Supplemental Materials
Unusual winding of helices under tension
Jian-Shan Wang1, 2, a, Yu-Hong Cui1, Takahiro Shimada2, Hua-ping Wu2, 3, Takayuki
Kitamura2
1
Tianjin Key Laboratory of Modern Engineering Mechanic, Department of Mechanics, Tianjin
University, Tianjin 300072, People’s Republic of China
2
Department of Mechanical Engineering and Science, Kyoto University, Nishikyo-ku, Kyoto
615-8540, Japan
3
Key Laboratory of E&M (Zhejiang University of Technology), Ministry of Education &
Zhejiang Province, Hangzhou 310032, People's Republic of China
a
Author to whom correspondence should be addressed. E-mail: wangjs@tju.edu.cn
1
1. Refined Cosserat rod model of chiral filament
We consider a slender chiral filament (rod) with complicated spatial configuration.
Following Whitman and DeSilva’s model, 1 the initial configuration CD of the rod
can be described by the position vector R( S ) and the direction vector Di ( S )
( i  1, 2,3 ) at any point on CD . Both R( S ) and Di ( S ) are the functions of the arc
length variable S of the non-deformed chiral rod. Similarly, the deformed
configuration cd of the chiral rod can be described by r ( s ) and d i ( s) at any point
on cd . r ( s ) and d i ( s) are the function of the arc length variable s of the chiral
rod. We define the following deformation measures zi and Fi to characterize the
deformation of chiral rods.1
zi  yi  Rˆ  Di , i  Fi  Fi 0 ,
(s1)
in which
yi  rˆ  di   r   di , Fi 
1
1
eijk dk  dˆ j   eijk dk  d j ,
2
2
(s2)
and, x̂  x S , x  x s , x̂   x and d i  d j   ij . The vectors d i are1
di  dii ei  ,
(s3)
where e is the orthonormal basis of a fixed rectangular Cartesian system and dii 
are expressed by the Euler angles as: 1
d11   sin sin   cos cos  cos , d12  cos sin   sin cos  cos ,
d13   sin  cos  , d 21   sin cos   cos sin  cos ,
d 22  cos cos   sin sin  cos , d 23  sin  sin  ,
(s4)
d31  sin  cos , d32  sin  sin , d33  cos .
Then, Fi can be expressed in terms of the Euler angles  ,  and  as
follows:1
2
F1  ˆ sin  ˆ sin  cos , F2  ˆ cos  ˆ sin  sin , F3  ˆ ˆ cos .
(s5)
Considering the stretch-twist and bend-shear coupling deformations, the chiral
filament with a spatial configuration has following strain energy function2
H
1
 E1 y12  E2 y22  E3 ( y3  1)2  A1 ( F1  F10 ) 2  A2 ( F2  F20 ) 2
2
 A3 ( F3  F30 )2  2 D1 y1 ( F1  F10 )  2 D2 y2 ( F2  F20 )
(s6)
 2 D3 ( y3  1)( F3  F30 )  ,
in which yi and Fi ( i  1  3 ) are the deformation measures of chiral filaments.
Fi 0 are the initial values of Fi of undeformed filament. The elastic constants
E1 =E2  Jk22 and E3 =k11πa 2 , J is the Timoshenko shear coefficient. The bending
rigidities A1  A2  π4 k11a 4 and the torsional rigidity A3  π2 k22a4 . Di are the
coupling elastic constants.
The internal forces  i and internal moments mi of chiral filament can be defined
as:
i 
H
H
, mi 
,
yi
Fi
(s7)
Eq. (s7) can be written as:
1  E1 y1  D1 ( F1  F10 ),  2  E2 y2  D2 ( F2  F20 ),
 3  E3 ( y3  1)  D3 ( F3  F30 ),
m1  D1 y1  A( F1  F10 ), m2  D2 y2  B( F2  F20 )
m3  D3 ( y3  1)  C ( F3  F30 ),
(s8)
(s9)
According the equilibrium of the forces and moments on the chiral rod, the
equilibrium equations can be obtained as1, 3
ˆi  eijk j Fk  fi 0  0, mˆ i  eijk (m j Fk   j yk ) + mi0 = 0,
3
(s10)
where x̂  x S , x̂   x , the internal forces  (s )   i d i and the internal
i 1
3
3
moments m ( s )   mi d i , and eijk is the permutation tensor.
i 1
2. Model of helix of chiral filament
The deformation measures for the chiral filament can be expressed by the
Euler angles  ,  and  . If the chiral filament has a helical configuration,
then
F1  0 , F2  ˆ sin  ,
F3  ˆ cos ,
F10  0 , F20  ˆ 0 sin  0
and
F30  ˆ 0 cos  0 , where   90   h , and 0 and  0 denote the Euler angles in
the initial configuration of the filament. 1,3 From Eq. (s6), the constitutive
relationship of straight chiral filament can be obtained as:
 3  E3 ( y3  1)  D3 F3 , m3  D3 ( y3  1)  A3 F3 ,
where
y3  1   ,
E3  π r2 k11 , D3  πr 2k12 , and
k11
and
(s12)
k22
are the
extensional and shear moduli of the filament, respectively. The material
chirality parameter is C f 
D32
k2
 12 , where and k12 is the coupling
A3 E3 k11k12
elastic constant.
For a slender filament, the shear deformation can be neglected, and only its
axial deformation need be considered:
y1  0, y2  0, y3   
F  D3ˆ
D
cos  3 ˆ 0 cos 0  1,
E3
E3
(s13)
where  denotes the elongation rate of the filament. The helical radius and
pitch of an undeformed helix can be expressed as R0 
cos  0
sin  0
and P0  2π
,
ˆ 0
ˆ 0
respectively. Similarly, the helical radius and pitch of a deformed helix are
R  y3
sin 
ˆ
and P  2πy3
cos 
, respectively. When the helix is stretched and
ˆ
twisted, the increment of the pitch Lp , is given as:
4
Lp 
 2π
D
2π  F  D3ˆ
cos2   3 ˆ 0 cos 0 cos   cos    cos0，

ˆ  E3
E3
 ˆ 0
and the twist of the helix is

 L0 sin 
R
 l0ˆ .
(s14)
(s15)
According to the equilibrium of forces of helix, the internal forces can be obtained
as  2  F sin  and  3  F cos  . The internal forces and moments of the helix
remain constant along the arc length of the filament, so that the second and third
formulations ( i  2,3 ) of Eq. (s10) are satisfied automatically. While i  1 , the
equilibrium equation of the internal moments in Eq. (s10) becomes
ˆ 1  m2 F3   2 y3  m3 F2   3 y2  0.
m
(s16)
Considering mˆ 1  0 and y2  0 , Eq. (s16) can be written as:
m2 F3   2 y3  m3 F2  0.
(s17)
Combining Eqs. (s9), (s13) and (s17) yields:
F sin  +
1 2
D
F sin  cos   3 F (ˆ sin 2 ˆ 0 cos  0 sin  )
E3
E3
D32 2
ˆ ˆ 0 cos 0 sin  )  ( B  G )ˆ 2 sin  cos 
(ˆ sin  cos  
E3
ˆ ˆ 0  0,
( B sin 0 cos   G cos 0 sin  )

where
the
effective
bending
rigidity
of
the
chiral
(s18)
filament
is
B  A1  A2  π4 k11a 4 , and the effective torsional rigidity of the chiral filament
π
is G  A3  2 k22a 4 .
The torque acting on the helix is
M  B (ˆ sin   ˆ 0 sin  0 )sin   G (ˆ cos   ˆ 0 cos  0 ) cos 
F  D3ˆ
D
 D3 (
cos   3 ˆ 0 cos  0 ) cos  ,
E3
E3
(s19)
where “  ” represents a right-handed material chirality ( C f  0 ) and “  ” a
5
left-handed material chirality ( C f  0 ) of a filament. Eq. (s19) shows that the
torque contributes from the bending, twisting and chirality induced
self-twisting of filament.
3. Elastic constants of helix
From Eqs. (s14), (s15), (s18) and (s19), we can have
ˆ 1
F

ˆ
 , a1
 a2
 a3
 0,
 l0



F

ˆ M
F

ˆ
b1
 b2
 b3

, c1
 c2
 c3
 0,







(s20)
where the coefficients ai , bi and ci are given in Appendix.
Then, we can obtain the effective elastic constants of helix as
Q12 
F c2a3  c3a2 1

,
Lp c1a2  c2a1 l0
1
M  c2a3  c3a2
c a ca
Q22 

b1  3 1 1 3 b2  b3  ,
  c1a2  c2a1
c1a2  c2a1
 l0
(s21)
Similarly, we have
ˆ
F
 ˆ
 0, f1
 f2

,
h
h
h 2π
F
 M
F

b1
 f3

, c1
 c2
 0,
h
h h
h
h
(s22)
where the coefficients f i are given in Appendix.
Then, we can obtain
Q11 
F
c2
ˆ
M c2b1  c1 f 3 ˆ

, Q21 

,
Lp c2 f1  c1 f 2 2π
Lp c1 f 2  c2 f1 2π
(s23)
References
1A.
G. Whitman and C. N. Desilva, J. Elasticity 4, 265 (1974).
2T.
J. Healey, Math. Mech. Solids 7, 405 (2002).
3J.
S. Wang, G. F. Wang, X. Q. Feng and Q. H. Qin, J. Phys.: Condens. Matter 24, 265303 (2012).
6
Appendix
The coefficients ai , bi , ci and f i are given as
a1  cos2  ,
a2   F sin 2  D3ˆ sin 2  D3ˆ 0 cos 0 sin   E3 sin  ,
1
a3   ( F cos2   D3ˆ 0 cos 0 cos   E3 cos  ).
ˆ
(A1)
D3
cos2  ,
E3
b2  ( B  C )ˆ sin 2  Bˆ 0 sin  0 cos   Cˆ 0 cos  0 sin 
b1 
 D3
F  D3ˆ
D2
sin 2  3 ˆ 0 cos  0 sin  ,
E3
E3
b3  B sin 2   C cos2  
D32
cos  .
E3
c1 
D
1
sin 2 F  3 ˆ 0 (sin 2  cos  0 sin  )  sin  ,
E3
E3
c2 
D
1
cos 2 F 2  cos  F  3 ˆ 0 (2 cos 2  cos  0 cos  )
E3
E3

(A2)
D32 2
ˆ 0 (cos 2  cos  0 cos  )  ( B  C )ˆ 02 cos 2
E3
(A3)
 ( B sin  0 sin   C cos  0 cos  )ˆ ,
2
0
D3
D2
sin 2 F  3 ˆ 0 (sin 2  cos  0 sin  )
E3
E3
 ( B  C )ˆ 0 sin 2
c3  
 ( B sin  0 cos   C cos  0 sin  )ˆ 0 .
f1 
1
cos 2  ,
E3
D
1
F sin 2  3 ˆ (sin 2  cos  0 sin  )  sin  ,
E3
E3
f3  Bˆ (sin 2  sin  0 cos  )  Cˆ (sin 2  cos  0 sin  ).
f2  
7
(A4)