Section Modulus calculation and optimization

AIM : Section Modulus Calculation Of Hood

To calculate the section modulus of the previously designed hood for analyzing its strength and also optimizing the design to see the difference in the bending strength of the hood.

Higher the section modulus of a structure, the more the resistive it becomes to bending.

Section modulus

It is the ratio of area moment of inertia about the neutral axis and to the distance of outermost fiber from neutral axis.

S=I/Y

Here,

S = Section modulus

I= Area moment of Inertia (Unit mm 4)

Y= Distance from Neutral axis to the outermost fiber (unit mm)

The Unit of a “section modulus” is (mm3)

Changing the position can show the difference in bending which is because of the centroid distance from extreme fiber end.

First Design Study

The Intersection of the design was taken and then section Inertia Analysis was done-

value of Y

value of y=876.2512/2=438.1256 mm

FOR CASE 1 MI CALCULATION

Moment of inertia(max) = 8.988265741×10^4×mm^4

Moment of inertia(min) = 2.628505858×10^4×mm^4

We take moment of inertia minimum:-

Y=438.1256 mm

S=I/Y

S=2.628505858×10^4×mm^4/438.1256 mm= 59.936 mm^3

Hood Design Optimization

By increasing the depth of the inner panel by 0.5 mm

CASE2

Here,

Moment of inertia(max) = 8.840460035×10^4×mm^4

Moment of inertia(min) = 2.641451185×10^4×mm^4

We take moment of inertia minimum:-

Y=438.1256 mm

S=I/Y

S=2.641451185×10^4×mm^4 /438.1256 mm=60.2898 mm^3

Conclusion

So, here the difference can be seen in the moment of inertia and the change in the section modulus can be seen by the change in the design(depth).

CASE 1 S = 2.628505858×10^4/438.1256 mm = 59.936 mm^3

CASE 2 S = 2.641451185×10^4/438.1256 mm = 60.289 mm^3

value of Y=876.2512/2=438.1256 mm