Consider the design of the bracket shown in the figure below (it is assumed that the anchors are sufficient and are not considered below).

My question concerns the detailing of the welded connection detail and the overall strength of the bracket. The design principle is that the welded connection shall have a capacity equal to atleast the moment capacity of the plate. Note that the bracket behaves as a cantilever so the point of maximum moment is the connection itself.

Steel partial factor $= 1.15$

Permissible stress in steel $= \frac{275 \text{ N/mm}^2}{1.15} = 239 \text{ N/mm}^2$

Elastic modulus $= \frac{bd^2}{6} = \frac{100 \times 10^2}{6} = 1666 \text{ mm}^3$

Moment capacity $ = 239 \times 1666 = 398174 \text{ Nmm} = 398 \text{ Nm}$

Load $F$ partial factor $= 1.6$

Maximum load $F_\text{max} = \frac{398 \text{ Nm}}{1.6 \times 0.5 \text{ m}} = 122.5 \text{ N} = 12.5 \text{ kg}$

The connection is then designed so that it is atleast as strong as the plate it is joining. There are three options shown in the picture below:

  1. closed corner (3 passes with a 6mm leg length E42 electrode)
  2. open corner (3 passes with a 6mm leg length E42 electrode)
  3. semi-open corner (2 passes with a 6mm leg length E42 electrode)

My first question is which of the three corner details is preferred for this design? Will all three details provide at least the moment capacity of the plate? Could you recommend an even better detail?

My secondary (bonus) question is whether my maths above is correct and that the bracket should conservatively be capable of supporting 12.5 kg (assuming a suitable anchoring detail)?

  • $\begingroup$ Are you also taking shear stress into account? $\endgroup$
    – Chuck
    Apr 26 '20 at 15:20
  • $\begingroup$ You missed the main point: What is the response of the base metal and filler metal to whatever welding process and procedures ( pre-heat, post heat, etc.) are chosen. $\endgroup$ Apr 26 '20 at 15:24
  • $\begingroup$ @Chuck are you referring to a combined shear/bending case? Does the issue of shear only affect weld details b) since the weld will take the shear? If there's an overlap will the shear not be transferred into the vertical plate and into the anchors? $\endgroup$
    – egg
    Apr 26 '20 at 16:24
  • 1
    $\begingroup$ Shear can only be transferred to the anchors after "passing through" the weld. $\endgroup$
    – Chuck
    Apr 26 '20 at 16:43

I haven't done a complete check of your calculations but I do have two comments:

  • You describe the thickness as 10mm yet use 100mm in the calculation of the section modulus, resulting in a capacity 100 times too great.
  • As Chuck has noted in a comment, you haven't accounted for the shear stresses. (With an elastic stress distribution, it is possible this won't impact the result, though.)

Your details 1 and 2 are one-sided fillet welds, and doing one-sided fillet welds like this is generally considered poor practice. They have three big problems:

  • It is very difficult to ensure full quality in the innermost fiber of the weld, which will have a significantly larger stress than the rest of the weld. Therefore, initial cracking/yielding in those fibers is likely, and that will weaken the weld.
  • Pitting corrossion is likely to occur if there is any possibility of humidity, resulting in a reduced lifetime.
  • With a 6mm leg length, you only have a throat thickness of about 4mm, resulting in the section modulus of the weld being $\frac{4^2}{10^2}=16\%$ of the section modulus of the plate, so you should not expect anything close to full capacity.

Your detail number 3 is a possibility but since the plates have an overlap of only 4mm (40% of the thickness), you should not assume full strength without a more detailed capacity calculation of the weld.

There are two obvious principles for doing full strength welds:

  • A butt weld with throat thickness equal to the plate thickness with a backing run.
  • A butt weld with throat thickness equal to half the plate thickness (5mm) on one side combined with a fillet weld with the same throat thickness on the other.
  • $\begingroup$ Great answer. Overestimating the capacity by a 100 fold was a blinder of an error - I've correct that now. So the capacity is only 12.5 kg. I will have a go at designing a better (two-sided) weld. Shall I do this is as a separate question or just edit my current question? $\endgroup$
    – egg
    Apr 26 '20 at 16:21
  • $\begingroup$ @egg - new question! Link to this one if you think it gives necessary context. $\endgroup$
    – Chuck
    Apr 26 '20 at 16:42

Your Answer

