Section Modulus calculation and optimization

Section Modulus calculation and optimization

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

Section modules (S): Section modulus is the first moment of area about the axis of bending for a beam cross-section. The strength of the beam sections depends mainly on the section modulus. It also defined as the ratio of moment of inertia to the distance between the neutral axis and the extreme end of the object.

Mathemetically it can defined as,

S=I/y

Euler Bernoulli’s Equation or (Bending stress formula) or Bending Equation:

`sigma`/y = M/I = E/R

Where `sigma` = Bending Stress
M = Bending Moment
I = Moment of Inertia
E = Modulus of elasticity
R = Radius of curvature
y = Distance of the fibre from NA (Neutral axis)

From this Euler Bernoulli's Equation we get Section Modulas concept and it's expression.

From the equation of Section Modulus, it's clear that the area moment of inertia of a beam cross sectional area measures the beam ability to resist bending.

So, we can conclude that from this concept, larger area moment of inertia gives a better section modulus.

Section modulus calculation of the hood: 

case 1: Existing model

Fig._1

Fig._2

Fig._3

From the above figure_2 we can see that moment of inertia (MOI) [max] = 8.885656925e+07[mm^4] and MOI [min] = 2.959721692e+05[mm^4]

From figure_3 we get the total distance = 875.4868 mm.

Now, y = 875.4868/2 = 437.7434 mm.

Then, Section Modulus (S) = I/y = 295972.1692/437.7434 mm^3 = 676.13165 mm^3

(considered mimimum MOI)

case 2: After modification ( offseting the inner panel 0.5mm [towards engine] )

Fig._4

Fig._5

From the above figure_5 we can see that moment of inertia (MOI) [max] = 8.869119130e+07[mm^4] and MOI [min] = 3.003551394e+05[mm^4]

Now, y = 875.4868/2 = 437.7434 mm.

Then, Section Modulus (S) = I/y = 300355.1394/437.7434 mm^3 = 686.14430 mm^3

(considered mimimum MOI)

case 3: After modification ( adding emboss )

Fig._6

From the above figure_6 we can see that moment of inertia (MOI) [max] = 8.870525479e+07[mm^4] and MOI [min] = 3.003814764e+05[mm^4]

Now, y = 875.4868/2 = 437.7434 mm.

Then, Section Modulus (S) = I/y = 300381.4764/437.7434 mm^3 = 686.20446 mm^3

(considered mimimum MOI)

Conclusion:

From the above case we can see the section modulus is increasing. In case 1, it is 676.13165 mm^3, in case 2, it is 686.14430 mm^3 and in 3rd Case it is 686.20446 mm^3.

It is clear that the section modulus increases as we add materials. The idea is that the section modulus is directly proportional to the area moment of ​​inertia. And increasing the sectional modulus increases the strength. We can conclude from this study that section modulus is important for vehicle design to increase vehicle strength.