0
$\begingroup$

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

$\endgroup$
2
  • 2
    $\begingroup$ 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 $\endgroup$ – blacksmith37 Apr 5 at 18:12
  • $\begingroup$ 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? $\endgroup$ – r13 Apr 5 at 18:44
1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$ – r13 Apr 5 at 20:53
  • $\begingroup$ @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. $\endgroup$ – kamran Apr 5 at 21:05
  • $\begingroup$ 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. $\endgroup$ – r13 Apr 5 at 21:08
  • 1
    $\begingroup$ @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. $\endgroup$ – kamran Apr 5 at 21:18
  • $\begingroup$ I wish I can learn something here :) Interesting problem. $\endgroup$ – r13 Apr 5 at 21:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.