If I have a piece of rubber between two steel plates, and four bolts (at equal preload) compressing the plates, will the total compressive force on the rubber be Preload x 4?
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2$\begingroup$ How do you define "preload"? The context you use that word can have a huge difference in this case. $\endgroup$– NMechCommented Oct 28, 2021 at 6:46
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$\begingroup$ The pressure won't be distributed evenly depending on stuff. $\endgroup$– DKNguyenCommented Oct 28, 2021 at 14:23
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1 Answer
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Let's say we have the two steel plates and rubber in place with the 4 bolts and nuts. Now we tighten the nuts while holding the steel plates by some spacer blocks between them not letting the sandwich assembly compress.
Now we apply torque turning the bolts until they are preloaded by a force Plbs.
If the bolts were free to move they would have shortened by;
$$\delta=\frac{Pl}{AE}$$
L is the length of the bolts and A its area. E is the Young modulus.
But as soon as we remove the spacer blocks the rubber plate will shrink allowing the bolts to relax significantly.
Let's say the stiffness of the rubber plate is 2/3 the sum of the 4 bolts.
Then we the sandwich will shrink by releasing some of the preloaded tension on the bolts until there is equilibrium, meaning the leftover tension in the bolts is balanced by compression force by the rubber.
In our case for each incremental shrinkage in the bolts, it will compress the rubber by $1/(2/3)=1.5\delta_s$
At equilibrium, the shrinkage of rubber is $2/3\delta_s\ (1/2+1/3)$ And the bolts have lost 0.66% of their preload.
This is assuming zero deflection on steel plates.
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$\begingroup$ Thank Kamran. At equilibrium, the rubber feels 0.33%P or 4x0.33%P? $\endgroup$ Commented Oct 28, 2021 at 16:08
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$\begingroup$ @astroball, We assumed P as the total preload of the 4 bolts. and A as the 4 bolts total area. So the rubber feels 0.33P. $\endgroup$– kamranCommented Oct 28, 2021 at 17:18