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

• fonseh
In summary: No, it's right.In summary, the homework statement is that 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.
fonseh

## 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 ?

## 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)

Nothing again ?

## 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.

fonseh
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 ?
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 ?
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.

fonseh
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 .

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 .
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 !

My advice: check out the derivation of 5-20

fonseh
BvU said:
You don't have to actually do it !

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 !

My advice: check out the derivation of 5-20
Can you explain about it ?

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 ?

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.

fonseh
BvU said:
174 came from
174 ? where is it ?

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

fonseh
BvU said:
And the correct number should be 184, not 174
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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Last edited:

## 1. What is the angle of twist?

The angle of twist is the amount of rotation that occurs in a thin wall with a closed section when a torque is applied.

## 2. How is the angle of twist calculated?

The angle of twist can be calculated by dividing the applied torque by the product of the shear modulus, polar moment of inertia, and the length of the thin wall.

## 3. What is the significance of the angle of twist?

The angle of twist is an important parameter in determining the structural integrity and stability of a thin wall with a closed section. It helps in understanding the behavior of the material under applied torque and predicting potential failure points.

## 4. What is stress in relation to thin wall with closed section?

Stress is the amount of force per unit area that is acting on the material of a thin wall with a closed section. It is induced by the applied torque and can cause the material to deform or fail.

## 5. How is stress calculated in thin wall with closed section?

Stress in a thin wall with a closed section can be calculated by dividing the applied torque by the product of the polar moment of inertia and the distance from the center of the thin wall to the outermost edge of the section.

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