[Engineer_Phil]

I have a design where a beam gets welded to a plate. The plate is not against a wall or foundation, but for simplicity I can say that the edges of the plate are fixed. I am concerned about the stress in the plate, and the deflection in the beam. I have access to solidworks simulation, but I want to know if anyone knows of an analytical solution or experimental solution that I can compare to. I don't like to just blindly trust a simulation.
I have attached an image of my setup. In the future I want to vary the beam cross section size and style.

 

[Windadct]

The process and quality of the weld is critical - welding is MUCH more organic than people recognize.

 

[Engineer_Phil]

We have qualified welding procedures and a good quality control program the welding should be fine. I am more concerned about how to determine the required thickness of the plate. The stress cannot exceed 20 kpsi.

 

[Windadct]

Do you mean the stress caused by the welding - or how much stress the weldament can withstand once complete?
For stress caused by the weld - there are different ways to make the weld, different processes and different electrode types. The determination of the HAZ will have a large impact in both the stress created by the process and how much the weld can take.
Probably the best is to get the datasheet for the electrodes your best welder would like to use.
I do think Solidworks has a simulator for welds - but you will need to know the parameters of the weld process and electrodes details.

 

[pongo38]

An elastic analysis will tell you how the stress changes due to the applied load, but not take account of the residual stresses from welding, and takes no account of stress redistribution after the first yield of the steel. Plastic theory or yield line theory will give a good indication of the collapse load, but give no information on the working load behaviour.

 

[Vadar2012]

If I was doing this, I'd trust the simulation more than any theoretical hand calculations. I've done something similar to this for work, but with something bolted to a corrugated steel sheet. To simplify the simlation, you can use simple statics to calculate the forces at the wall. Then just model only the plate in the simulation. What you're after is the stress on the plate around the beam and plate contact. An easy way to do this is make the contact surface a different element than the rest of the plate and turn it off when analysisng the results. This just gives you the stress on the plate.

In other words, simulation is the best way.

 

[Engineer_Phil]

Thanks for the help guys.

I guess I have no choice but to accept the Simulation results as Vadar2012 suggests.
I am going to look at getting some strain gauges to verify the Simulation results.

I hope the simulation is wrong though... otherwise the plate is going to have to be very very thick.

 

[Vadar2012]

Yeah, I found it was larger than I thought. Will be interesting to hear about the strain gauge results.

 

[Engineer_Phil]

Yeah, I found it was larger than I thought. Will be interesting to hear about the strain gauge results.

Did you run an FEA? I'd like to compare results if you did.

 

[Vadar2012]

Did you run an FEA? I'd like to compare results if you did.

Can't run one on yours, don't have any info.

 

[Engineer_Phil]

I made a simple drawing to give you the data.

A picture is worth a thousand words.

 

[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?

 

[RFreund]

What is the plate attached to? AISC give some guidance on welding plates to walls of HSS (hollow structural steel). Also Blodget's Design of Welded Structures also covers this.

 

[Engineer_Phil]

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?

I am not really concerned with residual stresses from welding at this point. I would weld the beam to the plate, then put the strain gauge on the back side of the plate directly behind the bottom of the beam. Then load the beam. And take the strain gauge measurement.

 

[Jupiter6]

Thanks for the help guys.

I guess I have no choice but to accept the Simulation results...

Accepting a simulations answer (especially Solidworks) without an approximate paper answer to corroborate it is never the way to do something correctly.

Check out Timoshenko's Theory of Plates and Shells. I don't think you'll find this to be as difficult as you think it is.

 

[PhanthomJay]

I would think that Roarke's "Stress and Strain" would give a solution, but i don't have a copy right now. But being very conservative by assuming simple supports along the short edges and free along the long side, I get a 5/8" thick plate required using a hand calc on the back of a matchbook. Looks like a 1/2 inch plate will do, as you have drawn. But someone needs to quality review my work
M_max in beam at beam/plate interface = 3(12) = 36 in-k
Load to short edge = M/L = 36/18 = 2 kips
M_max in plate at plate/beam interface = 2(6) = 12 in-k
S_req'd = M/allowable stress = 12/20 = 0.6 in^3
t_required = sq rt( 6S/b) = sq rt(6(.6)/10) = 0.6 in

Is a 5/8" plate too thick for you? i wouldn't cut it back too much...why fool with it.

 

[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.

 

[PhanthomJay]

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.

 

[Vadar2012]

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.

 

[nvn]

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.

 

[nvn]

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.

 

[nvn]

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 textbooks.