Section Modulus calculation and optimization

Objective:

To increase the section modulus of the hood by necessary changes.

SOFTWARE USED: NX CAD

Section modulus:

Section modulus is a geometric property for a given cross-section used in the design of beams or flexural members. Other geometric properties used in design include area for tension and shear, radius of gyration for compression, and moment of inertia and polar moment of inertia for stiffness. Any relationship between these properties is highly dependent on the shape in question. Equations for the section moduli of common shapes are given below.

Formula of section modulus:it may be define 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 netural axis and extreme end of the object (mm)

Area moment of inertia:

The area moment of inertia of a beams cross section area measure the beam ability to resist bending .the large 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.

Section modulus calculation of the hood design:

CASE 1:Initial hood design.

• Initial hood distance of embosse is 8.3419mm

• MOI(MAX):4.831306576e+06[mm^4]
• MOI(min):1.422109890e+04[mm^4]

• Y total distance/2 : 874.4393mm/2=437.21

As per section modulus equation,

S=I/Y

The distance between netural axis and extreme end of the object Y,Y total distance/2 : 874.4393mm/2=437.21

Therefore s=1.422109890e+04mm^4/437.21

S=32.5269296mm^3

CASE 2: After changes applied

The inner pannel is reduce from 8.3419-5.7771mm=2.5648mm

• After modified hood distance of embosse is 5.7771mm

• MOI(MAX):4.8856473358e+06[mm^4]
• MOI(min):1.42443587e+04[mm^4]

As per section modulus equation,

S=I/Y

The distance between netural axis and extreme end of the object Y,Y total distance/2 : 874.4393mm/2=437.21

Therefore s=1.42443587e+04mm^4/437.21

S=32.5801301mm^3

Conculsion:

• When comparing case 1and case 2, we conclude that value of case 2 is greater than case 1
• after decereasing the section area .the moment of inertia and section modulus increase
• which means strenght and load material also increase.