Calculating Maximum Bending Stress for a Beam - I'm Doing it Wrong!

[songtoan92]

bending stress in beam, help me !

Homework Statement

Homework Equations

singularity function

bending stress (max) = { (M max)/I } * ( a/2)

The Attempt at a Solution

after writing the singularity function, and compute reactions force I found that the maximum bending moment should be 12.04 kNm

I = 1/12 ( 0.1 * 0.14 ^ 3 - 2 * 0.06 * 0.04 ^ 3 ) (m ^ 4)

y/2 = 0.14 /2 = 0.07 m

therefore, the maximum bending stress is ~ 37.9 MPa

but this is totally wrong :( :(, can anyone help with

thank you very much

 

[Mapes]

Hi songtoan92, welcome to PF. Your maximum bending moment looks fine. Make sure to use the parallel axis theorem when subtracting the moment of inertia of the missing areas of the cross-section, though, since they're not centered on the centroid of the beam. Know what I mean?

 

[songtoan92]

Hi songtoan92, welcome to PF. Your maximum bending moment looks fine. Make sure to use the parallel axis theorem when subtracting the moment of inertia of the missing areas of the cross-section, though, since they're not centered on the centroid of the beam. Know what I mean?

Can you explain me more ?? I am a little bit confused now
Personally , I think , " I " is the 2nd moment of area, so considering the " H " cross-section of the beam I got I =... I think it does not a matter of moment of parallel axis theorem here ?

 

[Mapes]

Moment of inertia is another common name for second moment of area.
In any case, your calculation of I involves subtracting two areas of 60 mm x 40 mm. But the second moment of area of these regions is not simply $wh^3/12$ because their centroids do not coincide with the centroid of the beam. Rather, their centroids are 50 mm away from the beam centroid. So I is calculated as $I=wh^3/12+whr^2$, where $r=50\,\mathrm{mm}$. (This is called the http://en.wikipedia.org/wiki/Parallel_axis_theorem" .) Try this and see if you get the answer you're expecting.

 

[songtoan92]

Moment of inertia is another common name for second moment of area.
In any case, your calculation of I involves subtracting two areas of 60 mm x 40 mm. But the second moment of area of these regions is not simply $wh^3/12$ because their centroids do not coincide with the centroid of the beam. Rather, their centroids are 50 mm away from the beam centroid. So I is calculated as $I=wh^3/12+whr^2$, where $r=50\,\mathrm{mm}$. (This is called the http://en.wikipedia.org/wiki/Parallel_axis_theorem" .) Try this and see if you get the answer you're expecting.

So,...

I1 (of the outer rectangular) = 1 / 12 * 0.1 * 0.14 ^ 3 + 0.1 * 0.14 * 0.05 ^ 2
I2 (of the inner rectangular) = 1 / 12 * 0.06 * 0.04 ^ 3 + 0.04 * 0.06 * 0.05 ^ 2

and the I final = I1 - 2I2 ~ 4.5 * 10E-05

is that right ??

 

[Mapes]

You don't need to use $r=50\,\mathrm{mm}$ for the large rectangle. Its centroid coincides with the beam centroid.

 

[songtoan92]

You don't need to use $r=50\,\mathrm{mm}$ for the large rectangle. Its centroid coincides with the beam centroid.

I 've got it, thank you for your help :D