Section Modulus calculation and optimization
Aim
Use the Section from your Hood design and calculate the Section Modulus using the formula
S = I/y.
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
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.