Section Modulus calculation and optimization
Report On Section Modulus Calculation and optimization
Section Modulus: Section modulus is the property which shows how a body or a section behaves when a bending forces are applied in other word it is the property which provides resistance against bending.
Mathematically it is the ration Moment of Inertia of a Section to the distance between neutral axis and outer most point.
Significance of Section Modulus:
- It shows the strength of any section of beam.
- It provides resistance against bending.
- It helps to choose material if our crossectional dimention is fixed and we want higher resistance against bending.
The formula for calculation section modulus is:-
S= I / Y
Where,
S = Section Modulus
I = Moment of inertia (mm^4)
Y = Distance between neutral axes and outermost fiber (mm)
Steps to calculate elastic section modulus:-
- Step 1:- Find location of centroid of given cross section.
- Step 2:- Find moment of inertia about neutral axis
- Step 3:- Find "Y" from neutral axis.
- Step 4:- Find section modulus using the formula provided above.
There are many formulas for calculating Moment of Inertia (I) of some common shaped crossection which are mentioned below:-
Ex-
1. I (circle) = pi*D^4 / 64
2. I (Rectangle ) = b*d^3 / 12
For random shapes it is very difficult to calculate Moment of Inertia. But thankfully with help NX Software we can easily calculated Moment of Inertia of any random crosection.
Here we will be calculating the Section Modulus (S) of our Hood which we have designed in previous challenge.
We will take a crossection which passes vertically and exact middle of the hood assy.
As this crossection is not closed, we will project the outher most layer of the outer panel and inner most layer of the inner pannel and then with the help of sketch tool we will join this to make a close sectional area.
Now with the help of sectional analysis tool we can easily calculate Moment of Inertia (minimum) which is in this case found to be:-
[Note: We have taken the hollow section with thickness of 0.75mm]
I (MOI)-Min = 1.664 * 10^5 [mm^4]
And with the help of measurement tool , the distance between outer most fibers is := 488.914 mm
As the neutral axis is in the middle of the section , so the value Y will be :-
Y = 488.914/2 = 244.457 mm
Now the Section Modulus of the above section is :
S= I/Y
= 1.664 * 10^5 [mm^4] / 244.457 mm
= 680.69 [mm^3]
2nd Part (How to Increase Section Modulus)
As from section modulus formula we came to know that it is directly proportional to Moment of Inertia , It means in any way if we can increase the Moment of Inertia while keeping length from neutral axis constant then we might be able to increase the section Modulus.
So we will try to ofset the inner most boundry of inner panel by 2mm and then we will calculate the MOI as we done earlier.
Now we will join the ofseted boundry (downword direction) of Inner panel and the outer most boundry of Outer pannel.
After closing the boundry we will calculate the MOI with help of Section Inertia Analysis tool which is found to be.-
[Note: We have taken the hollow section with thickness of 0.75mm]
I (MOI)-Min = 1.864 * 10^5 [mm^4]
As we have taken the Y constant , so-
Y = 244.457 mm
Now the Section Modulus of the above section is :
S= I/Y
= 1.864 * 10^5 [mm^4] / 244.457 mm
= 762.506 [mm^3]
CONCLUSION: Now from the above two examples we came to know that if we want to increase the Section Modulus of any section of any object then we have to Increase its Moment of Inertia which is generally increasing the area of crossection or its boundry length while keeping its length from neutal axis to outermost fibre constant.
Also we can increase the section modulus by decreasing its length from outer most fibre to neutral axis while keeping MOI constant.
But in practical Changing the value of Y is not possible, so generally we try to modify MOI so that we can get the desired section modulus
Report Prepared By: Vikash Rai (Mechanical Engineer)