4. Buckling behavior in the CICC

1 tan αst

0 � ξst þ

� sin <sup>α</sup><sup>P</sup> 6 sin αst

þ GEτ<sup>P</sup> þ HEκ<sup>0</sup>

þ 6 GDτ<sup>P</sup> þ HDκ<sup>0</sup>

þ tan α<sup>p</sup> 

> 1 tan α<sup>p</sup>

0 � ξst þ 0 � ξ<sup>P</sup> þ tan αin þ

ð Þ EA stκ<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>

P

those cases with 200 N.

Figure 5.

88

<sup>ξ</sup>st � <sup>υ</sup>rp tan αst

<sup>þ</sup> tan <sup>α</sup><sup>p</sup> � <sup>υ</sup>RP

stξst þ GBτ<sup>P</sup> þ HBκ<sup>0</sup>

cos <sup>α</sup>PΔrin <sup>þ</sup> <sup>6</sup> GEτ<sup>p</sup> <sup>þ</sup> HEκ<sup>0</sup>

þ 6HE sin α<sup>P</sup> þ 6GE cos α<sup>P</sup> � 6 GEτ<sup>P</sup> þ HEκ<sup>0</sup>

rotation of cable can be evaluated [17].

� 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>

2 rin þ 2RP

> 1 rin þ RP

> > ξin þ

<sup>ξ</sup><sup>P</sup> <sup>þ</sup>

<sup>P</sup> � ð Þ EA <sup>P</sup>κ<sup>0</sup> P

tan α<sup>P</sup>

1 tan αin ξ<sup>P</sup> þ 0 � ξin þ

Nuclear Fusion - One Noble Goal and a Variety of Scientific and Technological Challenges

1 rin tan αin

ξ<sup>P</sup> þ 0 � ξin þ

<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

Δrin

The twist angle β<sup>0</sup> can be computed from the equation system (16). So, the

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

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> <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>

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

1 tan αst

1 rin þ 2RP

> 1 rin þ RP

1 tan α<sup>P</sup>

sin α<sup>P</sup> 6 sin αin

Δrin � tan αinε<sup>0</sup> � β<sup>0</sup> ¼ 0

ð Þ EA inκ<sup>0</sup>

P rP sin α<sup>P</sup>

Δrin � tan αstε<sup>0</sup> � β<sup>0</sup> ¼ 0

inξin þ GDτ<sup>P</sup> þ HDκ<sup>0</sup>

Δrin � tan αPε<sup>0</sup> � β<sup>0</sup> ¼ 0

P Δrin

P rP sin α<sup>P</sup>

ξP

(16)

## 4.1 Analytical model without the electromagnetic force

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

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.

The mechanical analysis of the strand microelement is shown in Figure 7. The equilibrium equations for the moments are as follows:

$$\frac{dM}{d\mathbf{x}} + N\frac{d\mathbf{y}}{d\mathbf{x}} - Q\_v = \mathbf{0} \tag{17}$$

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>M</sup> ¼ �EI d<sup>2</sup> <sup>y</sup>=dx<sup>2</sup> into Eq. (17), and making a substitution dQv=dx <sup>¼</sup> ky, we can get the differential equation of the rod with the spring constraints:

$$EI\frac{d^4y}{dx^4} - N\frac{d^2y}{dx^2} + ky = 0\tag{18}$$

#### Figure 6.

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 and εSlid.

Figure 7. Mechanical analysis for the strand microelement.

If the strand has no wrapping, one can get

$$EI\frac{d^4y}{d\mathfrak{x}^4} - N\frac{d^2y}{d\mathfrak{x}^2} = \mathbf{0} \tag{19}$$

wjx¼<sup>0</sup> ¼ wjx¼<sup>L</sup> ¼ 0

dx

This general solution can be divided into three situations with a variation of the

wi ¼ ai1cosh ð Þþ λ1xi ai2cosh ð Þþ λ2xi ai3sinh ð Þþ λ1xi ai4sinh ð Þ λ2xi (23)

wi ¼ ai<sup>1</sup> cosð Þþ λxi ai<sup>2</sup> cosð Þþ λxi ai3xi sin ð Þþ λxi ai4xi sin ð Þ λxi (24)

wi ¼ ai<sup>1</sup> cosð Þþ λ1xi ai<sup>2</sup> cosð Þþ λ2xi ai<sup>3</sup> sin ð Þþ λ1xi ai<sup>4</sup> sin ð Þ λ2xi (25)

dx <sup>¼</sup> <sup>L</sup>ε<sup>T</sup> � NL

Here, w<sup>0</sup> denotes the curvature of the strand transverse deflection. The radius of

<sup>ρ</sup> <sup>¼</sup> <sup>d</sup><sup>2</sup> w d2 x

Here, w stands for the strand transverse deflection. The corresponding maxi-

<sup>ε</sup>max <sup>¼</sup> <sup>D</sup> 2ρ

Here, D stands for the diameter of the strand. If the maximum strain is larger than 1%, the strand would be considered as cracking [10, 24–26]. Based on these equations, the coefficient aij can be calculated, as well as the buckling deflection, the

EA (26)

(27)

(28)

Here, λ, λ1, λ<sup>2</sup> and λ<sup>3</sup> are the eigenvalues, respectively; and aij is the constant

The axial compression strain can be neglected when the strand gets into the buckling state, as the compression force N is small. Therefore, the total compression

> 1 2 ð L

> > 0

<sup>w</sup><sup>0</sup> ð Þ<sup>2</sup>

� � � � �

<sup>x</sup>¼<sup>L</sup> ¼ 0

wi ¼ ai<sup>1</sup> cosð Þþ λxi ai<sup>2</sup> sin ð Þþ λxi ai3xi þ ai<sup>4</sup> (22)

(21)

dx <sup>x</sup>¼<sup>0</sup> <sup>¼</sup> dw

dw

EIk <sup>p</sup> , the general solution has the form as

Thus, the general solution of Eq. (18) is obtained as

external force P [21–23]:

1. If P≤2 ffiffiffiffiffiffiffi

2. If <sup>P</sup> <sup>¼</sup> <sup>2</sup> ffiffiffiffiffiffiffi

3. If P≥2 ffiffiffiffiffiffiffi

coefficients.

EIk <sup>p</sup> , it becomes

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

EIk <sup>p</sup> , it becomes

of the strand can be expressed as

curvature ρ has the form

mum strain is

91

� � � � �

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

where k denotes the stiffness of the wrapping.

The continuity of displacements, twist angles, bending moments, and shear forces across the coverage/free strand require

$$w\_i = w\_{i+1}$$

$$\frac{dw\_i}{dx\_i} = \frac{dw\_{i+1}}{dx\_{i+1}}$$

$$\frac{d^2w\_i}{dx\_i^2} = \frac{d^2w\_{i+1}}{dx\_{i+1}^2} \tag{20}$$

$$-\left[N\frac{dw\_i}{dx\_i} + \frac{d^3w\_i}{dx\_{i+1}^3}\right] = -\left[N\frac{dw\_{i+1}}{dx\_{i+1}} + \frac{d^3w\_{i+1}}{dx\_{i+1}^3}\right]$$

In Eq. (20) the subscript i represents the number of wrapping/free segments. The fixed boundary conditions at the coverage fraction have the form:

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

$$\begin{aligned} \left.w\right|\_{\mathbf{x}=0} &= w|\_{\mathbf{x}=L} = \mathbf{0} \\\\ \left.\frac{dw}{d\mathbf{x}}\right|\_{\mathbf{x}=0} &= \frac{dw}{d\mathbf{x}}\Big|\_{\mathbf{x}=L} = \mathbf{0} \end{aligned} \tag{21}$$

Thus, the general solution of Eq. (18) is obtained as

$$w\_i = a\_{i1}\cos\left(\lambda\mathbf{x}\_i\right) + a\_{i2}\sin\left(\lambda\mathbf{x}\_i\right) + a\_{i3}\mathbf{x}\_i + a\_{i4} \tag{22}$$

This general solution can be divided into three situations with a variation of the external force P [21–23]:

1. If P≤2 ffiffiffiffiffiffiffi EIk <sup>p</sup> , the general solution has the form as

$$w\_i = a\_{i1}\cosh\left(\lambda\_1 \mathbf{x}\_i\right) + a\_{i2}\cosh\left(\lambda\_2 \mathbf{x}\_i\right) + a\_{i3}\sinh\left(\lambda\_1 \mathbf{x}\_i\right) + a\_{i4}\sinh\left(\lambda\_2 \mathbf{x}\_i\right) \tag{23}$$

2. If <sup>P</sup> <sup>¼</sup> <sup>2</sup> ffiffiffiffiffiffiffi EIk <sup>p</sup> , it becomes

$$w\_i = a\_{i1}\cos\left(\lambda\mathbf{x}\_i\right) + a\_{i2}\cos\left(\lambda\mathbf{x}\_i\right) + a\_{i3}\mathbf{x}\_i\sin\left(\lambda\mathbf{x}\_i\right) + a\_{i4}\mathbf{x}\_i\sin\left(\lambda\mathbf{x}\_i\right) \tag{24}$$

3. If P≥2 ffiffiffiffiffiffiffi EIk <sup>p</sup> , it becomes

$$w\_i = a\_{i1}\cos\left(\lambda\_1 \mathbf{x}\_i\right) + a\_{i2}\cos\left(\lambda\_2 \mathbf{x}\_i\right) + a\_{i3}\sin\left(\lambda\_1 \mathbf{x}\_i\right) + a\_{i4}\sin\left(\lambda\_2 \mathbf{x}\_i\right) \tag{25}$$

Here, λ, λ1, λ<sup>2</sup> and λ<sup>3</sup> are the eigenvalues, respectively; and aij is the constant coefficients.

The axial compression strain can be neglected when the strand gets into the buckling state, as the compression force N is small. Therefore, the total compression of the strand can be expressed as

$$\frac{1}{2} \int\_{0}^{L} (w')^2 d\mathbf{x} = L\varepsilon\_T - \frac{NL}{EA} \tag{26}$$

Here, w<sup>0</sup> denotes the curvature of the strand transverse deflection. The radius of curvature ρ has the form

$$
\rho = \frac{d^2 w}{d^2 \mathbf{x}}\tag{27}
$$

Here, w stands for the strand transverse deflection. The corresponding maximum strain is

$$
\varepsilon\_{\text{max}} = \frac{D}{2\rho} \tag{28}
$$

Here, D stands for the diameter of the strand. If the maximum strain is larger than 1%, the strand would be considered as cracking [10, 24–26]. Based on these equations, the coefficient aij can be calculated, as well as the buckling deflection, the

If the strand has no wrapping, one can get

Mechanical analysis for the strand microelement.

Figure 7.

90

where k denotes the stiffness of the wrapping.

forces across the coverage/free strand require

� <sup>N</sup> dwi dxi þ d3 wi dx<sup>3</sup> iþ1

" #

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

Nuclear Fusion - One Noble Goal and a Variety of Scientific and Technological Challenges

y

The continuity of displacements, twist angles, bending moments, and shear

wi ¼ wiþ<sup>1</sup>

¼ d2 wiþ<sup>1</sup> dx<sup>2</sup> iþ1

In Eq. (20) the subscript i represents the number of wrapping/free segments.

<sup>¼</sup> dwiþ<sup>1</sup> dxiþ<sup>1</sup>

¼ � <sup>N</sup> dwiþ<sup>1</sup> dxiþ<sup>1</sup> þ d3 wiþ<sup>1</sup> dx<sup>3</sup> iþ1

" #

dwi dxi

d2 wi dx<sup>2</sup> i

The fixed boundary conditions at the coverage fraction have the form:

dx<sup>2</sup> <sup>¼</sup> <sup>0</sup> (19)

(20)

relationship between the radius of curvature, and the thermal compression strain εThermal or slid strainεSlid.

### 4.2 Analytical model with the electromagnetic force

As we know, the magnetic field is maintained at 13 T in the Tcs test procedure, and the electromagnetic force rises with the increasing the current. Therefore, the electromagnetic force is a cyclic load. The strand on the surface of the cable where the gap is formed is selected in this model. The EM force FEM is perpendicular to the Nb3Sn strand, and its direction points to the inner part of the cable. Thus, the strand at the lower loading side has the least lateral constraint that means the strand most likely to have a buckling behavior in the lateral direction. Therefore, the only thing that can prevent the buckling of the strands is the friction force against the cable. The friction factor can be written as a symbol μ.

In Figure 8, μ<sup>L</sup> and μ<sup>T</sup> are the axial friction factor and lateral friction factor, respectively; Ls and L denote the slipping length and buckling length, respectively; L þ 2Ls is the twist pitch of the first stage; FEM stands for the EM force, and then in the buckling area μTFEM is the lateral constraint on the strand; and P<sup>0</sup> and P are the compression force of the strand in the slipping and buckling area, respectively.

Accordingly, by the torque balance of the microelement, as shown in Figure 9. The equilibrium equation is as follows:

$$-Ely'' = Py + \frac{\mu\_T F\_{EM}}{2} \left(\frac{L}{2} - \infty\right) \left(\frac{L}{2} - \infty\right) - \frac{\mu\_T F\_{EM} L}{2} \left(\frac{L}{2} - \infty\right) \tag{29}$$

In Eq. (29) EI stands for the bending stiffness of the Nb3Sn strand; y stands for the deflection of the buckling. After simplifying formula (29), one can get

$$EI\mathbf{y}'' + P\mathbf{y} + \frac{\mu\_T F\_{\text{EM}}}{2} \left(\frac{L^2}{4} - \boldsymbol{\kappa}^2\right) = \mathbf{0} \tag{30}$$

The general solution [27] of formula (30) is given as

EI , n<sup>2</sup> <sup>¼</sup> <sup>P</sup>

undetermined coefficients. After applying y <sup>x</sup>¼ � <sup>L</sup>=<sup>2</sup>

Schematic view of the mechanical analysis for the microelement.

<sup>y</sup> <sup>¼</sup> <sup>m</sup> n4

If we substitute the boundary y<sup>0</sup> <sup>x</sup>¼ � <sup>L</sup>=<sup>2</sup>

sion force in the buckling area is <sup>P</sup> <sup>¼</sup> <sup>80</sup>:76EI=L<sup>2</sup>

ð Þ P<sup>0</sup> � P L EA �

and the length of the buckling area:

93

ð Þ P<sup>0</sup> � P L

<sup>n</sup><sup>4</sup> <sup>A</sup> cosð Þþ nx <sup>B</sup> sin ð Þ� nx

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

� cosð Þ nx cosð Þ nL=<sup>2</sup> � <sup>n</sup><sup>2</sup>x<sup>2</sup>

�

axial direction between the buckling and slipping segments; we can get that

The geometric compatibility equation of the strand can be expressed as

1 2 y02

Submitting Eq. (32) into Eq. (35), one can get the relation between axial force

EI � �<sup>2</sup>

dx <sup>þ</sup> <sup>μ</sup>LFEML<sup>2</sup>

S

<sup>L</sup><sup>7</sup> <sup>þ</sup> <sup>μ</sup>LFEML<sup>2</sup>

<sup>P</sup><sup>0</sup> � <sup>P</sup> <sup>¼</sup> <sup>μ</sup>LFEML

ð L=2

�L=2

EA � <sup>1</sup>:<sup>597</sup> � <sup>10</sup>�<sup>5</sup> <sup>μ</sup>TFEM

n<sup>2</sup>x<sup>2</sup> 2 þ

� ¼ 0, y<sup>0</sup>

n<sup>2</sup>L<sup>2</sup> <sup>8</sup> <sup>þ</sup> <sup>1</sup> � � (32)

� ¼ 0 to Eq. (32), Eq. (32) becomes

. There must be a balance in the

EA <sup>¼</sup> <sup>0</sup> (35)

S

EA <sup>¼</sup> 0 (36)

<sup>2</sup> <sup>þ</sup> <sup>μ</sup>LFEMLS (34)

y<sup>0</sup> ¼ tan ð Þ� nL=2 nL=2 ¼ 0 (33)

2 þ

�

We can deduce that nL ¼ 8:9868… from Eq. (33). Therefore, the axial compres-

� � (31)

n<sup>2</sup>L<sup>2</sup> <sup>8</sup> <sup>þ</sup> <sup>1</sup>

EI. Two boundary conditions are needed to solve the

j<sup>x</sup>¼<sup>0</sup> ¼ 0 to formula (31),

<sup>y</sup> <sup>¼</sup> <sup>m</sup>

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

In Eq. (31), <sup>m</sup> <sup>¼</sup> <sup>μ</sup>TFEM

one can get

Figure 9.

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

Figure 9. Schematic view of the mechanical analysis for the microelement.

The general solution [27] of formula (30) is given as

$$y = \frac{m}{n^4} \left[ A \cos \left( n\infty \right) + B \sin \left( n\infty \right) - \frac{n^2 \pi^2}{2} + \frac{n^2 L^2}{8} + \mathbf{1} \right] \tag{31}$$

In Eq. (31), <sup>m</sup> <sup>¼</sup> <sup>μ</sup>TFEM EI , n<sup>2</sup> <sup>¼</sup> <sup>P</sup> EI. Two boundary conditions are needed to solve the undetermined coefficients. After applying y <sup>x</sup>¼ � <sup>L</sup>=<sup>2</sup> � � ¼ 0, y<sup>0</sup> j<sup>x</sup>¼<sup>0</sup> ¼ 0 to formula (31), one can get

$$y = \frac{m}{n^4} \left[ \frac{-\cos\left(n\pi\right)}{\cos\left(nL/2\right)} - \frac{n^2\pi^2}{2} + \frac{n^2L^2}{8} + 1 \right] \tag{32}$$

If we substitute the boundary y<sup>0</sup> <sup>x</sup>¼ � <sup>L</sup>=<sup>2</sup> � � ¼ 0 to Eq. (32), Eq. (32) becomes

$$\mathbf{y}' = \tan\left(nL/2\right) - nL/2 = \mathbf{0} \tag{33}$$

We can deduce that nL ¼ 8:9868… from Eq. (33). Therefore, the axial compression force in the buckling area is <sup>P</sup> <sup>¼</sup> <sup>80</sup>:76EI=L<sup>2</sup> . There must be a balance in the axial direction between the buckling and slipping segments; we can get that

$$P\_0 - P = \frac{\mu\_L F\_{EM} L}{2} + \mu\_L F\_{EM} L\_S \tag{34}$$

The geometric compatibility equation of the strand can be expressed as

$$\frac{(P\_0 - P)L}{EA} - \int\_{-L/2}^{L/2} \frac{1}{2} \mathbf{y}'^2 d\mathbf{x} + \frac{\mu\_L F\_{EM} L\_S^2}{EA} = \mathbf{0} \tag{35}$$

Submitting Eq. (32) into Eq. (35), one can get the relation between axial force and the length of the buckling area:

$$\frac{(P\_0 - P)L}{EA} - \mathbf{1.597} \times \mathbf{10}^{-5} \left(\frac{\mu\_T F\_{\rm EM}}{EI}\right)^2 L^7 + \frac{\mu\_L F\_{\rm EM} L\_S^2}{EA} = \mathbf{0} \tag{36}$$

relationship between the radius of curvature, and the thermal compression strain

Nuclear Fusion - One Noble Goal and a Variety of Scientific and Technological Challenges

As we know, the magnetic field is maintained at 13 T in the Tcs test procedure, and the electromagnetic force rises with the increasing the current. Therefore, the electromagnetic force is a cyclic load. The strand on the surface of the cable where the gap is formed is selected in this model. The EM force FEM is perpendicular to the Nb3Sn strand, and its direction points to the inner part of the cable. Thus, the strand at the lower loading side has the least lateral constraint that means the strand most likely to have a buckling behavior in the lateral direction. Therefore, the only thing that can prevent the buckling of the strands is the friction force against the

In Figure 8, μ<sup>L</sup> and μ<sup>T</sup> are the axial friction factor and lateral friction factor, respectively; Ls and L denote the slipping length and buckling length, respectively; L þ 2Ls is the twist pitch of the first stage; FEM stands for the EM force, and then in the buckling area μTFEM is the lateral constraint on the strand; and P<sup>0</sup> and P are the compression force of the strand in the slipping and buckling area,

Accordingly, by the torque balance of the microelement, as shown in Figure 9.

<sup>2</sup> � <sup>x</sup> 

In Eq. (29) EI stands for the bending stiffness of the Nb3Sn strand; y stands for

L2 <sup>4</sup> � <sup>x</sup><sup>2</sup> 

the deflection of the buckling. After simplifying formula (29), one can get

2

EIy″ <sup>þ</sup> Py <sup>þ</sup> <sup>μ</sup>TFEM

� <sup>μ</sup>TFEML 2

L <sup>2</sup> � <sup>x</sup> 

¼ 0 (30)

(29)

4.2 Analytical model with the electromagnetic force

cable. The friction factor can be written as a symbol μ.

2

L <sup>2</sup> � <sup>x</sup> L

The equilibrium equation is as follows:

�EIy″ <sup>¼</sup> Py <sup>þ</sup> <sup>μ</sup>TFEM

Schematic of the analytical model with the EM force.

εThermal or slid strainεSlid.

respectively.

Figure 8.

92

Submitting Eq. (34) into Eq. (36) and eliminating the Ls, one can get P<sup>0</sup> ¼ P þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>1</sup>:<sup>597</sup> � <sup>10</sup>�<sup>5</sup> EAμLFEM <sup>μ</sup>TFEM EI � �<sup>2</sup> <sup>L</sup><sup>7</sup> � <sup>L</sup><sup>2</sup> <sup>4</sup> ð Þ μLFEM 2 r , in which <sup>P</sup> <sup>¼</sup> <sup>80</sup>:76EI=L<sup>2</sup> .

Consider the special case LS ¼ 0 that means the end of the strand is locked and the length of the buckling is equal to the twist pitch. In this case Eq. (36) can be simplified as follows:

$$P\_0 = 80.76 \frac{EI}{L^2} + 1.597 \times 10^{-5} EA \left(\frac{\mu\_T F\_{EM}}{EI}\right)^2 L^6 \tag{37}$$

region is about 0.4 m. During the EM cyclic loading, the cable can slide into the high-field region, which can accelerate the wire bending and leading the strand to

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

5. Mechanical behavior of the CICC caused by electromagnetic force

focus on the relationship between Tcs and axial strain of the cable.

be analyzed by using the thin rod model as shown in Figure 12.

The CICC qualification test samples show gradual degradation of the currentsharing temperature (Tcs) under several hundreds of EM cycles [29, 30], which leads to the Nb3Sn strand's bending or compressing deformation. In this section, we

It is known that 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 fourth petals and the sub-cables are wrapped with stainless steel tapes. Then, the wrapped cable were inserted into a stainless steel tube, which act as an amor, as illustrated in the cross-sectional view in Figure 11. The inner diameter of the CS armor is equal to 36 and 37 mm for TF, respectively. The side length of CS conductor is 49 mm, and the diameter of the TF conductor is of 40 mm. More than 1000 Nb3Sn strands were in the CICC conductors. In this section, the conductors can be simplified into a rope and the petals with circle cross section, as shown in Figure 11. Thus, the cable could

During the calculations, the parameters R, Rin, and twist pitch (h) are 6, 6, and 450 mm, respectively. According to the geometric relation h ¼ 2πrtan α, as shown

Figure 13(a) also displays the loads acting on the petal and the geometric relation of

Schematic illustration of the CS and TF CICC cross sections. The symbols Rin and R denote the radius of the He

(a) The deformation of the cable caused by the axial compression; (b) the cross section of the cable.

channel and petal, respectively; r is on behalf of the sum of Rin and R: r ¼ RIn þ R.

, where α represents the initial helix angle.

fracture. This process is shown in Figure 10(c).

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

and thermal mismatch

in Figure 13(b), we can get <sup>α</sup> <sup>¼</sup> <sup>80</sup>:5<sup>∘</sup>

the petal centerline.

Figure 11.

Figure 12.

95

Based on these equations, the critical buckling load P<sup>0</sup> can be calculated, as well as the relationship between the buckling length, bending stiffness, and the friction factor.

We can know that the strand buckling behavior is depending on the twist pitch of the first stage; the shorter the lay length, the lower the possibility of the strand buckling. The higher the wrap rigidity, the stronger the strand. Furthermore, if we fixed the coverage rate, and with a narrow wrap, which would lead to almost uniform lateral supporting, it can also prevent the strand slid into buckling. This mechanism of buckling is shown in Figure 10.

When there is no thermal load and EM force, the original strand in the CICCs is shown in Figure 10(a). When the temperature is cooled down, the axial strain caused by thermal mismatch and the lateral compression raised by the EM load are applied to the strands. We can see that the initial "straight" strand was keeping its shape by well lateral support. When the strand working in a large magnetic field (including self-field), it bears a huge EM force. A large void is generated, at the same time, some strands bending, which is shown in Figure 10(b). Additionally, this bending strain is not the only factor to make the strand fracture. For the SULTAN measurements [28], the samples is about 3.6 m long, and the high-field

Figure 10. Mechanism of the strand buckling behavior during the cooling and conducting processes.

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

region is about 0.4 m. During the EM cyclic loading, the cable can slide into the high-field region, which can accelerate the wire bending and leading the strand to fracture. This process is shown in Figure 10(c).
