Section Modulus calculation and optimization

1) Section modulus :

section modulus is the ratio of area moment of inertia about the neutral axis to the distance of outermost fiber from neutral axis.

it is done on the lower areas or smaller areas usually, where wanted to improve the strenth loaclly. it is comparative study with parent cars to improve the strength by increasing section modulus and it is inproved by many ways two of then mentioned below.

• larger the section modulus greater the strength of the part. so always larger section modulus is prefered.
• section modulus is improved by modifying the crossection or modifying the embosis or increasing number of embosis.

Section Modulus is calculated using the formula

S = I/y

Where,

S = Section Modulus

I = Area Moment of Inertia

y = distance between the neutral axis and the extreme end of the object

`I = (bd^3)/12`

case 1: Original section modulus

For the below section i have done section inertia analysis for frontal impact. this section is perpendicular to the frontal area.

For the above section the I = Area Moment of Inertia is found using the section inertia analysis in nx.

it is found that maximum moment of ineria = 9.022352760*10^7 mm^4

and minimum moment of inertia = 2.833366177*10^5 mm^4

out of these values always consider minimum moment of inertia for for factor of safety.

so,

I = Area Moment of Inertia = 2.833366177*10^5 mm^4

Now the y (distance between the neutral axis and the extreme end of the object) measured by

so,

y = distance between the neutral axis and the extreme end of the object = 879.7548/2

y = 439.8774 mm

we have, Section Modulus is calculated using the formula

S = I/y

S1 = 2.833366177*10^5 mm^4 / 439.8774 mm

S1= 644.126 mm^3

Section Modulus for case 1 is calculated as s1 = 644.126 mm^3

case 2 : modified section by offesting the inner panel by 0.5mm dounwords. ( modifying crossection )

the section modulus is improved by modifying crossection. I have offseted the inner panel curve by 0.5mm dounword so that my section modulus can improve. as shown below

so, for case 2

y2 = distance between the neutral axis and the extreme end of the object = 879.8760/2

y2 = 439.938 mm

For the above section the I = Area Moment of Inertia is found using the section inertia analysis in nx.

it is found that maximum moment of ineria = 9.019675057*10^7 mm^4

and minimum moment of inertia = 2.872239797*10^5 mm^4

out of these values always consider minimum moment of inertia for for factor of safety.

we have, Section Modulus is calculated using the formula

S = I/y

S2 = 2.872239797*10^5 mm^4/ 439.9380 mm

S2= 652.8737 mm^3

Section Modulus for case 1 is calculated as s2 = 652.8737 mm^3

so from the above two case study we can say that s2 > s1

i.e, s2 = 652.8737 is larger than the s1 = 644.126

so the hood section modulus is optimised. The section modulus is better for case 2 section than case 1 . so the case 2 section has more strength than compared to case 1 section.