Challenge

Section Modulus calculation and optimization

Objective:
Use the Section of Hood design and calculate the section modulus using the formula S = I/y
S = Section Modulus
I = Moment of Inertia
Y = distance between the neutral axis and the extreme end of the object

Create a new section that has improved the section modulus of the previous section

Consider Stainless steel AISI 302 – cold-rolled – y = 520 M Pa

Theory:
Section Modulus:
It is a geometric property for a given cross-section used in the design of beams or flexural members. Other geometric properties used in design include area for tension and shear, radius of gyration for compression, and moment of inertia and polar moment of inertia for stiffness.

There are two types of section moduli, the elastic section modulus and the plastic section modulus.

For general design, the Elastic section modulus is used, applying 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 also often used to determine the yield moment (My) such that My = S × σy, where σy is the yield strength of the material which is maximum allowable load in a mechanical component.

The Plastic section modulus is used for materials where elastic yielding is acceptable and plastic behavior is assumed to be an acceptable limit.

Procedure:
In the Hood design case we use Elastic section modulus. Model preparation as below.
Copied two parts of Hood outer panel and inner panel in another new part model which is named as Section modulus, then created the section geometry of the outer most geometry of hood outer panel and lower most geometry of hood inner panel and joined them with lines as we need closed profile to measure the inertia, then later using “Measure the Inertia” option in Catia noted the Moment of inertia ‘ I ’ value and then proceeded with the calculation to measure the allowable strength of the section 1.

See the below the images and calculation taken from Catia model of for your reference.

Section 1:

Calculation for Section 1:

Iox= 2.025e-006m4 =2025000 mm4 (IoxG value from above image)
Y= 430.17 mm (Measured value fom image 1)
S= I/Y = 2025000 mm4 /430.17 mm = 4707.44 mm3

Yield moment My = S × σy, where σy is the yield strength of the material which is 520 MPa or N/mm2 as mentioned in question.
M= S × σy = 4707.44 mm x 520 N/mm2 = 2447868.8 N mm = 2.4e6 N mm.

To increase the strength of the above section, I increased 3mm the distance between the outer panel geometry to inner panel geometry then repeated the above said procedures then we have got much better strength value. See below section 2 calculation part and images.

Section 2:


Calculation for Section 2:

Iox= 2.556e-006m4=2550000 mm4 (IoxG value from above image)
Y= 430.17 mm (Measured value fom image 1)
S= I/Y = 2550000 mm4 / 430.17 mm = 5927.8 mm3

Yield moment My = S × σy, where σy is the yield strength of the material which is 520 MPa or N/mm2 as mentioned in question.
M=
S × σy = 5927.8 mm3 x 520 N/mm2 = 3082456 N mm = 3.08 e6 N mm.

Conclusion:
Thus by increasing the distance between the Inner and outer panel of the Hood assembly we got the higher value of section modulus hence the strength also increased.

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