Section Modulus calculation and optimization

Problem Statement: -

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.

Consider Stainless steel AISI 302 – cold-rolled – y = 520 MPa

Theory: -

• Section modulus is a geometric property for a given cross-section used in the design of beams or flexural members.
• There are two type of section modulus

1.elastic section modulus

1. plastic section modulus

• For general design, the elastic section modulus is used, applying up to the yield point for most metals and other common materials.
• The elastic section modulus is defined as S = I / y,

I = is the second moment of area (or area moment of inertia, not to be confused with moment of inertia)

Y = is the distance from the neutral axis to any given fibre. It is often reported using y = c,

C = is the distance from the neutral axis to the most extreme fibre, as seen in the table below.

• It is also often used to determine the yield moment (My) such that My = S × σy, where σy is the yield strength of the material.
• Higher the Section Modulus higher will be the Resistance to Bending.

Procedure: -

• By using intersection command, we have found out the shape of inner and outer panel of hood.
• After getting intersection curve we have to extract the outer most curve of both panel.
• Connect them by using line
• Convert the closed sketch to surface using fill command and measure the 2D inertia of section.

For improvising section modulus, we will increase the offset distance by 0.5 mm in negative y direction which will result in significantly increasing in section modulus.

Comaparision Data:-

Conclusion: -

• 5 % Increase in section modulus is found.