Appendix A: Buckling Equations
The total energy of the pipe can be expressed as
  U W .
(A-1)
The elastic potential energy is
(A-2)
U  U a  Ub .
Eq. A-3 describes the elastic compressive deformation energy of the coiled tubing. Its variation
can be expressed as shown in Eq. A-4.
L
du 
1 
(A-3)
U a    F  z  a  dz ,
20
dz 
L
 U a  FL ua  L   
0
dF  z 
dz
 ua  z  dz .
(A-4)
The elastic compressive bending energy is
L
EI 2
EI
Ub 
k dz 

2 0
2
2
4
L
EI 2  d 2   d  

k

k
dz

r







  dz
0
2 0  dz 2   dz  


L
2
r
2
2
2
2
L
 dr d d 2  d  2 d 2 r 
EI  d 2 r 
 d   dr  

 EI  r  2

dz


4





    dz
2
2 
2 0  dz 2 
 dz  dz 
 dz   dz  
 dz dz dz
0 

,
(A-5)
L
E 4
 RP  rP4  is the bending stiffness in N. m 2 ; here, rp is the inner radius of the coiled
4
tubing, in m , and R p is the outer radius, also in m .
where EI 
 U b   U b , r   U b , ,
(A-6)
  d 
 d 4 r
 d 2 
d d 3 
d  dr  d   
 r rdz  EI   4  6  
  3 2   4
    rdz , (A-7)
3
dz 
dz dz 
dz  dz  dz   
dz 
dz


0 
0




2
4
L
 U b , r  EI  
2
L
3
3
L
L
 d 4
 2

 d 3r d 3 d 2 r d 2 
d  d   

 dr  d 3  d   
 2 

r dz  4EI    3  
   r  dz  4 EI   3
   r dz
4
2
2 
dz
dz
dz
dz
dz
2
dz
dz
dz
dz
dz






0
0
0


 
 
 

L
 U b,  EI  
. (A-8)
The total variation of U is given by
3
L
L
dF  z 
 d 3r d 3 d 2 r d 2 
 dr  d 3  d   
 U a  z  dz  4 EI   3

r dz  4 EI    3  
   r dz
2
2 
dz
dz dz 2 dz dz 
0
0
0
 dz  dz  dz   
L
U  U a  Ub  
2
3
2
4
L 4
L
 d 4
 d 2 
d  d    2
d  dr  d   
d d 3 
d r
 d 

 r rdz
 EI   4  2 
r

dz

EI

6

rdz

EI

3






 2  4





4


dz  dz   
dz  dz  dz   
dz 
dz dz 3 
 dz 
0
0
0
 dz
 dz


L
. (A-9)
The work done by the force in the X direction
Let WG2 represent the work done by gravity in the X direction; its variation is expressed as
shown in Eq. A-10. The work done by the normal contact force WN and its variation is given by Eq.
A-11.
L
L
L
0
0
0
WG2    q sin   rc  r cos   dz,  WG2  q sin   cos  rdz  q sin   r sin  dz .
L
L
0
0
WN    N  rc  r cos   dz ,  WN    N  rdz .
(A-10)
(A-11)
The work done by the force in the Y direction
Let Wx, f2 represent the work done by the lateral friction force. Its variation is given by Eq. A-13.
  z
W f2  
 
  z 0
0
 z
f 2 Nrddz 
 
  z  0 0
L
 z
0
0
f 2 Nrddz    sign  
1

f 2 Nrddz .
(A-12)
L
 W f    f 2 Nrsign    dz .
(A-13)
2
0
The work done by the force in the Z direction
For a given axial displacement profile ua ( z ) , the work done by the force along the Z direction is
denoted by WFz , a . Similarly, for the displacements defined by ub ( z ) , WFz ,b denotes the work. These
quantities and their variations are expressed as follows:
WFz  WFz , a  WFz ,b ,
WFz , a 
ua ( L )

0
L ua ( z )
FL dua ( L)  
0

0
L
(A-14)
L ua ( z )
f1  z  N  z  dua ( z )dz  
0

q cos  dua ( z )dz ,
(A-15)
0
L
 WFz , a  FL ua ( L)   f1  z  N  z   ua ( z )dz   q cos  ua ( z )dz ,
0
(A-16)
0
where FL represents the axial force at the loading end, and
WFz ,b 
ub ( L )

0
L ub ( z )
FL dub ( L)  
0

0
L ub ( z )
f1  z  N  z  dua ( z )dz  
0

q cos  dua ( z )dz ,
L
z
 dub  z  
 dub  z  
dzdz   q cos    
dzdz .
dz
0
0


0
0
 dz 
By interchanging the order of integration, this expression can be rewritten as follows:
L
z
 WFz ,b  FL ub ( L)   f1  z  N  z    
L
x
L x
L L
L
L
0
0
0 0
0 s
0
s
 H ( x) G  s  dsdx    H ( x)G  s  dsdx    H ( x)G  s  dxds   G  s   H ( x)dxds ,
L

(A-17)
0
L
  dub ( z ) 
 du ( z ) 
dz   F  z    b dz .

 dz 
0
  dz 
L
 WFz ,b    FL    f1 N  q cos   dz   
z

The axial force is defined as
0
(A-18)
(A-19)
(A-20)
L
F  z   FL    f1 ( z ) N ( z )  q cos   dz .
(A-21)
z
Let ub ( z ) denote the axial displacement induced by buckling or lateral bending.
ub ( z ) 
2
2
2
2
z
dub ( z ) 1  dr 
1  dr 
 d  
 d  
2
2

r
z
dz
,


r
z
   
    ,
 
 
2 0  dz 
dx
2  dz 
 dz  
 dz  
2
 du ( z )  dr  dr 
 d 
 d  ,
2 d
  b       r r 

 r

dz  dz 
 dz 
 dx  dz  dz 
(A-22)
(A-23)
d  2 d 
d  dr 
 dub ( z ) 
 d 
 dz    dz  Fr dz   dz   Fr  dz   rdz    dz  F dz   rdz . (A-24)
dz








0
0
0
0
The total variation of  is
   U  W .
(A-25)
 ua ( z) ,  r , and  can respectively be expressed as
L
L
 WFz ,b   F  z   
L

dF  z 
dz
0
2
L
L
L
0
0
L
 ua  z  dz   f1  z  N  z   ua ( z )dz   q cos  ua ( z )dz  0 ,
2
4
2
L
L 4
 d 2 
d d 3 
d  dr  d   
d r
 d 

EI  

3

4
r

rdz

EI

6
 2 

0  dz 4 dz  dz  dz     rdz
dz 
dz dz 3 
dz 


0 





,
L
L
L
0
0
0
q sin   cos  rdz   N  rdz   
d  dr 
 d 
 F   rdz   Fr 
  rdz  0
dz  dz 
 dz 
0
L
2
2
(A-26)
(A-27)
3
3
L 4
L
L
3
d  d    2
d 
 dr  d   d   
EI   4  2 
   r  dz  4 EI    3  
   r dz   f 2 Nrsign    dz
dz  dz   
0
0
0
 dz
 dz  dz  dz   
. (A-28)
L
L
L
3
2
2
 d r d 3 d r d  
d  2 d 
4 EI   3

 r dz    Fr
  dz  q sin   r sin  dz  0
dz 2 dz 2 dz 2 
dz 
dz 
0  dz
0
0
The buckling equations can thus be obtained using the principle of virtual work.
dF  z 
(A-29)
 f1  z  N  z   q cos  ,
dz
4
2
  d 2  2
d d 3  d  
 d 

N  EIrc 3  2   4


q
sin

cos


Fr
(A-30)
c



 ,
dz dz 3  dz  
 dz 
  dz 

2
d 4
d  d 
 d  d 
EIr
 6 EIrc2 
 rc2  F

  q sin  rc sin   f 2 Nrc sign    0 .
4
2
dz
dz
dz
dz


 dz 
2
2
c
(A-31)
Appendix B: Work and Elastic Deformation Energy
We assume that the coiled tubing is in continuous contact with the wall. In other words, r is a
constant ( r  rc ). Therefore, the elastic deformation energy and its variation are, respectively,
L
Ub 
EI 2
EI
k dz 
2 0
2
L
2
2
  kr  k  dz 
0
2
4
L
EI 2  d 2   d  

r


 
  dz ,
2 0  dz 2   dz  


(B-1)
L
L
L
2
L
 d 2
3
 d 3  
 d 4
d 2  d  


 d    d  
 U b  EIr  2      3    2       EIrc2   4  6 2     dz . (B-2)
dz  dz   0  dz
dz  dz  
 dz
0
0

 dz 
 0 


Similarly, the work done by the axial load and its variation can be expressed as
2
c
WFz ,b 
ub ( L )

0
L ub ( z )
FL dub ( L)  
L
0

q cos  dua ( z )dz ,
(B-3)
d  d 
d 
d  d 
2 

   rc2   F
 dz  rc  F
  dz .
dz  dz 
dz  dz 
 dz
0
0
(B-4)
0
 WFz ,b,  rc2  F

L ub ( z )
f1  z  N  z  dua ( z )dz  
0
0
L
0
L
Appendix C: Derivation of the Potential Energy
When the effect of the lateral friction is considered, the total dimensionless potential energy is
defined as
2
4
2
  



 d  
1 L  d 2   d 
1 L
1 L





2
d


sign

f
nd

d




1  cos   d  , (C-1)






0 2
2m L 0  d  2   d  
m L 0
 L 0
 d   

where n( ) represents the dimensionless normal contact force on the coiled tubing, and
2
4
2
 d 2 
 d 
d d 3  d 
(C-2)
n  3 2   4

  m cos   2 
 .
3
d d  d 
 d 
 d 
When the coiled tubing enters a state of helical buckling, it can be assumed that the configuration
of the pipe will take the following form:
(C-3)
 ( , t )  p(t ) ,
where
p will range from 0 to
ph . By substituting Eq. C-3 into Eq. C-2, we obtain
n  2 p 2  p 4  m cos  .
3
(C-4)
For a given  , d   dp , and
(  )
(  )
ph
1 
2
 p 4 ) dp  m sin ph   ph2  ph4  ph   m sin ph . (C-5)
5 
3
0
0
0
The integer k represents the number of helical turns of the coiled tubing with a given length
sign( )  nd 

nd 
 (2 p
2
L :
L 
where  h 

ph
2k
 2k h ,
ph
(C-6)
denotes the length of one half -helix turn.
L
  
0
0
 sign   
 h   
ndd  2k 
0

0
1 
4k   2 2 1 4 
2
2
ndd  k  ph2  ph4  ph  h2 
  ph  ph   L  m L , (C-7)
3
5
p
2
3
5





h
2
  d   2  d  4
 d  
4
2



2
0  d 2   d   d   d   ph  2 ph   L ,


L
2
(C-8)
L
 1  cos  d  
L
.
(C-9)
0
Thus, we arrive at the total dimensionless potential energy for the helical buckling of the coiled
tubing by substituting Eqs. C-6 to C-9 into Eq. C-1.
 1  f2  4   f2 1  2 2 f2
(C-10)
h  

 ph   3m  m  ph    1 .
 2m 10m 


Appendix D: Solution for Sinusoidal Buckling
If buckling occurs, axial friction becomes dominant. Therefore, for post-buckling analysis, we
can assume that the lateral friction can be neglected, f 2  0 , and f1  f . Then, Eq. 8 can be
simplified to
2
 d  d 2
d 4
d 2

6

2
 m sin   0 .
(D-1)


2
d 4
d 2
 d  d
Once buckling is induced, the wavelength of the sine wave will remain unchanged. However, the
amplitude will increase with increasing axial load. Therefore, during the initiation of sinusoidal
buckling, the solution to Eq. D-1 can be assumed to have the following form:
  , t   a  t  sin    a 2  t  g 2  , t   a 3  t  g 3  , t     a 4  .
(D-2)
Substituting Eq. D-2 into Eq. D-1 yields
 4

d 2 g 2  , t 

3a 2 
2 d g 2  , t 
a m  1 
2
 mg 2  , t  
 sin    a 
4
2
2 
d

 d 

.
(D-3)
 d 4 g3  , t 

d 2 g 3  , t 
3
4
a 
2
 mg3  , t   sin(3 )    (a )  0
4
2
d 2
 d 

The solution to Eq. D-3 can be expressed as follows:
2 1  m 
(126  2m)a sin   (1  m)3sin(3 )
a
, g 2  , t   0, g3  , t  
.
(D-4)
3
2m 2  124m  126
Because the term a 3 (t ) g3 ( , t ) is on the order of 103 , it can be ignored in practical engineering
applications. Therefore, the solution to Eq. D-1 can be approximated as
2 1  m 
 ( , t )=
sin( ) .
(D-5)
3
3
4
Appendix E: Solution for Helical Buckling
The coiled tubing will transition from a sinusoidal shape to a helical buckling shape when the
axial force exceeds the critical load for helical buckling.
2
 d  d 2
d 4
d 2

6

2
 m sin   0 .
(E-1)


2
d 4
d 2
 d  d
Because buckling occurs for m  1 , the solution to Eq. E-1 can be assumed to be of the form
 ( )   0 ( )  m1 ( )   (m2 ) ,
(E-2)
m sin   m sin  0   (m2 ) .
Substituting Eqs. E-2 and E-3 into Eq. E-1 results in
(E-3)
d 40
d 20
 d0  d 20

6

2
0,
(E-4)


2
d 4
d 2
 d  d
2
d 41
 d0   d1 
d 


(E-5)
2
 1  3 
  sin  0  0 .
 
4
d   
d
 d    d  


If we can solve Eq. E-4 for  0 , then subsequently solving Eq. E-5 for 1 will be simple. Thus,
let us now focus on seeking a solution to Eq. E-4 for  0 for various boundary conditions. Integrating
Eq. E-4 results in
2
 d 3
  d  2  d
  d  2  d 
d 3 0
0
0
0


2
1




0,



2
 
1   0   0  .
 
1
1
3
d 3
 d 
  d    d 
  d    d    0
(E-6)
For the free end at   0 , under the natural boundary conditions (Eq. 19), we arrive at 1  0 .
2
d0 d 3 0
d  d  1 d  d  


,

 can be obtained by applying




d  3 d   d   2 d  d   
d
the chain rule. Therefore, Eq. E-6 can be rewritten as
2
1 d  d  
3
(E-7)

   2  2  1  0 ,
2 d  d   
Otherwise, 1  0 . For  
2
 d 
4
2
(E-8)

    2  2 1   2 ,
d



 d 2

4
2
(E-9)
 2  
    2  2 1  .
 d  
  0
For simplicity, this article discusses only the case of a length of coiled tubing with a free end at
  0 . Therefore, the value of 1 is zero. Fig E-1 presents the phase diagrams of Eq. E-8 for various
constant values of 2 . The black-colored curves represent the case of  2  0 .
3

2
3
2
1
-2
0
-2
0.5
0
-1
-1
-2
-3
Fig E-1. Phase diagram of   vs. 
5
0
1

1
2
The red curve in Fig E-1 , corresponding to 0  2  1 , assumes the shape of a closed loop. This
means that Eq. E-4 has a periodic solution. This is not the main result of this analysis and corresponds
to the sinusoidal buckling considered in our previous study. For 2  1 (the green curves), there exists
no value of  for which    0 can be obtained. This indicates that there is no solution that satisfies
d 20
 0 when
d 2
  1 . In other words,   1 is a specific solution that satisfies the boundary conditions for a
the boundary conditions for a pinned or free end. For 2  1 (the blue curve),   
pinned or free end. Because one end of the coiled tubing is free
configuration of the massless coiled tubing can be expressed as
    c0 .
 1  0 ,
the helical buckling
(E-10)
For coiled tubing with two free ends, c0 is an arbitrary constant. However, c0  0 when one
end is a pinned end. The expression 1  0 corresponds to the other types of boundary conditions at
  0 . However, for two pinned ends, the boundary conditions are a special case of 1  0 . For any
integer k , the solution  
2k 
L
 is suitable not only for the two-pinned-end boundary conditions
but also for the buckling equation given by Eq. E-4. Given an integer k ,  ( L )  2k indicates that
the end at    L will not undergo horizontal displacement. Substituting 0    c0 into Eq. E-5
yields
2
d 41
2 d 1

2
1

3

(E-11)

 d 2  sin   c0   0 .
d 4
The solution to Eq. E-11 is given by
sin(  c0 )
1   2
.
(E-12)
 (7 2  2)
For the boundary conditions for pinned or free ends, the solution is
sin(  c )
 ( )    c0  m 2 2 0
(E-13)
 (7  2)
Compared with the linear term, the perturbation term is very small. Therefore, we can ignore this
term for practical engineering applications.
6