Section Modulus calculation and optimization

 SECTION MODULUS OF AN HOOD

• Section modulus is a direct measure of strength of any  sectional area of an object about an reference axis(neutral axis).
• Strength is nothing but, capable of supporting greater loads, so higher the strength it has a high capable of supporting  greater loads.
• This strength can measured directly from Section modulus formula, and defined by below formula.
• Section modulus 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 end 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, higher the value of moment of inertia , have higher resistance to deform under certain load, hence we can say that moment of inertia directly proportional to the strength of the material , and strength of the material can be calculated with the help of section modulus.
• If we vary the  neutral axis towards positive or negative side the overall strength of the body will decrease , because when nuetral axis line lies on the middle of the section only have higher strength at particular point, so,  here 'Y'  value was inversely proportional to section modulus.

Thats why formula represented as above.

By using the same principle i going to calculate the strength of an hood panel.

First Design Study

The Intersection of the design was taken and then section Inertia Analysis was done

Moment of inertia(max) = 8.19398×10^5×mm^4

Moment of inertia(min) = 5.34031×10^3×mm^4

Y=485.7599/2=242.8799 mm

S=I/Y

S=5.34031×10^3/242.87= 21.98 mm^3

Hood Design Optimization

By increasing the depth of the inner panel by 0.5 mm

Here,

Moment of inertia(max) = 8.16698×10^5×mm^4

Moment of inertia(min) = 5.51216×10^3×mm^4

Y=485.7599/2=242.8799 mm

S=I/Y

S=5.51216×10^3/242.87=22.69 mm^3

CONCLUSION:-

The strength of the material is directly proportional to the moment of inertia, hence increase in sectional area there is have increase in overall mass of an object ,hence have increase in strength.