Starian = (length/initial-length) -1 Let forces are acting on a spring at both ends and no end is fixed. Each node can move. for example ... F1 o----k------o F2 How much force (which forc) is will stretch the spring? What will be starin corresponding to that force ? As i under stand..... 1: If both F1 & F2 are in same direction then only nodes will move in that direction. There will be no stretch. It will be just pulling the spring & only position of nodes is shifted. 2: If one of the 2 forces is zero. Then there will be no stretch and only shifting. 3: If left end is fixed then F2 = -K * dL (for spring) 4: If Both forces are in opposite direction then spring will stretch. What is starain when F1=F2 , F1>F2 & F1<F2 ? ? Is my understanding is correct ? May I apply same technique to understand String or Rigid beam? What will happand if i use the multiple springs like F1 o----k1---o----k2-----o-----k3----o F2 Any comments or mathematical relations(prefered) will be helpful for me. Any link/site regarding it is alao Ok. Regards Alam -- Learning never ends --

quote:

---

Let forces are acting on a spring at both ends and no end is fixed. Each node can move. For example ...
F1 o----k------o F2
How much force (which forc) is will stretch the spring?

---

Well... Both forces will stretch the spring. You need two forces to stretch the spring. If one of the forces is 0, then as you said there will be no stretch.

quote:

---

What will be starin corresponding to that force ?

---

The elongation of a spring is directly proportional to the force that acts upon it:
Elongation = constant · Force
And Elongation is DeltaLength:
StretchedLength = UnloadedLength + Elongation
I don''t remember a lot about strain though... And my books are at school

quote:

---

As i under stand.....
1: If both F1 & F2 are in same direction then only nodes will move in that direction. There will be no stretch. It will be just pulling the spring & only position of nodes is shifted.
2: If one of the 2 forces is zero. Then there will be no stretch and only shifting.
3: If left end is fixed then F2 = -K * dL (for spring)

---

Because if the left end is fixed (i.e. to a wall) and you pull on the string, then the wall exerts an equally great force (giving the F1) as you upon the string (but of opposite direction/sign compared to F2). The wall holds the spring back.
And there you have the equation again:
Elongation = constant · Force

quote:

---

4: If Both forces are in opposite direction then spring will stretch. What is starain when F1=F2 , F1>F2 & F1

---

Do you just want to know these 2d cases (where the forces are parallel to the spring)? It''s more complicated to explain it in 3d.
If F1 and F2 is equally strong, well… We’ve covered that one before…
If F1 is bigger than F2… Imagine dividing F1 into two parts. One is as big as F2 (opposite sign), the other the ’surplus’. The balanced pair of forces will stretch the string while the surplus force will accelerate the spring to the left.
The case when F2 is bigger than F1 is really just the same, but the spring will start moving to the right instead.

quote:

---

Is my understanding is correct ?
May I apply same technique to understand String or Rigid beam?

---

I don''t really see how you could apply this on rigid beams. Could you elaborate? But for strings, you’re right on. Hugo Elias has a tutorial on strings, which are really just a lot of interconnected springs. I don''t know if you''ve seen it.

quote:

---

What will happand if i use the multiple springs like
F1 o----k1---o----k2-----o-----k3----o F2

---

Speaking freely... I guess that you''re trying to implement some sort of spring/string system... This is the way I''d do it. Known are the position of each spring''s end points, the spring constant and the ”mass” of each spring end point (or node, as you call them).
Every node or end point is connected to two other neighbor nodes via a spring.
pseudo-code

for every spring{Calculate distance to neighbor point 1Force1 = constant · distance1Calculate distance to neighbor point 2Force2 = constant · distance2SumForce = Force1 + Force2MovingEndPointPosition += MovingEndPointVelocity · deltaTime + SumForce/MassOfEndPoint · deltaTime · deltaTime / 2MovingEndPointVelocity += SumForce/MassOfEndPoint · deltaTime} 

(Note: the above is not 100 % usable. You really should calculate the forces and movements of all nodes before actually moving/updating the position of any node)
Post again if you have any questions!
/Ksero

Thanks for ur kind response.

F1 o----k------o F2
Well... Both forces will stretch the spring. You need two forces to stretch the spring. If one of the forces is 0, then as you said there will be no stretch.
***
Yes both will stretch. I specifically want to know in terms of F1 & F2 forexample (F1+F2) or (F1-F2) so on .....?

*****-----
Because if the left end is fixed (i.e. to a wall) and you pull on the string, then the wall exerts an equally great force (giving the F1) as you upon the string (but of opposite direction/sign compared to F2). The wall holds the spring back.
And there you have the equation again:
Elongation = constant · Force
-------
How this recation force by wall be related to Elongation? Is it equal to F2(opposite direction)?


-----------------------------------------------------------------4: If Both forces are in opposite direction then spring will stretch. What is starain when F1=F2 , F1>F2 & F1
-----------------------------------------------------------------
Do you just want to know these 2d cases (where the forces are parallel to the spring)? It''s more complicated to explain it in 3d.
If F1 and F2 is equally strong, well… We’ve covered that one before…
If F1 is bigger than F2… Imagine dividing F1 into two parts. One is as big as F2 (opposite sign), the other the ’surplus’. The balanced pair of forces will stretch the string while the surplus force will accelerate the spring to the left.
The case when F2 is bigger than F1 is really just the same, but the spring will start moving to the right instead.
****
Yes i am talking about axial force/stretch and taking force along string.
You mentioned that ballance pair of forces is stretching the string.
Is it menas Elongation = constant* Blanace force ?
Then what will be this force in terms of F1 & F2(if F1>F2 then it will be 2*F2 ?) ?

*****------
I don''t really see how you could apply this on rigid beams. Could you elaborate? But for strings, you’re right on. Hugo Elias has a tutorial on strings, which are really just a lot of interconnected springs. I don''t know if you''ve seen it.
****
In above tutorial they defined the term xv & yv. I think that are velocity in x & y direction. But can''t understand how he derive this relation ?
X1 = Distance between node(i-1) and Node(i)
X2 = Distance between node(i+1) and Node(i)
xv = X1/(Length1 * Ext1)+ X2/(Length2 * Ext2) ??????

-----------------------------------------------------------------

What will happand if i use the multiple springs like
F1 o----k1---o----k2-----o-----k3----o F2
****-------
Speaking freely... I guess that you''re trying to implement some sort of spring/string system... This is the way I''d do it. Known are the position of each spring''s end points, the spring constant and the ”mass” of each spring end point (or node, as you call them).
Every node or end point is connected to two other neighbor nodes via a spring.

*****
Yes i going to make spring-system. But i m going towards opposite diection. I know the forces F1 & F2 and the intiial position of each node. Forces r same as case of only one string, now divided to N-strings. My aim is to find the elaongation in each string.
But i don''t know the force at each node. If that is known then elongation can be found?
What will be force at each node (it is the sharing F1 & F2 how?)?

****--------
pseudo-code
*****
Is this code is for each node ? (not for each spring)

I hope our discussion will be continue...
Regards

Alam
-- Learning never ends --

Do you want to examine a static system that has reached equilibrium, or do you want to simulate a dynamic system with moving springs?
In systems that have reached equilibrium, F1 = F2. If you disturb this, then the system will eventually settle down into a new balanced state. Before that, the elongation will vary with time.

Perhaps you should think that elongation has nothing whatsoever to do with forces. Rather, it’s about where the spring’s endpoints (nodes) are placed in space. The endpoints can be moved, and are influenced by several forces. The string stretches enough so that it begins in one node and ends in the other.
While thinking about this, I had a revelation. You need to separate the contraction force of the spring (which is always proportional to elongation) and the outside pulling force.

(note: when drawing this, the spring was supposed to be moving leftwards, hence the “pull” and “brake”-force)
The two contraction forces are *always* equally long (force = elongation/constant. It doesn’t matter which force, as both the forces are equally long).
Then you look at each node and sums the forces applied to it. Then a=F/m. and v=v0 + a·t.

If you have several springs:

You see that the springs pull each other out. The contracting force of the left spring pulls on the right spring. The contracting force of the right one pulls on the left one.

quote:

---

F1 o----k------o F2
Well... Both forces will stretch the spring. You need two forces to stretch
the spring. If one of the forces is 0, then as you said there will be no
stretch.
***
Yes both will stretch. I specifically want to know in terms of F1 & F2
forexample (F1+F2) or (F1-F2) so on .....?

---

Assuming the system has reached equilibrium, F1=F2. As the forces are equally big, it doesn’t matter what force you choose for inserting into the formula (when I was studying this in school, I too wondered why you didn’t take both forces into calculation, like F1 + F2... But eventually, I figured this out). As I said, if the system has not reached equilibrium, the stretch will be a function depending on time, eventually settling down in a balanced state.

quote:

---

*****-----
Because if the left end is fixed (i.e. to a wall) and you pull on the
string, then the wall exerts an equally great force (giving the F1) as you
upon the string (but of opposite direction/sign compared to F2). The wall
holds the spring back.
And there you have the equation again:
Elongation = constant · Force
-------
How this recation force by wall be related to Elongation? Is it equal to
F2(opposite direction)?

---

Yes it is, Most of the time (when the system is in an equilibrium).

quote:

---

Yes i am talking about axial force/stretch and taking force along string.
You mentioned that ballance pair of forces is stretching the string.
Is it menas Elongation = constant* Blanace force ?
Then what will be this force in terms of F1 & F2(if F1>F2 then it will be
2*F2 ?) ?

---

Actually, in hindsight, I don’t believe that my analogy with the balanced pair of forces was that good. Sorry for misleading you...
If you want to calculate (i.e. not simulate) the equilibrium elongation, then check out the picture above… When the system is in balance, the acceleration of the left node is equal to the acceleration of the right node. If the masses of the endpoints are equal, that means that the forces that act upon the endpoints are equal. So we’ve got (forces pointing left are positive):
Pull Force – Contraction Force = Contraction Force – Brake Force
(Pull Force + Brake Force)/2 = Contraction Force = Elongation/constant

quote:

---

In above tutorial they defined the term xv & yv. I think that are velocity
in x & y direction. But can''t understand how he derive this relation ?
X1 = Distance between node(i-1) and Node(i)
X2 = Distance between node(i+1) and Node(i)
xv = X1/(Length1 * Ext1)+ X2/(Length2 * Ext2) ??????

---

Actually, as he hasn’t put any parenthesis in that expression, I think he means:

xv = Ext1 · X1/Length1+ Ext2 · X2/Length2 

He normalises the vectors to the next & previous node (X/length) so that he gets the direction that the force pulls toward. Then he scales this by the elongation, giving the force(s) (proportional to elongation).
The “massless” code is not really physically correct (but it might look OK...). He scales xv and then just adds it to the x-coordinate. He has another version, where he stores the velocities of all points, giving an accurate simulation.

quote:

---

****-------
Speaking freely... I guess that you''re trying to implement some sort of
spring/string system... This is the way I''d do it. Known are the position
of each spring''s end points, the spring constant and the "mass" of each
spring end point (or node, as you call them).
Every node or end point is connected to two other neighbor nodes via a
spring.
*****
Yes i going to make spring-system. But i m going towards opposite diection.
I know the forces F1 & F2 and the intiial position of each node. Forces r
same as case of only one string, now divided to N-strings. My aim is to
find the elaongation in each string.
But i don''t know the force at each node. If that is known then elongation
can be found?
What will be force at each node (it is the sharing F1 & F2 how?)?

---

I don’t quite see how this is different from the algorithm I proposed. Is your aim only to find/calculate the elongation? In equilibrium? With gravity?

quote:

---

Is this code is for each node ? (not for each spring)

---

Right... I''m afraid I made a mistake there. It should have read "for each node"
Go ahead and reply when you come up with further questions!
/Ksero

I had trouble loading the pictures I posted... Here are links to them instead... I don''t know if they load OK in your browser...
Image 1, a single spring analyzed
Image 2, two springs affecting each other

I think the idea of contracting force is more logical. May be it is called as internal force of string or Tensile force (Not sure about terminology). So total force at each node is
Total force = Pull/Break force - contracting force[sign should be selected carefully]
Contracting force = (F-Pull + F_Break)/2
It is only true when F_pull = F_Break [Equlibrium]

Waht i am thinking is some sort of static situation[ considering only one step of time]. Intially string is equlibrium. No forces & no elongation. Then i apply forces F1(pull) & F2(break). F1 & F2 may or may not be equal. Now the situation is imballance[no equlibrium]. I want to know the elongation[new position of nodes] that make equlibrium. Suppose neglect what happand before reaching equlibrium [dynamic behavior].
Let constant force is applied, initially nodes are moving, elongation is produced. Finally nodes will stop at some position until force is further increased. This is the equlibrium.
Note: forces are constant, let neglect vibrations.
Waht will be elongation now ?
Elongation will be proporional to what amount of force[F1? or F2? or ??)
What when we divide it into spring/string system?

In case of hanging spring, only one force is applied. So contracting force = pulling force. At fixed edge, same is the reaction force. Complete system is in balance and so
F = Constant * Elongation
This case is fine. What about the previous one?

Bye

Alam
-- Learning never ends --