# Aircraft structure (thin walled closed section beam)

1. Jun 9, 2007

### sohaib

Figure shows the cross section of a single cell, thin walled
beam with the horizontal axis of symmetry. The direct stresses are carried
by the booms B1 to B4, while the walls are effective only in carrying shear
stresses. Assuming that the basic theory of bending is applicable, calculate
the position of the shear center S. The shear modulus G is the same for all walls.

Cell Area = 135000 mm^2
Boom Areas B1 = B4 = 450 mm^2, B2 = B3 = 550 mm^2

Wall Length(mm) Thickness(mm)
12,34 500 0.8
23 580 1.0
41 200 1.2

Ans: 197.2 mm from vertical through booms 2 and 3.

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2. Jun 9, 2007

### Aero Stud

And so far you have done... ? If you know how to solve one of these you know how to solve them all as I remember.

3. Jun 9, 2007

### sohaib

i was looking for the approach i should take to solve the problem... iam not here to find out the entire solution because i know u (aero stud) can't solve it anyway...

4. Jun 9, 2007

### Aero Stud

LOL are you studying engineering or psychology ? lol
Those are the rules - what's the point of just telling you the answer, the shear centre is where the moment is zero so use that for every section and it's moment and you can find it. I studied this stuff a year ago, I think maybe a tiny bit did sink in... lol You could at least google it, you lazy shrink...

5. Jun 10, 2007

### sohaib

U are the most lazy person i know in this world and have so much
free time that u are writing such big replies rather than answering the question. So sick and so pathetic...

Anywayz, i have figure out the answer myself of my question no thanks to u.... Anywayz, iam off to solving the next question...
lolz... ;)

6. Jun 10, 2007

### sohaib

I recently asked the above mentioned question on this forum....

I have figured out the answer myself to this question... So iam posting the way to solve this problem so it might be helpful to anyone who is having trouble solving it...

First of all see that the shear center S lies on the horizontal axis of symmetry, the x axis. Therefore what we have to do is to apply an arbitrary shear load Sy through S. The internal shear flow distribution is given by eq. 9.80 (aircraft structure, thg megson). and since we have
Ixy = 0
Sx =0
and tD=0
Calculate Ixx.
Calculate qb23, qb34, qb41

note that qb23 = 0
The value of the shear flow at the cut is obtained by using eq. 9.47 .
Calculate qs,0

Now take moments about O we have,
SyEs = 2x0.61x10^-3Syx500x100 + 2.86x10^-3Syx200x500

Solve this, we have
Es = 197.2 mm