Section Modulus calculation and optimization
Section Modulus:
Section Modulus is done to improve the strength of weaker section in the part. A beam that has a greater section modulus will have the more strength carrying capacity then that which has less section modulus.Section modulus directly reresents the strength of the given part or object.
`sigma`b/y =MOI/I =E/R
Section Modulus : S= I/y
where ,
`sigma`b - `bending stress
y= Max distance from CG to outermost surface mm
M- Bending moment N-mm , a moment from a vertical force on a beam.
I- moment of innertia mm^3 -, measure of the resistance of a body to angular acceleration about a given axis that is equal to the sum of the products of each element of mass in the body and the square of the element's distance from the axis
E-Modulus of rigidity GPa - ,It is the ratio of shear stress to shear strength.
R-radius of gyration mm -,Radius of gyration is the distance from the center of mass of a body at which the whole mass could be concentrated without changing its moment of rotational inertia about an axis through the center of mass.
Figure 1 shows the curve before optimization
Fig 1-1
fig1-2:
Figure 2 shows the images after optimizing the weaker section curve to 0.5mm
Fig2-1
Fig 2-2
Fig-3- to get the max distance frog CG.
Before Optimization
Given: MOI(Max)-8.088*10^5 mm^4
MOI(Min)-4.087*10^3 mm^4
Section Modulus : S= I/y
I- MOI
y- Distance from the axis.
Total Distance from CG of the axis:-877.03 ~877mm
Therefore y= 877/2 =438.5mm ,i.e the maximum distance from the centre point to the outermost surface.
S(max)= Imax/y = 8.088*10^5/438.5 = 1844.46mm^3
After optimizing
Given: MOI(Max)-8.11*10^5 mm^4
MOI(Min)-3.94*10^3 mm^4
S(max)= Imax/y = 8.11*10^5/438.5 = 1849.48mm^3
Therfore from the above results it is clear after optimizing we have the higher value of section modulus which results in less bending stresses developed then that before optimizing the curve. By increasing the area, MOI increases and therefore , the load capacity also increases.