# Angle of twist / stress of thin wall having closed section

1. Jan 3, 2017

### fonseh

1. The problem statement, all variables and given/known data
why the mean area enclosed within the boundary of the centerline of tube thickness is (0.035)(0.057) ? Is it wrong ?

Second question , why for the ds it's 2(57) and 2(35) respctively ? what is ds actually ?

2. Relevant equations

3. The attempt at a solution

for the first question , Shouldnt it be (0.040-0.005-0.005) x ( 0.060 -0.003 -0.0030 = 0.054 ?

File size:
44.3 KB
Views:
29
File size:
35.3 KB
Views:
26
File size:
26.5 KB
Views:
28
2. Jan 3, 2017

### BvU

No, it's right. Mean height is 54+3/2+3/2 and mean width is 30+5/2+5/2.
He's following the circumference of the mean area: the dashed line in figure 5-30 (d)
Nothing again ?
That's not an attempt, that's a question !
Anyway, that would be following the inside of the tube, which is not the intention here.

3. Jan 3, 2017

### fonseh

Do you mean the area enclosed by the hollow position + area of half thickness of the frame ?

4. Jan 3, 2017

### fonseh

ok , i understand why it's 57mm and 35 mm , because it's the mean area enclosed within the boundary of the centerline of the tube thickness

why the thickness shouldnt be 3/2 and 5/2 mm now ? Because for the area of 57x 35 mm , the thickness of the 'frame' has been halved , right ?

5. Jan 3, 2017

### BvU

I think the book is perfectly clear. He's following
which has a certain length and encloses a certain area. The tube thickness itself is of course not halved. (however, the expressions developed -- such as for 5-20 -- are valid only for thin walls)

6. Jan 4, 2017

### fonseh

why The tube thickness itself is not halved ??? We can see that the figure in 7.30d , the thickness has been halved , right ?
for the original frame , we have the inner length of 60-3-3 = 54m and 40-5-5 = 30m , for the mean area it's 54+(3/2) + (3/2) = 57 , and 30+(5/2)+(5/2) = 35 , so the thickness for the mean area is 3/2 and 5/2 respectively , right ? Which the thickness has been halved ?

7. Jan 5, 2017

### BvU

The dashed line is simply an average circumference. If you cut the pipe lengthwise you get a long sheet with a certain width and a certain thickness (that varies with lateral position, so it's kept inside the $\int ds$). No halving.

See post #5 and check out the derivation of 5-20.

8. Jan 5, 2017

### fonseh

CIRCUMFERENCE ? I didnt see any circle here

9. Jan 5, 2017

### BvU

10. Jan 5, 2017

### fonseh

why the thickness shouldnt be the same through out the length of the pipe ?

11. Jan 5, 2017

### BvU

If you cut it lengthwise and fold it flat, thickness is the same throughout the length of the pipe. That's in the problem statement. You get a long slab of approximately 174 mm wide with two sections of 3 mm thickness and two of 5 mm thickness.

My estimate was/is that you could have figured that out for yourself. Idem circumference/perimeter .

12. Jan 5, 2017

### fonseh

Do you mean cut the beam at 2 locations( cut off a portion of the beam) and hold it flat ???
How to fold it flat ? I couldnt imagine it at all

13. Jan 5, 2017

### BvU

You don't have to actually do it !

My advice: check out the derivation of 5-20

14. Jan 5, 2017

### fonseh

it's not given in the book

15. Jan 5, 2017

### fonseh

Can you explain about it ?

16. Jan 5, 2017

### BvU

17. Jan 5, 2017

### fonseh

it's not the thickness of the wall bounded by the centerline ? it's still the same thickness before the centerline is drawn ?

18. Jan 5, 2017

### BvU

Yes to the latter. The shear stress is distributed over the entire thickness (5-18).
I drew the centerline so you could see where the 174 came from; not to confuse you.

19. Jan 5, 2017

### fonseh

174 ? where is it ?

20. Jan 5, 2017

### BvU

With the browser find function CTRL-F you can see that it is in post #11.

My estimate was/is that you could have figured that out for yourself