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 Modulus (S):-
The section modulus (S) is geometric property of the cross section used for designing beams and flexural members. It does not represent anything physically.
Formula of Section Modulus: It may be defined as the ration of total moment resisted by the section to the stress in the extreme fibre which is equal to yield stress.
There fore
S = I/Y
where
S=section modulus(mm^3),
I= Moment of Inertia of the section about the bending axis. (mm^4),
Y= Distance between the neutral axis and extreme end of the object (mm).
Importance:-
• It is significant for beam and flexural member design.
• More the section modulus, the more the member is resistant to the bending of the beam.
• It is easier to calculate stresses in the beam and an exact measure of the strength of steel.
• If two beams are made of the same material and comparing the “section modulus” of two beams, the beam with bigger section modulus will be tougher and more capable to withstand larger loads.
Area momentum of inertia:-
The moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis that passes through the centre of mass and the product of its mass and the square of the distance between the two lines.The area moment of inertia of a beams cross sectional 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 = Youngs 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 fibres of the beams are subjected to maximum tension.
Material Data: Stainless steel AISI 302 – cold rolled
Where yield strength of stainless steel is 520MPa = 520 N/mm^2
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 materia.
This could be shown mathematically with the formula:-
shape factor for a rectangular section is 1.5.
So here we are Calculating Elastic section Modulus
Case 1:-
to find a section modulus of the hood, and optimizing the section modulus value using NX_cad.
section modulus of the hood 1:-
Calcultion:-
Moment of inertia(max) = 4.811416923e+06[mm^4]
Moment of inertia(min) = 1.418790217e+04[mm^4]
=>1.418790217x10^4[mm^4]
=>14187.90217mm^4
Distance between the neutral axis and extreme end of the object(Y)=H%2=>440.24835mm
S=I/Y
=14187.90217mm^4/440.24835mm
s=32.227mm^3
Hood Design Optimization:-
By increasing the depth of the inner panel by 0.5 mm
calculation:-
Moment of inertia(max) = 4.804316768e+06[mm^4]
Moment of inertia(min) = 1.438075719e+04[mm^4]
=>1.438075719x10^4[mm^4]
=>14380.75719mm^4
Distance between the neutral axis and extreme end of the object(Y)=H%2=>440.24835mm
S=I/Y
=14380.75719mm^4/440.24835mm
s=32.665mm^3
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