Section Modulus calculation and optimization

OBJECTIVE:

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

Section Modulus:

The section modulus of the cross-sectional shape is of significant importance in designing. It is a direct measure of the strength of the object. An object that has a larger section modulus than another will be stronger and capable of supporting greater loads.

CASE 1

Minimum moment of inertia (M1) = 6.881 X 10^7 m^4 = 6.881 X 10^5 mm^4

Maximum moment of inertia (M2) = 1.538X10^4 m^4 = 1.538X10^8 mm^4

So, Inertia I= 6.881 X 10^5 mm^4

Y is the distance between the neutral axis to the outermost of the hood panel

Y=Distance/2 = 478.314/2

Y= 239.157 mm

Section Modulus = I/Y

= 6.881 X 10^5 /239.157

S.M =28771.8946132 mm^3

CASE 2

The hood is offset for a distance of 1mm without changing the overall shape of the hood and section modulus is calulated.

Mimum moment of inertia (M1) = 7.62 X 10^7 m^4 = 7.62 X 10^5 mm^4

Maximum moment of inertia (M2) = 1.632X10^4 m^4 = 1.632X10^8 mm^4

So Inertia I= 7.62 X 10^5 mm^4

Y is the distance between the neutral axis to the outermostof the hood

Y=Distance/2 = 478.314/2

Y= 239.157 mm

Section Modulus = I/Y

= 7.62 X 10^5 /239.157

S.M = 31861.9149764 mm^3

CONCLUSION:

On comparing the results of case 1 and case 2 its seen that case 2 has a higher section modulus because of its increase in gap beyween the inner panel and outer panel. So case 2 is more stronger than case 1.