Section Modulus calculation and optimization
Section Modulus: The section modulus (S) is geometric property of the cross section used for designing beams and flexural members. It may be defined as the ratio of total moment resisted by the section to the stress in the extreme fibre which is equal to yield stress.
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)
Neutral axis: When beam is subjected to pure bending, then 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. The Mean axis of the fibre neither has tension nor compression. This is neutral axis.
Before Optimization:
For practice purpose, the Hood was taken for analysis. Here the cross section of the hood (from top face of the outer hood panel to the bottom face of the inner panel) is considered. These are ensured to be a closed section such as the gaps are connected with the lines. While analysis, the following result occurs.
Calculation part:
MOI (max) = 4.757960857e+06 (mm^4)
MOI (min) = 1.466515894e+04 (mm^4)
As per the Section modulus equation,
S = I/Y
Here for I = MOI (min) value is considered
Y = Total distance /2, Total distance =876.8271mm
Y = 876.8271 / 2
= 438.41355 mm
Therefore, S = 14665.15894 mm^4/438.41355 mm
S = 33.4505 mm^3
After Optimization:
Now the same Hood was taken for analysis. Here the cross section of the hood (from top face of the outer hood panel to the offsetted bottom face of the inner panel) is considered. These are ensured to be a closed section such as the gaps are connected with the lines. The offset of inner panel bottom face is done for 0.5mm. Now the inner volume is increased. While analysis, the following result occurs.
Calculation part:
MOI (max) = 4.749156680e+06 (mm^4)
MOI (min) = 1.498554330e+04 (mm^4)
As per the Section modulus equation,
S = I/Y
Here for I = MOI (min) value is considered
Y = Total distance /2, Total distance =876.8271mm
Y = 876.8271 / 2
= 438.41355 mm
Therefore, S = 14985.54330 mm^4/438.41355 mm
S = 34.1813 mm^3
Conclusion:
Hence when comparing case 1 & case 2, The section modulus value of after optimization is higher than the before optimization. This implies if the section areaincreases , then the moment of inertia & section modulus also increases which means the material strength also increases.