Section Modulus calculation and optimization

                                     Challenge - SECTION MODULUS OF THE HOOD

Objective: To find the section modulus of the hood and optimization to improve the overall section modulus.

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

The formula of Section Modulus: It may be defined as the ratio of the total moment resisted by the section to the stress in the extreme fiber 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 beam's cross-sectional area measures the beam's 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 = Young's modulus

R = Radius of gyration

Neutral axis: When the beam is subjected to pure bending the topmost fibers of the beam are subjected to maximum compression, while the bottom most fibers of the beams are subjected to maximum tension.

Material Data: Stainless steel AISI 302 – cold rolled

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

 

Calculations:

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 1.1

          

                                                               Fig 1.2

                                                     

                                                               Fig 1.3

MOI (max) = 5.097517577e+06 mm^4

MOI (min) = 1.569042653e+04 mm^4

As per the section modulus equation,

S = I / Y

I = MOI (min) value is considered

Y = Total distance /2, Total distance =882.6661mm

Y = 882.6661/ 2

   = 441.33305 mm

Therefore, S = 15690.42653 mm^4 / 441.3305 mm

                   S = 35.552 mm^3

CASE 2: After optimization

The distance between the hood inner panel and the outer panel is decreased by 10 mm and section inertia analysis is done as same as done in case 1 for a closed curve.

      

                                                             Fig 1.4

       

                                                             Fig 1.5

        

                                                             Fig 1.6

MOI (max) = 5.201653635e+06 (mm^4)

MOI (min) = 1.666018422e+04 (mm^4)

As per the section modulus equation,

S = I/Y

I = MOI (min) value is considered

Y = Total distance /2, Total distance =882.6661mm

Y = 882.6661/ 2

   = 441.33305 mm

Therefore,  Section Modulus (s) = I / Y = 16660.18422 / 441.33305 = 37.749 [mm^3]

                    S = 37.749 mm^3

 

Conclusion:

Hence when comparing case 1 & case 2,

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

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