Section Modulus calculation and optimization
SECTION MODULUS OF THE HOOD
Objective: To find the section modulus of the hood and optimisation to improve the overall section modulus.
Section Modulus: The section modulus (S) is the geometric property of the cross-section used for designing beams and flexural members. It does not represent anything physically.
The 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.
There fore
S = I/Y, Where
S = Section modulus (mm^3)
I = Moment of Inertia of the section about the bending axis. (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-sectional area measures the beam's 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 = Youngs modulus
R = Radius of gyration
Neutral axis: When the beam is subjected to pure bending the top fibres of the beam are subjected to maximum compression, while the bottom-most fibres of the beams are subjected to maximum tension.
Material Data: Stainless steel AISI 302 – cold rolled
yield strength of stainless steel is 520MPa = 520 N/mm^2
Calculation:
CASE 1: Before Optimization
The Hood which was already created is imported and the section inertia analysis is done.
We have to make sure that the curve obtained is a single/closed curve.
MOI(Max): 8.717502642e+05(mm^4)
MOI(Min): 3.809744446e+03(mm^4)
As per the section modulus equation,
S = I/Y
I = MOI (min) value is considered
Y = Total distance /2, Total distance =482.27mm
Y=482.27/2=241.135mm
Therefore,S=38097.44446mm^4/241.135mm=157.9921mm^3
CASE 2: After optimization
The hood inner panel’s curve is offset by 0.5mm outwards and section inertia analysis is done the
same as case 1 for a closed curve.
MOI(Max):8.765109064e+05(mm^4)
MOI(Min):3.810533999e+03(mm^4)
As per the section modulus equation,
S = I/Y
I = MOI (min) value is considered
Y = Total distance /2, Total distance =482.27mm
Y=482.27/2=241.135mm
S=38105.33999mm^4/241.135mm=158.0249mm^3
Conclusion:
Hence when comparing case 1 & case 2,
1. The section modulus value of case 2 is higher than case 1
2. As we increase the section area, the moment of inertia & section modulus increase which means strength and load material also increases.