Section Modulus calculation and optimization

SECTION MODULUS OF THE HOOD

Aim

To calculate the section modulus of the 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: 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 ration of total moment resisted by the section to the stress in the extreme fibre which is equal to yield stress.

There fore

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 bending. The larger the moment of inertia less will be the bend.

Where,

M = Bending moment

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 stainless steel is 520MPa = 520 N/mm^2

Calculation:

CASE 1: Before Optimization

The Hood which was already created is imported and the section inertia analysis is done.

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

MOI (max) = 9.056472367e+07 (mm^4)

MOI (min) = 2.877755730e+05 (mm^4)

As per the section modulus equation,

S = I/Y

I = MOI (min) value is considered

Y = Total distance /2, Total distance =876.1544mm

Y = 876.1544 / 2

= 438.077 mm

Therefore, S = 287777.557 mm^4/438.077 mm

S = 656.910 mm^3

CASE 2: After optimization

MOI (max) = 9.054106651e^07 mm^4

MOI (min) = 2.923447773e^05 mm^4

As per section modulus equation,

S = I / Y

I = MOI (min) value is considered

Y = Total distance /2, Total distance =876.1544 mm

Y = 876.1544 / 2

= 438.0772 mm

Therefore, S = 292344.7733 mm^4 / 438.0772 mm

S = 667.336 mm^3

Conclusion:

Hence when 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 increases which means strength and load material also increase

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)