Section Modulus calculation and optimization

SECTION MODULUS HOOD :

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

                  * The section modulus is geometry property of the cross section used for designing beams and flexural members.it does not reprsent anything physically.

                        S=I/Y

           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 objet(mm)

Area moment of inertia:

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

                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 beam maximum tension .

      Material data : stainless steel AISI 302 -cold rolled

  where yield strength of stainless steel is 520Mpa = 520 N/mm^2

Case : 1 Before optimization:

      

                                                             Fig 1.1

      

                                                             Fig 1.2

                      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 1.3

          

                                                               Fig 1.4

                 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