Suppose a section of steel angle is used to support a structural beam, which would otherwise have no available bearing, as in the following sketch.

enter image description here

Assuming that appropriate connection/fixing on the side marked L1 is not an issue, how should the load-bearing capacity of the angle correctly be calculated?


Ordinarily, one can assume a cantilever is fixed at the wall, and deflection only occurs within L2, giving a boundary condition of zero slope. But here, angular deflection also occurs within the steel corner, so that's no longer the case, and it probably also depends on the radius/thickness of that corner as well as thickness of L2.

I don't know what property/ies to refer to for this, or the needed calculation, in real world steel angle.

Other information:

  • The beam spreads its load uniformly along the length of the angle (Z-axis), rather than a point load at some place along the angle, and the load is static not dynamic. It is supported close to the inner corner of the angle.
  • We can ignore the beam itself, and its own bending. This question is purely about the angle as bearing, when supporting the end of the beam.
  • For simplicity, assume it is fixed to the vertical element (weld or bolt) along the side L1, sufficiently that the primary mode of failure is distortion of the position of leg L2, or local distortion around the join of L1 and L2, rather than L1 or L2 developing "saddle like" distortion etc.
  • A suitable scale for this might be t around say 5-10% of L1 or L2, and as with all simplified beam/deflection workings, assuming elastic behaviour and smaller scale distortions.
  • $\begingroup$ every cantilever bends "a little bit" at the mount. The simple method tells you if the deflection is enough to worry about. There is nothing different from this angle than from a plate embedded in the wall. $\endgroup$
    – Tiger Guy
    Commented Dec 28, 2021 at 15:18

3 Answers 3


If you assume the bottom part of the angle as a longer section made up of two parts with two I's, one much larger than the other, it will give you a better aproximation.

lets call the angle thickness, t and it's side L.

then you have a cantilver beam of length

$$legth= t+L$$

That is composed of two sections and is loaded with a uniform loading only over the L part.

Or you can consider the entire lower part of the angle as a cantilver beam but it is suspended from the upper leg und solve for compatibility of definition s.

  • $\begingroup$ Ahhhaaa - thank you! That gives me a way to think about it, very helpful. @r13 also took a similar approach, I see. $\endgroup$
    – Stilez
    Commented Dec 28, 2021 at 17:36

enter image description here

Assume $L_1$ is perfectly welded to the steel post, there are two potential failure modes - 1) $L_2$ (similar to cantilever) fail in bending, and 2) failure of the weldment due to combined stresses (shear and bending stresses).

  • Determine $t$:

  • $M = R*b$

Assume effective bearing width = $2*b$, $45^o$ spread from the center of bearing.

  • $S = \dfrac{2*b*t^2}{6} = \dfrac{b*t^2}{3}$

  • $f_b = \dfrac{M}{S} = \dfrac{3*R*b}{b*t^2} = \dfrac{3R}{t^2}$

$f_b$ must equal to or less than the allowable bending stress $f_a = 0.6*f_y$, thus

  • $t_{min} = \sqrt{\dfrac{3R}{0.6f_y}} = \sqrt{\dfrac{5R}{f_y}}$

The next step is to check the vertical weldment, which is subjected to the beam end reaction $R$ and the bending moment $R*b$.

  • Assume the weld group consists of two fillet line with unit width along the entire leg length $L_1$,

  • $f_v = \dfrac{R}{2*L_1}$

  • $f_t = \dfrac{M}{S_w}$, $S_w = L_1^2/3$ is the section modulus of the weld group.

  • The weld stress, $f_w = \sqrt{f_v^2 + f_t^2} \le f_a$ (allowable stress of the weld metal). From here, the weld size $t_w$) can be determined.

Finally, $t$ must greater or equal to $t_w$. My personal preference is $t_{min} \ge t_w +\dfrac{1}{16}$.

  • $\begingroup$ I hadn't thought to decompose into 2 plates and a join. That makes a lot of sense, thank you. I'll hold off marking it as "the answer" until others have a chance to comment. $\endgroup$
    – Stilez
    Commented Dec 28, 2021 at 17:37
  • $\begingroup$ Angle has a fillet in between the legs that will interfere with the beam and its connection. In practice, unless limited by space below, you should flip the angle letting the vertical leg point down. It will make the connection more stable, and provide better seating for the beam. The analytical/design procedures remain the same. $\endgroup$
    – r13
    Commented Dec 28, 2021 at 17:54
  • $\begingroup$ Interesting, because I thought it'd be more stable that way up, pressing into the post not pulling away from it. I see what you mean, though $\endgroup$
    – Stilez
    Commented Dec 28, 2021 at 19:14

simplest approach (cantilever beam)

Assuming that:

  • the angle is fixed on the $L_1$ there are no worries about detacking
  • the load on the $L_1$ part is known and uniform then the easiest way to determine the maximum load is to consider the horizontal part as a cantilever beam with a uniform load.

enter image description here

Figure: Cantilever beam with uniform load sourceEngineering toolbox)

In that case

property equation
maximum moment $M_{max} = -\frac{qL^2}{2}$
maximum deflection $d_{max} = -\frac{qL^4}{8EI}$
maximum stress (at root a) $\sigma_{max} = \frac{M\cdot t}{2\cdot I} = \frac{qL^2\cdot t}{4\cdot I} $


  • E is the Young's modulus of steel (E=200 GPa)
  • I is the second moment of area of the horizontal part of the cross-section $I = \frac{t^3 b}{12}$
  • b is the width of the steel angle.


So what you need to check is that the deflection $\delta_{max}$ is less than the permitted, and that the maximum observed stress $\sigma_{max}$ is less than the allowable.

Other considerations

Although the above should give you adequate approximations, there are some cases that some of the assumptions above will not hold. E.g.:

  • if the beam which is resting on the angle is not rigid enough then the distribution of the loads cannot be considered uniform
  • if the span of the beam over the gap is small compared to the length, again the distribution cannot be considered as uniform.

There are probably other asterisks, but these are all I can think of now.

  • $\begingroup$ I guess my main question is that the cantilever approach assumes a fixed end at "A" (zero slope at boundary). But in this case, the end at "A" is a steel corner, and can deflect, to > 90 degrees (and will to some tiny extent). Hence should not assume zero slope boundary? We need to consider angular deflection at "A", don't we? Based on thickness/curvature at the corner/root? $\endgroup$
    – Stilez
    Commented Dec 28, 2021 at 10:48
  • $\begingroup$ Depending on the value of $t$ you might need to. However if the thickness is less than 1/10 of the $L_2$ there is little significance (especially for low loads). $\endgroup$
    – NMech
    Commented Dec 28, 2021 at 10:56
  • 1
    $\begingroup$ Btw, you might want to update the question with that information. I.e. focus that you are worried about the slope at A. The remaining problem (beam over a gap) is not adequately defined, and distracts to other considerations. $\endgroup$
    – NMech
    Commented Dec 28, 2021 at 10:58
  • $\begingroup$ Updated, thanks! $\endgroup$
    – Stilez
    Commented Dec 28, 2021 at 11:10

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