Angle of twist / stress of thin wall having closed section

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SUMMARY

The discussion centers on calculating the mean area enclosed within the centerline of a thin-walled tube, specifically addressing the dimensions of 0.035 m and 0.057 m. Participants clarify that the mean height and width are derived from the dimensions of 54 mm and 30 mm, respectively, adjusted for half the thickness of the frame. The term "ds" refers to the differential arc length along the mean area circumference, which is crucial for understanding shear stress distribution in thin-walled structures.

PREREQUISITES
  • Understanding of thin-walled tube mechanics
  • Familiarity with shear stress distribution concepts
  • Knowledge of differential calculus, particularly arc length calculations
  • Ability to interpret engineering diagrams, such as those in textbooks
NEXT STEPS
  • Study the derivation of shear stress equations in thin-walled tubes, specifically equation 5-20
  • Learn about the implications of mean area calculations in structural engineering
  • Explore the concept of differential arc length (ds) in engineering applications
  • Review engineering diagrams and their significance in understanding material properties
USEFUL FOR

Mechanical engineers, structural analysts, and students studying materials science or mechanical design will benefit from this discussion, particularly those focusing on the analysis of thin-walled structures and stress distribution.

fonseh
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Homework Statement


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 ?

Homework Equations

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 ?[/B]
 

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fonseh said:

Homework Statement


why the mean area enclosed within the boundary of the centerline of tube thickness is (0.035)(0.057) ? Is it wrong ?
No, it's right. Mean height is 54+3/2+3/2 and mean width is 30+5/2+5/2.
Second question , why for the ds it's 2(57) and 2(35) respctively ? what is ds actually ?
He's following the circumference of the mean area: the dashed line in figure 5-30 (d)

Homework Equations

Nothing again ? :smile:

The Attempt at a Solution


[/B]
for the first question , Shouldnt it be (0.040-0.005-0.005) x ( 0.060 -0.003 -0.0030 = 0.054 ?
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.
 
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BvU said:
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 ? :smile:
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.

Do you mean the area enclosed by the hollow position + area of half thickness of the frame ?
 
BvU said:
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 ? :smile:
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.
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 thicknesswhy the thickness shouldn't 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 ?
 
I think the book is perfectly clear. He's following
fonseh said:
centerline of tube thickness
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)
 
BvU said:
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)
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 ?
 
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.
 
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BvU said:
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.
CIRCUMFERENCE ? I didnt see any circle here
 
  • #10
BvU said:
and a certain thickness (that varies with lateral position, so it's kept inside the ∫ds∫ds\int ds). No halving.
why the thickness shouldn't be the same through out the length of the pipe ?
 
  • #11
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 :rolleyes: .
 
  • #12
BvU said:
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 :rolleyes: .
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 couldn't imagine it at all
 
  • #13
You don't have to actually do it ! :smile:

My advice: check out the derivation of 5-20
 
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  • #14
BvU said:
You don't have to actually do it ! :smile:

My advice: check out the derivation of 5-20
it's not given in the book
 
  • #15
BvU said:
You don't have to actually do it ! :smile:

My advice: check out the derivation of 5-20
Can you explain about it ?
 
  • #16
upload_2017-1-5_13-48-38.png
 
  • #17
BvU said:
it's not the thickness of the wall bounded by the centerline ? it's still the same thickness before the centerline is drawn ?
 
  • #18
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.
 
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  • #19
BvU said:
174 came from
174 ? where is it ?
 
  • #20
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
 
  • #21
It is the length of the dashed line in figure 5-30 (d), the same line.
 
  • #22
And the correct number should be 184, not 174 o:) :biggrin:
 
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  • #23
BvU said:
And the correct number should be 184, not 174 o:) :biggrin:
here's what i found for the t , from the figure , we can notice that it's not the thickness of the frame as you stated , rather , it's a width from the centerline to the outer surface

P/s : Here's the 3 consecutive pages of the notes .
 

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