3. Rotation analysis of the CICC

2.2 The tensile stiffness of the higher stage strand

higher-level strand can be expressed as

T<sup>k</sup> ¼

κp2 κb2 κt2

3 7 7

equilibrium equations can be expressed as

<sup>5</sup> <sup>¼</sup> <sup>T</sup>2T<sup>1</sup> <sup>T</sup><sup>T</sup>

1

8 >>><

>>>:

0 0

Ft0

Space line of the triplet and single wire (a) and 2D view of the triplet and single wire (b).

respectively.

Figure 2.

84

Based on the equivalent modulus and thermal expansion of the triplet, the space and the 2D view of the triplet and single wire are shown in Figure 2(a) and (b),

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The conversion relationship between the local coordinates of the triplet and the

� cos θ<sup>k</sup> � sin θ<sup>k</sup> 0

> 9 >>>=

> >>>;

: (5)

: (6)

(7)

sin θ<sup>k</sup> sin α<sup>k</sup> � cos θ<sup>k</sup> sin α<sup>k</sup> cos α<sup>k</sup>

� sin θ<sup>k</sup> cos α<sup>k</sup> cos θ<sup>k</sup> cos α<sup>k</sup> sin α<sup>k</sup>

T1T<sup>2</sup>

Fp2

Fb2 Ft2

0 0 cos α<sup>1</sup> r1

sin α<sup>2</sup>

The curvature and torsion of the secondary stage strand can be given by

0 0 cos α<sup>2</sup> r2

According to the geometric compatibility of the secondary-stage strand, the deformation of the triplet is equal to the tangential strain of the secondary-stage strand, and the torsion of the triplet is equal to the twist angle of the secondary cable. The axial loads and torque of the secondary-stage strand can be obtained. The

In the CICC conductor manufacture process, they twist a superconducting cable and penetrate it into the stainless steel tube. However, due to the friction between the superconducting cable and the stainless steel armor, the drag force of the cable is as high as several tons during the cable penetration. The friction force of the pipe leads to the axial elongation of the superconducting cable, accompanied by the untwisting of the cable, which causes the cable pitch to increase. This makes that the pitch is much larger than the ITER requirement [20]. Therefore, the untwist behavior of the cable must be controlled [17].

In this section, the untwist model is described. The large-scale cable is considered, e.g., ITER TF, CS, and CFETR CSMC. The components of the final cable include petals, central cooling spiral, and wrap, as shown in Figure 3. The model ignores the friction between the jacket and the cable, only modeling de-twists of the cable under the insertion force FInsert.

The cable is divided into three parts in the model: central cooling spiral, six petals, and the wrap. The twist direction of wrap and cooling spiral is left and with the reverse direction for the petal. The torsion constraint is free for the cable when there is undering the uniaxial tension. Therefore, the boundary conditions can be set as F ¼ F0; M ¼ 0. The force of the whole cable is from those acts on wrap, subcables, and central cooling spiral, which can be described as follows:

$$\begin{aligned} F &= F\_{in} + F\_p + F\_{st} = F\_{0}, \\ M &= M\_{in} + M\_p + M\_{st} = \mathbf{0}. \end{aligned} \tag{9}$$

In Eq. (9), F is the insertion force in the axial direction for the cable. Fst, Fp, Fin are the forces loading on the stainless steel wrap, petals, and inner cooling spiral,

Figure 3. The dimensions and parameters of a large-scale cable (e.g., CFETR CSMC).

He � inlet :

8 < :

Wrap � stiffness :

refers to the cross section of each component. EPI

twist, respectively. ξ<sup>P</sup> is the axial strain of petal.

<sup>x</sup>τ<sup>x</sup> þ Hxκ<sup>0</sup>

<sup>x</sup>τ<sup>x</sup> � Txκ<sup>0</sup>

Fx ¼ Tx sin α<sup>x</sup> þ N<sup>0</sup>

Mx ¼ Hx sin α<sup>x</sup> þ G<sup>0</sup>

ΔRP ¼ �νRPξP, v is Poisson's ratio of the petal.

x,

x,

<sup>x</sup> cos αx,

rin ¼ Rin; rP ¼ Rin þ RP; rst ¼ Rin þ 2RP,

The contact force loading on the petal can be written as follows:

6XP sin α<sup>P</sup>

�Tstrst cos αst � Tinrin cos αin

reaction force between inner He-inlet spiral and petal.

From the above Eqs. (10)–(15), we can obtain

From Eqs. (9)–(12), one can get

F ¼ F<sup>0</sup> ¼ 6 TP sin α<sup>P</sup> þ N<sup>0</sup>

M ¼ 0 ¼ 6 HP sin α<sup>P</sup> þ G<sup>0</sup>

87

Δrin ¼ ΔRin; ΔrP ¼ ΔRin þ ΔRP; Δrst ¼ ΔRin þ 2ΔRP:

where x stands for anyone of in, p, st, which represent inner He-inlet spiral, petal, and stainless steel wrap, respectively; Rp denote the twist radius of petal, and

> <sup>¼</sup> Xst sin αst

<sup>P</sup> cos α<sup>P</sup>

where XP stand for the line pressure between the petal and inner He-inlet spiral; Xst is the uniform force between the stainless steel wrap and petal; and Xin is the

� � <sup>þ</sup> Tst sin <sup>α</sup>st <sup>þ</sup> Tin sin <sup>α</sup>in

� �

<sup>P</sup> cos α<sup>P</sup> þ TPrP cos α<sup>P</sup> � N<sup>0</sup>

<sup>x</sup> cos α<sup>x</sup> þ Txrx cos α<sup>x</sup> � N<sup>0</sup>

� Xin sin αin

torsion stiffness of petals, respectively. Δκ<sup>0</sup>

subcables, and wrap can be obtained as

Xx ¼ N<sup>0</sup>

N0 <sup>x</sup> ¼ G<sup>0</sup>

Petal :

DOI: http://dx.doi.org/10.5772/intechopen.82349

Gin ¼ 0; G<sup>0</sup>

The Mechanical Behavior of the Cable-in-Conduit Conductor in the ITER Project

8 < :

GP ¼ 0; G<sup>0</sup>

TP ¼ EPAPξ<sup>P</sup>

8 < :

Tin ¼ EinAinξin

<sup>P</sup> ¼ EPI x PΔκ<sup>0</sup>

Gst ¼ 0; G<sup>0</sup>

In the equation systems, E refers to Young's modulus of each component. A

The balance equations and the compatible equations of central cooling spiral,

Tst ¼ EstAstξst

x

in ¼ 0; Hin ¼ 0

P; HP ¼ GPI

st ¼ 0; Hst ¼ 0

z <sup>P</sup>Δτ<sup>P</sup>

P, GPIP are the bending and

xrx sin αx:

PrP sin α<sup>P</sup>

P, Δτ<sup>P</sup> are the changes in curvature and

(11)

(12)

(13)

(14)

(15)

Figure 4. Force distribution (A) and spatial relationship (B and C).

respectively; M represents the torque of the whole cable. Mst, Mp, Min denote the torques of stainless steel wrap, petals, and inner cooling spiral.

The force distribution and spatial relationship are shown in Figure 4. Gx, G<sup>0</sup> x, Hx are sectional moment components of the thin rod. Nx, N<sup>0</sup> x, Tx are the sectional force components. Xx is the contact force. α<sup>x</sup> is the spiral angle of each component. rx is the distance between the centroid and the center of the cable, and Lx is the twist pitch.

We simplified the central cooling spiral into a single helical thin plate. The change of spiral angle, radius, and the axial strain are defined as Δαst, <sup>Δ</sup>rst rst , ξst, respectively. Then, the geometrical equations of the He-inlet spiral, six petals, and stainless steel wrap can be deduced by Costello-Velinsky theory [19]; we can obtain

$$He-inlet: \begin{cases} \varepsilon\_0 = \xi\_{in} + \frac{\Delta\alpha\_{in}}{\tan\alpha\_{in}}, \\\\ \beta\_0 = \frac{\xi\_{in}}{\tan\alpha\_{in}} + \frac{1}{\tan\alpha\_{in}}\frac{\Delta r\_{in}}{r\_{in}} - \Delta a\_{in}, \end{cases} \\\\ \text{Petal}: \begin{cases} \varepsilon\_0 = \xi\_P + \frac{\Delta\alpha\_P}{\tan\alpha\_P}, \\\\ \beta\_0 = \frac{\xi\_P}{\tan\alpha\_{P\!}} + \frac{1}{\tan\alpha\_{P\!}}\frac{\Delta r\_{P}}{r\_{P}} - \Delta a\_{P}, \end{cases} \\\\ \text{Wrap}-stififfness: \begin{cases} \varepsilon\_0 = \xi\_{\it t} + \frac{\Delta\alpha\_{\it t}}{\tan\alpha\_{\it t}}, \\\\ \beta\_0 = \frac{\xi\_{\it t}}{\tan\alpha\_{\it t}} + \frac{1}{\tan\alpha\_{\it t}}\frac{\Delta r\_{\it t}}{r\_{\it t}} - \Delta a\_{\it t}. \end{cases} \end{cases} \tag{10}$$

In the equations, the axial deformation and torsion angle of the cable are expressed as ε0, β0. The only axial tension is considered for central cooling spiral and wrap. Therefore, the equations can be updated as

The Mechanical Behavior of the Cable-in-Conduit Conductor in the ITER Project DOI: http://dx.doi.org/10.5772/intechopen.82349

$$\begin{aligned} \left. He - inlet : \begin{cases} G\_{in} = \mathbf{0}; G\_{in}^{\prime} = \mathbf{0}; H\_{in} = \mathbf{0} \\\\ T\_{in} = E\_{in} A\_{in} \tilde{\xi}\_{in} \end{cases} \right. \\\\ \left. \begin{aligned} \left. \mathbf{G} p = \mathbf{0}; G\_{P}^{\prime} = E p\_{P}^{\mathbb{E}} \Delta \kappa\_{P}^{\prime}; H\_{P} = G p\_{P}^{\mathbb{E}} \Delta \tau\_{P} \\\\ T\_{P} = E\_{P} A\_{P} \tilde{\xi}\_{P} \end{cases} \right. \\\\ \left. Wrap - siffness : \begin{cases} \mathbf{G}\_{\mathcal{U}} = \mathbf{0}; \mathbf{G}\_{\mathcal{U}}^{\prime} = \mathbf{0}; H\_{\mathcal{U}} = \mathbf{0} \\\\ T\_{\mathcal{U}} = E\_{\mathcal{U}} A\_{\mathcal{U}} \tilde{\xi}\_{\mathcal{U}} \end{cases} \end{aligned} \end{aligned} \tag{11}$$

In the equation systems, E refers to Young's modulus of each component. A refers to the cross section of each component. EPI x P, GPIP are the bending and torsion stiffness of petals, respectively. Δκ<sup>0</sup> P, Δτ<sup>P</sup> are the changes in curvature and twist, respectively. ξ<sup>P</sup> is the axial strain of petal.

The balance equations and the compatible equations of central cooling spiral, subcables, and wrap can be obtained as

$$\begin{aligned} N\_{\boldsymbol{x}}' &= G\_{\boldsymbol{x}}' \boldsymbol{\tau}\_{\boldsymbol{x}} + H\_{\boldsymbol{x}} \boldsymbol{\kappa}\_{\boldsymbol{x}}', \\\\ X\_{\boldsymbol{x}} &= N\_{\boldsymbol{x}}' \boldsymbol{\tau}\_{\boldsymbol{x}} - T\_{\boldsymbol{x}} \boldsymbol{\kappa}\_{\boldsymbol{x}}', \\\\ F\_{\boldsymbol{x}} &= T\_{\boldsymbol{x}} \sin a\_{\boldsymbol{x}} + N\_{\boldsymbol{x}}' \cos a\_{\boldsymbol{x}}, \\\\ M\_{\boldsymbol{x}} &= H\_{\boldsymbol{x}} \sin a\_{\boldsymbol{x}} + G\_{\boldsymbol{x}}' \cos a\_{\boldsymbol{x}} + T\_{\boldsymbol{x}} r\_{\boldsymbol{x}} \cos a\_{\boldsymbol{x}} - N\_{\boldsymbol{x}}' r\_{\boldsymbol{x}} \sin a\_{\boldsymbol{x}}. \\\\ r\_{\dot{m}} &= R\_{\dot{m}}; r\_{P} = R\_{\dot{m}} + R\_{P}; r\_{\boldsymbol{x}} = R\_{\dot{m}} + 2R\_{P}, \\\\ \Delta r\_{\dot{m}} &= \Delta R\_{\dot{m}}; \Delta r\_{P} = \Delta R\_{\dot{m}} + \Delta R\_{P}; \Delta r\_{\boldsymbol{x}} = \Delta R\_{\dot{m}} + 2\Delta R\_{P}. \end{aligned} \tag{13}$$

where x stands for anyone of in, p, st, which represent inner He-inlet spiral, petal, and stainless steel wrap, respectively; Rp denote the twist radius of petal, and ΔRP ¼ �νRPξP, v is Poisson's ratio of the petal.

The contact force loading on the petal can be written as follows:

$$\frac{\mathsf{G}X\_P}{\sin a\_P} = \frac{X\_{\mathfrak{s}}}{\sin a\_{\mathfrak{s}}} - \frac{X\_{\mathfrak{m}}}{\sin a\_{\mathfrak{m}}} \tag{14}$$

where XP stand for the line pressure between the petal and inner He-inlet spiral; Xst is the uniform force between the stainless steel wrap and petal; and Xin is the reaction force between inner He-inlet spiral and petal.

From Eqs. (9)–(12), one can get

$$F = F\_0 = \left(T\_P \sin a\_P + N\_P' \cos a\_P\right) + T\_{st} \sin a\_{st} + T\_{in} \sin a\_{in}$$

$$M = 0 = \left(H\_P \sin a\_P + G\_P' \cos a\_P + T\_P r\_P \cos a\_P - N\_P' r\_P \sin a\_P\right) \tag{15}$$

$$-T\_{st} r\_{st} \cos a\_{st} - T\_{in} r\_{in} \cos a\_{in}$$

From the above Eqs. (10)–(15), we can obtain

respectively; M represents the torque of the whole cable. Mst, Mp, Min denote the

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The force distribution and spatial relationship are shown in Figure 4. Gx, G<sup>0</sup>

components. Xx is the contact force. α<sup>x</sup> is the spiral angle of each component. rx is the distance between the centroid and the center of the cable, and Lx is the twist

We simplified the central cooling spiral into a single helical thin plate. The

Δαin tan αin ,

þ

1 tan α<sup>P</sup>

ε<sup>0</sup> ¼ ξst þ

<sup>β</sup><sup>0</sup> <sup>¼</sup> <sup>ξ</sup>st tan αst þ

In the equations, the axial deformation and torsion angle of the cable are expressed as ε0, β0. The only axial tension is considered for central cooling spiral

1 tan αin

> ΔrP rP

Δαst tan αst ,

Δrin rin

� ΔαP,

1 tan αst Δrst rst

� Δαst:

� Δαin,

respectively. Then, the geometrical equations of the He-inlet spiral, six petals, and stainless steel wrap can be deduced by Costello-Velinsky theory [19]; we

change of spiral angle, radius, and the axial strain are defined as Δαst, <sup>Δ</sup>rst

ε<sup>0</sup> ¼ ξin þ

<sup>β</sup><sup>0</sup> <sup>¼</sup> <sup>ξ</sup>in tan αin

> 8 >>>><

> >>>>:

Δα<sup>P</sup> tan α<sup>P</sup> , x, Hx

x, Tx are the sectional force

rst , ξst,

(10)

torques of stainless steel wrap, petals, and inner cooling spiral.

are sectional moment components of the thin rod. Nx, N<sup>0</sup>

Force distribution (A) and spatial relationship (B and C).

8 >>>><

>>>>:

ε<sup>0</sup> ¼ ξ<sup>P</sup> þ

<sup>β</sup><sup>0</sup> <sup>¼</sup> <sup>ξ</sup><sup>P</sup> tan α<sup>P</sup> þ

He � inlet :

8 >>>><

>>>>:

Wrap � stiffness :

and wrap. Therefore, the equations can be updated as

Petal :

pitch.

Figure 4.

can obtain

86

4. Buckling behavior in the CICC

DOI: http://dx.doi.org/10.5772/intechopen.82349

L<sup>1</sup> þ L<sup>2</sup> þ L<sup>3</sup> ¼ L.

<sup>M</sup> ¼ �EI d<sup>2</sup>

Figure 6.

and εSlid.

89

4.1 Analytical model without the electromagnetic force

The Mechanical Behavior of the Cable-in-Conduit Conductor in the ITER Project

It is known that all the ITER CS and TF coils are wounded by CICCs, which made up of five-stage sub-cables formed around a central cooling tube. The petals and the sub-cables are wrapped with stainless steel tapes. Then, let the wrapped cable inserted into a stainless steel tube, which act as an amor. We assume that the total length of free segment of the superconducting strand on the surface of the cable is L (equal to the twist pitch), and set the fixed constraints on both sides, the wrap band as a uniform spring constraint. The schematic of this analytical model is illustrated in Figure 6. The lengths of spring constraint on both sides are equal to L<sup>1</sup> and L3, respectively. The length of the free fraction of the strand is of L2; we can get

Since the coefficient of thermal expansion of stainless steel between 923 and 4.2 K is approximately twice that of the Nb3Sn strand, then the superconducting cable is in compression at the end of the cooling. The thermal shrinkage of the cable is denoted by εThermal. Other than this, the strands of the cable can be squeezed into another side of the conduit by the large electromagnetic force; then, there will generate a large void on the other side of the conduit. Due to the gap, the friction force between the strands and the stainless steel armor decrease significantly. As there is no lateral restraint by wrap or friction, the surface strands around the void will show bending deformation by the thermal mismatch. In addition, the strand will slide into the high magnetic field region that will aggravate its bending behavior. εSlid is the stand for this slid strain. Therefore, the total compression strain of the

strand ε<sup>T</sup> is the sum of εSlid and εThermal. In this case, ε<sup>T</sup> ¼ εThermal þ εSlid.

dM dx <sup>þ</sup> <sup>N</sup> dy

get the differential equation of the rod with the spring constraints:

EI <sup>d</sup><sup>4</sup> y dx<sup>4</sup> � <sup>N</sup> <sup>d</sup><sup>2</sup>

equilibrium equations for the moments are as follows:

The mechanical analysis of the strand microelement is shown in Figure 7. The

In Eq. (17), Qv, M, N denote the vertical shear force, the bending moment, and the compression force along the axial direction, respectively. After submitting

<sup>y</sup>=dx<sup>2</sup> into Eq. (17), and making a substitution dQv=dx <sup>¼</sup> ky, we can

y

Schematic of the strand model ignores the EM force. L, L1, L<sup>2</sup> and L<sup>3</sup> denote the length of the twist pitch, left spring constraint, free segment, and right spring constraint, respectively. k denotes the rigidity of the bandaging. EI is the bending stiffness of the strand. ε<sup>T</sup> is the total compression strain, which is equal to the sum of εThermal

dx � Qv <sup>¼</sup> <sup>0</sup> (17)

dx<sup>2</sup> <sup>þ</sup> ky <sup>¼</sup> <sup>0</sup> (18)

1 tan αst þ tan α<sup>p</sup> <sup>ξ</sup>st � <sup>υ</sup>rp tan αst 2 rin þ 2RP ξ<sup>P</sup> þ 0 � ξin þ 1 tan αst 1 rin þ 2RP Δrin � tan αstε<sup>0</sup> � β<sup>0</sup> ¼ 0 0 � ξst þ 1 tan α<sup>p</sup> <sup>þ</sup> tan <sup>α</sup><sup>p</sup> � <sup>υ</sup>RP tan α<sup>P</sup> 1 rin þ RP <sup>ξ</sup><sup>P</sup> <sup>þ</sup> <sup>0</sup> � <sup>ξ</sup>in <sup>þ</sup> 1 tan α<sup>P</sup> 1 rin þ RP Δrin � tan αPε<sup>0</sup> � β<sup>0</sup> ¼ 0 0 � ξst þ 0 � ξ<sup>P</sup> þ tan αin þ 1 tan <sup>α</sup>in <sup>ξ</sup>in <sup>þ</sup> 1 rin tan αin Δrin � tan αinε<sup>0</sup> � β<sup>0</sup> ¼ 0 � sin <sup>α</sup><sup>P</sup> 6 sin αst ð Þ EA stκ<sup>0</sup> stξst þ GBτ<sup>P</sup> þ HBκ<sup>0</sup> <sup>P</sup> � ð Þ EA <sup>P</sup>κ<sup>0</sup> P <sup>ξ</sup><sup>P</sup> <sup>þ</sup> sin α<sup>P</sup> 6 sin αin ð Þ EA inκ<sup>0</sup> inξin þ GDτ<sup>P</sup> þ HDκ<sup>0</sup> P Δrin þ GEτ<sup>P</sup> þ HEκ<sup>0</sup> P ε<sup>0</sup> <sup>þ</sup> <sup>0</sup> � <sup>β</sup><sup>0</sup> <sup>¼</sup> <sup>0</sup> ð Þ EA st sin αstξst þ 6GBτ<sup>P</sup> cos α<sup>P</sup> þ 6HBκ<sup>0</sup> <sup>P</sup> cos α<sup>P</sup> þ 6ð Þ EA <sup>P</sup> sin α<sup>P</sup> <sup>ξ</sup><sup>P</sup> <sup>þ</sup> ð Þ EA in sin <sup>α</sup>inξin þ 6 GDτ<sup>P</sup> þ HDκ<sup>0</sup> P cos <sup>α</sup>PΔrin <sup>þ</sup> <sup>6</sup> GEτ<sup>p</sup> <sup>þ</sup> HEκ<sup>0</sup> P cos <sup>α</sup>Pε<sup>0</sup> <sup>þ</sup> <sup>0</sup> � <sup>β</sup><sup>0</sup> <sup>¼</sup> <sup>F</sup><sup>0</sup> � ð Þ EA st cos αstrstξst þ 6HB sin α<sup>P</sup> þ 6GB cos α<sup>P</sup> þ 6ð Þ EA <sup>P</sup>ð Þ rin þ rP cos α<sup>P</sup> � 6 GBτ<sup>P</sup> þ HBκ<sup>0</sup> P rP sin α<sup>P</sup> ξP � ð Þ EA in cos αinrinξin þ 6HD sin α<sup>P</sup> þ 6GD cos α<sup>P</sup> � 6 GDτ<sup>P</sup> þ HDκ<sup>0</sup> P rP sin α<sup>P</sup> Δrin þ 6HE sin α<sup>P</sup> þ 6GE cos α<sup>P</sup> � 6 GEτ<sup>P</sup> þ HEκ<sup>0</sup> P rP sin α<sup>P</sup> <sup>ε</sup><sup>0</sup> <sup>þ</sup> <sup>0</sup> � <sup>β</sup><sup>0</sup> <sup>¼</sup> <sup>M</sup><sup>0</sup> (16)

The twist angle β<sup>0</sup> can be computed from the equation system (16). So, the rotation of cable can be evaluated [17].

The experimental and numerical results are shown in Figure 5. First, it is easily found that the cabling tension has less impact on cable rotation. Taking the wrapping tension with 200 N, for example, there is no deviation between two different cables with cabling tension 200 N and 800 N, respectively. This result is in good agreement with the numerical model results. Second, the untwisting of the cable was mainly controlled by wrapping tension. Therefore, we can reduce the rotation significantly by increasing the wrapping tension. When insertion force is about 40 kN, the untwisting angle of cables with 600 N wrapping tension is about half of those cases with 200 N.

Figure 5. Rotated angle per meter as a function of the force: Numerical and experimental results.
