Engineering Asked by Mech_Engineer on September 1, 2021
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?
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.
Answered by kamran on September 1, 2021
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