Section Modulus calculation and optimization
What is Section Modulus?
- Ratio of total moment resisted by the section to the stress in the extreme fiber.
- It gives us an idea about how much load the given structure can take.
- If the section modulus increases the moment of resistance increases
- and hence its load-bearing capacity.
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Product : Face...Fill.1...hood
Date : Saturday, 10 April 2021 03:29:34
Author : Admin
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Component | Sub-Component | Area[m2] | Gx[mm] | Gy[mm] | Gz[mm] | M1[m4] | M2[m4] | IoxG[m4] | IoyG[m4] | IxyG[m4] |
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Face...Fill.1...hood | | 0.010 | 5.000 | 743.067 | -908.551 | 1.14114e-06 | 0.000162454 | 1.141e-006 | 1.625e-004 | 6.776e-021 |
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The formula for section modulus
S=l/y
S - Section Modulus
l - Moment of inertia value from the inertia analysis
y - Distance from the neutral axis to outermost fiber
I=1.141*10^6m^4
Y=488.961/2=244.4805mm
S=1.141*10^6mm^4/244.4805mm
=4667.03mm^3
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Product : Face...Fill.2...hood
Date : Saturday, 10 April 2021 03:35:58
Author : Admin
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Component | Sub-Component | Area[m2] | Gx[mm] | Gy[mm] | Gz[mm] | M1[m4] | M2[m4] | IoxG[m4] | IoyG[m4] | IxyG[m4] |
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Face...Fill.2...hood | | 0.011 | 5.000 | 742.941 | -909.128 | 1.19306e-06 | 0.000167592 | 1.193e-006 | 1.676e-004 | 1.403e-018 |
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The formula for section modulus
S=l/y
S - Section Modulus
l - Moment of inertia value from the inertia analysis
y - Distance from the neutral axis to outermost fiber
I = 1.19306*10^6mm^4
y = 244.4805mm
S = 1.19306*10^6mm^4/244.4805mm
=7896.744mm^3
Conclusion:
So decreasing the distance between the 2 panels leads to increment in section modulus, so resistance to bending also increases, the strength of the hood also increase