# Finding Neutral Axis & Second Moment on a T-beam

1. Sep 12, 2005

### Struggling

hi im having trouble with finding the neutral axis and second moment of a T-beam.

here are the measurements of the beam:

i understand to find the neutral axis you divide the beam into 2 rectangles but iam having problem finding the thickness of the bottom rectangle of the t-beam. how can i find the neutral axis if i cant find the width of the bottom rectangle, because i cant find the area of it. iam totally stuck.

also calcutale the second moment of area

for a member with a rectangular cross section bent about the z axis,
Izz =(1/12)bd^3
to determine the second moment of area of a cross section made of a number of different shapes the parallel axis theorem is used.
Iaa = Izz + Ay^2

what i did was:

Izz = (1/12)6.4(38.1)^3
= 29496.72

but it doesnt make sense i was going to insert this into Iaa= Izz + Ay^2 to get the second moment of the vertical.

can someone please help

2. Sep 12, 2005

### mathmike

the bottom thickness is 6.4 just as the vertical is

3. Sep 12, 2005

### Struggling

how did you get that?
or is it just assumed that it is in my case because both beams are of the same length?

4. Sep 12, 2005

### mathmike

no i did not assume, it is usually the case that the webs of a beam are the same thickness. In this case since the bottom thickness is not given it should be of the same thickness.

5. Sep 12, 2005

### Struggling

ARGH i still cant get it this is what i had
rectangle 1 = 38.1 x 6.4
rectangle 2 = 31.7 x 6.4

from these notes:

i got this:

(Ycent x A) = (sumof)Ai Yi
Rectangle 1
(Ycent x 446.72) = 243.84(3.2)
= 1.746
Rectangle 2
(Ycent x 446.72) = 202.88(19.05)
= 8.651

theyre sposed to be the distances from the arbitary axis to the centroid but common sense tells me its wrong.
please help

6. Sep 12, 2005

### Struggling

anyone at all?

7. Sep 12, 2005

### Kazza_765

Ok, I'm not sure I understand the method you are using. The way I do it is to consider the distance from the bottom edge of the lowest rectangle to the central axis. Then we have the formula:

$$A_1*d_1 + A_2*d_2=A_{1+2}*d_0$$

where A1 is the area of rectangle 1, d1 is the distance from the bottom of the lower rectangle to the centroid of rectangle 1, and d0 is the distance of the central axis.

$$(38.1*6.4)(3.2)+(31.7*6.4)(22.25)=(38.1*6.4+31.7*6.4)d_0$$

And we find that the neutral axis is 11.85mm above the lowest point, or 5.45mm above the intersection of the two rectangles.

Once you know the location of the neutral axis you should be able to find the second moment of area using the parallel axis theorem.

8. Sep 14, 2005

### Struggling

thanks for the help but i still am not able to find the second moment of area.

Iaa = Izz + Ay^2

iam lacking a detailed description of how to use the formula.
i understand that

Iaa = parrallel axis
Izz = Centroid Axis
A = Area
y = distance between Iaa and Izz

but i have no clue how to do it. my first attempty went something like this
Neutral axis = 11.85mm from the bottom axis (as kazaa stated above)
Iaa = 11.85+202.88(10.4)^2
Iaa = 21955.35
which obviously isnt the answer so i tried
Izz = (1/2)bd^3
Izz = (1/12)6.4x31.7
Izz = 16.9

Iaa = 16.9 + 202.88(5.05)^2
Iaa = 26145.33

can anyone help me out with what iam supposed to use or direct me to a detailed explanation, i have searched many pages on yahoo etc but still fail to find an explanation i can relate to and the only text books i have are very breif on it.

9. Sep 15, 2005

### Kazza_765

To use the parallel axis theorem, consider each rectangle seperatly, so we'll call the top one A, and the bottom one B. For each rectangle, calculate the second moment of area about its own neutral axis, and then add to that A*y^2 where y is the distance between the rectangles neutral axis, and the overall neutral axis.

So for rectangle A, the second moment of area is
$$\frac{bd^3}{12}$$
And the distance between rectangle A's neutral axis and the overall neutral axis is
$$y=6.4+\frac{31.7}{2}-11.85$$
$$y=10.4$$

So A's contribution to the total second moment of area is:
$$I_A=\frac{bd^3}{12}+(10.4^2)bd$$

And I was about to put all that in my calculator, but I just realised I left it at uni after a physics prac today But anyway, you should be able to work it out from there.

10. Sep 15, 2005

### Struggling

oh alright well wat i ended up doing this morning was:

Izz = (1/12)6.4x31.7^3 = 16989.34
Iaa = (1/12)6.4x38.1^3 = 29496.72

Iaa = Izz +Ay^2
therfore

(Square Root)Iaa-Izz/A = Y
(square root) 29496.72 - 16989.34/ 446.72 = Y
Y = 5.29mm

and i just tried the forumla you gave me and i got and answer of 43363.07???

Ia = (6.4)(31.7)^3/12 + (10.4^2)(38.1)(6.4)
Ia = 16989.34+26373.73
Ia = 43363.07

sooo confused argghhh

ur help is much appreciated kazaa thank you

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