Section Modulus calculation and optimization

Section Modulus

The section modulus is defined as S = I / y, where I is the second moment of area

(or area moment of inertia, not to be confused with moment of inertia)

and y is the distance from the neutral axis to any given fibre. It is often reported using y = c,

where c is the distance from the neutral axis to the most extreme fibre,

It is also often used to determine the yield moment (M_y) such that

M_y = S ⋅ σ_y, where σ_y is the yield strength of the material.

Moment of inertia

The moment of inertia is a measure of how resistant an object is to changes in its rotational motion. T

he moment of inertia of the cross-section of a body is its resistance to changes in its rotation.

It depends on how far each part of the body's mass is from its center.

Calculations Useing Hood

S = Section Modulus

I = Moment of Inertia

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

First analyse the base section

Case - 1

where ,

max MOI = 9.010442045e+07[mm^4] = 90104420.45 mm^4

min MOI = 2.673811115e+05[mm^4] = 267381.11 mm^4

y = H/2, = 879.4966/2, y = 439.74

section modulus S = I/y

For Max MOI S = I max/y

= 90104420.45/439.74

= 204903.85 mm^3

For Min MOI S = I min/y

= 267381.11/439.74

= 608.04 mm^3

AFTER DESIGN MODIFICATION

Case - 2

where ,

max MOI = 8.957084368e+07[mm^4] = 89570843.68 mm^4

min MOI = 2.579612130e+05[mm^4] = 257961.21 mm^4

y = H/2, = 879.4966/2, y = 439.74

section modulus S = I/y

For Max MOI S = I max/y

= 89570843.68/439.74

= 203690.46 mm^3

For Min MOI S = I min/y

= 257961.21/439.74

= 586.62 mm^3

Conclusion

Here we can see, while we change our design MOI is changing as well as section modulus