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 and '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.

Here the moment of inertia was calculated by using the NX-software obtained values are

I_MAX:8.152381372X10⁵mm⁴

I_min:5.428652574x10³mm⁴

Total length of 'Y' value was486.1938mm

So,at neutral axis we need 'Y' value:Y/2=486.1938/2=243.0969mm

As per the section modulus equation,

S = I/Y

I = MOI (min) value is considered

1)Section modulus(S):-I/Y=5.428652574X10^3mm⁴/243.0969mm

=2.233X10³mm³

CASE 2: After optimization

The hood inner panel’s curve is offset by 0.5mm outwards and section inertia analysis is done same as case 1 for a closed curve.

MOI (max) = 8.271630048e^05 mm^4

MOI (min) = 6.915281504e^03 mm^4

As per section modulus equation,

S = I / Y

I = MOI (min) value is considered

Y = Total distance /2, Total distance =875.5680mm

Y = 486.1938/2=243.0969mm

Therefore, S = 6.915281504e^03 mm^4 / 243.0969mm

S = 2.84x10³mm³ mm^3

Conclusion:

Hence when comparing case 1 & case 2,

1. The section modulus value of case 2 is higher than the case 1

2. As we increase the section area, the moment of inertia & section modulus increases which means strength and load material also increase