I need to make a flat pattern for a cylinder made from a steel plate 44mm thick laminated with explosion cladding stainless steel plate of 8mm total plate thickness is 52mm.

How I usually work is take the mean diameter of the cylinder (Inside diameter + thickness ) * PI. To get the developed length of the cylinder. resulting in use of Kfactor 0.5

But now there are materials bonded to each other I fear the Kfactor of 0.5 might not be correct.

Additional the laminated plate 8mm is on the inside diameter side.

The plate will be rolled into shape.

What Kfactor should I use in this case or is 0.5 a good Kfactor?

  • $\begingroup$ What will you use for heads ? Will it have nozzles or only small penetrations. ? I presume you will have a longitudinal weld seam,; you need to have a plan for weld metal dilution. For titanium clad vessels we welded a strip of titanium over the pressure weld. $\endgroup$ Jul 21, 2020 at 0:40

1 Answer 1


Stainless still's modulus of elasticity is 28*10^6 psi, compared to 29*10^6 of steel.

So in order to calculated the neutral axis, centroid of bend you can the following:

$$ A_{stainless} * 28/29 =A_{convert\ to\ stell} $$ $$So\quad 8*28/29 =7.724mm$$

$$ Y_{Neutraal} = \frac{(44+7.724/2) * 7.724 + 44*22)}{44+7.724}= 25.862$$

$$\frac{ 25.862}{26} *0.05 =0.4973 \ Kfactor $$

As we see the difference is lees than 0.003 so you may even assume Kfactor of 0.5.


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