Challenge

Section Modulus calculation and optimization

SECTION MODULUS:

It is the ratio of area moment of inertia about the neutral axis and to the distance of outer most fiber from the neutral axis.

Z=I/Y HERE,Z= SECTION MODULUS,I= AREA MOMENT OF INERTIA,Y=DISTANCE FROM NETRUAL AXIS TO THE OUTER MOST FIBER.

Section modulus is the direct measure of the strength of the steel. Bending a steel section that has a larger section modulus than another will be stronger and harder to bend. Section modulus is a geometric property for a given cross-section used in the design of flexural members.

The more section modulus, the more the member is resistant to bending of the beam. If two beams are made of the same material and comparing the “section modulus” of two beams, the beam with bigger section modulus will be tougher and more capable to withstand larger loads.

Intersection of hood:

Section modulus of part one with 0.75:

Minimum value(I)=2.727e-007m4

=2.727e007* 10^12 mm^4

=2.727e007mm^4

Y=DISTANCE FROM NETRUAL AXIS TO THE OUTER MOST FIBER:

Y=879.427/2

=439.71

Hence Section modulus is,S=I/Y

=2.727e007/439.71

=6201.830mm^3

NOTE:Here,i have made changes of sketch of hood intersection for increase the moi,section modulus .the changes are ,instead of flat/origional intersectionsection i have increases the embosses and the curve shape diagrams.

Section modulus of part one with 0.75:(OPTIMAIZATION)

Minimum value(I)=2.911e-007m4

=2.911e007*10^12 mm^4

  • =2.911e007mm^4

Y=DISTANCE FROM NETRUAL AXIS TO THE OUTER MOST FIBER:

  • Y=Distance =879.427

=879.427/2

=439.71

Hence Section modulus is,S=I/Y

=2.911e007/439.71

=6620.245mm^3

Section modulus of part one with 0.5:

Minimum value(I)=1.832e-007m4

=1.832e007* 10^12 mm^4

=1.832e007mm^4

Y=DISTANCE FROM NETRUAL AXIS TO THE OUTER MOST FIBER:

  • Y=Distance =879.487

=879.427/2

=439.71

Hence Section modulus is,S=I/Y

=1.832e007/439.71

=4166.39mm^3

Section modulus of part with 0.5:

Minimum value(I)=1.96e-007m4

=1.96e007* 10^12 mm^4

=1.96e007mm^4

  • Y=Distance =879.427

=879.427/2

=439.71

Hence Section modulus is,S=I/Y

=1.96e007/439.71

=4457.64mm^3

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