Kfactor for laminated steel plates rolling

Question

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?

kamran · Answer

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:

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

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

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

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

Kfactor for laminated steel plates rolling

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