Section Modulus calculation and optimization

Aim:

To find the section modulus of the hood to improve the overall section modulus.

Objective:

By using the section from Hood design and calculate the section modulus using the

formula

S= I/y.

Section Modulus:

The section modulus (S) is geometry property of the cross section used for designing beams and flexural members. It does not represent anything physically.

Formula of section Modulus: It may be defined as the ratio of total moment resisted by the section to the stress in the extreme fibre which is equal to yield stress.

Which is equal to yield stress.

Therefore

S=I/y, Where

S= Section modulus (mm^3)

I = Moment of inertia of the section to the stress in the extreme end of the object (mm^4)

Y= Distance between the neutral axis and extreme end of the object (mm).

Area Moment of Inertia:

The area moment of inertia of a beam cross section area measures the beam ability to resist bending. The larger the moment of inertia less will be the bend.

Where,

M = Bending Moment

I = Area moment of Inertia

E = Young modulus

R = Radius of gyration

Neutral Axis: When beam is subjected to pure bending the top most fibres of the beam are subjected to maximum compression, while the bottom most fibre of the beams are subjected to maximum tension.

Calculation of Moment Of Inertia:

Case 1: Before Optimization

The hood which was already created is imported and the section of inertia analysis is calculated.

(MAX) = 8.992483998e+07 (mm^4)

I (MIN) = 2.665186461e+05 (mm^4)

As per the section modulus equation,

S = I/Y

I = MOI (min) Value is considered

Y = Total distance / 2, Total distance = 877.1375 mm

Y = 877.1375 / 2

= 438.568 mm

Therefore, S = 266518.646 mm^4 /438.568 mm

S = 607.701 mm^3

Case 2: After Optimization

The hood inner panel curve is offset by 0.6mm outwards and section inertia analysis is calculated.

I (MAX) = 8.972477787e+07 (mm^4)

I (MIN) = 2.714461720e+05 (mm^4)

As per the section modulus equation,

S = I/Y

I = MOI (min) Value is considered

Y = Total distance / 2, Total distance = 877.1375 mm

Y = 877.1375 / 2

= 438.568 mm

Therefore, S = 271446.172 mm^4 /438.568 mm

S = 618.937 mm^3

Conclusion:

Resulting by 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 increase which means strength and load of the material also increase. Which the moment of inertia directly proportional to the strength of the material.