Simple Angle Bracket Sizing and Stress Analysis
In this post, we will go through the various steps involved in analyzing a simple ‘L’ angle bracket, in other words, performing ‘angle bracket sizing’.
We already touched on typical shear/tension clips in other blog posts listed below, you can also click on the related blog posts images below this post for other similar posts:
- https://www.stressebook.com/stress-engineering-interview-questions-part-3/
- https://www.stressebook.com/classical-hand-calculations-in-structural-analysis/
- https://www.stressebook.com/shear-clip-freebody-diagram-fbd/
Click here to access pdf versions of the latest blog posts…
Known Values:
In general, sheet metal angle bracket sizing from scratch involves a few different steps.
First of all, we already know the following:
- Applied loading (applied at the top leg, P = 100 lb)
- The bracket is 0.050″ thick (t), it is a bent sheet metal part made from aluminum alloy 2024-T3 sheet
- Note that the standard ‘gage’ of Aluminum alloy sheet at this thickness level is actually 0.05082″, but lets use 0.050″ for simplicity
- The bracket is secured with two pan-head style NAS623-3-2 (#10, nominal grip of 0.125″, 0.275″ thread engagement, nominal length of 0.4″) screws at the bottom leg used with appropriate washers
- It is loaded at the top leg using the same fasteners and washers
- The critical section is analyzed as a pure bending section
Bend Radius Requirements:
However, when we are dealing with sheet metal, the following characteristics play into the minimum recommended bend radius at the transition of the bends in the part:
- The sheet raw material (Al 2024 alloy sheet)
- Heat treatment or temper level (T3 temper)
- The thickness (0.050″) (note: these thicknesses are usually to the third decimal, so its good practice to use trailing zeros up to three decimals)
For our angle bracket, the minimum recommended bend radius is 0.19″. You can use the following links to read more about this topic:
In Addition, you can read more about tension clips in Chapter D3.3 of this book: Analysis And Design Of Flight Vehicles Structures.pdf

Figure 1 shows the general features of the part as well as its main dimensions.
In this post, our main focus will be on the following tricks of the trade:
Angle Bracket Sizing – Freebody Diagram (FBD):
Figure 2 below illustrates the following:
- The applied loading ‘P’ assumed uniformly distributed
- The critical section we are going to analyze, Section A-A
- Section A-A is a section cut that passes through the center line vertical plane of the fastener holes
- The reason we pick this section is that it has the lowest net cross sectional area in bending, and the largest moment arm from the vertical leg
- The three red surfaces in the figure below make up the net cross sectional area to be analyzed
- The portion of the bottom leg beyond Section A-A is not important for this check
- This is because it is assumed to react the heel-toe load
- This produces the counter acting moment required to balance the applied moment
- This allows the development of the full applied moment at Section A-A

Angle Bracket Sizing – Analysis:
As we can see in Figure 2, the left and right sections are 0.66″ long each, the mid section is 1.31″ long.
Net Area Moment of Inertia:
Ixx = 2*(0.66″)*(0.050″)^3/12 + 1.31″*(0.050″)^3/12 = 2.74E-05 in^4
Section max fiber distance:
c = t/2 = 0.050″/2 = 0.025″
Applied Load:
P = 100 lb
From Figure 1, we calculate the moment arm from load P to Section A-A:
Moment Arm:
y = 2.0″ – 0.9″ = 1.1″
Applied Moment:
Mxx = P*y = 100*1.1 = 110 in-lb
Section A-A Bending Stress:
fb = Mxx*c/Ixx =110 in-lb*0.025″/2.74E-05 in^4 = 100,365 psi
From MMPDS Table 3.2.4.0(b1), the A-Basis properties of aluminum alloy 2024-T3 sheet, per AMS 4037 (0.010 – 0.128 thick) are as follows:
Ftu = 63 ksi (LT, lowest grain direction value)
Fitting Factor FF = 1.15
M.S. (Angle Bracket, Section A-A Bending, 100 lb Ultimate Load) = [Ftu / (FF)*fb] – 1 = 63,000 / (1.15*100,365) – 1 = -0.46 (fail)
Angle Bracket Sizing – Redesign and Resizing:
Clearly, this bracket is undersized for the applied load. So what can we do now?
There are a few different options:
- Firstly, thicken the bracket
- However this could mean a larger bend radius, so watch out for fastener washers impinging on the start of the bend
- Secondly, change material of the bracket to one that has a higher ultimate tensile strength
- But for an application like this, we typically just increase the thickness
- Thirdly, reduce the moment arm but keep the same thickness
- Although this could also mean fastener washers could impinge on the start of the bend
Refer to this page and look at the Aluminum Gage Chart:
In this case, let us choose the next thickness (or ‘gage’) level from 0.050″, which is 0.064″ (rounded down from 0.06408″).
Note: Assuming we made sure there is no interference between the washer and the new bend radius for the thicker part, the bending moment remains unchanged.
Net Area Moment of Inertia:
Ixx = 2*(0.66″)*(0.064″)^3/12 + 1.31″*(0.064″)^3/12 = 5.745E-05 in^4
Section max fiber distance:
c = t/2 = 0.063″/2 = 0.032″
Section A-A Bending Stress:
fb = Mxx*c/Ixx =110 in-lb*0.032″/5.745E-05 in^4 = 61,271 psi
M.S. (Angle Bracket, Section A-A Bending, 100 lb Ultimate Load) = [Ftu / (FF)*fb] – 1 = 63,000 / (1.15*61,271) – 1 = -0.11 (fail)
Angle Bracket Sizing – Plastic Bending, Ultimate Check:
In summary so far, even the 0.064″ gage thickness does not seem to cut it. However, there is one more trick up my sleeve that I will discuss with you.
This concept is called Cozzone’s plastic bending modulus of rupture, Fb.
You can read more about this concept here: Chapter CD3.1 of this book: Analysis And Design Of Flight Vehicles Structures.pdf
Without going into too much detail, for a rectangular Section A-A in this example:
Fb = fm + fo*(k-1.0)
Where:
Fb = Plastic Bending Modulus of Rupture
fm = Ultimate Tensile Strength = 63 ksi
fo = Yield Stregth = =Fty = 42 ksi (LT) (From the same MMPDS table)
k = Shape Factor = 1.5 (rectangular)
Fb = fm + fo*(k-1) = 63000 + 42000*(1.5-1.0) = 84,000 psi
M.S. (Angle Bracket, Section A-A Plastic Bending, 100 lb Ultimate Load) = [Fb / (FF)*fb] – 1 = 84,000 / (1.15*61,271) – 1 = +0.19 (OK)
Angle Bracket Sizing – Limit Check:
In addition, since we showed this bracket good for ultimate load (limit load*1.5) using plastic bending, we also need to analyze the bracket for limit loading against the material yield strength. Therefore, the first step is to determine the limit load.
Limit Load:
P_lim = P/1.5 = 100/1.5 = 66.67 lb
Applied Limit Moment:
Mxx_lim = P_lim*y = 66.67*1.1 = 73.34 in-lb
Section A-A Limit Bending Stress:
fb_lim = Mxx_lim*c/Ixx =73.34 in-lb*0.032″/5.745E-05 in^4 = 40,851 psi
M.S. (Angle Bracket, Section A-A Bending, 66.67 lb Limit Load) = [Fty / (FF)*fb] – 1 = 42,000 / (1.15*40,851) – 1 = -0.11 (fail)
Since the bracket fails under limit loading, it is evident that just showing it good using plastic bending under ultimate loading is not acceptable. As a result, we need to go back to the drawing board and increase the thickness by probably one more gage, or use a better material, or product form. Following that, we repeat all the above steps.
Alternatively, if the bend radius becomes too large as you increase the thickness, then you may need to switch from a sheet metal bracket to an extrusion. The extrusions do not have this issue with bend radius vs thickness.
More importantly, I suggest you do this on your own as practice work.
Conclusions:
In conclusion, for this unique angle bracket, we learnt many things:
- How to properly size it
- Select material and appropriate bend radius
- Select the material gage
- Balance freebody diagrams
- Calculate ultimate and limit moments
- Calculate section properties, and bending stresses
- Cozzone plastic bending
- Ultimate and limit checks
- And finally write margins of safety
This bracket may be a simple bracket, but the process detailed above is essential for most components in Aerostructures.
I hope you enjoyed reading through this blog post and learnt from it. Feel free to leave your comments below.
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