jerryv
Mechanical
- Nov 10, 2000
- 22
I need a formula that help me calculate sag in a steel wire due to it’s own weight over a given length.
Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations TugboatEng on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
Wire Sag 3
- Thread starter jerryv
- Start date
- Status
-
1
- #2
- Thread starter
- #3
Look in Mark's Standard Handbook for the Catenary curve. The curve represents a "curve in which a flexible chain or cord of uniform density will hang when supported by the two ends". This means that no stiffness is assumed, hence no Modulus of Elasticity.
I have used this to also model straps that are restraining uniformly distributed loads.
A short span of steel wire with no tension will act like a beam, while a long span will act like a catenary. Something in between will be half and half. The point where one becomes another depends on the thickness of the wire relative to the span.
STF
I have used this to also model straps that are restraining uniformly distributed loads.
A short span of steel wire with no tension will act like a beam, while a long span will act like a catenary. Something in between will be half and half. The point where one becomes another depends on the thickness of the wire relative to the span.
STF
-
2
- #5
EnglishMuffin
Mechanical
The modulus of elasticity is a factor ! The equations you need can be found in Roark's Formulas for Stress and Strain.
ymax = l*(3*w*l/(64EA))^(1/3)
P = w*l^2/(8*ymax)
Where ymax is the maximum sag
l is the unstretched length
w is the weight/unit length
E Youngs modulus
A Wire area
P axial tension
Rich2001 is correct in saying that the curve is a catenary
One interesting point is that the cables on suspension bridges do not hang in catenaries, but parabolas (because of the horizontal road weight).
ymax = l*(3*w*l/(64EA))^(1/3)
P = w*l^2/(8*ymax)
Where ymax is the maximum sag
l is the unstretched length
w is the weight/unit length
E Youngs modulus
A Wire area
P axial tension
Rich2001 is correct in saying that the curve is a catenary
One interesting point is that the cables on suspension bridges do not hang in catenaries, but parabolas (because of the horizontal road weight).
EnglishMuffin
Mechanical
Actually, more correctly should have said that P is the horizontally applied stretching force at the ends.
EnglishMuffin
Mechanical
I should further add that although the curve is in reality a catenary, the two equations I have given appear to be derived on the basis that the curve is a parabola, which is a good approximation if the depth of sag is no greater than 1/8th of the span.
It appears to me that you can use the first equation to determine what the sag would be in a cable that had zero inital tension (this obviously depends on E).
You can then use the second equation to derive the induced tension P. I should note that according to a second reference that I have for the second formula (Blevins formulas for natural frequency), P is not horizontal but is the assumed mean tension in the wire. This tension is actually not constant along the wire, but can be assumed so for shallow sags.
It further appears that one can use the second equation for any tension P.
Blevins also gives this equation:
ymax/L = (5/24)^(1/2)*(1-(1-18/5*(Lc-L)/L)^(1/2))^(1/2)
- where L is the span as before, and Lc is the unstressed cable length. You can use this is conjunction with the formula for P. That should give you everything you need to reduce your sag as required..
It appears to me that you can use the first equation to determine what the sag would be in a cable that had zero inital tension (this obviously depends on E).
You can then use the second equation to derive the induced tension P. I should note that according to a second reference that I have for the second formula (Blevins formulas for natural frequency), P is not horizontal but is the assumed mean tension in the wire. This tension is actually not constant along the wire, but can be assumed so for shallow sags.
It further appears that one can use the second equation for any tension P.
Blevins also gives this equation:
ymax/L = (5/24)^(1/2)*(1-(1-18/5*(Lc-L)/L)^(1/2))^(1/2)
- where L is the span as before, and Lc is the unstressed cable length. You can use this is conjunction with the formula for P. That should give you everything you need to reduce your sag as required..
Line sag (ITT Ref. Data with supports at same elev.):
H:=WL^2/8S
S:=WL^2/8H
L: distance between support
S: Sag at mid-span
H: tension (horzontal component of tension)
W: weight of cable per unit length
(parabolic approx.)
The modulus of elasticity is only important where cable stiffness is sufficient to justify its modeling as a beam rather than a cable.
H:=WL^2/8S
S:=WL^2/8H
L: distance between support
S: Sag at mid-span
H: tension (horzontal component of tension)
W: weight of cable per unit length
(parabolic approx.)
The modulus of elasticity is only important where cable stiffness is sufficient to justify its modeling as a beam rather than a cable.
EnglishMuffin
Mechanical
Hacksaw :
Your equation for tension is identical to mine.
I do not agree with your final statement however.
My first reply should probably have read "modulus of elasticity can be a factor, depending on how the problem is formulated".
I provided three equations, which all relate to either extensible or inextensible strings (or cables). None of them have anything to do with beams. Usually, the term "beam" relates to a bar which can resist bending and/or shear.
The first equation applies to the case where an extensible string (or cable) of initial length LC is fixed between two supports distant L apart, and LC = L.
In this case, ymax = L*(3*w*L/(64EA))^(1/3)
The sag is directly related to E
The second equation applies to the case of an inextensible string, where LC > L
In this case,ymax = L*(5/24)^(1/2)*(1-(1-18/5*(LC-L)/L)^(1/2))^(1/2)
This equation could easily be extended to the case of an extensible string (or cable), in which case you would replace LC with LC*(1+P/(A*E)) where P is given by the next equation - you would then have to iterate to find the solution, but in most cases it would hardly be worth it.
The third equation, which is the same as yours, gives the tension in terms of the sag - and is valid for all cases, as I stated in my second post -including either of the above cases, or the case where the tension has been increased to a higher value.
P = w*l^2/(8*ymax)
So it all depends how the problem is formulated. If the tension is specified, all you need is eqn 3.
If the length of the string (or cable) is specified, then you need eqn 1 or 2 as well.
Your equation for tension is identical to mine.
I do not agree with your final statement however.
My first reply should probably have read "modulus of elasticity can be a factor, depending on how the problem is formulated".
I provided three equations, which all relate to either extensible or inextensible strings (or cables). None of them have anything to do with beams. Usually, the term "beam" relates to a bar which can resist bending and/or shear.
The first equation applies to the case where an extensible string (or cable) of initial length LC is fixed between two supports distant L apart, and LC = L.
In this case, ymax = L*(3*w*L/(64EA))^(1/3)
The sag is directly related to E
The second equation applies to the case of an inextensible string, where LC > L
In this case,ymax = L*(5/24)^(1/2)*(1-(1-18/5*(LC-L)/L)^(1/2))^(1/2)
This equation could easily be extended to the case of an extensible string (or cable), in which case you would replace LC with LC*(1+P/(A*E)) where P is given by the next equation - you would then have to iterate to find the solution, but in most cases it would hardly be worth it.
The third equation, which is the same as yours, gives the tension in terms of the sag - and is valid for all cases, as I stated in my second post -including either of the above cases, or the case where the tension has been increased to a higher value.
P = w*l^2/(8*ymax)
So it all depends how the problem is formulated. If the tension is specified, all you need is eqn 3.
If the length of the string (or cable) is specified, then you need eqn 1 or 2 as well.
EnglishMuffin
Mechanical
Thanks hacksaw - very flattering and thanks for the star!
Did you realise there is another thread on this subject with different answers ?
thread507-60466
Did you realise there is another thread on this subject with different answers ?
thread507-60466
I have nothing to add to this post's technical content, but I would like to provide an editing tip: you can create subscripts and superscripts by using the code described in the Process TGML link below in Step 2 Options of the reply area. For example the equation given by EnglishMuffin could look like this:
ymax = L &[ignore]middot[/ignore]; (5/24)0.5 &[ignore]middot[/ignore]; (1 - (1 - 18/5 &[ignore]middot[/ignore]; (LC-L)/L)0.5)0.5
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
ymax = L &[ignore]middot[/ignore]; (5/24)0.5 &[ignore]middot[/ignore]; (1 - (1 - 18/5 &[ignore]middot[/ignore]; (LC-L)/L)0.5)0.5
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
![[hammer]](/data/assets/smilies/hammer.gif "[hammer] [hammer]")
I really shot from the hip in my first post.
EnglishMuffin
Mechanical
Thanks CoryPad :
I'm spending too much time writing and not enough reading !
I'm spending too much time writing and not enough reading !
- Thread starter
- #15
jerryv
Mechanical
- Nov 10, 2000
- 22
Thanks EnglishMuffin for all your input. I tried to put your formula into an Excel spreadsheet and I am having some difficulties with the results. It appears that no matter what diameter I use the Ymax stays the same. Any thoughts.
Wire Sag
Ymax: 7.223390595 Max. Sag l*(3*w*l/(64*E*A))^(1/3)
l: 960 Is the Unstretched Length
Wire dia. 0.015
w: 5.01869E-05 Is the weight/unit length (steel) (PI*((Wire Dia/2)^2)*0.284)
E: 30000000 Youngs Modulus
A: 0.000176715 Wire Area PI*((Wire Dia./2)^2)
P: 0.800390857 Axial Tension w * I^2 /(8*ymac)
Wire Sag
Ymax: 7.223390595 Max. Sag l*(3*w*l/(64*E*A))^(1/3)
l: 960 Is the Unstretched Length
Wire dia. 0.015
w: 5.01869E-05 Is the weight/unit length (steel) (PI*((Wire Dia/2)^2)*0.284)
E: 30000000 Youngs Modulus
A: 0.000176715 Wire Area PI*((Wire Dia./2)^2)
P: 0.800390857 Axial Tension w * I^2 /(8*ymac)
Here is the first set of circumstances that JerryV and i are attempting to solve.
A band of 7 clamped cables are pulled 960" from a locked multi-cable "precision" metered source which can be assumed fixed. The clamped group (to negate cable to cable shifting) is placed under 5.25 lbs. of tension. It is observed that one cable is the tightest, with a slight sag. The next cable is 1/2" from the first at the mid point (ymax), the next is 1" from the first (1/2 from the second), fourth is 1.5", fifth is 2.0", sixth is 2.5", and the seventh is 3.0". This creates a 1/2" gap between each cable.
These gaps translate to approximately .000725" in longer lenght per .5" gap. (this was determined using a basic arc from the 960" end points, this approx is probably debatable as well).
Given .015" diameter cable. .284 lb/in3. and P = w*l^2/(8*ymax). Assumption that the tight wire sees all of the 5.25 tension.
It was determined that the top wire sag was .01096".
There is debate as to why E is not necessary here.
It seems to me that there would be a variety of regimes for consideration. Because the wire is not in tension to the point where elongation would occur, then i would not think E is relative in the "Undeformed Regime".
I would think that the sag of the first cable can be used as the sag of the group. If additional tension is added to reduce the sag, at what point can it be determined that the "short" cable is under tension and entering a "elastic reginme" where E becomes a factor and elongation can be calculated to the accumulated point that all of the end are tight and of the same length as the long end.
Your thoughts are appreciated, valued, and applied.
My general thought is how the stated equations relate to this specific scenario.
A band of 7 clamped cables are pulled 960" from a locked multi-cable "precision" metered source which can be assumed fixed. The clamped group (to negate cable to cable shifting) is placed under 5.25 lbs. of tension. It is observed that one cable is the tightest, with a slight sag. The next cable is 1/2" from the first at the mid point (ymax), the next is 1" from the first (1/2 from the second), fourth is 1.5", fifth is 2.0", sixth is 2.5", and the seventh is 3.0". This creates a 1/2" gap between each cable.
These gaps translate to approximately .000725" in longer lenght per .5" gap. (this was determined using a basic arc from the 960" end points, this approx is probably debatable as well).
Given .015" diameter cable. .284 lb/in3. and P = w*l^2/(8*ymax). Assumption that the tight wire sees all of the 5.25 tension.
It was determined that the top wire sag was .01096".
There is debate as to why E is not necessary here.
It seems to me that there would be a variety of regimes for consideration. Because the wire is not in tension to the point where elongation would occur, then i would not think E is relative in the "Undeformed Regime".
I would think that the sag of the first cable can be used as the sag of the group. If additional tension is added to reduce the sag, at what point can it be determined that the "short" cable is under tension and entering a "elastic reginme" where E becomes a factor and elongation can be calculated to the accumulated point that all of the end are tight and of the same length as the long end.
Your thoughts are appreciated, valued, and applied.
My general thought is how the stated equations relate to this specific scenario.
EnglishMuffin
Mechanical
jerryv - no I think you are quite correct.
You will find this formula in Roark, for a cable with a distributed load on it, and I suppose the only time the wire diameter would actually come into it would be if the weight per unit length were independent of the wire - this would apply in the case of an axially stiff wire with a heavy coating on it, that itself had a very low youngs modulus.
You will find this formula in Roark, for a cable with a distributed load on it, and I suppose the only time the wire diameter would actually come into it would be if the weight per unit length were independent of the wire - this would apply in the case of an axially stiff wire with a heavy coating on it, that itself had a very low youngs modulus.
EnglishMuffin
Mechanical
ZackS
As far as the equations are concerned, if you know the tension, you know ymax, and vica versa. It doesn't matter whether the cable is axially stretched or not, or what the value of E is. The tension depends purely on the loaded geomety and the weight per unit length. Think about the equilibrium of a small element of cable. Bear in mind that all these equations are approximate, based on a parabola rather than a catenary - which is the true shape - but the error is minute with these small sags. They also assume that the tension is constant along the length, which is also not true exactly - but again, the error is minute for small sags. I will study your problem and see whether I understand what you have - in the meantime, someone will probably beat me to it - I'm going to the local Mexican restaurant.
As far as the equations are concerned, if you know the tension, you know ymax, and vica versa. It doesn't matter whether the cable is axially stretched or not, or what the value of E is. The tension depends purely on the loaded geomety and the weight per unit length. Think about the equilibrium of a small element of cable. Bear in mind that all these equations are approximate, based on a parabola rather than a catenary - which is the true shape - but the error is minute with these small sags. They also assume that the tension is constant along the length, which is also not true exactly - but again, the error is minute for small sags. I will study your problem and see whether I understand what you have - in the meantime, someone will probably beat me to it - I'm going to the local Mexican restaurant.
EnglishMuffin
Mechanical
ZackS :
Let me see if I understand this correctly - I'm a little slow on the uptake, and without a drawing its tough to be sure that I am visualizing it correctly. Can I assume we are effectively talking about this :
The first wire (the shortest) is stretched between two points 960 inches apart. The sag of this wire is .011". Six other wires are connected between these two points, having sags from the horizontal of 1/2", 1", 1.5", 2",2.5", 3".
There is a horizontal force of 5.25 lbf applied between the points.
I am not clear as to whether the .011" is measured, required, or calculated.
I am not clear whether the 5.25 lbf is measured or required.
I am also not clear what it is you want to calculate. The sags, if known, enable the tension at the end of each cable to be computed. By figuring the slopes at the ends we can compute the necessary total axial force from superposition. It should come out to the measured value - if known.
First - can I get the above clarified ?
And I would not assume at the outset that most of the 5.25 lbf is resisted by tension in the first wire, but it may be so.
Let me see if I understand this correctly - I'm a little slow on the uptake, and without a drawing its tough to be sure that I am visualizing it correctly. Can I assume we are effectively talking about this :
The first wire (the shortest) is stretched between two points 960 inches apart. The sag of this wire is .011". Six other wires are connected between these two points, having sags from the horizontal of 1/2", 1", 1.5", 2",2.5", 3".
There is a horizontal force of 5.25 lbf applied between the points.
I am not clear as to whether the .011" is measured, required, or calculated.
I am not clear whether the 5.25 lbf is measured or required.
I am also not clear what it is you want to calculate. The sags, if known, enable the tension at the end of each cable to be computed. By figuring the slopes at the ends we can compute the necessary total axial force from superposition. It should come out to the measured value - if known.
First - can I get the above clarified ?
And I would not assume at the outset that most of the 5.25 lbf is resisted by tension in the first wire, but it may be so.
EnglishMuffin
Mechanical
Actually - we'd beter cut to the chase here
You state the following :
Diameter of cable : .015"
Density :.284 lb/in3
Cable length : 960 in
This means that the weight per inch of the cable is 5.02*10^-5
Which means that for a sag of 0.01" you would need a tension of :
P = 5.02*10^-5*960^2/(8*.01) = 578 lbf
Where are you getting the 5.25 lbf from ?
You state the following :
Diameter of cable : .015"
Density :.284 lb/in3
Cable length : 960 in
This means that the weight per inch of the cable is 5.02*10^-5
Which means that for a sag of 0.01" you would need a tension of :
P = 5.02*10^-5*960^2/(8*.01) = 578 lbf
Where are you getting the 5.25 lbf from ?
- Status
Similar threads
- Locked
- Question
- Question