Section Modulus calculation and optimization

Aim

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

Objective

     By optimizing the hood,the section modulus is calculated using the formula i.e..,

         S=l/y

     where, S=Section modulus,

                l=Moment of inertia,

                y=Distance between the neutral axis.

Section modulus

      Section modulus is a geometric property for a given cross-section used in the design of beam and flexural members. By finding the section modulus we can find the resistance to remain unchanged when the force is applying to the object or substances.

Formula

     It may be defined as the ratio of total moment resisted by the section to the stress in the extreme fibre which is equal to yield stress. 

      (S=l/y).

Moment of inertia

     Moment of inertia is defined as the quantity expressed by the body resisting angular acceleration which is the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.

Moment of Inertia is also known as the angular mass or rotational inertia.

The SI unit of moment of inertia is kg m².

Moment of Inertia is expressed as I = m × r²

where,

              m = Sum of the product of the mass.

              r = Distance from the axis of the rotation.

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 fibre of the beams are subjected to maximum tension.

Section modulus

Case 1

         

Here the moment of inertia is calculated by using the NX-software 

   MOI (Max):9.558802x10^7mm^4

   MOI (Min):3.138893x10^5mm^4

Total distance is 889.2816mm

Calculation,

Neutal axis we need is 'y' i.e.., y/2=889.2816/2=444.6408mm.

MOI for max,

9.558802x10^7mm^4/444.6408mm=214.9784x10^3mm^3,

S=214.9784x10^3mm^3.

MOI for min,

3.138893x10^5mm^4/444.6408mm=705.939mm^3

S=705.939mm^3.

Case 2

Here the moment of inertia is calculated by using the NX-software,

     MOI (Max):9.610205x10^7mm^4.

     MOI (Min):3.357282x10^5mm^4.

MOI for max,

9.610205x10^7mm^4/444.6408mm=216.1341x10^3mm^3.

S=216.1341x10^3mm^3.

MOI for min,

3.357282x10^5mm^4/444.6408mm=755.054mm^3.

S=755.054mm^3.

Similarities difference between the two cases of study are,

For maximum is 1.1561mm^3.

For minimum is 49.115mm^3.

Conclusion,

               Thus i have offsetted the inner and outer panel by 1mm towards outside.The section module of case 2 is higher than of case 1.

Section modulus is successfully determined and further increased the sectional area.