Section Modulus calculation and optimization

Section modulus using the formula S = I/y by the help of CATIA V5 in the below link:-

SECTION MODULUS CALCULATION AND OPTIMIZATION

The plastic section modulus is used for materials where elastic yielding is acceptable and plastic behaviour is assumed to be an acceptable limit. Designs generally strive to ultimately remain below the plastic limit to avoid permanent deformations.

Here, Hood inner and outer is used for calculation the section modulus using the formula i.e. S= I/Y using cad tool CATIA V5.

Where, S = Section Modulus

I = Moment of inertia

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

It gives us an idea about how much of load the given structure can take in starting and where it is going to act. If the section modulus increases the moment of resistance increases and its load bearing capacity also increase.

By calculating and optimization of section modulus, it is easy to do modification to the object design so that strength of the object is to be done or it make easy to compare.

Calculation: -

Intersection Part of Hood: -

Measure Inertia: -

Here, From the above data

M1( Minimum):- 4.194* 10^-7 m^4

4.194*10^5 mm^4

M2 ( Maximum):- 1.326*10^-4 m^4

1.326*10^8 mm^4

For Calculation, considering minimum value because it can be act on body in starting. i.e. 4.194*10^5 mm^4

Distance between the neutral axis and extreme end of the object

i.e. Y = 477.97/2 = 238.985 mm

Then

S= I/Y

S = 1.75* 10^3 mm^3

Optimization Data: -

• It is obtained when inner hood boundary is offset by 0.5 mm in downward direction.

Here, From the above data

M1( Minimum):- 4.453* 10^-7 m^4

4.453*10^5 mm^4

M2 ( Maximum):- 1.371*10^-4 m^4

1.371*10^8 mm^4

Here, Minimum value is I = 4.453*10^5 mm^4

Y = 477.97/2 = 238.985 mm

S= I/Y

S = 1.8* 10^3 mm^3

• It is obtained when inner hood boundary is offset by 0.5 mm in upward direction.

Here, From the above data

M1( Minimum):- 3.947* 10^-7 m^4

3.947*10^5 mm^4

M2 ( Maximum):- 1.28*10^-4 m^4

1.28*10^8 mm^4

Here, Minimum value is I = 3.947*10^5 mm^4

Y = 477.97/2 = 238.985 mm

S= I/Y

S = 1.4* 10^3 mm^3

Conclusion

When I increase the area of Hood I got more section modulus value i.e. S = 1.8* 10^3 mm^3 so, in this condition hood has more strength.

THANK You