Section Modulus calculation and optimization
Section Modulus Calculation and Optimization:-
In this project i have done section modulus of the hood and optimization to improve the overall section modulus of the Hood.
Objective:-
I have used the Hood design and calculate the section modulus using the formula S= I/y.
Section Modulus:-
The section modulus (S) is geometry property of the cross section used for designing beams.It does not represent anything physically,Mainly it helps to increase the strength of the component
higher the section modulus higher the strength lower the section modulus lower the strength of the component
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 nature of the component.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
Calculation of the section modulus:-
Case 01:- "Total Distance of the Hood is 880.2487 mm"
As per the thickness of the component I have took to consider for a section modulus.
I (MAX) = 8.965418623e+07 (mm^4)
I (MIN) = 3.337274983e+05 (mm^4)
As per the section modulus equation,
S = I/Y
I = MOI (min) Value is considered
Y = Total distance / 2, Total distance = 880.2487 mm
Y = 880.2487 / 2
= 440.12435 mm
Therefore, S = 333727.4983 mm^4 / 440.12435 mm
S = 758.257mm^3
Case 02:- "Total Distance of the Hood is 880.2487 mm"
In this case the depth of the inner panel is increased to 1mm.
I (MAX) = 8.962756697e+07 (mm^4)
I (MIN) = 3.411968171e+05 (mm^4)
As per the section modulus equation,
S = I/Y
I = MOI (min) Value is considered
Y = Total distance / 2, Total distance = 880.2487 mm
Y = 880.2487 / 2
= 440.12435 mm
Therefore, S = 341196.8171mm^4 / 440.12435 mm
S = 775.2282 mm^3
Conclusion:
The section modulus of the case 1 is S = 758.257 mm^3
The section modulus of the case 2 is S = 775.228 mm^3
Higher the Section modulus gives the higher the strength to the material. so the Case 2 is having the higher when compared to the Case 1
So our conclusion is Case 2.