Section Modulus calculation and optimization
Section Modulus:
The section modulus (S) is geometry property of the cross section used for designing beams and flexural members. It does not represent anything physically.
Formula of section Modulus: It may be defined as the ratio of total moment resisted by the section to the stress in the extreme fibre which is equal to yield stress.
Which is equal to yield stress.
Therefore
S=I/y, Where
S= Section modulus (mm^3)
I = Moment of inertia of the section to the stress in the extreme end of the object (mm^4)
Y= Distance between the neutral axis and extreme end of the object (mm).
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: HOOD _W/O OPTIMIZATION
I (MAX) = 4.871626006e+06 (mm^4)
I (MIN) = 1.51966019e+04 (mm^4)
As per the section modulus equation,
S = I/Y
I = MOI (min) Value is considered
Y = Total distance / 2,
Total distance = 875.9330 mm
Y = 875.9330 / 2
Y= 437.9665 mm
Therefore, S = 15196.6019 mm^4 /437.9665 mm
S = 34.6980 mm^3
CASE 2: HOOD_AFTER OPTIMIZATION [1]
The Lines are offsetted for 1 mm and the Section modulus is generated.
I (MAX) = 4.854281345e+06 (mm^4)
I (MIN) = 1.1554953239 e+04 (mm^4)
As per the section modulus equation,
S = I/Y
I = MOI (min) Value is considered
Y = Total distance / 2,
Total distance = 875.9330 mm
Y = 875.9330 / 2
Y= 437.9665 mm
Therefore, S = 15549.53239 mm^4 /437.9665 mm
S = 35.5039 mm^3
Compared to Case 1, Case 2 has higher Section Modulus and there has been a increase on 2.32% in section modulus.
CASE 3: HOOD_AFTER OPTIMIZATION [2]
The Lines are offsetted for 1 mm and the Section modulus is generated.
I (MAX) = 4.842084198e+06 (mm^4)
I (MIN) = 1.581804840e+04 (mm^4)
As per the section modulus equation,
S = I/Y
I = MOI (min) Value is considered
Y = Total distance / 2,
Total distance = 875.9330 mm
Y = 875.9330 / 2
Y= 437.9665 mm
Therefore, S = 15818.04840 mm^4 /437.9665 mm
S = 36.1170 mm^3
Compared to Case 1, Case 3 has higher Section Modulus and there has been a increase on 4.08% in section modulus.