Challenge

Section Modulus calculation and optimization

OBJECTIVE:-

S = I/y

S = Section Modulus

I = Moment of Inertia

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

SECTION MODULAS:-

  • Ratio of minimum moment of inertia of componant to the distance between the neutral axis and the extreme end of the object is called as section modulas
  • section modulas is calculation for strenght of componant and plasticity of compnante.

AS PER DESIGN BEFORE MEASURING SECTION MODULAS:-

MOI (max) = 4.814703602e+07 (mm^4)

MOI (min) = 1.560769698e+04 (mm^4)

As per the section modulus equation,

S = I/Y

I = MOI (min) value is considered

Y = Total distance /2, Total distance =876.2742mm

Y = 876.2742/ 2

= 438.1371 mm

Therefore, S = 1.560769698e+04 mm^4/438.1371 mm

S = 35.622 mm^3

AFTER MEASURING SECTION MODULAS:-

HOOD INNER PANNEL OFFSET TO 0.5MM

MOI (max) =7.191111842e+06 (mm^4)

MOI (min) = 2.473454269e+04 (mm^4)

As per the section modulus equation

S = I/Y

I = MOI (min) value is considered

Y = Total distance /2, Total distance =876.2742mm

Y = 876.2742/ 2

= 438.1371 mm

Therefore, S = 1.560769698e+04 mm^4/438.1371 mm

S = 56.4538 mm^3

CONCLUSION:-

  1. In first case the section modulas is minimum so we get 0.5 mm offset of inner pannel in second case and we get more section modulas as first case .
  2. More section modulas is good for componant.
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