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OBJECTIVE: To Calculate the section modulus for the previously designed hood and optimized hood. To Come up with the new section that has improved section modulus of the previous section. MATERIAL OF THE HOOD: Stainless steel AISI 302 – cold-rolled – y = 520 MPa SECTION MODULUS: Section modulus is the geometric property…
Bharath P
updated on 07 Dec 2022
OBJECTIVE:
MATERIAL OF THE HOOD:
SECTION MODULUS:
There are two types of section modulus:
For general design, the elastic section modulus is used, applicable up to the yield point for most metals and other common materials.
The elastic section modulus is defined as S = I / y, where I is the second moment of area (or area moment of inertia, not to be confused with moment of inertia) and y is the distance from the neutral axis to any given fibre. It is often reported using y = c, where c is the distance from the neutral axis to the most extreme fibre, as seen in the table below. It is also often used to determine the yield moment (My) such that My = S ⋅ σy, where σy is the yield strength of the material.
Cross-sectional shape | Figure | Equation | Comment |
---|---|---|---|
Rectangle | ![]() |
S = b h 2 6 |
Solid arrow represents neutral axis |
doubly symmetric I-section (major axis) | ![]() |
S x = B H 2 6 − b h 3 6 H S x = I x y with y = H 2 |
NA indicates neutral axis |
doubly symmetric I-section (minor axis) | ![]() |
S y = B 2 ( H − h ) 6 + ( B − b ) 3 h 6 B |
NA indicates neutral axis |
Circle | ![]() |
S = π d 3 32 |
Solid arrow represents neutral axis |
Circular hollow section | ![]() |
S = π ( r 2 4 − r 1 4 ) 4 r 2 = π ( d 2 4 − d 1 4 ) 32 d 2 |
Solid arrow represents neutral axis |
Rectangular hollow section | ![]() |
S = B H 2 6 − b h 3 6 H |
NA indicates neutral axis |
Diamond | ![]() |
S = B H 2 24 |
NA indicates neutral axis |
C-channel | ![]() |
S = B H 2 6 − b h 3 6 H |
NA indicates neutral axis |
The plastic section modulus is used for materials where elastic yielding is acceptable and plastic behavior is assumed to be an acceptable limit. Designs generally strive to ultimately remain below the plastic limit to avoid permanent deformations, often comparing the plastic capacity against amplified forces or stresses.
The plastic section modulus depends on the location of the plastic neutral axis (PNA).The PNA is defined as the axis that splits the cross section such that the compression force from the area in compression equals the tension force from the area in tension. So, for sections with constant, and equal compressive and tensile yielding stress, the area above and below the PNA will be equal, but for composite sections, this is not necessarily the case.
The plastic section modulus is the sum of the areas of the cross section on each side of the PNA (which may or may not be equal) multiplied by the distance from the local centroids of the two areas to the PNA:
Z P = A C y C + A T y T
The Plastic Section Modulus is not the 'First moment of area'. Both relate to the calculation of the centroid, but Plastic Section Modulus is the Sum of all areas on both sides of PNA (Plastic Neutral Axis) and multiplied with the distances from the centroid of the corresponding areas to the centroid of the cross section, while the First moment of area is calculated based on either side of the "considering point" of the cross section and it is different along the cross section and depends on the point of consideration.
Description | Figure | Equation | Comment |
---|---|---|---|
Rectangular section | ![]() |
Z P = b h 2 4 |
A C = A T = b h 2 |
Rectangular hollow section | Z P = b h 2 4 − ( b − 2 t ) ( h 2 − t ) 2 |
where: b = width, h = height, t = wall thickness | |
For the two flanges of an I-beam with the web excluded[7] | Z P = b 1 t 1 y 1 + b 2 t 2 y 2 |
where:
b 1 , b 2 |
|
For an I Beam including the web | Z P = b t f ( d − t f ) + 0.25 t w ( d − 2 t f ) 2 |
[8] | |
For an I Beam (weak axis) | Z P = ( b 2 t f ) / 2 + 0.25 t w 2 ( d − 2 t f ) |
d = full height of the I beam | |
Solid Circle | Z P = d 3 6 |
||
Circular hollow section | Z P = d 2 3 − d 1 3 6 |
The plastic section modulus is used to calculate the plastic moment, Mp, or full capacity of a cross-section. The two terms are related by the yield strength of the material in question, Fy, by Mp = Fy ⋅ Z. Plastic section modulus and elastic section modulus are related by a shape factor which can be denoted by k, used for an indication of capacity beyond elastic limit of material. This could be shown mathematically with the formula :-
k = Z S
Shape factor for a rectangular section is 1.5.
Significance of Section Modulus:
We know that stress (σ) in a beam or structure is given by
σ=M⋅yIσ=M⋅yI...................(1)
Where I = Moment of Inertia
Y = distance between the neutral axis and the extreme end of the object
M = the moment about the neutral axis
Section Modulus is given by
S=IyS=Iy.......................(2)
From equations (1) and (2),
σ=MSσ=MS........................(3)
Thus from equation (3), it can be said that higher the section modulus, higher will be the resistance to bending.
# DESIGN 1: Section Modulus of Initial Hood Design
From the above analysis, following minimum and maximum moment of inertia are obtained
Imax = 95091198.37 mm4
Imin = 301754.6017 mm4
y = 895.2/2 mm
=447.6 mm
Smax = Imax / y = 95091198.37 / 447.6 = 212,446.8238 mm3
Smin = Imin / y = 301754.6017 / 447.6 = 674.1613 mm3
# DESIGN 2: Section Modulus of Optimized Hood design
What changes I made and Why?
Section Modulus can also be represented as
S = Σ miAi2 / y where A is the area of the section and M is the mass of the body
From the above equation, it can be observed that the section modulus of the body is directly proportional to the sectional area of the body.
Here the section area is increased by reducing the embossing depth and offsetting the outer panel by 3 mm. This increases the value of section modulus.
From the above analysis, following minimum and maximum moment of inertia are obtained
Imax = 94875533.26 mm4
Imin = 309456.7634 mm4
y = 895.2/2 mm
=447.6 mm
Smax = Imax / y = 94875533.26 / 447.6 =211,964.9983 mm3
Smin = Imin / y = 309456.7634 / 447.6 = 691.3689 mm3
CONCLUSION:
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