Section Modulus calculation and optimization
SECTION MODULUS
Section moduluds is a geometric property for a given cross section used in the design of beams or flexural members. It does not represent anything physically. To define section modulus, it may be defined as the ratio of total moment resisted by the section to the stress in the stress in the extreme fiber which is equal to yield stress.
There are two types of section moduli
1. Elastic section modulus
For general design, the elastic sexction modulus is used, applying up to the yield point fot most metals and another common materials.The elastic section modulus is desfined as S=I/y,where I is the second moment of inertia(area moment of inertia, not to be confused with the moment of inertia)and y is the distance from the neutral axis to any given fiber. It is often reported using y=c, where c is the distance from the neautral axis to the most extreme fiber. It is also aften used to determine the yield moment(My) such that My=S x σy,where σy is the yield strength of the material.
2. Plastic section modulus
the plastic section modulus is used for material where elastic yielding is accepltable and plastic behavior is used for materials where elastic yielding is acceptable and plastic behavior is assumed to be an accepltable limit. Design generally strive to ultimately remain below the plastic limit to avold permanent deformations, often comparing the plastic capacity aganist amplified forces or stresses. The plastic section modulus depends on tje location of the plastic neutral axis(PNA). The PNA is defined as the axis that splits the cross-sectio such that the comparission forces from the area in compression equals the tension force from the area in tension . so, for section with constant yielding stress,the area above and below the PNA will be equal but for composite sections,this is not necessarily the case.
But here i have to calculate the elastic section modulus.The elastic section modulus is given by:
S=I/y
S=section modulus
I=moment of inertia
y= distance between the neutral axis and the extreme end of the object
it is basically from the bending equaltion. the bending moment equation is given by
M/I=σ/y=E/R
where M= bending moment
σ = fiber stress at distance 'y' from the centroidal/neutral axis.
R= radius of curvature of the bent beam.
E= young's modulus of the material of the beam
MI=σY
Mσ=IY
Mσ=S
Here in the equation, we can see that the section modulus (S) is iversely proportional to the bending stress(σ ) which means thatr if the section modulus of any sectio is increased then the comonent will have mpre resistance to bending stresses so basically the strength of the component will be increased.
section modulus equation for different shapes
HOOD SECTION MODULUS CALCULATION
PROCEDURE
1. At first we sholud convert CATProduct file in to CATPart
before generating CATPart file hide Latche and Hinges.
2.now select the Hood_assembly on the tree and enter OK.
now we created a CATPart file and new tab is opened.
3. Now select the intersection commond tool,select the first element as plane(yz),for second element drag and select the hood and click OK.
4.now the intersection curve is created and hide inner panel and outer panel.by using Multi extract command extract outer most curve and take tangent contunity on dilauge box as shown in the figure.
5.after hide the intersection curves on the tree as shown in above figure.
now join the lines
and by using join command join all the curves.
4.create a point on the curve to trim join the both curves(inner and outer hood curve ). as shown in below figures.
5.create the line by joining the point on both sides.then join all the curves by using join command.
use fill tool to create 2D object for further calculations
case1: now 2D object is created and next step is measuring moment of inertia.
measure inertia ---> measure inertia 2D ---> select the object ---> OK
now we get the values of moment of inertia.
so we got two values M1 and M2
M1 =1.077e-006m^4 and
M2 = 1.541e-004m^4.
here catia shows the values in meter (m) so we need to conver it in to milimeters(mm).
after converting the values to mm we get as, we need to consider the lowest value that is
M1=1.077x10^6mm^4
8.next step is to calculate y which is distance between the neutral axis and the exxtreme end of the object.
for calculating y i will measure the distance between the extreme ends of the outer most panel and then divide that value by 2 to get the value of y.
to measure that distance we have to draw the lines at to ends of the panels as shown in figure and measure the distance.
so the distance is 478.203mm. by dividing it by 2 we get 239.1015mm
there for the value of y=239.1015mm.
section modulus=1.077x10^6mm^4/239.1015mm.
section modulus = 4.505x10^3mm^3
Optimization
case 2: for optimization we need to offset the inner panel curve. for that hide the surface and lines and unhide the multiple extract and select the curve to offset, enter the offset distance as 0.5mm and enter ok and join the curves by line and follow the same procedure and calculate the section modulus..
moment of inertia for offset 0.5mm and y remains same
we need to consider the lowest value that is M1
M1= 1.126e-006m^4---> 1.126x10^6mm^4
section modulus=1.126x10^6mm^4/239.1015mm.
Section modulus = 4.171x10^3mm^3
case 3: moment of inertia for offset 1mm and y remains same
M1= 1.176e-006m^4--->1.176x10^6mm^4
section modulus=1.176x10^6mm^4/239.1015mm.
Section modulus = 4.919x10^3mm^3
case 4: now i will change the sketch and check for the moment of inertia
we need to consider the lowest value that is M1
M1=1.19e-006m^4--->1.19x10^6mm^4
section modulus=1.19x10^6mm^4/239.1015mm.
section modulus=4.978x10^4mm^3
Result :-
The values of the section modulus are calculated and tabulated in below table:
| case | off set distance(mm) | moment of inertia (mm^4) | distance from the neutral axis and the extreme end of the object(y) mm | Section modulus(S)mm^3 |
| 1 | - | 1.077x10^6 | 239.1015 | 4.505x10^3 |
| 2 | 0.5 | 1.126x10^6 | 239.1015 | 4.71x10^3 |
| 3 | 1 | 1.176x10^6 | 239.1015 | 4.919x10^3 |
| 4 | different c/s area | 1.19x10^6 | 239.1015 | 4.978x10^4 |
as we can see in the above table that initial value of section modulus i got was 4.505x10^3 mm^3but after optimization it iwe got values as 4.978x10^3 mm^3 which shows that the strength of the coponent in that section increased
Conclusion
The above calculation shows that case 4 optimization has greater section modulus than cas 1,2,3.
All these calculations are essential while designing to know the strength and load carrying capacity of the given component.
It is quite evident that increasing the more number of embooses has significant impact on increasing the section modulus which in turn its strength also But it depends on the designer to choose the best option from the calculaion that gives satisfactory result