Section Modulus calculation and optimization

THEORY:

SECTION MODULUS:

It is a geometrics property for 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 stiffnes.

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 difined as S=l/y , where l 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 neurtal 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 point strength of the MATERIAL which is maximum allowable load in a mechanical component.

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

PROCEDURE:

In the hood design case we use elastic section modulus . section modulus . Model prepartion as below

Copied two parets 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 option in Catia noted the moment of inertia l value and then proceeded with the calculation to measure the allowable strength of the section 1.

see the below the image and calculation taken from catia model of for your reference.

SECTION 1:

CALCCULATION FOR SECTION 1:

lox = 2.025e-006m4=2025000mm4

Y= 430.17mm(measured value )

S=l/y= 2025000mm4/ 430.17mm=4707.44mm3

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= 4707.44mm×520N/mm2= 2447868.8N mm=2.4e6N mm.

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

SECTION2:

lox=2.556e-006m4=2550000mm4(loxG value)

Y= 430.17mm (measured value)

s=l/y =2550000mm4/430.17mm=5927.8mm3

yield mpmentMy=S×σy, where σy is the yield strength of the materials which is 520Mpa or N/mm2 as mentioned in question.

M=S×σy= 5927.8mm3520N/mm2=3082456Nmm=3.08e6 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 .