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  1. Home/
  2. Bharath P/
  3. Section Modulus calculation and optimization

Section Modulus calculation and optimization

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…

  • DESIGN
  • Bharath P

    updated on 07 Dec 2022

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 of a given cross-section used in the design of flexural members. Section modulus depends only on the cross-section shape of the beam or structure.
  • For symmetric sections, the value of section modulus is the same for both above and below of centroid of the cross-section. An asymmetric section has two values for the top and bottom of the centroid, the maximum section modulus (Smax) and minimum section modulus (Smin).

 

There are two types of section modulus:

  • Elastic section modulus(S).
  • Plastic section modulus(Z).

 

Elastic 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.

Section modulus equations[3]
Cross-sectional shape Figure Equation Comment
Rectangle Area moment of inertia of a rectangle.svg S = b h 2 6 S={\cfrac  {bh^{2}}{6}} Solid arrow represents neutral axis
doubly symmetric I-section (major axis) Section modulus-I-beam-strong axis.svg S x = B H 2 6 − b h 3 6 H {\displaystyle S_{x}={\cfrac {BH^{2}}{6}}-{\cfrac {bh^{3}}{6H}}}

S x = I x y {\displaystyle S_{x}={\tfrac {I_{x}}{y}}},

with y = H 2 {\displaystyle y={\cfrac {H}{2}}}

NA indicates neutral axis
doubly symmetric I-section (minor axis) Section modulus-I-beam-weak axis.svg S y = B 2 ( H − h ) 6 + ( B − b ) 3 h 6 B {\displaystyle S_{y}={\cfrac {B^{2}(H-h)}{6}}+{\cfrac {(B-b)^{3}h}{6B}}}[4] NA indicates neutral axis
Circle Area moment of inertia of a circle.svg S = π d 3 32 {\displaystyle S={\cfrac {\pi d^{3}}{32}}}[3] Solid arrow represents neutral axis
Circular hollow section Area moment of inertia of a circular area.svg S = π ( r 2 4 − r 1 4 ) 4 r 2 = π ( d 2 4 − d 1 4 ) 32 d 2 S={\cfrac  {\pi \left(r_{2}^{4}-r_{1}^{4}\right)}{4r_{2}}}={\cfrac  {\pi (d_{2}^{4}-d_{1}^{4})}{32d_{2}}} Solid arrow represents neutral axis
Rectangular hollow section Section modulus-rectangular tube.svg S = B H 2 6 − b h 3 6 H S={\cfrac  {BH^{2}}{6}}-{\cfrac  {bh^{3}}{6H}} NA indicates neutral axis
Diamond Secion modulus-diamond.svg S = B H 2 24 S={\cfrac  {BH^{2}}{24}} NA indicates neutral axis
C-channel Section modulus-C-channel.svg S = B H 2 6 − b h 3 6 H S={\cfrac  {BH^{2}}{6}}-{\cfrac  {bh^{3}}{6H}} NA indicates neutral axis

Plastic section modulus

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 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 Area moment of inertia of a rectangle.svg Z P = b h 2 4 Z_{P}={\cfrac  {bh^{2}}{4}}[5][6] A C = A T = b h 2 {\displaystyle A_{C}=A_{T}={\cfrac {bh}{2}}} , y C = y T = h 4 {\displaystyle y_{C}=y_{T}={\cfrac {h}{4}}}
Rectangular hollow section   Z P = b h 2 4 − ( b − 2 t ) ( h 2 − t ) 2 {\displaystyle Z_{P}={\cfrac {bh^{2}}{4}}-(b-2t)\left({\cfrac {h}{2}}-t\right)^{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 Z_{P}=b_{1}t_{1}y_{1}+b_{2}t_{2}y_{2}\, where:

b 1 , b 2 b_{1},b_{2} =width, t 1 , t 2 t_{1},t_{2}=thickness, y 1 , y 2 y_{1},y_{2} are the distances from the neutral axis to the centroids of the flanges respectively.

For an I Beam including the web   Z P = b t f ( d − t f ) + 0.25 t w ( d − 2 t f ) 2 Z_{{P}}=bt_{f}(d-t_{f})+0.25t_{w}(d-2t_{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 ) Z_{{P}}=(b^{2}t_{f})/2+0.25t_{w}^{2}(d-2t_{f}) d = full height of the I beam
Solid Circle   Z P = d 3 6 Z_{P}={\cfrac  {d^{3}}{6}}  
Circular hollow section   Z P = d 2 3 − d 1 3 6 Z_{P}={\cfrac  {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 k={\cfrac  {Z}{S}}

Shape factor for a rectangular section is 1.5.

 
 

Significance of Section Modulus:

  • Section modulus is an important factor for designing beams and flexural members.
  • Higher the section modulus, higher will be the resistance to bending.
  • Easier to calculate strength and stresses in beams.
  • If two beams are made of the same material and the section modulus of two beams are being compared and vice versa, the beam with larger section modulus will be tougher and more capable to withstand larger loads.

 

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:

  • By comparing the values we can say that section modulus of the optimized hood (#Design 2) is increased by 2.52 %.
  • MOI increases with the increase of section area.
  • MOI increases with the increase of section length.
  • If we increase the gap between the inner and outer panel of hood eventually section length will increase thus MOI will increase.

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