While this is about a home improvement project, I figure the core part is interesting physics.

My situation is this: I am hanging 500 lbs of drywall (two layers weighing about 250 lbs each) on 5 rows of metal furring channels (hat channels) which are mounted onto RSIC (sound insulation clips) which are screwed into studs with drywall screws. There are 25 total clip assemblies across all 5 rows of channels. This wall is suspended (ie, mounted solely on channels and none of the edges touch the walls, floor, or ceiling). Here's a diagram of a single clip assembly:


Below is a diagram of my actual clip assembly array. The wall is 13' 10" long and 8' tall. The vertical black lines represent the studs and are spaced apart at about 16" (it's not exact because I had to add a couple studs and the frame joins up with another frame at one point in the wall), the red dots are the clip assemblies, and the horizontal grey lines are the hat channels that are mounted onto the clips. Ignore the green dots and yellow arrows.

clip array

The problem: Since drywall screws are not as strong as wood screws which I should have used, I am worried that the 25 screw/clip assemblies may not be strong enough to bear the weight of 500 lbs. To test the strength of a single screw, I mounted a single clip into a dummy stud and ended up being able to hang 120 lbs of dumbells on it. This mock assembly has been up for over a month and shows no signs of being on the verge of snapping. I could probably hang another 30 lbs on it before it gave away.

So, using simple math, it seems I would be able to say, "Well, if one screw can support 120 lbs then 25 screws can support 3,000 lbs!" Of course, I am sure the math is not so simple. I am sure there is a curve there somewhere where adding more screws certainly allows for greater weight capacity, but the weight capacity deteriorates the more screws and more weight you add even if the number of screws and amount of weight was proportionate.

So is it really as easy as using simple math to solve this problem or does it require something more advanced?

  • $\begingroup$ I assume that your test standing included and rail and all, or is that a untested possible point of failure? In any case, I imagine the problem is indeterminate without knowing the full geometry: how are the screws placed in relation to one another and to the sheets of dry wall; how many channels on each sheet and in what geometry; and so on. With enough data it becomes a basic statics problem suitable to freshmen physics or engineering students to find the screws with the maximum load on them. $\endgroup$ – dmckee --- ex-moderator kitten Nov 1 '12 at 0:46
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    $\begingroup$ The problem is that it's hard to be sure the model fits reality. If you could ensure that the load perfectly distributed on all the screws then it's just 25*120lb. If the weight is all on one tightened screw which then fails dropping the load onto the next screw and so on - then it can support about 120lbs. Quite a few large buildings have failed on not understanding this! $\endgroup$ – Martin Beckett Nov 1 '12 at 1:51
  • $\begingroup$ Thanks for the comments. Updated my answer with a diagram of my entire clip assembly. $\endgroup$ – oscilatingcretin Nov 1 '12 at 9:03
  • $\begingroup$ So, Martin, you're saying that, assuming the clips are spaced in such a way that the weight is evenly distributed, calculating how much weight these 25 screws can collectively support is as easy as taking the weight a single screw can support and multiplying that by the number of screws in the assembly? $\endgroup$ – oscilatingcretin Nov 1 '12 at 10:14
  • $\begingroup$ FYI while this isn't the type of question we particularly want to encourage on the site, it isn't off topic either. $\endgroup$ – David Z Nov 1 '12 at 17:36

As others have pointed out in the comments, it is not really trivial to apply any model to reality, especially since we don’t know much about reality. However, we can make a few educated guesses and estimations and see how well everything fits together.

First, assuming perfect weight distribution, we see that five screws would probably be sufficient to support your wall. And while the assumption is probably wrong, we can be relatively sure that a safety margin of 500% is a good start.

Second, we can look at how much a single screw has to support according to your diagram. An upper boundary for the area of the wall supported by a single screw seems to be four ‘rectangles’, corresponding to about $$\frac{4}{44} \approx 0.09 \hat{=} 45.5\textrm{ lb }\hat{=} \frac{1}{3}\textrm{ screwweight}_{\textrm{max}}\quad.$$ That still looks rather good, doesn’t it?

Third, we could check if there are any screws that, if they are removed, leave another screw with many more ‘neighbouring’ tiles. As far as I can see, the maximum would still be about six (corresponding to approx. $\frac{1}{2}\textrm{ screwweight}_{\textrm{max}}$).

I would hence tend to say that you are fine, but there are many, many problems that could possibly arise (not to mention that I am not a construction engineer and roughly followed http://xkcd.com/793/).

  • It appears that you are building some sort of sound-proofing. Not to take into account possible issues with vibrations (and hence faster wear of the screws) seems silly.
  • Depending on how and where you fix things to the wall, you might have to deal with ugly resonances, both between the two walls and within the drywall. Without knowing the speed of sound in drywalls, it is difficult to make any estimates here.
  • Everything else I didn’t think of.
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  • $\begingroup$ Yes, there are more things to consider like sudden structural settling. I can take a hammer and snap a drywall screw if I whack it really hard whereas a wood screw would just bend. My question assumes total structural stability. Being that I am no physicist, what does your equation/expression/whatever it is equate to? What's it saying in plain, layman's English terms? $\endgroup$ – oscilatingcretin Nov 1 '12 at 11:37
  • $\begingroup$ Oh, I was just relating the area of the drywall (made up of 44 rectangles) to those closest to one particular screw (4). Then the weight of the wall closest to a particular screw is $\frac{4}{44} \times \textrm{ total mass of wall}$. This happens to be $45.5\textrm{ lbs}$, which is equal to $\frac{1}{3}$ of the weight a single screw can support. $\endgroup$ – Claudius Nov 1 '12 at 13:05

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