Angle of twist / stress of thin wall having closed section

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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
 
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 ?
 
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: .
 
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
 
You don't have to actually do it ! :smile:

My advice: check out the derivation of 5-20
 
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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
 
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 ?
 
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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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
 
It is the length of the dashed line in figure 5-30 (d), the same line.
 
And the correct number should be 184, not 174 o:) :biggrin:
 
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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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