Section Modulus calculation and optimization

Aim : To calculate the section modulus and optimization of hood.

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.

Section Modulus :

• The section modulus (S) is geometric property of the cross section used for designing beams and flexural members.
• It is defined as the ratio of the moment of inertia of a section about its centroidal axis and to the
  distance of the extreme layer from the neutral axis.

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.

• S = I/y

  Where,

  S = Section Modulus (mm^3)

  I = Moment of Inertia of the section about bendind axis (mm^4)

  y = distance between the neutral axis and the extreme end of the object (mm)

Case 1 : Befor Optimization

MOI (max) = 8.446716248e+05 (mm^4)

MOI (min) = 6.476093924e+03 (mm^4)

As per the section modulus equation,

S = I/Y

I = MOI (min) value is considered

Y = 484.3055 mm

Therefore, S = 647609.392 mm^4/484.305 mmtion

S = 1337.19 mm^3

Case 2 : After Optimization

MOI (max) = 8.504567145e+05 (mm^4)

MOI (min) = 9.17797198e+03 (mm^4)

As per the section modulus equation,

S = I/Y

I = MOI (min) value is considered

Y = 484.3055 mm

Therefore, S = 917797.198 mm^4/484.305 mmtion

S = 1895.06 mm^3

Conclusion:
1) compare case 1 & case 2.

2) The section modulus value of case 2 is higher than the case 1 .
3) As we increase the section area, the moment of inertia & section
modulus increases, which means strength and load material also increase.