OBJECTIVE:

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

SECTION MODULUS:

The section modulus (S) is geometric 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 of inertia resisted by the section to the stress in the extreme fiber which is equal to the yield stress.

Therefore

S = I / y

S = Section modulus (mm^3)

I = Moment of inertia of the section about the bending axis (mm^4)

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

AREA OF MOMENT OF INERTIA:

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

M = moment of inertia

I = area of moment of Inertia

E = Young’s modulus

R = Radius of gyration

NEUTRAL AXIS:

When the beam is subjected to pure bending the top most fibers of the beam are subjected to the maximum compression, while the bottom most fibers of the beam are subjected to a maximum tension.

Material of data: Stainless steel AISI 302- cold rolled

Where yield strength of stainless steel is 520 Mpa = 520N/mm^2

CALCULATION:

• CASE 1: Before optimization

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

We have to make sure that the curve obtained is a single and closed curve

MOI (max) = 9.227734630e+07 (mm^4)

MOI (min) = 2.704507237e+05 (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 = 875.5680 / 2

= 437.784 mm

Therefore S = 270450.7237mm^4 / 437.784mm

S = 617.7720 mm^3

• CASE 2: After optimization

The hood inner panel’s curve is offset 0.50mm outwards and section inertia analyses is done and results are as shown below.

MOI (max) = 9.34547637e+07 (mm^4)

MOI (min) = 2.726612015e+05 (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 = 875.5680 / 2

= 437.784 mm

Therefore S = 272661.2015mm^4 / 437.784mm

S = 622.821 mm^3

CONCLUSION:

Hence by comparing case 1 and 2

1. Section modulus value 2 is greater than case 1
2. As we increase the section area, the moment of inertia and section modulus increases which means strength and load material also increase.