Section Modulus calculation and optimization
SECTION MODULUS OF AN HOOD
- Section modulus is a direct measure of strength if any sectional area of an object about an reference axis(neutral axis).
- Strength is nothing but, capable of supporting greater loads. so higher the strength it has a high capable of supporting greater loads.
- This strength can measured directly from section modulus formula, and defined by below formula.
- Section modulus represented with a symbolof 's'.
S=l/y
where
'S' Section Modulus
'I' Moment of inertia
'y' is the Distance from the reference line(neutral axis) to one of the extreme end of an Object.
- Moment of inertia is nothing but offering resistance against of moment. bending or twisting, i.e if we applied any external load on the boby, how much resistance its offering to tend to deform.
- So, higher the value of moment of inertia. have higher resistance to deform under certain load, hence we can say that moment of inertia directly proportional to the strength of the material , and strength of the material can be calculated with the help of section modulus.
- If we vary the neutral axis towards positive or negative side the overall strength of body will decause when nuetral axis line lies on the middle of the section only have higher strength at particular point ,so, here 'y'value was inversely proportional to section modulus.
Case 1:
Maximum Moment of Inertia =4.759968X10^6 mm^4
Minimum Moment of Inertia =1.345675X10^4 mm^4
The Total Distance of the Hood Y=876.4365mm
Distance between the Neutral Axis to the Extreme End=876.4365/2=438.218 mm
To find Section Modulus of HOOD section, we have to calculate
=4.759968X10^6 mm^4/438.218 mm
=10.862X10^3 mm^3(Maximum Strength)
=1.345675X10^4 mm^4/438.218
=0.030X10^3 mm^3(Minimum Strength)
Case 2:
Maximum Moment of Inertia =4.74523X10^6 mm^4
Minimum Moment of Inertia =1.3998257X10^4 mm^4
The Total Distance of the Hood Y=876.4365mm
Distance between the Neutral Axis to the Extreme End=876.4365/2=438.218 mm
To find Section Modulus of HOOD section, we have to calculate
=4.74523 X10^6 mm^4/438.218 mm
= 10.828X10^3 mm^3(Maximum Strength)
=1.3998257 X10^4 mm^4/438.218
= 0.031X10^3 mm^3(Minimum Strength)
Conclusion
Hence when comparing case 1 and case 2
1. The section modulus value of case 2 is higher than the case 1
2. The strength of material is directly proportional to the moment of inertia, Hence increase in sectional area there is have increase in overall mass of an object, so that increase in strength also.