2
$\begingroup$

We are evaluating a design in which a fall-arresting safeguard would be subject to a shock load in the event of a catastrophic failure of a different component, which is required to meet compliance with our design standard. We believe that the equation below (found in Machinery's Handbook, p.279 of the 29th Ed.) is appropriate for approximating the stresses due to this type of loading:

$$ p = p_s\left(1+\sqrt{1+\frac{2h}{y}}\right) $$

Where:

  • $p$ = stress (psi) due to shock caused by impact of a moving load
  • $p_s$ = stress (psi) resulting when moving load is applied statically
  • $h$ = distance (in) that load falls before striking member
  • $y$ = deflection (in) resulting from static load

Our problem is that in sample calculations of real-world scenarios, we are finding that the resultant stress is far higher than the ultimate tensile strength of the material. My intuition is telling me that the transient nature of the loading means that tensile strength alone is not an appropriate constraint but I'm unable to find any literature to back up that hypothesis. Additionally, our design standards (CMAA 70 and 74) don't seem to indicate any special allowed stress specifically for shock loading.

To be clear, since this is a safe-guard in case of catastrophic failure, we are not concerned with investigating fatigue rupture. If this safeguard is ever activated once, the product is rendered unable to function and replacement of the entire assembly is necessary.

So my question is, what strength is appropriate to use (if not ultimate tensile) as a measure of acceptable dynamic stresses due to shock load as a result of falling and what reputable sources are there to cite this information? Otherwise, are we incorrect to be using the above formula to estimate stresses due to shock load as a result of falling?

Here is a sample calculation:

Inputs:

  • $p_s = \frac{Wl}{Z} = \frac{1102.5lbf\times1in}{\frac{4in\times(0.375in)^{2}}{6}} = 11760 psi$
  • $h = 0.5 in$
  • $y = \frac{Wl^3}{3EI} = \frac{1102.5lbf\times(1in)^3}{3\times(2.90\times10^7psi)\times\frac{4in\times(0.375in)^{3}}{12}} \approx 7.21x10^{-4} in$

Result: $p \approx 4.50x10^5 psi$. Without any special factor on strength, comparing this to $UTS=65000psi$ for the steel we are using, this would lead one to believe that this shock would cause this safe-guard to fail (to put it lightly).

Here is a loading diagram (rounding 0.375 to 0.38) to help illustrate the loading scenario which would produce the numbers above: loading

$\endgroup$
  • $\begingroup$ Are the numbers correct? ps is awfully high and y is really low. Is the mechanism actually so stiff that it doesnt defelct by 10^-4 in under the static pressure already? $\endgroup$ – mart Sep 9 at 15:01
  • $\begingroup$ I'm afraid the numbers appear to be correct for this sample case. I guess it's a matter of perspective, but that static stress doesn't seem unusually high for me for your average steel. Keep in mind that this specific member's sole purpose is to be very stiff as it's only function is to engage in the event of a failure to prevent dangerous load drop. I'll try and whip up a diagram showing the member and loading if you want to verify. $\endgroup$ – Luke Sep 9 at 15:11
  • $\begingroup$ That would be good and might help to answer the question. $\endgroup$ – mart Sep 9 at 15:15
  • $\begingroup$ @mart Image is added, if it's difficult to read please see full res here: i.imgur.com/ZDIziJ9.png $\endgroup$ – Luke Sep 9 at 15:26
  • $\begingroup$ w is the load that causes this static stress? where does the value for y given in the question come from? $\endgroup$ – mart Sep 9 at 15:39
1
$\begingroup$

Your static deflection is way too low.

let's consider a 1-inch length of your angle for a basic check. Assuming steel E=29000ksi

$$I=1*0.38^3/12=0.00457 in^4 $$ $\delta = \frac{wL^3}{(3EI)} = \frac{11760*1^3}{3EI}=0.0295 in$

Therefore

$$ p = p_s\left(1+\sqrt{1+\frac{2h}{y}}\right) =11760(1+\sqrt{1+\frac{2*0.5}{0.0295}}=11760(1+6.899)=92786 psi \ \text{no good}$$

Let's look at it from the energy angle.

The stiffness of each one-inch length of the angle is $$K=[\frac{Ebh^3}{4L^3}]=\frac{29000000*1*0.38^3}{4*1}=397.822 ksi$$

$$mgh_ \text{potential e}= 1/2 Kx^2_\text{strain e} \rightarrow 11760*9.8*0.5=1/2*397822*x^2 \\x_{deflection}^2=\frac{11760*9.8}{397822}=0.29in\ \text{no good, large angle}$$

The angle is not okay.

| improve this answer | |
$\endgroup$
  • $\begingroup$ I appreciate the effort in your answer but you have 2 large mistakes in your deflection calculation: 11760psi is a stress, not a force, so it is being used inappropriately. The static load is 1102.5 lbf, please see loading diagram. Additionally, in the loading diagram, see that the angle is 4" long. We are assuming that the 1102.5 lbf load is distributed along the entire length of the angle, 1 inch from where it is fixed. So I = 4*0.38^3/12 ~= 0.018in^4 $\endgroup$ – Luke Sep 10 at 13:20
  • $\begingroup$ I also noticed several large errors in your energy equation. K is off because the angle is 4" long, not 1" as already mentioned. For m you are again using a stress in psi when it should be a mass, and then for g you are using 9.8 which is the metric constant, where we are using imperial units here. I don't mean to sound rude but there are a large number of issues here so it is surprising to me that someone chose to upvote this answer as-is. $\endgroup$ – Luke Sep 10 at 13:41
  • $\begingroup$ One other error: you also forgot to take the square root of the final result for the energy equation... $\endgroup$ – Luke Sep 10 at 13:51
  • $\begingroup$ yes, I did not get the square root of x because it is obvious. as for my errors in reading the length of the cantilever my apologies if the diagram was dark for me. as for the load i did not see it. if you edit the question we can revise the answer. $\endgroup$ – kamran Sep 10 at 15:53
  • $\begingroup$ Those two issues were honestly the least of my concerns. To be quite frank, your answer is sort of littered with problems. With that said, I did edit my answer to include details on how I arrived at my input numbers. I included units throughout so if you choose to edit your answer I'd appreciate if you did the same, as that accounts for about 75% of the errors in your answer. $\endgroup$ – Luke Sep 10 at 17:42

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.