Section Modulus calculation and optimization
Aim -
calculation of section modulus of hood design and also come up with improved section modulus
Use the Section from your Hood design and calculate the Section Modulus 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
Introduction
SECTION MODULUS
- Section modulus is the ratio of total moment resisted by the section to the stress in the extreme fibre.
- Section modulus give us idea about how much of load the given structure can take.
- If the section modulus increases the moment of resistance increases and hence its load bearing capacity.
- S=I/Y ,
Where S= section modulus
I=moment of inertia of a section
Y=distance from neutral axis to any given fibre.
| Cross-sectional shape | Figure | Equation | Comment |
|---|---|---|---|
| Rectangle |  |
Solid arrow represents neutral axis | |
| doubly symmetric I-section (major axis) | 
|
NA indicates neutral axis | |
| doubly symmetric I-section (minor axis) |  [4] |
NA indicates neutral axis | |
| Circle |  [3] |
Solid arrow represents neutral axis | |
| Circular hollow section |  |
Solid arrow represents neutral axis | |
| Rectangular hollow section |  |
NA indicates neutral axis | |
| Diamond |  |
NA indicates neutral axis | |
| C-channel | |
NA indicates neutral axis |
,
Calculation:
CASE 1: Before Optimization
The Hood which was already created is imported and the section inertia analysis is done.
We have to make sure that the curve obtained is a single/closed curve.
MOI (max) = 4.770798694e+06 (mm^4)
MOI (min) = 1.540393782e+04 (mm^4)
As per the section modulus equation,
S = I/Y
I = MOI (min) value is considered
Y = Total distance /2, Total distance =875.5680mm
Y = 878.9458 / 2
= 439.4729 mm
Therefore, S = 154039.3782 mm^4/439.4729 mm
S = 350.50939022633 mm^3
CASE 2: After optimization
The hood inner panel’s curve is offset by 0.5mm outwards and section inertia analysis is done same as case 2 for a closed curve.
MOI (max) = 4.819629018e+06 mm^4
MOI (min) = 1.690673290e+04 mm^4
As per section modulus equation,
S = I / Y
I = MOI (min) value is considered
Y = Total distance /2, Total distance =875.5680mm
Y = 878.9458 / 2
= 439.4729 mm
Therefore, S = 169067.3290 mm^4 / 439.4729 mm
S = 384.70478839537 mm^3
Conclusion:
Hence when comparing case 1 & case 2,
1. The section modulus value of case 2 is higher than the case 1
2. As we increase the section area, the moment of inertia & section modulus increases which means strength and load material also increase