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 geometric 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 ration 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 beams cross sectional 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 = Youngs 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 fibres of the beams are subjected to maximum tension.
Calculation:
Frontal crash:
Itration-1:
Itration-2:
Front crash 2mm offset:
Side crash:
Itration-1
Itration-2
side crash 2mm offset:
Section view:
Observation:
| SECTION MODULUS DATA | FRONT CRASH ORIGINAL | FRONT CRASH 2MM | SIDE CRASH ORIGINAL | SIDE CRASH 2MM |
| Imin(in meters) | 1.772*10^6 | 2.172*10^6 | 1.044*10^6 | 1.252*10^6 |
| Imax(in meters) | 6.501*10^8 | 7.74*10^8 | 1.454*10^8 | 1.682*10^8 |
| I min(in mm) | 1772000 | 2172000 | 1044000 | 1252000 |
| I max(in mm) | 650100000 | 774000000 | 145400000 | 168200000 |
| Total distance(in mm) | 878.998 | 878.646 | 485.15 | 484.927 |
| Y=Total dis/2 (in mm) | 439.499 | 439.323 | 242.575 | 242.4635 |
| S min= I min/y(in mm) | 4031.86355 | 5522.17897 | 4303.82356 | 5163.66381 |
| S max= I max/y (in mm) | 1479184.25 | 1761801.68 | 599402.247 | 693712.662 |
| S avg=(Smin+Smax)/2(in mm) | 741608.055 | 883661.93 | 301853.036 | 349438.163 |
sample calculation:
Itration-1
Frontal crash original:
Imin(in meter)=1.772*10^6 m
Imax(in meter)=6.501*10^8m
Imin (in mm)= 1772000 mm
Imax (in mm)= 650100000 mm
Total distance = 878.998mm
Y=total distance/2 =439.499 mm
S=I/Y
Smin=Imin/y
Smax=Imax/Y
Smin = 1772000/439.499 =4031.86355 mm
Smax = 650100000/439.499 = 1479184.25 mm
S avg =(Smin + Smax)/2
=(4031.86355+1479184.25)/2
=741608.055 mm
Itration-2
Frontal crash 2 mm offset:
Imin(in meter)=2.172*10^6
Imax(in meter)=7.74*10^8
Imin (in mm)= 2172000
Imax (in mm)= 774000000
Total distance =878.646
Y=total distance/2 =439.323
S=I/Y
Smin=Imin/y
Smax=Imax/Y
Smin = 2172000/439.323
Smax = 774000000/439.323
S avg =(Smin + Smax)/2
=(5522.17897+1761801.68)/2
=883661.93
By comparing the both the cases
741608.055 < 883661.93
Conclusion:
The section modulus value for the offset range of 2 mm of the inner panel of the hood has the maximum value when compared to the original dimension. I conclude that section modulus increases the stress will be decreased. If we design in this method we can achieve the structure.