# Homework Help: Tensile strength of a cord winding

1. Jan 4, 2014

### strive

1. The problem statement, all variables and given/known data

Hi

Can someone please tell me how to set equations for the following problem:

A cord of length L is wound on a hollow bobbin (inner radius r and outer radius R). Cord on cord friction (static) coefficient (Ctc) and cord on bobbin (Ctb) friction (static) coefficient is given, cord elasticity modulus (E) is given, cord radius is given and cord tensile strength (sigma) is given. (any other necessary data can be obtained)

What is the max pressure on the bobbin internal radius (reaching cords braking point) and what is the necessary force on the cords ends to hold the winding from unwinding?

(The bobbin is ideally elastic, thus it only serves to give shape to the cord winding and to equally distribute the pressure.)

2. Relevant equations

3. The attempt at a solution

I presume that first winding thickness (denoted t) must be obtained from cord length and its diameter and from bobbins outer diameter (this is done).

If this winding was a solid wall the max internal pressure would be simple to calculate by:

p=t*sigma/r (presuming bobbin wall thickness is small enough to be ignored.

But this is a cord. Each loop (of the winding) is in contact with several others. Should I presume the friction (static) between these contacts acts like a binder in composite materials?
If so how do I calculate the actual strength of the wall (SIGMAwall) in order to use it in the above equation?

Also as the cord has a circular cross section this leaves some empty space in the windings cross section and this leaves space for elastic deformation of the cord (of its cross-sectional shape), thus I predict some tension during winding of the cord might help to distribute the load uniformly among the cord layers.
Should I presume the force on the cord ends would be equal to this force subtracted by the sum of the friction (static) force on the last layer of the winding?

p.s.: regarding my mathematical/physical background – I am a last year student of mechanical/aerospace engineering (undergraduate)

2. Jan 6, 2014

### nvn

strive: Check out this page. I have not been able to figure out anything useful for you yet. I keep getting his "p" symbols mixed up. I currently think we need some kind of derivative with respect to d(theta), perhaps dp/d(theta)?, but I am not sure. I am familiar with computing hoop stress on a cylinder due to uniform pressure, but I am not yet familiar with computing hoop stress due to a varying pressure around the perimeter, if the pressure varies. I am having trouble visualizing yet how a varying hoop stress in the bobbin would even be possible, or how to compute it, if it varies. If you have figured out a few things yet, and you want to let us know, someone might see something and it might generate new ideas.

Last edited: Jan 6, 2014
3. Jan 7, 2014

### strive

The pressure around the perimeter of the bobbins wall varies only minutely, but the pressure further away from the bobbin (throughout the winding) varies with X (X being the distance from the center of the circle). The bobbin is only there to give initial circular shape to the winding (imagine a spool of thread and the thread takes the entire load from the pressure (the pressure is uniformly distributed over the inside radius of the spool - the spool only serves to transmit that pressure to the winding and thus the spool takes no load)).

Right now my postulate is that each layer of the winding can only take as much force as there is friction between it and the layer beneath it, as otherwise slip would occur and the bottom layer of the cord would be loaded beyond its breaking point.

However I am not certain how to calculate the friction between layers as each layer adds additional compression load (towards the center of the circle/bobbin) thus increasing max friction force between the two layers bellow.

It seems to me that I need to obtain dFfriction/dX, but I am beginning to think this is an iterative problem, as a continuous function cannot be obtained*, rather this dFfriction/dX changes at each 360° interval (thus with each new layer).

Calculating in this manner I think it would be best done by starting from the most top layer (if I choose the winding tension force equal to cords tensile strength this should yield max wall tensile strength) and progressing down. But I don’t know how to calculate the friction between layers, I have been looking at the capstan equation (here: http://en.wikipedia.org/wiki/Belt_friction) but I’m not sure it is directly applicable.

*A continuous function can be obtained by using the Archimedean spiral equation but in that case I don’t know how to obtain dFfriction/dX.

I have been searching through the internet and books for weeks to find a similar problem but I failed to find one.
If I take the capstan equation for a reference I wander if the friction is so high that the wall (winding) tensile strength is simply equal to 0.99*cord_tensile_strength*A/Ag
(“A” being the wall cross-section area and “Ag” being the area of the gaps in that same cross-section).

I will check the book.

4. Jan 7, 2014

### nvn

Then why does post 1 say, "What is the maximum pressure on the bobbin internal radius (r)?" But perhaps you meant to say what is the hoop stress on the bobbin inner radius (r), instead of "pressure," because air pressure inside the bobbin is 0 MPa (gauge), atmospheric pressure. Therefore, one of the questions given in post 1 seems to say, determine the hoop stress on the bobbin inner radius (r).

How many layers of cord are assumed to be on the spool? And how many turns per layer?

Are r and R the inner and outer radii of the spool steel wall? Or is r the radius to the centerline of the first (inner) cord layer, and R the radius to the centerline of the last (outer) cord layer? Or is r the inner radius of the first cord layer, which would be the outer radius of the steel spool?

Last edited: Jan 7, 2014
5. Jan 7, 2014

### nvn

strive: I went ahead and assumed R = steel spool outer radius. The spool axial length relative to t is a1/t, where t = cord diameter. For simplicity, always ensure a1/t is an integer. Then, r1 = R + 0.5*t + (nn - 1)*t = R + (nn - 0.5)*t, where nn = number of cord layers = 1 - floor[L*t/(2*pi*r1*a1)]. For simplicity, for now, ensure nn = 1.

For an example, I assumed t = 2.50 mm, R = 50 mm, a1 = 12.5 mm, L = 1400 mm, Ctb = 0.20, Stu = cord tensile ultimate strength = 550 MPa, E = 200 000 MPa, A = cord cross-sectional area = 4.9087 mm^2, T2 = cord applied tensile load = Stu*A = 2700 N, and T1 = cord holding tensile force.

Therefore, for nn = 1, then r1 = R + (nn - 0.5)*t = 51.25 mm. Check, nn = 1 - floor[1400*2.5/(2*pi*51.25*12.5)] = 1 - floor(0.8695) = 1 - 0 = 1 layer. And theta = L/r1 = 1400/51.25 = 27.3171 rad.

Compute cord elongation, delta = [T2*r1/(E*A*Ctb)]*[1 - e^(-Ctb*theta)] = [2700*51.25/(200 000*4.9087*0.20)]*[1 - e^(-0.20*27.3171)] = 0.7018 mm. Compute corrected cord length, L = L + delta = 1400.7018 mm. Compute corrected theta, theta = L/r1 = 1400.7018/51.25 = 27.3308 rad.

Therefore, T1 = T2/[e^(Ctb*theta)] = 2700/[e^(0.20*27.3308)] = 2700/236.5501 = 11.41 N.

The problem continues similarly, no matter how many cord layers there are. For nn > 1, cut the cord at the end of each layer, then solve each layer as a separate problem, starting with the last (outer) layer. T1 at the beginning of a layer becomes T2 at the end of the preceding (underlying) layer. For each layer, compute r1. For nn = 1, use Ctb; for nn > 1, use Ctc.

For steel cable, you can just ignore the elongation (i.e., skip paragraph 4 altogether), because the elongation makes a negligible difference in the answer.

Last edited: Jan 7, 2014
6. Jan 8, 2014

### strive

Thank you, this is extremely helpful. I think I truly understand the problem now.

I apologize for initial complications with the bobbin, apparently I did not understand the problem well enough to describe it clearly.
I will now try to expand this to a 3D problem with slight elastic deformation. As the principle is the same it should not be a problem.

Thank you very much

7. Jan 8, 2014

### nvn

strive: I made a typographic mistake in the equation that uses floor() in post 5. However, you do not need that equation anyway, so just throw it out. Also, unlike what I said in the third sentence of post 5, parameter a1 can be any value you desire; i.e., a1/t does not need to be an integer.

In post 5, I forgot to take into account the fact that cords in layers after the first layer sit in "valleys" formed by the underlying layer cords. Therefore, the revised r1 equation is listed below, where ii = cord layer number = (1, 2, ..., nn), and nn = outermost cord layer number. Also, the number of cord turns in each full layer is listed below. Notice, every even-numbered full layer contains one less turn than odd-numbered full layers, due to the "valleys," which cause staggering of the cord.

r1ii = R + 0.5*t + (ii - 1)*0.8660*t
turnsii = (a1/t) - 1 + mod(ii,2)​

You can now compute the length of cord in each full layer:

L0ii = turnsii*2*pi*r1ii

You can now compute the summation of all L0ii values; i.e., summation(L0ii). Then compute the length of cord in the outermost layer:

Ltemp = L0nn - [summation(L0ii) - L]​

You can now replace L0nn with Ltemp, which is the length of unstrained cord in the outermost layer:

L0nn = Ltemp​

You can now compute the radians in each cord layer:

theta1ii = L0ii/r1ii

Now that you have the radians in each layer, theta1ii, you can compute T1 and T2 at the ends of each layer, as described in post 5, paragraph 6. After that, you could compute the elongation in each layer, if you wish, and the summation of elongation, then add the total elongation to L, then repeat the above calculations (in post 7), to obtain the corrected number of radians in the outermost layer. Then compute, again, T1 and T2 at the ends of each layer.

For steel cable, you could just skip computing the elongation, because it probably makes a negligible difference in the answer.

Last edited: Jan 8, 2014
8. Jan 9, 2014

### SlideRuler

Yarn on Bobbin Forces

In yarn mechanics, its convenient to use the normal force per length "n" rather than the pressure.
If a yarn is wound on a bobbin of radius "R at tension "T",
n=T/R
(This can easiloy be proved by writing the force equilibrium equation of a section of arc.)

The total normal force "N" on the bobbin will be the length of yarn "L" times the normal force per length "n" - assuming that the yarns are wound perpendicular to the bobbin axis.
N=nL
N=TL/R
The pressure "p" is the total normal force divided by the bobbin area "A". If "H" is the axial wind length
A = 2 pi R H
p = N/A
Combining the equations gives
p = TL / (2 pi R^2 H)

Complicating factors solved in the literature are: multiple layers, angled yarns, deformable yarns, deformable bobbins,.....

Last edited: Jan 9, 2014
9. Jan 9, 2014

### Staff: Mentor

Hi SlideRuler. Welcome to Physics Forums!!!!
Very nice analysis.

10. Jan 9, 2014

### nvn

strive: Post 1, in the above quote, states a cord is wound on a bobbin, but with what cord tensile force during the winding?

(1) Is the cord tensile force T2 during the entire winding process?

(2) Or is tensile force T2 applied after the essentially-unstrained cord has been wound onto the bobbin?​

Notice, posts 5 and 7, and the links in posts 2 and 3, address item 2, whereas post 8 addresses item 1. Two different problems. So it depends on which problem you are trying to address in your question. Is it item 1 or item 2?

11. Jan 10, 2014

### strive

Hi

Item 2 is (rather was) initially intended, but as i mentioned in post 1 some pretension will mean less cord is necessary. Thus i need to solve both items and compare the results.

This is sort of what i am currently working with. (i hope the picture explains my thoughts)
Can you tell more about complicating factor: multiple layers? Or where to find some material on it?

My point is that each layer of winding exerts pressure (acts with a force F) on underlying layers. This increases the force of static friction with each added layer. This means that at some point the force of friction in a layer equals the breaking force of the cord (ultimate tensile strength of the cord). At this point slip can no longer occur thus that layer together with each underlying layer sort of acts like a uniform wall (at this point I am casting aside the “porous” structure of the “wall”).
Of course without pre-tension force there can be no friction force… thus this would describe “item 1”.

However if the end of the cord in the outermost layer is statically fixed (loaded with “the breaking force”) and the winding was wound with absolutely 0 pre-tension force (setting aside that this is mechanically impossible) when the pressure (“p” in the picture) increases the cord is again compressed and the result is the same as in “item 1”.
The only problem here is that because nothing is perfect the cord layers would not be compacted evenly thus uniform “wall” strength would not be achieved (but this is ignored for now).

Please tell me if my postulate about the “uniform wall” is incorrect or somehow flawed.
So far I have been trying to determine at which point the friction force and breaking force equal.

I have this (but it can’t be correct):

dc=1; % cord diameter [mm]
sig=33; % cord material tensile strength [Pa]
ctc=0.2; % cord on cord friction coefficient [/]
ctb=0.16; % cord on bobbin friction coefficient [/]
vpt=0.8; % cord pretension safety factor [/]

Fpt=((dc^2)/4)*(sig/(1e6))*vpt % cord winding pre-tension force[N]

Ff1r=Fpt/exp(2*ctc) % tensile force on the cord at the end of layer 1 (outermost layer) [N]
Ff1=Fpt-Ff1r % friction force in cord layer 1 [N]
Fc1=Ff1 % useful load on the cord layer 1 [N]

Ff2r=(2*Fpt)/exp(2*ctc) % tensile force on the cord at the end of layer 1 (outermost layer) [N]
Ff2=(2*Fpt)-Ff2r % friction force in cord layer 2 [N]
Fc2=Ff1+Ff2 % useful load on the cord layer 2 [N]

Ff3r=(3*Fpt)/exp(2*ctc) % tensile force on the cord at the end of layer 1 (outermost layer) [N]
Ff3=(3*Fpt)-Ff3r % friction force in cord layer 2 [N]
Fc3=Ff1+Ff2+Ffr % useful load on the cord layer 2 [N]

The pattern of the last 6 rows is repeated until condition “Fpt = Ff(index)” is achieved.

(note: 2 radians are taken by default as I am currently calculating only for 1 row – sort of a 2D problem)
(note 2: values of variables are arbitrary as the problem is theoretical)

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Last edited: Jan 10, 2014
12. Jan 11, 2014

### nvn

Last edited: Jan 11, 2014
13. Jan 12, 2014

### strive

Yes.

Thus it should state:

Ff(ii)r = (ii*Fpt) / exp(2*pi*ctc)

ii – layer number (outermost being 1)

Sorry about that... apparently i forgot to type it in the first line and then copy-pasted it 3 times.

Last edited: Jan 12, 2014
14. Jan 15, 2014

### SlideRuler

Winding Tension.
In a commercial winding process, the yarn tension is controlled before it touches the bobbin.

Yarn Tension within Bobbin.
Many layers are usually wound on a bobbin. The tension of the outer layers reduces the tension of the inner layers - often to the point of compression and yarn buckling. Friction is NOT important.

15. Jan 15, 2014

### Staff: Mentor

Am I correct in saying that the reduction in tension of the inner layers is the result of the radial compressive stress from the outer layers and the associated Poisson effect?

16. Jan 16, 2014

### strive

Well that is definitely the main factor. However I wonder if it is the only one.

SlideRuler, could we get some actual case values so we can check?

But regarding friction: if you linearize the problem (just straighten the circle/hoop), the friction actually acts as glue between individual plates, thus it spreads load from 1 plate (the innermost layer) to all other adjacent plates (layers).
Unless I have incorrectly linearized…?
Take a plywood bobbin as an example. Is it not the same?