Section Modulus calculation and optimization
AIM:
- To find the section modulus of the hood and optimization to improve the overall section modulus.
OBJECTIVE:
- Use the Section from your Hood design and calculate the Section Modulus using the formula
S = I/y.
SECTION MODULUS:
It is the ratio of area moment of inertia about the neutral axis and to the distance of outermost fiber from the neutral axis.
S = I/y
Where,
S = Section Modulus
I = Moment of Inertia
y = distance between the neutral axis and the extreme end of the object .
AREA MOMENT OF INERTIA:
The area moment of inertia of a beam cross section area measures the beam ability to resist bending. The larger 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.
NEUTRAL AXIS:
- When beam is subjected to pure bending the top most fibres of the beam are subjected to maximum compression, while the bottom most fibre of the beams are subjected to maximum tension.
- This picture contains the following scene, Diagram of a beam on its narrow edge supported at both ends. A man is walking on a straight beam. Diagram of a beam resting on its flat side supported at both ends. A man is walking on the straight beam and its bending in the middle.
CASE 1:-(BEFOR OPTIMIZATION)
- The hood which was created early is imported and the section inertia analysis is done.
- We have to check the curve obtained is closed/single curve.
I (max) = 9.068467353e+07
I (min) = 2.753184571e+05
Total distance = 876.8173mm
Total distance/2 = 438.40865mm
As per the section modulus equation,
S = I/Y
Therefore s1 = 2.753184571e+05/438.40865
s1 = 627.9757mm^3 .
CASE 2:(After optimization)
I (max) = 9.042777040e+07
I (min) = 2.807563089e+05
Total distance = 876.8173mm
Total distance/2 = 438.40865mm
As per the section modulus equation,
S = I/Y
Therefore s2 = 2.807563089e+05/438.40865
s2 = 640.384 mm^3 .
CONCLUSION:
Resulting by comparing case 1 & case 2,
- The section modulus value of case 2 is greater than the case 1 (s2>s1).
- As we increase the section area, The moment of inertia and section modulus increase which means the strength and load material increases.
- Hence the moment of inertial is directly proportional to strength of material.