Section Modulus calculation and optimization

Question:-

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

S = Section Modulus

I = Moment of Inertia

y = distance between the neutral axis and the extreme end of the object

Come up with a new section that has improved the section modulus of the previous section and mention what change you made that has increased the strength.

Answer:-

Section Modulus:-

*It is a geometry property for a given cross-section used in the design of beams or flexural members. Section modulus can also be defined as the ratio of total moment resisted by section to the stress in the extreme fibre which is equal to yield stress. Also, section modulus is the direct measure of strength of a beam. Higher the section modulus higher will be the resistant to bending.

Section modulus is indicated as follow

S=I/Y

I =moment of inertia

Y= distance of neutral axis to top or bottom of fibre

Unit

Unit of section modulus is mm^3

RESULT OBTAINED AS FOLLOW

1

Imin = 9.9619*10^4 (mm^4)

Y= 483.2097/2=241.6085

S1= Imin / Y

= 9.9619*10^4

241.6085

=412.3158 (mm^3) ------------------------------------------------------ (1)

2.

Imin =2.78663*10^5 (mm^4)

Y= 483.2097/2=241.6085

S2= = Imin / Y

= 2.78663*10^5

241.6085

=1153.37 (mm^3) ------------------------------------------------ (2)

MODIFIED SECTION:-

By incorporating more emboss and increasing in area can increase our moment of inertia. Because according to formula of moment of inertia more the distance of area from neutral axis greater will be its inertia so keeping that thing.

CONCLUSION:-

From result (1) & (2) its conclude that by increasing number of embosses we get more section of inertia.