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I have 2 plates that are fillet welded to each other. How can I determinate the max force the weld can take before it cracks?

The plates are only welded on one side because the other side is not reachable.

I added a image to clarify.

enter image description here

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  • $\begingroup$ Where is the force applied? is it parallel with the weld or is there an angle involved? ie where are the attachment points? $\endgroup$ – Solar Mike Jul 30 '18 at 11:55
  • $\begingroup$ It is parallel with the weld. The black plate is being pulled down. $\endgroup$ – Mech_Engineer Jul 30 '18 at 12:00
  • $\begingroup$ I'm clarifying how much you already know... Do you know how to calculate the strength of a weld in general? Is the load direction what is causing issues? $\endgroup$ – hazzey Jul 30 '18 at 12:49
  • $\begingroup$ I would like to know how to calculate the strength of a weld, but in this case I need to know how much weight I can attach to a existing structure. The load direction is not the issue, this issue is that I can't calculate it :D $\endgroup$ – Mech_Engineer Jul 30 '18 at 12:52
  • $\begingroup$ It depends on the weld technique, filler metal, weld conditions ( preheat, etc.) . and base metal. $\endgroup$ – blacksmith37 Aug 30 '18 at 16:46
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Depending on the type of electrode and the properties of the base material, we get a wide range of possible answers.

Typical electrodes tensile strength is ranging between 60 -70- 80- 90 ksi. They ultimately fail at shear which we call f_w at ultimate level.

And we consider factor of safety of 75% let's assume a= width of the weld and L =length of the weld.

$$φ Rn = 0.75 f_w \times 0.707\times a \times L$$

So to simplify, you plug in 3mm=1/6 inch and 11mm = 15/32 inch length of the weld in the formula and input the f_w from the electrode's spec sheet.

At the same time you need to check the base material shear strength and factor of safety which is usually 0.60.

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For fillet weld only axial tensile/compressive load is possible. Due to this load shear will produce on weld joint. We know that shear stress is equal to (applied load/minimum cross-section area). Minimum cross section is 0.707XSXL

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