Section Modulus calculation and optimization

Answer:

Use the Section from your Hood design and calculate the section modulus using the formula S = I/y

S = Section Modulus

I = Moment of Inertia

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

Section modulus is a direct measure of the strength of any sectional area of an object about a reference axis(neutral axis).
Strength is nothing but, capable of supporting greater loads, so the higher the strength it has the high capacity of support greater
loads.
This strength can be measured directly from the Section modulus formula, and defined by the below formula.
Section modulus is represented with a symbol of 'S'.

S=I/Y
where I ='Moment of inertia

'Y' is the distance from the reference line(neutral axis) to one of the extreme ends of an object.

Moment of inertia is nothing but offering resistance against of moment, bending or twisting, i.e if we applied any external load
on the body, how much resistance its offering to tend to deform. So, the higher the value of the moment of inertia, have higher the resistance to deform under a certain load, hence we can say that moment of inertia is directly proportional to the strength of the material, and the strength of the material can be calculated with the help of section modulus.

If we vary the neutral axis towards the positive or negative side the overall strength of the body will decrease, because when
The neutral axis line lies in the middle of the section and only has higher strength at a particular point, so, here 'Y' value was inversely proportional to section modulus.

By using the above principle going to calculate the strength of a hood panel.

here the moment of inertia is calculated for the obtained values

I _max = 7.587*10^7 m^4

I_min = 1.489*10^4 m^4

y = distance from neutral axis 537.704/2 = 268.852 mm

section modulus, s= I/Y

SECTION MODULUS ( S) = I/Y = 7.587*10^7 M^4 / 268.852

= 282.199*10^3 mm^3 (maximum strength)

SECTION MODULUS (S) = I/Y = 1.489*10^4 /268.852

= 0.055*10^3 mm^3 ( minimum strength)

so the strength of the material varies from the 0.055*10^3 mm^3 to 282.199*10^3 mm^3

case 2:

Now the length of the inner panel is offset by the 5 mm and by increasing the cross-sectional area as shown below,

section modulus, s= I/Y

SECTION MODULUS ( S) = I/Y = 7.977*10^7 MM^4/268.852

= 296.705*10^3 mm^3 ( maximum strength)

SECTION MODULUS (S) = I/Y = 1.539*10^4/ 268.852

= 0.057*10^3 mm ^3 (minimum strength).

so the strength of the material varies from the 0.057*10^3 mm^3 to 296.705*10^3 mm^3

when we calculate both the max and the min strength values we know that when I offset the length of the section by 5 mm the max and the min values of the cross-section increased

conclusion :

The strength of the material is directly proportional to the moment of the inertia hence the increase in cross-section area has increased the overall mass of an object hence the strength of the material is increased