Section Modulus calculation and optimization

AIM:

TO FIND THE SECTION MODULUS FROM THE HOOD DESIGN AND OPTIMATIONS NEEDS TO BE DONE IN ORDER TO IMPROVE SECTION MODULUS .

SECTION MODULUS :

The section modulus of the cross-sectional shape is of significant importance in designing beams. It is a direct measure of the strength of the beam. A beam that has a larger section modulus than another will be stronger and capable of supporting greater loads.

It is the ratio of Moment of Inertia (I) of the beam cross-section about the neutral axis to the distance (ymax) of extreme fiber from the neutral axis.

Section modulus ( s ) = I / Ymax

MOMENT OF INERTIA (I )  : Newton's first law explains that ,if higher the mass , the force required to move the object is also high due to offering more resistance in order to move the object, This resistance we called it as a MOMENT OF INERTIA.

Section modulus is a geometric property for a given cross-section used in the design of beams or flexural members, it is also referred to as the Polar Modulus or the Torsional Sectional Modulus

In terms of bending stress, the section modulus is the ratio of bending moment to bending stress for steel. If  steel has a high section modulus it will be harder to bend and can withstand a high moment without having high bending stres

CASE 1 : BEFORE OPTIMISATION 

In this , we have taken the hood assembly , then we have taken the intersection curve from the inner and outer pannel . then section inertia analysis done.

we , measure the distance total , by MEASURE DISTANCE.

from the above section inertia analysis , 

I max = 8.476029203e^005 (mm^4)

I min = 8.265615427e^003 (mm^4) = 166.01  (mm^4)

TOTAL Y distance = 486.8161 mm

Ymax= Y/2 = 243.40805 mm

To find section modulus of  HOOD section, we have to calculate

S   =  I min  / Y = 0.68280 (mm^3)

S = I max/Ymax= 5.16 (mm^3)

case 2 : After optimisation

 new section that has an improved Section Modulus compared to the previous one , as in this case we have taken offset curve in the inner pannel section , hence area has increased in this section , now we will optimise this .

Imax =    8.50525e^005 (mm^4)

I min =    8.767668e^003 ( mmm^4)  = 176.103

TOTAL DISTANCE = 486.8161 mm

Ymax = Y/2 = 243.40805

To find section modulus of  HOOD section, we have to calculate

S = I min / Ymax = 0.7234 (mm^3)

S = Imax/Ymax= 5.185 (mm^3)

CONCLUSION:-

As , before optimisation , the section modulus values comes oout to be is 0.68280 (mm^3), then after optimisation , the section modulus comes out to be 0.7234 (mm^3). Therefore we can conclude that strength of the beam is incread.

ALSO, here we have consider the minimum values of inertia , because, when we safe the mimimum sterngth , maximum strength will automatically safe . 

The  strength of the beam ( section modulus )  is directly proportional to the moment of inertia, and inversly propotional to the the max distance that is from the centroid , hence by increasing the area , we saw that , section modulus or strength of the beam is increased.