# force required to bend .375 thick aluminum tab

We are trying to determine the force that was needed to bend an aluminum tab on a piece of our aircraft equipment.

The tab is deflected 2 inches and is 3 inches in length. load was at the tip (3 in moment arm)

Material is .375 in thick, 1.5 width and as mentioned was 3 inches long, with the load hitting the end. load was cantilever. 6061 aluminum

• Heat - treatment will make a difference. The most accurate number would be to get an equal part and measure the force necessary to bend it – blacksmith37 Apr 5 at 18:12
• I am getting lost in your description - The tab is "deflected 2 inches" and is 3 inches in length. Can you provide a better description or a sketch? – r13 Apr 5 at 18:44

correct me if, with dimensions. You have a cantilevered tab with a thickness of 0.375" and 1.5" width and 3" length with 2 inches deflection. so the tab is bent by an angle $$\alpha=41.8$$

The force must be smaller than section plastic hinge moment,

$$F > \sigma Y* M_{P \ of \ tab}*3"cos41.8$$

$$M_{P \ of \ tab}= b \frac{H^2}{4}= 1.5\frac{ 0.375^2}{4}= 0.0527inch^4$$

Let's pick 42ksi as the yield strength of aluminum,

$$F> 0.0527"^4*3*cos 41.8"*42 ksi = 4940lbs$$

The projection of the tab on the x-axis counts. the cos41.8 factor is for that.

• If this is the case, upon permanent deflection of 2", the extreme fiber of the plate has already stressed to yield by a force with unknown magnitude. Thus, this problem is to find the additional force required to stress the entire cross-section into yield. This problem is not as simple as it is seen. – r13 Apr 5 at 20:53
• @r13, I assumed they have removed the initial source of bending, so the beam is deformed permanently and I ignored the hardening or any residual stresses after removal of that force, which is I believe what the OP is asking. otherwise, why stop at two-inch deflection. A moment greater than the section plastic moment will keep rotating it. – kamran Apr 5 at 21:05
• I understand your assumption, don't mean to criticize, just unsure about it. Maybe the energy method is the better approach for this problem, but I am not familiar with that either. – r13 Apr 5 at 21:08
• @r13, onother assumption i made is at the end the beam we have a small curve terminating into a horizontal boundary angle. i am a private pilot and have seen many similar tabs and fish eye hooks for anchorage of the plane bent. – kamran Apr 5 at 21:18
• I wish I can learn something here :) Interesting problem. – r13 Apr 5 at 21:27