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Author: neil moran
Time: 01/31/03 14:23 PST
This is a reply to message no. 10466 by ddelaiar
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Current Topic:
Press Fit Pressure
Hi Dan,

Sorry I think I misunderstood your problem, it is the pressure between the head and the tube due to the interference fit that you are trying to arrive at. In that case I shall try again.

Let us assume that the head is made from steel with internal and external diameters of say 2” and 2.25” respectively and Es = 30 x 10^6 lb/in^2. The tube is made from copper with internal and external diameters of say 1.875” and 2.005” respectively and Ec = 10 x 10^6 lb/in^2.

Now the thickness of the steel head St =  (2.25 – 2.00) / 2 = 0.125” and that of the copper tube Ct =  (2.000 – 1.875) / 2 = 0.0625”

Since there are no applied external forces, the sum of the internal forces must = 0

Or the ((stress in steel x area) + (stress in copper x area)) = 0

Which gives us 0.125Ss + 0.0625Sc = 0 (note that lengths cancel out), therefore Sc = -2.0Ss

Now from our college days we remember that E = Stress / Strain and Strain = Extension / Original length. Also, at the interface between the head and the tube, the circumferential strains in the copper and steel must be the same ie. Strain(steel) = Strain(copper). We now have all we need to do the calculations.

(Ss / Es + Sc / Ec) = Extension / Original length

(Ss / 30 x 10^6 + -2.0Ss / 10 x 10^6) = Pi x 0.005 / Pi x 2.000

Solving gives Circumferential stress in the steel  Ss = 10714 lb/in^2

And Circumferential stress in the copper  Sc = 2 x 10714 = 21428 lb/in^2

From this we can get the radial pressure at the interface using the thin cylinder formula – Stress = P x r / t

Using steel values P = Ss x St / radius = 10714 x 0.125 / 2.000 = 670 lb/in^2

And check using copper values P = Sc x Ct / radius = 21428 x 0.0625 / 2.000 = 670 lb/in^2

Phew, got there at last.

Hope this helps and I have got my maths right.

Regards,
Neil Moran
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