Section Modulus calculation and optimization

Significance

It is also often used to determine the extreme fiber elastic stresses as well as the yield moment (My) of elastic sections such that My=S×σy, where σy is the yield strength of the material. It is a commonly used property in the design of flexural members of steel, post-tensioned concrete, and timber sections.

Cross-sectional shape	Figure	Equation	Comment
Rectangle		{\displaystyle S={\cfrac {bh^{2}}{6}}}	Solid arrow represents neutral axis
doubly symmetric I-section (major axis)		{\displaystyle Sx={\cfrac {BH^{2}}{6}}-{\cfrac {bh^{3}}{6H}}} {\displaystyle Sx={\tfrac {Ix}{y}}}, with {\displaystyle y={\cfrac {H}{2}}}	NA indicates neutral axis
doubly symmetric I-section (minor axis)		{\displaystyle Sy={\cfrac {B^{2}(H-h)}{6}}+{\cfrac {(B-b)^{3}h}{6B}}}[4]	NA indicates neutral axis
Circle		{\displaystyle S={\cfrac {\pi d^{3}}{32}}}[3]	Solid arrow represents neutral axis
Circular hollow section		{\displaystyle 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		{\displaystyle S={\cfrac {BH^{2}}{6}}-{\cfrac {bh^{3}}{6H}}}	NA indicates neutral axis
Diamond		{\displaystyle S={\cfrac {BH^{2}}{24}}}	NA indicates neutral axis
C-channel		{\displaystyle 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 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: Zp = Acyc + Atyt

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Description	Figure	Equation	Comment
Rectangular section		{\displaystyle Z_{P}={\cfrac {bh^{2}}{4}}}[5][6]	{\displaystyle A_{C}=A_{T}={\cfrac {bh}{2}}} , {\displaystyle y_{C}=y_{T}={\cfrac {h}{4}}}
Rectangular hollow section		{\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]		{\displaystyle Z_{P}=b_{1}t_{1}y_{1}+b_{2}t_{2}y_{2}\,}	where: {\displaystyle b_{1},b_{2}} =width, {\displaystyle t_{1},t_{2}}=thickness, {\displaystyle 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		{\displaystyle Z_{P}=bt_{f}(d-t_{f})+0.25t_{w}(d-2t_{f})^{2}}	[8]
For an I Beam (weak axis)		{\displaystyle Z_{P}=(b^{2}t_{f})/2+0.25t_{w}^{2}(d-2t_{f})}	d = Full height of the I Beam
Solid Circle		{\displaystyle Z_{P}={\cfrac {d^{3}}{6}}}
Circular hollow section		{\displaystyle 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)

SECTION MODULUS CALCULATON AND OPTIMIZATION :

Hood Main Design, first case:

Calculation of Section Modulus of main Hood:

S = section modulus.

I = moment of inertia.

Y = Distance between the neutral axis and the extreme end of the object.

S = ?

I (min) = 1.583961374 x 10^4 [mm^4]

Y = 881.898568 / 2

Y = 440.9492

FORMULA,

S=I/Y

Moment of Inertia (MOI), Imin = 1.58396137 * 10^4[mm^4]

=15839.6137 / 440.949[mm]

=35.92164[mm^3]

Moment of Inertia (MOI), Imax = 4.873385030 * 10^6[mm^4]

= 4873385.03 / 440.949 [mm]

= 11052.03783 [mm^3]

SECOND CASE,

HOOD INNER PANEL OFFSETTED 5 MM TO DOWNWARDS.

Moment of Inertia (MOI), Imin = 1.605521867 * 10^4[mm^4]

= 16055.21867 /440.949[mm]

= 36.410602[mm^3]

Moment of Inertia (MOI), Imax = 4.865732120 * 10^6[mm^4]

= 4865732.12 / 440.949[mm]

= 11034.68228[mm^3]

Conclusion:

Thus, the section modulus of the Front Hood of a car was calculated and compared with the optimized hood and found the better result.

second case optimized hood resulted in the increased strength.