Section Modulus calculation and optimization

Section modulus:-

It is a ratio of total moment resisted by the section to the stress in extreme fibre.If section modulus increasesthen moment of resistance increases an hence its load bearing capacity.

Section modulus(S) = I/Y

Where,

I=moment of inertia

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

Intersection of a hood

For the above section the output values are

I min = 1.983*10^-006*10^12

I max = 7.626*10^-004*10^12

Y = 439.7015 mm

Therefore,

S min = I/Y

= 1.983*10^-006*10^12/(879.403/2)

= 1983000/439.7015

S min = 4509.87 mm^3

S max = 7.626*10^-004*10^12/439.7015

= 762600000/439.7015

S max = 1734358.423 mm^3]

After that made a small modification in section which makes which increase the strength i.e, increase the thickness of lower panel to 0.5mm.

So that we get a values as

I min = 2.0902*10^-006*10^12

I max = 7.925*10^-004*10^12

Y = 439.7015 mm

Therefore,

S min = I/Y

= 2.0902*10^-006*10^12/(879.403/2)

= 2092000/439.7015

S min = 4757.77 mm^3

S max = 7.925*10^-004*10^12/439.7015

= 792500000/439.7015

S max = 1802359.1 mm^3