SECTION MODULUS CALCULATION AND OPTIMIZATION OF THE HOOD

Objective:

To find the section modulus of the hood and optimisation 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 the section modulus can be defined as the ration of total moment resisted by the section to the stress in the extreme fibre which is equal to yield stress.

Therefore,

S=I/Y

Where

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 moment of inertia:

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

Where,

M=Bending modulus I=Area moment of inertia E=Youngs 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 fibres of the beams are subjected to maximum tension.

Material data: stainless steel AISI 302-cold rolled

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

Calculation:

Case 1: Before optimization

The hood which was already created and imported then the section inertia analysis is done. We have to make sure that the curves are obtained single connected and closed curves.

Fig:1.1

Fig:1.2

Fig:1.3

MOI (max)=4.902893054e+06(mm^4) MOI (min)=1.47041006e+04(mm^4)

As per section modulus equation, S=I/Y

I=MOI (min) value is considered Y=Total distance/2

Total distance=876.9748mm Y=876.9748/2 Y=438.4874mm

Therefore, S=14704.1006 mm^4/438.4874mm

S=335.336 mm^3

Case 2:

The hood inner panel curve is offset to 0.5mm outwards and section inertia analysis is done as followed for the case 1.

Fig:1.4

Fig:1.5

Fig:1.6

MOI (max)=4.927119534e+06 (mm^4) MOI (min)=1.520688727e+04 (mm^4)

As per section modulus equation, S=I/Y

I=MOI (min) value is considered Y=Total distance /2 Y=876.8004/2

Y=438.4002mm

Therefore, S=15206.88 mm^4/ 438.4002mm

S=346.872 mm^3

Conclusion:

Hence by comparing case 1 and case 2 we can conclude that 1.The section modulus value of case 2 is greater than the case 1.

2.As by increasing the section area, the moment of inertia and section modulus increases which means strength and load material also increases.