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I have a uniform load applied to the underside of a plastic sheet which is to be resisted by the combination of a bolt, washer, and nut as depicted below (the other end of the bolt is secured and will not budge).

Diagram of nut, bolt, washer, and ASA plastic.

The thickness of the Luran S 777k ASA (Acrylonitrile Styrene Acrylate) plastic is currently unknown, but to get the ball rolling let's just say it is 1/8" thick (EDIT: confirmed). The plastic will bear directly on the square washer.

For modeling purposes, I have decided to assume the plastic is totally fixed/rigid and that the load is being applied to the bolt. In this model, it is the washer that is bearing on the fixed plastic.

I expect that the limiting factor of this system is going to be the pull-through capacity of the plastic as (I assume) the washer will tear it apart way before the washer or the bolt fail. How can I determine the pull-through capacity of this system? Is my assumption that the pull-through capacity of the plastic is the limiting factor valid?


My idea is to simply obtain the shear resistance of the plastic material online somewhere, and calculate the pull-through capacity like so:

  • Shear strength of the material is $\tau_{\text{ASA}}=??$
  • Load-resisting cross-sectional area: $$A=4s_{\text{washer}}\cdot t_{\text{plastic}}$$
  • Therefore the pull-through capacity of washer bearing on plastic is: $$P_t=\tau_{\text{ASA}}\cdot 4s_{\text{washer}}\cdot t_{\text{plastic}}$$

However, I'm concerned there may be other considerations at work here that I need to take into account. Is it really this simple?


Thanks to hazzy for the simple but reassuring answer, and the points he made.

It turns out that the limiting consideration for this particular problem seems to be fatigue/cyclic bending of the plastic around the washer, since in actuality the load is being applied at either end of the plastic section (instead of uniformly as I proposed in the model above).

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    $\begingroup$ Failure for shear in the plastic would occur at the perimeter of the washer, so if it's 2" square, your area resisting shear would be the perimeter times the thickness (8" * 1/8" = 1 Square Inch.) If the nut if more than snug tight, the preload will add some shear stress before there is any uplift on the plastic sheet.I'm sure someone who knows about fracture mechanics of plastics will have more to add - I mostly work with metal. I don't know if asymmetric loading causing a crack to start and grow is a concern. $\endgroup$ – Ethan48 Feb 12 '15 at 3:13
  • $\begingroup$ Is Luran S 777k ASA a brittle or ductile material? Are there supports close to the washer or are they far away? $\endgroup$ – hazzey Feb 12 '15 at 3:24
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    $\begingroup$ There is one effect you are not accounting for. Since your washer is square, you will have stress concentrations at the corners (assuming a uniform upward force on the sheeting). In addition, because the wasmer is metal, the corners may be sharp, which could cut into the sheeting. $\endgroup$ – regdoug Mar 30 '15 at 0:43
  • $\begingroup$ @regdough That's a good point. I think, though, that the way the washer is actually being loaded (as opposed to the simplification I gave in the question) will somewhat alleviate this concern. $\endgroup$ – Rick supports Monica Mar 30 '15 at 3:37
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Short Answer

Yes, it should be as simple as taking the perimeter of your washer multiplied by the thickness of the plastic. The shear stress over this area should be your maximum tension on the bolt.

Longer Justification

The washer should be considered to be rigid in relationship to the plastic. This will set the minimum area of shear as the perimeter of the washer.

Since the plastic likely exhibits an amount of plastic deformation before failure, the actual load at failure will likely be larger than the value calculated by the simple method. Much like designing a concrete slab or block shear in a bolted connection, there are many different possible failure planes, but there is a minimum conservative failure plane.

The design should take into consideration this minimum possible failure plane. If this is assumed, then the design will be conservative.

The data sheet of the material shows that there isn't a large amount of strain at yield (3.3%). This is less than most structural metals, so you shouldn't have to worry about the effects of large deformations on the calculations.

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  • $\begingroup$ I was actually going to start looking at this problem again tomorrow morning, and this is helpful. Two things, though: 1. the yield stress and strain listed is in tensile mode of failure, with nothing given for shear. There's probably not a good way to convert the tensile yield capacity to shear, is there? 2. I think the controlling concern is going to turn out to be something I didn't mention in the question: fatigue failure due to cyclic loading. Unfortunately I'm having trouble finding a cyclic loading curve for this particular plastic (have contacted Styrolution, waiting for a call). $\endgroup$ – Rick supports Monica Mar 30 '15 at 3:31
  • $\begingroup$ There are a couple other problems. The moduli given are also tensile only (no shear moduli, and I doubt I can assume the stuff is isotropic with G = E/2(1+ν)). The tensile failure numbers are at 23 C, whereas the actual application is outdoors and extremes could be more like -30 C. $\endgroup$ – Rick supports Monica Mar 30 '15 at 3:42
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    $\begingroup$ @RickTeachey The cold would certainly be an issue. I also wonder if failure in bending between supports would control over pull-through. $\endgroup$ – hazzey Mar 30 '15 at 12:55
  • $\begingroup$ The moment arm is only 14 inches (135 lb load - the section modulus is around 2 in^3). Yield failure doesn't control, but I think I've decided that fatigue in bending does - about 10 MPa for 10^7 cycles according to the data I just got this morning. $\endgroup$ – Rick supports Monica Mar 30 '15 at 16:57
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You are KIND OF RIGHT !

IN THIN PLASTIC DEFORMATION ONLY

OD of steel washer = OD

Thickness of Material (Pastic) = Tp

Shear Area = (OD+2.5*Tp)*pi()*Tp

Shear force is Shear Area * MPa = KN force.

The plastic deformation forms a "bulb" effect same goes for concrete piles in the ground ! the diameter x 2.5 is the effective area for the end of the pile. This gets adjusted should the depth be less than 2.5 x the diameter.

Garret Krampe

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  • $\begingroup$ Can you explain why the 2.5 Factor applies in the case of plastic or provide a source that backs it up? $\endgroup$ – Rick supports Monica Jul 22 '16 at 12:42

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