Calculate weld stress of a fabricated clevis 2

[flydan27]

I have a fabricated clevis with 8mm fillet weld all round the two lugs. The force is 20mT axial but might see some shear. How do you calculate the weld tensile/shear shear stress of this clevis ? I have attached some phots above and started with the calculation.

 

[flydan27]

 

[BrianE22]

Blodgett's Design of Welded Structures is such a good book for welding calculations. I'd advise getting a copy of it. (Sorry - no direct help, I make too many mistakes!)

 

[desertfox]

Hi flydan

Firstly you need to get the allowable shear stress for the weld and the material which should be a fairly similar number for both, fillet welds usually fail in shear and not tension. Secondly we really need a sketch of the Clevis and see exactly where the welds are because depending how the force acts relative to the fillet weld throats the allowable stresses can vary.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[flydan27]

Here is the clevis with the 8mm fillet weld all round the lugs

 

[flydan27]

This is s355 plate , the allowable shear stress of weld, is this a figure on a data sheet?

 

[BrianE22]

I doubt you'll find "allowable stress" on the data sheet. I'd ask a welder what filler he/she would use. You might find some data on the filler. If not, then assuming the weld is stronger than the parent material you could start out with the yield stress for s355 plate. That would give you a starting point for figuring the allowable stress.

BTW, your first post says load is 20 mT. It's not really millitons is it?

 

[desertfox]

Hi flydan27

Usually the shear stress maximum is about 40% of the yield stress or at least that’s the figure I used when I did fillet weld stresses but I thought you might have some better information. However it would appear to me that the welds are to highly stressed, if I take 40% of the 355 I only get a maximum of 145 N/mm^2 without any safety factor.

Is the length of weld 249mm for one half the clevis or for both sides? Also is the 20 tonnes divided across the clevis?

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[flydan27]

Thanks for that. Lets just say we go with the 145 N/mm^2, with the figures provided, what would be the method for calculating the stress of the weld for this clevis ? are you able to provide a worked example ?

 

[NRP99]

This is ok for axial force but welds are more weak in shear. The shear force will cause some bending of legs and bending stress in welds. And hence you need to consider shear and then combined shear and axial/bending stresses.

Check Shigley's Mechanical engineering design. But Blodgett's book is best resource.

 

[dik]

In Canada and the US (I think) the allowable weld load is predicated on the shear strength of the weld and the yield ultimate strength of the base material. Something like the attached:

I can include the SMath program. You can download the executable from the link on the *.pdf file. It's an excellent program for design.

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[flydan27]

The length of weld is the total , Added the perimeter length around each lugs . The load 20mt will be divided any the two Clevis .

 

[IRstuff]

Kudos for using Mathcad, if that's what's shown ;-)

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[dik]

It's SMath... my favourite. I only use MathCAD for opening other MathCAD files.

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[desertfox]

Hi flydan27

The weld area you are using is correct it appears to me you are using the tensile yield stress to evaluate your weld stress which is in my opinion incorrect and based on my quick calculation you need a larger weld to increase the safety margin, I estimate you need at least a 10mm leg length fillet weld.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[flydan27]

Ok cheers , I’ll go with 10mm weld then . Have the the correct weld areas formula ? And the shear stress formula for this application ?

 

[dik]

mt? Megatonne, millatonne? I usually write my SMath stuff in both units since the clients that I design for may be using either. I'm too old, and even if I work in Metric, I think in Imperial.

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[flydan27]

20 metric tonne

 

[desertfox]

Hi flydan

Just change the terminology from tensile stress to allowable shear stress which needs to be less than 142N/mm^2, so divide by the 1.33 you used previously and then size the weld accordingly.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[flydan27]

How about this ,