Section Modulus calculation and optimization
AIM -
To calculate the section modulus of Section from Hood design and improve the Section Modulus to increase the strenght of the hood.
Introduction -
Section Modulus is a geometric property used to calculate the bending stresses in a structural member.
Mathematically,It is defined as the ratio of the moment of inertia of a section about its centroidal axis and to the distance of the extreme layer from the neutral axis.
the formula is
S = I/y
Where,
S = Section Modulus
I = Moment of Inertia
y = distance between the neutral axis and the extreme end of the object
Types of Section Modulus
There are two types of section modulus:
- Elastic Section Modulus
- Plastic Section Modulus
Elastic Section Modulus
Elastic section modulus is calculated when we are to use the material behavior till the yield point.
For general design purposes, we use elastic section modulus.
Generally, elastic section modulus is denoted by S.
So,
S = I/y
Where,
S = Section Modulus
I = Moment of Inertia
y = distance between the neutral axis and the extreme end of the object
Elastic Section Modulus of Different Section
There are following different elastic section modulus:
Rectangular Section
As we know the moment of inertia of the rectangular section,
I = bd³/12
And the distance from the neutral axis,
ymax = d/2
So,
S = I/ymax
S = (bd³/12)/(d/2)
After calculating them,
Section modulus of rectangular section,
S = bd²/6
Hollow Rectangular Section
As we know the moment of inertia of the hollow rectangular section,
I = (BD³/12) - (bd³/12)
ymax = D/2
After putting these values,
S = I/ymax
(BD³/12) - (bd³/12)/(D/2)
S = (BD³ - bd³)/6D
Solid Circular
For solid Circular moment of inertia,
I = (π/64) × d⁴
ymax = d/2
Then,
S = {(π/64) × d⁴}/(d/2)
After calculating then find values of section modulus of solid circular
S = (π/32) × d³
Hollow Circular
The moment of inertia of hollow Circular,
I = (π/64) × (D⁴ - d⁴)
And,
ymax = D/2
After putting these values,
S = (π/64) × (D⁴ - d⁴)/(D/2)
S = π/32 × {(D⁴ - d⁴)/D}
Symmetric I & C Section
Moment of inertia for the I & C section,
I = (BD³/12) - {(B - b)d³}/12
ymax = D/2
So,
S = [(BD³/12) - {(B - b)d³}/12]/(D/2)
S = {BD³ - (B - b)d³}/6D
Diamond
Elastic section modulus for diamond is,
S = bd²/24
Plastic Section Modulus
The plastic section modulus is used for materials where the elastic yield is acceptable and the plastic behavior is considered to be an acceptable range.
It depends on the location of the plastic neutral axis. The PNA is the axis that bisects the cross-section such that the compressive force from the region in compression is equal to the tension force from the region under tension. So, for sections with a constant yield stress, the area above and below the PNA will be equal, but for mixed sections, this is not necessarily the case.
Mathematically,
The plastic section modulus is the sum of the areas of cross-sections on each side of the plastic neutral axis that may or may not be equal and multiplied by the distance from the local centroid of the two regions to the plastic neutral axis.
S = (Ac × yc) + (Aₜ + yₜ)
Where,
Ac = Area of cross-section under compression.
Aₜ = Area of cross-section under tension.
yc = Distance from the plastic neutral axis to the center of gravity of the areas under compression.
yₜ = Distance from the plastic neutral axis to the center of gravity of areas under tension.
Since on the neutral axis, the area under compression is equal to the area under tension.
So,
Ac = Aₜ
And,
yc = yₜ
Plastic Section Modulus of Different Section
There are following section modulus for different section:
Rectangular Section
As we know,
on the neutral axis, the area under compression is equal to the area under tension.
So,
Ac = Aₜ
And,
yc = yₜ
Ac = Aₜ = bd/2
yc = yₜ = d/4
So,
Sp= (Ac × yc) + (Aₜ + yₜ)
Sp= (bd/2 × d/4) + (bd/2 × d/4))
After calculating
Sp= bd²/4
Similarly, we can calculate the section modulus of the rectangular hollow section,
Sp= (bd²/4) - (b- 2t) × {(b/2) - t}²
Circular Section
As we know, on the neutral axis, the area under compression is equal to the area under tension.
So,
Ac = Aₜ = πd²/8
yc = yₜ = 2d/3π
Sp= (πd²/8 × 2d/3π) + (πd²/8 × 2d/3π)
After calculating,
Sp= d³/6
Similarly, we can calculate the section modulus of the circular hollow section,
Circular hollow section,
Sp= (D³ - d³)/6
So here you have to know all aspects related to the section modulus.
PROCEDURE -
For calculating the section modulus first I have to take an intersect of the outer panel and the inner panel of the hood with a plane.
NOW , after getting intersection curve use the sketch command and project curve command fill the obtained using intersection curve .
remmenber to use the outermost part of the intersection curve.
use section inertia analysis command from search bar
and calculate the inertia of the section.
from calculation we get the required data.
here,
Imax =4.8361*10000
y= 880.1948/2
= 440.019
so,
S=I/Y
=48361/440.019
S=109.9 MM3
now , we just measure the distance between inner and outer panel by 0.5mm distance we get the new value of I.
from calculation we get the required data.
here,
Imax =4.83509*10000
y= 880.1948/2
= 440.019
so,
S=I/Y
=48350/440.019
S=109.8816MM3
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