Section Modulus calculation and optimization

OBJECTIVE:

The goal of the project is to discover the section modulus of the proposed front hood using the formula S=I/Y and optimise it to achieve a better section modulus than the preceding section.

Introduction:

The geometric property of a cross section of a part is called section modulus. It is a direct representation of a part's strength at the specified cross section. It's the ratio of the area moment of inertia to the distance between the neutral axis and the part's furthest point. Its measurement unit is mm^3. The ratio of the section's total moment withstood to the stress in the extreme fibre.It provides us an indication of how much weight a building can support. The moment of resistance, and hence the load bearing capacity, increases as the section modulus increases.

Then the value can be used to find out the section modulus using the formula given

S=I/y

where,

S - Section Modulus

I- Moment of inertia value from the inertia analysis

Y- Distance from Neutral axis to outermost fiber

It is also often used to determine the yield moment (My) such that,

My = s×σy

where,

σy- Yield Strength of the material (515 MPa = 515 N/mm2mm2 for stainless steel)

using the above formula, the section inertia analysis was carried out for the Hood along the X-direction at the center plane as the cross section.

Two ways of increasing S value is increasing the Area Moment of Inertia value and decreasing the y distance

Analysis:

case 1: Before modification

Y=440mm (apporx)

Maximum Moment of Inertia I=6.038631567 X `10^7``mm^4`

Minimum Moment of Inertia I=1.916048251 X `10^5``mm^4`

For Maximum Moment of Inertia,

S=`6.038631567 X `10^7``mm^4`/440 mm

S= 137241.626 `mm^3`

My =s×σy

My =137241.626 `mm^3` X 515

= 7.069 X `10^7` Nmm

For Minimum Moment of Inertia,

S=1.916048251 X `10^5``mm^4`/440

S=435.46 `mm^3`

My = 435.46 `mm^3`×515

My = 224264.709Nmm

Case 2: After modification

Maximum Moment of Inertia I=9.214432634 X `10^7``mm^4`

Minimum Moment of Inertia I=3.1123136 X `10^5``mm^4`

For Maximum Moment of Inertia,

S=`9.214432634 X`10^7``mm^4`/440 m`

S= 209418.927 `mm^3`

My =s×σy

My =209418.927 `mm^3` X 515

`= 10.785 X `10^7` Nm`

For Minimum Moment of Inertia,

S= 3.1123136 X `10^5``mm^4`/440

S=707.344 `mm^3`

My = 707.344`mm^3`×515

My = 364282.16Nmm

Conclusion:

As a result, the section modulus of a car's front hood was determined. As a result, the stiffness and resistance of the hood will rise as the sectional area of the hood grows.