Section Modulus calculation and optimization

Section modulus:-

Section modulus is a direct measure of a strength of any cross sectional area of an object about an reference axis.

The formula of section modulus is given by,

S=I / Ymax

Where,
I = Moment of inertia (mm^4)
Y = Distance between neutral axes and outermost fiber (mm)

Area moment of Inertia:

The area moment of inertia of a beams cross sectional area measures the beam ability to resist bending.  The larger the moment of inertia less will be the bend.

Where,

M = Bending moment

I = Area moment of Inertia

E = Youngs modulus

R = Radius of gyration

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.

Material Data: Stainless steel AISI 302 – cold rolled

  Where yield strength of stainless steel is 520MPa = 520 N/mm^2

Calculation:

CASE 1: Before Optimization

The Hood which was already created is imported and the section inertia analysis is done.

We have to make sure that the curve obtained is a single/closed curve.

                                                                   fig 1

                                                                    fig 2

MOI (max) = 8.928754301e+07 (mm^4)

MOI (min) = 3.196609661e+05 (mm^4)

As per the section modulus equation,

S = I/Y

I = MOI (min) value is considered

Y = Total distance /2, Total distance =875.5680mm

Y = 875.5680 / 2

   = 437.784 mm

Therefore, S = 319660.966 mm^4/437.784 mm

                    S = 730.179 mm^3

CASE 2: After optimization

The hood inner panel’s curve is offset by 0.5mm outwards and section inertia analysis is done same as case 1 for a closed curve.

                                                                     fig 3

                                                                 fig 4

MOI (max) = 8.910657812e^07 mm^4

MOI (min) = 3.264954768e^05 mm^4

As per section modulus equation,

S = I / Y

I = MOI (min) value is considered

Y = Total distance /2, Total distance =875.5680mm

Y = 875.5680 / 2

   = 437.784 mm

Therefore, S = 326495.4768 mm^4 / 437.784 mm

                   S = 745.792 mm^3

Conclusion:

Hence when comparing case 1 & case 2,

1. The section modulus value of case 2 is higher than the case 1

2. As we increase the section area, the moment of inertia & section modulus increases which means strength and load material also increase