Section Modulus calculation and optimization

Title : 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

Objective : In this challenge we need to find the section modulus of the hood we have created and how to improve it which will result in increasing its strength.

Tool used :

Intersection curve

sketch tools

section moment of inertia

Description :

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

• It is the ration of moment of inertia and the distance of the topmost or the bottommost fiber from the neutral axis S = I/y

• S = Section Modulus

  I = Moment of Inertia

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

• It represents the strength of the section, Greater the value of S, greater will be the strength of the section.
• The section modulus depends upon the moment of inertia which again depends not only on the total area of the cross-section but also on the disposition of the area with respect to neutral axis.

Steps followed :

• First using intersection curve command a intersection is defined on the hood by selecting its outer and inner panel.
• The obtained intersection curve is then joined at both ends to form a close profile.
  • This close profile formed is then analsed by the command section moment of inertia which gives the value of MOI min = 1.9276992 × 10^-7 m⁴
  • Y is the distance between the neutral axis and the extreme end of the object which is found to be 488.91 mm / 2 = 244.45 mm
  • So section modulus S = I/Y as given above
  • S = 1.9276992 × 10-7 / 244.45 = `7.885e-10 mm3`
• Now to improve the Moment of inertia without changing thickness of the sheet we will offset the inner pannel for 2mm and then calculate the value to check its deviations from the orginal value.
• So now outer and inner curve of the hood is projected and modified with sketch command to the offset of 2mm from the original position and the new sketch is obtained.
• On this new sketch moment of inertia is calculated and the value found to be is MOI min = ``2.14077053 × 10-7 m4`
• So, new S = `2.14077053 × 10-7 m4 / 244.45` = `8.757e-10 mm3`

Conclusion:

we found out that with increasing the area of cross section the moment of inertia increases which in result increases the section modulus S of the component from 7.885e-10 mm3 to 8.757e-10 mm3 which means just by increasing the area we can increase the strength of the component without changing the thickness of the material.

Thankyou