# How to determine impact load factor on small beams

For small beams, the following equation explodes to high values, which seems unreasonable:

$$n = 1 + \sqrt{1+\frac{2h}{\delta}}$$, where $$n = \dfrac{F}{W}$$, with $$F$$ the maximum force on the beam and $$W$$ a suddenly applied weight.

If it helps, I'm evaluating the following case: for a given work in height, the worker will be anchored to a structure. In order for him to be protected from a fall, I'm designing eyebolts for him to anchor himself with his lanyard. In this case, the height of the structure is much higher than the length of the eyebolt.

The problem is that if I consider the eyebolt as a fixed-free beam, then its displacement $$\delta = \dfrac{PL^3}{3EI}$$ will lead to a very small number, which, as I mentioned before, will lead to a high value of n.

I suppose the equation for suddenly applied load doesn't apply here, right? If so, any tips for how to work around my problem?

For reference, see the "eyebolt" dimensions below: Consider it a plate of 1/4". I don't think those are called eyebolts, though, and I'm unsure what's the correct translation.

• As a special heights rope access trained person, the length of a lanyard is seldom more than 50cm stinky because the forces become extreme at longer lengths. A 100kg body falling that far causes quite considerable impact on the lanyards and they are designed usually to withstand 5000kgf. May 11 '20 at 21:59

That equation does not distinguish between large and small beams. It is as valid in this case as any other.

But this will only be a problem if $$h \gg 0$$. Remember that $$h$$ isn't the height of the beam, but the "drop height". How much slack would be expected on the lanyard? If it's small, it won't be a problem. At an extreme, if you can assume no slack on the lanyard, you get $$h = 0$$, which gives the absolute minimum of $$n = 2$$.

Also, your equation for the deflection is incorrect. $$\delta = \dfrac{PL^3}{3EI}$$ is valid for cantilevers with perpendicular loads. You're dealing with a vertical cantilever resisting a vertical applied force. That applied force is offset from the beam's neutral line, so it will generate a concentrated bending moment.

If we ignore second-order effects, the deflection on a cantilever due to a concentrated bending moment at its end is equal to

$$\delta = \dfrac{ML^2}{2EI}$$

Looking at the eyebolt, your original deflection equation is more reasonable. In such cases, you're probably right: the given equation for $$n$$ is likely unusable. The problem here is that the eyebolt won't behave as a beam: beams usually have a span/height ratio of $$\ell/d > 4$$; the eyebolt has $$\ell/d = 1$$. Also, the eyebolt has a giant hole in the middle, which will make it behave even less like a beam. This is especially the case if the eye isn't welded closed (has a gap between the two sides of the loop, like the left one on this image).

Therefore, to calculate the eyebolt, you'll probably have to rely on FEM.

After further consideration, I think there is a simple model you can use to calculate the static deflection.

The solution is to think of the eyebolt not as a cantilever with span $$\ell$$ and cross-section $$b \times h$$ (which would always be wrong, since that's not what an eyebolt cross-section looks like), but as a looped beam with a circular cross-section equal to the eyebolt's ring's diameter. This will lead to a much smaller moment of inertia and therefore greater deflection.

Using this eyebolt as an example (it was the first one I found, absolutely not a recommendation, no idea if it's suitable for this particular use case!): We get a vertical deflection of 1.671e-6 m, which gives us an $$n = 425$$... On the bright side, that's one third of what you'd get using a simple cantilever ($$n = 1394$$).

Making the model more accurate should further increase deflections and therefore lower $$n$$. For example, I put a fixed support on the right side of the loop. A more accurate model would likely have a small horizontal span representing the actual bolt.

• I'm evaluating for both the eyebolt and for the column. Your equation for deflection refers to the displacement in the columns, correct? What about for the eyebolt itself? I considered it as a cantilever with a perpendicular force. Try these numbers: $50mm$ for length, $100kg$ (or $980N$ for the dropped mass, section of $50mmx6,5mm$, resulting in $3.08e-07m$ of displacement. For height I considered a $0.3m$ fall, resulting in $n = 1394$! May 11 '20 at 18:39
• @CaioGuimaraes Ah, I see. Then yes, the equation probably isn't valid because the eyebolt itself probably won't behave as a beam. Beams are usually assumed to have span/height ratios of $\ell/d > 4$. Also, the eyebolt has a giant hole in the middle, which will change it's behavior even further away from that of a beam.
– Wasabi
May 11 '20 at 18:54
• do you have any tips on how to work in this situation? Or any literature I might look into? For the column, using the formula you proposed, n was a much more reasonable number (although still too high for my tastes... oh well). May 11 '20 at 19:01
• @CaioGuimaraes no suggestions on literature, but in terms of actually calculating it, use FEM. It'll probably be easier than whatever funky equations you get from the literature, and it'll be more accurate.
– Wasabi
May 11 '20 at 19:11
• I figured as much. Currently I'm unavailable to simulate it on FEM, but I'll try it. Thanks! May 11 '20 at 19:13

A close approximation is to equate the potential energy of the weight of the worker falling by the amount of the slack, h by the strain energy of the eyeball beam.

Assuming he drops clean with no entanglement with the rope into the eyeball and then by ignoring the weight of the eyeball, it bends the eyeball down in a cantilever deflection.

$$mgh= \frac{3\delta^2 EI}{2L^3} \\ m (9.8*h)=\frac{3\delta^2 EI }{2L^3}$$

L is your 50 mm, mg is his weight. watch out for consistency of dimensions.

If you want to be a bit pickier you could consider the strain energy of the column and the flexibility of the rope.

Edit

Therefore your dynamic loading factor is $$n=\frac{h }{\delta}$$

• Hey @kamran, But then I'll fall on the same issue that $\delta$ is too small. Also, could you elaborate on the physical meaning behind $n = \frac{h}{delta}?$ Isn't n supposed to be the ratio between dynamic and static delta? May 12 '20 at 13:01
• The strain energy of the cantilever is kx^2/2. X is delta and k is 3EI/L^3.. the man falls h and then stops after small delta so the acceleration factor is h/delta. Or the tension in the rope during deceleration. May 12 '20 at 13:32
• let's say we attach a spring scale to the rope. during the deceleration, it will measure the tension in the rope (perceived weight of the man) h/delta. dynamic load factor. mg gains V= sqrt(2gh) now this V decelerates to zero, 0 over delta. so the deceleration is h/d May 12 '20 at 15:57
• I see. A deleted my comment because my math was wrong. Correcting, and for further reading: using your result and still comparing the eyebolt to a beam (let's say... its a beam, in another situation), I obtained the ratio between dynamic delta and static delta as \$n = \sqrt{\frac{2kh}{P}} which is different from my original equation, but results in a much lower ratio. May 12 '20 at 16:02