Section Modulus calculation and optimization

SECTION MODULUS OF THE HOOD:

Objective:

To find the section modulus of the hood and optimisation to improve the overal 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 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.

MOl(max)=8.211e+008mm4

MOl(min)=2.238e+006mm4

AS per the section modulus equation,

s=l/y

l=MOl(MIN) value is considered

y=total distance/2

y=880.586/2=440.293

section modulus(s)=l/y=22,380/440.293=50.82[mm^3]

from the section 1 using minimum moment of inertia and extreme distance of layer , we got section modulus(s)=50.82[mm^3]

section 2:

IN the section modulus to increase of inner panel distance from the previous section 1 off of the inner panel 2mm

MOl(max)=8.465e+008mm4

MOl(min)=2.407e+006mm4

AS per the section modulus equation,

s=l/y

l=MOl(MIN) value is considered

y=total distance/2

y=880.586/2=440.293

section modulus(s)=l/y=24,070/440.293=54.66[mm^3]

from the section 2 using minimum moment of inertia and extreme distance of layer , we got section modulus(s)=54.66[mm^3]

Conclusion:

Hence when comparing case 1& case 2,

1. The section modulus value of case 2 is higher than the casel

2. As we increase the section area, the moment of inertia & section modulus increases which means strength and load material also increase