Section Modulus calculation and optimization
Aim:
To find the section modulus of the hood and optimization to improve the overall section modulus.
Objective:
Use the section from your Hood design and calculate the section modulus using the formula S= I/y.
Section Modulus:
Section modulus is a geometric property for a given cross-section used in the design of beams or flexural members, it is also referred to as the Polar Modulus or the Torsional Sectional Modulus. Other geometric properties used in design include area for tension and shear, radius of gyration for compression, and moment of inertia and polar moment of inertia for stiffness. Any relationship between these properties is highly dependent on the shape in question. Equations for the section moduli of common shapes are given below. There are two types of section moduli, the elastic section modulus and the plastic section modulus.
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.
Which is equal to yield stress.
Therefore
S=I/y, Where
S= Section modulus (mm^3)
I = Moment of inertia of the section to the stress in the extreme end of the object (mm^4)
Y= Distance between the neutral axis and extreme end of the object (mm).
Neutral Axis:
When beam is subjected to pure bending the top most fibres of the beam are subjected to maximum compression, while the bottom most fibre of the beams are subjected to maximum tension.
Changing the position can show the difference in bending which is because of the centroid distance from extreme fiber end.
in this position there is more material from axis to the specific direction, to resist the load which makes less bend and improves the strength.
Area Moment of Inertia:
The area moment of inertia of a beam cross section area measures the beam ability to resist bending. The larger the moment of inertia less will be the bend.
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 fibres of the beam are subjected to maximum compression, while the bottom most fibre of the beams are subjected to maximum tension.
Elastic section modulus
For general design, the elastic section modulus is used, applying up to the yield point for most metals and other common materials.
The elastic section modulus is defined as S = I / y, where I is the second moment of area (or area moment of inertia, not to be confused with moment of inertia) and y is the distance from the neutral axis to any given fibre. It is often reported using y = c, where c is the distance from the neutral axis to the most extreme fibre, as seen in the table below. It is also often used to determine the yield moment (My) such that My = S × σy, where σy is the yield strength of the material.
Plastic section modulus:
The plastic section modulus is used for materials where elastic yielding is acceptable and plastic behavior is assumed to be an acceptable limit. Designs generally strive to ultimately remain below the plastic limit to avoid permanent deformations, often comparing the plastic capacity against amplified forces or stresses.
The plastic section modulus depends on the location of the plastic neutral axis (PNA). The PNA is defined as the axis that splits the cross section such that the compression force from the area in compression equals the tension force from the area in tension. So, for sections with constant yielding stress, the area above and below the PNA will be equal, but for composite sections, this is not necessarily the case.
The plastic section modulus is the sum of the areas of the cross section on each side of the PNA (which may or may not be equal) multiplied by the distance from the local centroids of the two areas to the PNA:
the Plastic Section Modulus can also be called the 'First moment of area'
The plastic section modulus is used to calculate the plastic moment, Mp, or full capacity of a cross-section. The two terms are related by the yield strength of the material in question, Fy, by Mp=Fy*Z. Plastic section modulus and elastic section modulus are related by a shape factor which can be denoted by 'k', used for an indication of capacity beyond elastic limit of material. This could be shown mathematically with the formula :-
CASE 1
Design Study
The Intersection of the design was taken and then section Inertia Analysis was done.
I = MOI (min) Value is considered
Moment of inertia MOI (max) =9.065027704 x 10^7mm
Moment of inertia MOI (min) =2.921342694 x 10^5mm
Y = Total distance / 2, Total distance = 880.06 mm
Y=880.06/2=440.03 mm
S1= I/Y
S1=2.921342694 x 10^5/440.03
S1=663.90 mm^3
CASE 2
HOOD DESIGN OPTIMIZATION:
Increasing the depth of inner panel.
Moment of inertia(max) =9.048219230x10^7 mm^4
Moment of inertia(min) =2.958380529x10^5 mm^4
Y=880.06/2=440.03 mm
S2= I/Y
S2=2.958380529x10^5/440.06
S2=672.31 mm^3
Therefore, S2>S1
Conclusion:
Resulting by comparing case 1 & case 2,
- The section modulus value of case 2 is higher than the case 1
- As we increase the section area, the moment of inertia & section modulus increase which means strength and load material also increase. Which the moment of inertia directly proportional to the strength of the material.
So, here the difference can be seen in the moment of inertia and the change in the section modulus can be seen by the change in the design(depth).