Section Modulus calculation and optimization

SECTION MODULUS;

Section modulus is a geometric property for a given cross-section used in the design of beams or flexural members.Other geometric properties used in design include area for tension and shear, radius of gyration for compression, and second moment of area and polar second moment of area for stiffness.

Any relationship between these properties is highly dependent on the shape in question. Equations for the section moduli of common shapes are given below.

There are two types of section moduli, the elastic section modulus and the plastic section modulus. The section moduli of different profiles can also be found as numerical values for common profiles in tables listing properties of such.

For general design, the elastic section modulus is used, applicable up to the yield point for most metals and other common materials.

The elastic 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 fiber. It is often reported using y = c, where c is the distance from the neutral axis to the most extreme fiber, as seen in the table below. It is also often used to determine the yield moment (My) such that My = S ⋅ σy, where σy is the yield strength of the material.

SECTION MODULUS OF HOOD;

                                                                CROSS SECTION OF HOOD

                                                           SECTION INERTIA ANALYSIS

Formula for section modulus is; S = I / Y

I = 1.251997990e+05 mm^4

Y = 243.7076 mm

S = 1.25199799e+05 / 243.7076

 S = 513.7289312 mm^3

SECTION MODULUS OF OPTIMISED CURVED;

In the optimised curved, we offseted the base by 1.5mm downwards and created a new curved for cross section and analysed it.

                                                                OPTIMISED CURVED

        SECTION INERTIA VALUE OF OPTIMISED CURVED

Formula for section modulus is; S = I / Y

I = 1.383478464e+05 mm^3

Y = 243.7076 mm

S = 1.25199799e+05 / 243.7076

 S = 567.679656 mm^3

Thus it is clear from the section modulus that by increasing the cross section the moment of inertia increases.