Section Modulus calculation and optimization

Objective: To se the Section from Hood design and calculate the Section Modulus. Come up with a new section that has an improved Section Modulus compared to the previous one.

Section Modulus: The section modulus (S) is geometric property of the cross section used for designing beams and flexural members. It does not represent anything physically.

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.

There fore

S = I/Y, Where

S = Section modulus (mm^3)

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

Y = Distance between the neutral axis and extreme end of the object (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

Case 1: Calculation on initial design: This calculation is done on the initial design. We take the intersection curve along the assembly and make sure that it is a closed curve to calculate the section modulus.

First we find the total length of the section:

Total length of section = 936.0106mm
Then we find the moment of inertia using section Inertia Analysis option.

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 = 936.0106mm

Y = 936.0106/2

= 468.0053mm

Therefore, S = 319660.966 mm^4/468.0053

= 682.70 mm^3

Case 2 Optimized Design: In order to improve the section modulus, cross section area has to be increased. To obtain that we have offsetted the inner panel by 5 mm.

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 = 936.0106mm

Y = 936.0106/2

= 468.0053mm

Therefore, S = 326495.4768 mm^4 / 468.0053 mm

S = 697.632 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.