Question

As shown in the given figure, a rubber cylinder A with a diameter of 50 mm and a Poisson's ratio of 0.4 is compressed in a steel rigid cylinder B by a force F = 47 kN. Determine the pressure between rubber and steel (horizontal). Friction between rubber and steel is neglected. steel B rubber A

          As shown in the given figure, a rubber cylinder A with a diameter of 50 mm and a Poisson's ratio of 0.4 is compressed in a steel rigid cylinder B by a force F = 47 kN. Determine the pressure between rubber and steel (horizontal). Friction between rubber and steel is neglected.

steel
B

rubber
A
        
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as shown in the given figure a rubber cylinder a with d 50 mm diameter and 04 of poissons ratio is compressed in a steel rigid cylinder b by a force f47 kn determine the pressure between rub 84436

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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As shown in the given figure, a rubber cylinder A with a diameter of 50 mm and a Poisson's ratio of 0.4 is compressed in a steel rigid cylinder B by a force F = 47 kN. Determine the pressure between rubber and steel (horizontal). Friction between rubber and steel is neglected. steel B rubber A
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Transcript

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00:01 Hello, in the question we have given that a rubber cylinder of length l and a cross -section area a is compressed inside a steel cylinder by a force f that applies a uniformly distributed pressure to the rubber.
00:17 So, we have to find out the following quantities.
00:19 So, the first one we have to find out the lateral pressure.
00:22 So, in order to calculate the lateral pressure, we will find the strain in x because strain in x is zero because this rubber is enclosed inside a steel.
00:32 So, we will use that and we will calculate.
00:36 So, we get this epsilon x is equal to sigma x divided by e minus v by e times sigma y plus sigma z.
00:45 Now, here we have used the formula of this e that is young's modulus, we know it is stress upon strain.
00:54 So, from here strain will be equal to sigma by e.
00:58 So, now we calculate, we will calculate in the, so when we will calculate in the x direction the strain, so we will get this formula.
01:07 So, that is a pretty long calculation.
01:09 So, i have directly adopted the formula over here.
01:12 So, by calculating i get this epsilon x as i mentioned it is zero, there is no strain in x, sigma x is minus p and sigma y is minus f upon e because pressure is applied in the y direction and sigma z is minus p because it is uniformly distributed.
01:31 So, plugging the values, i get this equation that is zero is equal to minus p by e minus v by e times minus f by e minus p.
01:42 So, now a bit simplification, i am taking this to this side.
01:45 So, i get p by e is equal to, so minus minus become plus that is vf divided by ea, then minus minus plus, so vp divided by e.
01:55 So, now i can take this p terms on one side.
02:00 So, i get p by e minus v, p divided by e is equal to vf divided by e times a.
02:06 So, this e gets cancelled, we can take common and cancel it off.
02:10 So, by taking common i get p times 1 minus v is equal to vf divided by a.
02:17 So, from here i get the lateral pressure as vf divided by a times 1 minus v.
02:24 Now, in the second part, they are asking us to calculate the shortening del of rubber.
02:28 So, we will calculate the strain in y because the pressure is, the force is applied in the y direction.
02:38 So, there will be strain only in the y.
02:42 So, we get epsilon y is equal to this sigma y divided by e minus v by e times sigma x plus sigma z.
02:55 Now, as we have mentioned above, so we will plug the value of sigma y, sigma x and sigma z...
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