Hydraulic tensioner pressure conversion in a fretting testing rig

Hello all

I have a problem I hoped some of you guys might be able to shed some light on.

First off, I have a rig like crudely drawn in the attachment. In it is a block clamped between the two plates using 4 bolts that go through the rig. The bolts have nuts on the left side and hyrdaulic tensioners on the right side.

The block has jagged edges that are in contact with the plates, the area of these jagged edges are 1290.3 mm^2 on each side, so both sides add up to 2580.6 mm^2, or 2.5806*10^-3 m^2

The smallest cross-sectional area of the bolts are 364 mm^2

The area of the hydraulic tensioners (as inscribed on them) is 20.27 cm^2, or 2.027*10^-3 m^2

I am required to excert a compression pressure of 15 000 psi on this block using the hydraulic tensioners. This means I have to find the pressure input needed in the hydraulic tensioners.

What I did was as follows:

15 000 psi = 103 421 359 Pa

I decided to take this pressure and multiply it with the area to be clamped to get the total compressive force needed:
since 1 Pa = 1 N/m^2 => 103 421 359 N/m^2 * 2.5806*10^-3 m^2 = 266893.3 N

Since there are 4 bolts, I then divided this force by 4 to get 66723.33 N as the compressional force needed to be done by each bolt.

I then took this force and divided by the area of the tensioner to get the pressure force needed to be excerted by the tensioner in order to get the required compressional force in the bolt.

66723.33 N / (2.027*10^-3 m^2) = 32.92 MPa per tensioner.

Now, apparently this is wrong, can anyone tell me where I went wrong in this? Or point me in the right direction to where I could find an answer? 😀

Thanks in advance

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