Section Modulus calculation and optimization

SECTION MODULUS :-

SECTION MODULUS IS THE AREA MOMENT OF INERTIA DIVIDED BY THE MAXIMUM DISTANCE FROM THE CENTROID TO THE OUTER FIBER OF A CROSS SECTION. IT MEASURE THE STRENGTH OF ANY AREA OF AN OBJECT.,GREATER THE VALUE OF SECTION MODULUS, GREATER WILL BE THE STRENGTH OF THE SECTION.THE SECTION MODULUS DEPENDS UPON THE MOMENT OF INERTIA WHICH AGAIN DEPENDS NOT ONLY ON THE TOTAL AREA OF THE CROSS-SECTION BUT ALSO ON THE DISPOSITION OF THE AREA WITH RESPECT TO NEUTRAL AXIS.

S = I/Y

WHERE,

S = SECTION MODULUS

I = MOMENT OF INERTIA

Y = DISTANCE BETWEEN THE NEUTRAL AXIS AND THE EXTREME END OF THE OBJECT

1^st CALCULATE OF THE SECTION MODULUS

SECTION FROM HOOD DESIGN

MOI(max): 8.990374857e+07[mm^4]

MOI(min): 2.861171832e+05[mm^4]

TOTAL HOOD DISTANCE = 876.4579 mm

DISTANCE BETWEEN THE END OF THE HOOD AND NEUTRAL AXIS (Y)= 438.22895 mm

SECTION MODULUS -> S = I/Y

MOI(min) I/Y = 2.861171832/438.22895

= 0.006528943

S = 0.006528943 x 10^5 mm3

= 652.8943 mm3

THIS IS THE MOMENTUM OF INERTIA MINIMUM VALUE FOR 1st CALCULATE OF THE SECTION MODULUS.

2^nd CALCULATE OF THE SECTION MODULUS

SECTION FROM OPTIMIZED HOOD DESIGN

MOI (max) : 9.081169082e+07[mm^4]

MOI (MIN) : 3.078434032e+05[mm^4]

TOTAL HOOD DISTANCE = 876.1055 mm

DISTANCE BETWEEN THE END OF THE HOOD AND NEUTRAL AXIS (Y)= 438.05275 mm

SECTION MODULUS -> S = I/Y

MOI(min) I/Y = 3.078434032/438.05275

= 0.0070275418

S = 0.0070275418 x 10^5 mm3

= 702.75418 mm3

THIS IS THE MOMENTUM OF INERTIA MINIMUM VALUE FOR 2^nd CALCULATE OF THE SECTION MODULUS.