- #1

masterchiefo

- 212

- 2

## Homework Statement

The profiled steel I above is topped by a " profiled C ".

Determine the moments of inertia of the structure composed by the axes

central x and y through its geometric center C.

Picture representing the problem:

## Homework Equations

Profiled C and I information:

## The Attempt at a Solution

Section

*All numbers are coming from the table for this first part.*Profiled I: A= 14400mm

^{2}/y=0 /x=0 /y*A=0 /x*A=0

Profiled C: A= 2897mm

^{2}/y=16.10 /x=0 /y*A=46641.7 /x*A=0

Total A = 17297mm

^{2}

/Y∑A = ∑/yA /Y*(17297) = 46641.7 /Y =26.9652mm

For the Profiled I :

Ix' = /Ix +A/Y

^{2}=554*10

^{6}+ 14400*26.9652

^{2}

= 564.47*10

^{6}

For the Profiled C:

Height of profiled I = 460, half = 230

Now to get height of mass center of I to mass center of section C I do 230-(22.8-16.10)

16.10 is the distance from mass center of section C to the top of the section, but I need mass center to bottom of that section so 22.8-16.10.

Ix' = /Ix +A*height

^{2}= 0.949*10

^{6}+(2897)*(230-(22.8-16.10))

^{2}

=145.402*10

^{6}

Total

145.402*10

^{6}+ 564.47*10

^{6}

= 709.872*10

^{6}

Last edited: