The resulting E' is in the units of ksi, but it is coupled to the thickness of the gauged deck in the first place. So when inputting the equivalent thickness into RAM, should it be 1" thick, or should it be the thickness of the deck that is assumed in the calculation? Effectively, the E' (a material property) is tied to the thickness of the deck in this calculation. This seems more like a flexibility property. Also, is there a necessary G equivalent for RAM?
Just to make sure, using 22 gauge deck in this calculation means I use 22 gauge thickness in RAM with the resulting E' or I use unit thickness (1") instead?
For in plane stiffness what matters is E' and t (the product of the two). I often advise people to make the thickness more than the gage, while reducing E', so that the out of plane stiffness is greater (since this is related to t^3). You don't enter G directly, but it's obtained from Hooke's law give E and Poisson's ratio.
Understood, thank you Seth. To reiterate, the E' that is calculated using this approach means to enter in the same plate thickness as the gauge thickness, but because this doesn't capture the out-of-plane stiffness that is present from the deck geometry you are saying to arbitrarily proportion components in the product t*E' to adjust for this.
Yes, the equation for the G is given in the example. By not giving the option to enter it you are making sure it is not inadvertently defined twice, also meaning isotropic properties. This is fine, just good for me to know.