Section Modulus calculation and optimization

INPUT DATA:

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

DEFINITIONS OF SECTION MODULUS:

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

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

• In 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 (M_y) such that M_y = S × σ_y, where σ_y 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.

CALCULATION:

In mean time we are calculating a two values from the hood design:

CASE 1

Iox= 2.556e-006m4=2550000 mm4
Y= 430.17 mm
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
M= S × σy = 5927.8 mm3 x 520 N/mm2 = 3082456 N mm = 3.08 e6 N mm.

CASE 2

Iox= 2.025e-006m4 =2025000 mm4
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.

Conclusion:

By the end of the two cases which increasing the distance between the Inner and outer panel of the Hood assembly we got the higher value of section modulus hence we can increase the strength also.