Section Modulus calculation and optimization
AIM : To
i)Use the Section from the created Hood design and to calculate the Section Modulus of that particular section using the formula
S = I/y
Where,
S = Section Modulus
I = Moment of Inertia
y = distance between the neutral axis and the extreme end of the object
ii) Come up with a new section that has an improved Section Modulus compared to the previous one and to mention what changes you made which resulted in the increased strength.
Brief explanation about Section Modulus :
Section Modulus : Section Modulus is defined as the ratio of the moment of inertia of a section about the neutral axis to the distance of Outermost layer from the neutral axis .
- It is denoted by 'Z' or 'S' .
Hence mathematically section modulus is given by Z=I/Y
Where I= moment of Inertia about neutral axis
Y=Distance of the outermost layer from the neutral axis
- Note:
1. The neutral axis of a section always passes through its centroid.
2. In case of symmetrical sections such as circular, square or rectangular, the neutral axis passes through its geometrical centre and the distance of extreme fibre from the neutral axis is y = d / 2,
where d is the diameter in case of circular section or depth in case of square or rectangular section.
3. In case of unsymmetrical sections such as L-section or T-section, the neutral axis does not pass through its geometrical centre.
In such cases, first of all the centroid of the section is calculated and then the distance of the extreme fibres for both lower and upper side of the
section is obtained. Out of these two values, the bigger value is used in bending equation
- From Bending equation , M/I=σ/Y=E/R
where M=Bending moment acting at the given section
Y=Distance of the outermost layer from the neutral axis
σ=Bending stress
I=Moment of Inertia of the crossection about the neutral axis
E=Young's Modulus of the material
R=Radius of the curvature
The stress σ will be maximum when Y is maximum. Hence the above bending equation can be written as M/I=σ">σmax/Ymax
and M=σ">σmax *I/Ymax
but σ">I/Ymax =Z
Therefore M=σ">σmax * Z
σ">In the above equation ,M = maximum bending moment (or the moment of resistance offered by the section) .
σ">Hence ,Moment of resistance offered by the section is maximum when section modulus is maximum . Hence,Section Modulus represents the
σ">strength of the section
- the higher is the value of Section Modulus (Z) for cross-section of a particular shape, the higher is the bending moment carrying capacity of the beam for a given
maximum bending stress.
CALCULATIONS :
Case 1: Before Optimization
Taking a section of Hood using Intersection curve command and performing ' Section Inertial Analysis ' .
From the above analysis
Imax= 9.04398*10^7 mm^4
Imin = 2.92*10^5 mm^4
and calculated total Length of the section = 881 mm
Now Y= 0.5* Total length =0.5*881 = 440.5mm
Considering only Imin to see the minimum value for which the design fails
Section Modulus= S1 = Imin/Y = 2.92*10^5 mm^4/ 440.5mm =662.88 mm^3
Therefore S1= 662.88 mm^3 .
Case 2 : After Optimization
By increasing the depth of inner panel by 0.5mm (using offset command ) , the crossectional area of the hood increases which further increases Section Modulus
From the above improved section analysis
Imax= 9.05411*10^7 mm^4
I min = 2.964*10^5 mm^5
and calculated total Length of the section = 881 mm
Now Y= 0.5* Total length =0.5*881 = 440.5mm
Considering only Imin to see the minimum value for which the design fails
Section Modulus= S2 = Imin/Y= 2.964*10^5 mm^5 /440.5mm = 672.871 mm^3 .
Therefore S2 = 672.871 mm^3 .
Conclusion :
After the calculations of the above 2 cases ,the section modulus S1 and S2 are differ by 9.991 mm (~ 10mm)
>The section modulus value of case 2 is higher than the case 1
Thus, Strength of material is directly proportional to the moment of interia, and thus increasing the sectional area also increases the strength of component.
Thus sectional modulus for hood design is calculated and optimized design is also done to increase the section modulus of the hood.