Section Modulus calculation and optimization

Objective:-

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

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 ratio of total moment resisted by the section to the stress in the extreme fibre which is equal to yield stress.

Therefore,

S = I/Y

where S = section modulus 

I - moment of inertia of the section about the bending axis.(mm^4)

Y - Distance between the neutral axis and the extreme end of the object (mm)

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

Where,

M - Bending moment 

I - area moment of inertia 

E - Young modulus 

R - Radius of gyration 

Neutral axis:- when beam is subjected to pure bending the top most fibre of the beam are subjected to maximum compression while the bottom most fibre of the beam are subjected to maximum tension 

Material data :- stainless steel AISI 302 - cold rolled 

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

Calculation 

CASE 1:- Before optimisation 

The hood whichwas already created is imported and the section inertia analysis is done 

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

 

MOI (max) = 5.646612896e+08[mm^4]

MOI (min) = 1.604103590e+06[mm^4]

As per the section modulus equation 

S = I/Y 

I = moment of inertia min is considered 

Y = total distance from the neutral axis 

i.e Y = total distance /2 

Y = 875.5680/2

Y = 437.784 mm 

Therefore 

S= 1604103.5/437.784

S = 3664.143 mm^3

CASE 2 : After optimisation 

The hood inner pannel curves is offset by 0.5 mm outwards and section inertia is done as same as case 1 for a closed surface 

 

MOI (max) = 5.934075155e+08[mm^4]

MOI (min) = 1.690964905e+06[mm^4]

As per the section modulus equation 

S = I/Y 

I = moment of inertia min is considered 

Y = total distance from the neutral axis 

i.e Y = total distance /2 

Y = 875.5680/2

Y = 437.784 mm 

Therefore 

S= 1690964.905/437.784

S = 3862.94 mm^3

CONCLUSION:-

Hence when comparing case 1 and case 2 

1) The section modulus value of cas2 is higher then the case 1 

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