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Calculating weld stress under bending load using Shigley?(4)

Hello, Does anyone have experience calculating the stress in a welded connections that are under a bending load? Specifically using the techniques outlined in the Shigley Mechanical Engineering Design. These techniques represent the weld as line.
On page 397 of Mechanical Engineering Design (5th edition), Shigley has a table that lists the Unit Second Moment of Inertia for a weld treated as a line. My understanding of the use of this table is to mulitply the Iu in the table by .707(h) to get the actual I, h being the weld size. The .707 is the proportion of the leg length to the thickness of the fillet at the throat. The .707 is initially described on page 392 (equation 97) dealing with J for a torsional load on a welded connection. This is the first use of that type of table (treating the weld as a line). The sample problems is Shigley all use this .707 in these types of problems. Unfortunately the homework problems in Shigley for bending welds don't list answers.
I am preparing for the PE exam and have done some practice problems that use different techniques. When I use the Shigley techniques as described above, I get different answers than those in the practice problems. I donâ€™t know if I am applying the Shigley table properly.
Any help is appreciated, Shaggy
BTW, I have crossposted this question in the Mechanical Engineering Other Topics.


Shaggy, Maybe you'll find some clarifications or examples in the following websites... http://files.aws.org/wj/supplement/Weaver/ARTICLE2.pdfhttp://www.mhprofessional.com/downloads/products/0071428674/0071428674_Section13.pdf (Section 13.3) Like you said the factor 0.707 is the relation of the leg size to the throat size (i.e. throat size = Sin 45° x leg size OR throat size = leg size/√2). But since your problem is about treating the weld as a line, I don't understand why your section modulus (Sw) or polar moment of inertia (Jw) should have any reference to the thickness of the weld. The Sw or Jw should only have the length or the weld or distance between welds. The 0.707 factor only applies to the weld thickness 

Hi Shaggy Haven't got Shigley but a similar treatment can be found in Schaum's outline series for machine design. sw = M/Zw and further sw=f= M/Zw where s=stress in standard design formula w= leg size of weld f= force per inch of weld when weld is treated as a line Zw= section modulus of the weld (in^2) when weld is treated as a line Here's a site which might help: http://www.mech.uwa.edu.au/DANotes/welds/fillets/fillets.html#geometricPropertiesregards desertfox 

Gentlemen, thank you both for the information (a star to both). I am still a little fuzzy however.
desertfox, If I am understanding your set of formulas correctly (and they are consistent with the links that you both provided), then:
f = M/Zw (units are lbf/in) and then as you say Stress = f/w = M/(Zw*w) (units are lbf/in^2)
I guess the question I have is at this point. The shigley reference uses a w of .707*weld height and the other references simply use weld height. Which is right? Essentially, is w the weld height or the throat length?
I suppose for the purposes of the PE exam, I can use both and pick the answer that most closely matches one of the choices, but I would rather understand the concepts.
Shaggy 

Lion06 (Structural) 
10 Apr 07 11:06 
shaggy The 0.707 is most likely to account for a fillet weld leg size. The effective throat of a fillet weld is 0.707*weld leg. The other references that just use a 1.0 are expecting you to compare the actual value to the allowable (which for a fillet weld is 0.707*weld leg). The method I would use is this. Blodgett has all of the section modulus values for various weld configurations (in in^2) in "Design of Welded Structures" in Chapter 7. Take the total shear divided by the total weld length. That is the direct shear value. Take the moment divided section modulus as calculated from Blodgett to get the maximum stress in the weld due to bending. Square both of the preceeding values and take the square root of that. That will give you the maximum shearing stress in your weld group. Compare that to your allowable weld stress (based on your effective throat). If you don't have blodgett's book, let me know the type of weld configuration you have and I will give you formula for the section modulus. 

Hi Shaggy
The .707 is to get the throat length, by multiplying this figure with the leg length of the weld, so imagine a weld with a leg length of say .5" the throat length would be .707 * .5" which is the w you base your stresses on. Actually the .707 is the cosine of 45 degrees as usually all fillet welds are assumed to be at 45 degrees.
regards
desertfox 

Lion06 (Structural) 
10 Apr 07 12:55 
desertfox This 0.707 ONLY applies to fillet welds. If you have a BTCP10 or flare bevel weld (or any other kind of weld that is not a fillet), the effective throats are different.
Shaggy Be careful how you apply this and make sure you understand exactly where they are coming from. 

".707 * .5" which is the w you base your stresses on." desertfox
This was basically the answer I was looking for. The sample problem that I was working on was for a fillet weld and they did not use the throat length to calculate the stress, they used the weld height (resulting in a lower stress by a factor of .707). I thought their method was wrong and your statement agrees with that assertion.
StructuralEIT, Thank you for the added info. All of the problems that I would expect to see on the PE exam (mechanical) should be fillets. I will look into getting a copy of the Blodgett book. Not in time for the test, but just for future reference.
Shaggy 

Hi StructuralEIT
I assumed that were only talking about fillet welds as there is no other type of weld mentioned in any of the posts
regards
desertfox 



