Section Modulus calculation and optimization

S = Section Modulus

I = Moment of Inertia

y = distance between the neutral axis and the extreme end of the object

S=I/y

Distance from the center of the panel to the end of the panel

For the initial design

Section Modulus= I/y

For the MOI (Max) = 4.84618x10^06[mm^4] / 440.1[mm] = 11.01154x10^3 mm^3

For the MOI (Min) = 1.50226x10^04[mm^4] / 400.1[mm] = 34.13451mm^3

Decrease the distance between the panel by 1mm

For the MOI (Max) = 7.89411x10^06[mm^4] / 440.1[mm] = 17.93708x10^3 mm^3

For the MOI (Min) = 2.19308x10^04[mm^4] / 440.1[mm] = 49.83140mm^3

Increase the distance between the panel by 1mm

For the MOI (Max) = 4.83005x10^06[mm^4] / 440.1[mm] = 10.97489x10^3 mm^3

For the MOI (Min) =1.54852x10^04[mm^4] / 440.1[mm] = 35.18563 mm^3

Conclusion - If the distance between the outer and inner panel is reduced the strength can be increased by a significant amount