Section Modulus calculation and optimization

Aim:

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

Objective:

Use the section from your Hood design and calculate the section modulus using the formula S= I/y.

Section Modulus:

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

Which is equal to yield stress.

Therefore

S=I/y, Where

S= Section modulus (mm^3)

I = Moment of inertia of the section to the stress in the extreme end of the object (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 beam cross section area measures the beam ability to resist bending. The larger the moment of inertia less will be the bend.

Where,

M = Bending Moment

I = Area moment of Inertia

E = Young 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 maximum compression, while the bottom most fibers of the beams are subjected to maximum tension.

Calculation:

Case 1: Before Optimization

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

I (MAX) = 1.542709062e+07 (mm^4)

I (MIN) = 1.20372805e+05 (mm^4)

As per the section modulus equation,

S = I/Y

I = MOI (min) Value is considered

Y = Total distance / 2, Total distance = mm

Y = 485.4562 / 2

= 242.7281 mm

Therefore, S =1.20372805e+05 (mm^4) /242.7281 mm

S = 495.9162mm^3

Case 2: After Optimization

The hood inner panel’s curve is offset by 0.5mm outwards and section inertia analysis is done the same as case 1 for a closed curve.

I (MAX) = 1.540905889e+07 (mm^4)

I (MIN) = 1.245216358e+05 (mm^4)

As per the section modulus equation,

S = I/Y

I = MOI (min) Value is considered

Y = Total distance / 2, Total distance = 484.0709 mm

Y = 484.0709 / 2

= 242.03545 mm

Therefore, S =1.245216358e+05 (mm^4) / 242.03545 mm

S = 514.4768mm^3

Conclusion:

Resulting by comparing case 1 & case 2,

1. The section modulus value of case 2 is higher than the case 1
2. As we increase the section area, the moment of inertia & section modulus increase which means strength and load material also increase. The moment of inertia is directly proportional to the strength of the material.