Section Modulus calculation and optimization
AIM:
Use the Section from your Hood design and calculate the section modulus.
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.
What is Section Modulus:
It is the ratio of moment of inertia to the distance from the neutral axis and the extreme end of the object.
S = (I/y)mm^3
S = Section Modulus
I = Moment of Inertia
y = distance between the neutral axis and the extreme end of the object
Section modulus depends only on the cross section shape of the beam. Cross section shapes like rectangular, square, circular, I section and T, composite section etc.
For symmetric sections, the value of “section modulus” is for both above and below of centroid of the fiber.
An asymmetric section has two values for top and bottom of the fiber due to different distance from centroid to the top and bottom of the material, the maximum “section modulus” is S max and the minimum section modulus is S min.
Importance:
- It is significant for beam and flexural member design.
- More the section modulus, more the member is resistant to bending of the beam.
- It is easier to calculate stresses in beam and an exact measure of strength of steel.
- If two beams are made of same material and comparing the “section modulus” of two beams, the beam with bigger section modulus will be tougher and more capable to withstand larger loads.
Types of section module,
- Elastic section modulus (S)
- Plastic section modulus (Z).
Material Used for Hood,
AISI302 - Cold Rolled - Stainless Steel, y = 520Mpa
Hood:
Section Modulus Calculation in NX CAD:
Design Approch 1:
For above section the section modulus is calculated by using 'Section inertia analysis' command.
Where,
Maximum moment of inertia I(max)= 9.2385e+07 mm^4
Minimum moment of inertia I(min)= 2.8768e+05 mm^4
Y = 439.75 mm
S=I(min)/Y = 2.8768e+05/439.75 = 654.189 mm^3
Design Approch 2:
By offsetting the complete cross-section in outer direction by 0.3mm,
Where,
Maximum moment of inertia I(max)= 9.2627e+07 mm^4
Minimum moment of inertia I(min)= 2.9049e+05 mm^4
Y = 439.75 mm
S=I(min)/Y = 2.9049e+05/439.75 = 660.579 mm^3
Conclusion:
It is clearly seen that by offsetting the enitre cross-section by 0.3mm outer side will increase the section modulus of the Hood, which results in increasing the strength of Hood.