# Pressure caused by clamping base and cover

1. Feb 10, 2010

### anlouk

Hello, i've posted this on the pyshics forum but no-one has replied so I thought I might try this one.

I have been asked to solve the following problem at work. Unfortunately I haven’t done anything like this since college (over 12 years) and am a bit confused now.

We have an assemblied unit, consisting of an aluminium base and cover with a PCA clamped between. The unit is clamped using 27 screws (M2 SS Pan TORX SEMS) with a torque value of 0.35Nm. The electrical engineer wants to place some compressive material into a milled pocket in the base, which will allow for tolerance build up in an area of the PCA and therefore needs to know the pressure caused by clamping the unit together.

I was given the following formula by another of the electrical engineers:
T = (C x D x P x A) / n

T = torque per screw (Nm)
C = torque coefficient
D = nominal screw size (m)
P = desired pressure (force per unit area) (N/sq m)
A = surface area (sq mm) – is this unit correct should this not be sq m?
n = no of screws

Could anyone confirm if this is the correct formula?
Also I can only find the torque coefficient for aluminium/steel (0.45 dry) and not stainless steel, does anyone know what this value is?
Will this affect the answer greatly as I worked out the formula using the aluminium/steel value (0.45 dry).

Depending on the surface area unit (surface area of unit is 8604.8788mm²) I am getting the pressure to be either:

Using surface area in mm² = 1.220238Nm² = 1.220238 Pa
Using surface area in m² = 1220238Nm² = 1220.238 kPa

Both values seem to be a bit extreme and when comparing to the data sheet of the compressive material, the graphical range for the pressure in kPA is in units of 100 upto 700kPa.

I would appreciate any advice given. Thank You.