5. Mechanical behavior of the CICC caused by electromagnetic force and thermal mismatch

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

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

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 in Figure 13(b), we can get <sup>α</sup> <sup>¼</sup> <sup>80</sup>:5<sup>∘</sup> , where α represents the initial helix angle. Figure 13(a) also displays the loads acting on the petal and the geometric relation of the petal centerline.

Figure 11.

Submitting Eq. (34) into Eq. (36) and eliminating the Ls, one can get

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EI � �<sup>2</sup>

<sup>L</sup><sup>2</sup> <sup>þ</sup> <sup>1</sup>:<sup>597</sup> � <sup>10</sup>�<sup>5</sup>

<sup>L</sup><sup>7</sup> � <sup>L</sup><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

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

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

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

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

<sup>4</sup> ð Þ μLFEM

EA <sup>μ</sup>TFEM EI � �<sup>2</sup>

2

, in which

L<sup>6</sup> (37)

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

EAμLFEM <sup>μ</sup>TFEM

<sup>P</sup><sup>0</sup> <sup>¼</sup> <sup>80</sup>:<sup>76</sup> EI

mechanism of buckling is shown in Figure 10.

P<sup>0</sup> ¼ P þ

factor.

Figure 10.

94

<sup>P</sup> <sup>¼</sup> <sup>80</sup>:76EI=L<sup>2</sup>

r

simplified as follows:

<sup>1</sup>:<sup>597</sup> � <sup>10</sup>�<sup>5</sup>

.

Schematic illustration of the CS and TF CICC cross sections. The symbols Rin and R denote the radius of the He channel and petal, respectively; r is on behalf of the sum of Rin and R: r ¼ RIn þ R.

Figure 12. (a) The deformation of the cable caused by the axial compression; (b) the cross section of the cable.

Substituting Eqs. (40) and (43) into Eq. (41), the axial strain of the petal ξ which

<sup>1</sup> � 2 sin <sup>2</sup><sup>α</sup>

Δα

<sup>1</sup> � 2 sin <sup>2</sup><sup>α</sup>

" # � � <sup>Δ</sup><sup>α</sup> (46)

<sup>1</sup> <sup>þ</sup> <sup>υ</sup> <sup>þ</sup> 2 sin <sup>2</sup>

� <sup>1</sup>

<sup>1</sup> � 2 sin <sup>2</sup><sup>α</sup>

<sup>1</sup> <sup>þ</sup> <sup>υ</sup> <sup>þ</sup> 2 sin <sup>2</sup>

� �

" #

<sup>1</sup> � 2 sin <sup>2</sup><sup>α</sup>

" #

" # � � <sup>Δ</sup><sup>α</sup>

<sup>1</sup> � 2 sin <sup>2</sup><sup>α</sup>

<sup>1</sup> <sup>þ</sup> <sup>υ</sup> <sup>þ</sup> 2 sin <sup>2</sup>

ε<sup>0</sup> (50)

<sup>1</sup> <sup>þ</sup> <sup>υ</sup> <sup>þ</sup> 2 sin <sup>2</sup>

� �

<sup>1</sup> <sup>þ</sup> <sup>υ</sup> <sup>þ</sup> 2 sin <sup>2</sup>

α � �Δ<sup>α</sup> (44)

tan <sup>α</sup> (45)

α

tan <sup>α</sup> (47)

α

� 1

α

� 1

<sup>r</sup> can be

Δα

(49)

α

Δα

(48)

sin α cos α <sup>4</sup>ð Þ <sup>r</sup>=<sup>R</sup> <sup>2</sup>

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

The deformed configuration of the petal in Figure 13 yields

cos α sin α þ

θ � θ <sup>h</sup> <sup>¼</sup> <sup>r</sup> r

sin α

r þ

compression strain obeys: ε ¼ �ε0; then, Eq. (46) and Eq. (48) become

cos α sin α þ

> r þ

The relation between axial strain <sup>ε</sup><sup>0</sup> and transverse strain <sup>ε</sup>Trans <sup>¼</sup> <sup>Δ</sup><sup>r</sup>

<sup>ε</sup>Trans <sup>¼</sup> <sup>C</sup><sup>4</sup>

sin <sup>α</sup> <sup>þ</sup> sin <sup>α</sup> cos <sup>α</sup> <sup>4</sup>ð Þ <sup>r</sup>=<sup>R</sup> <sup>2</sup>

<sup>1</sup>�2 sin <sup>2</sup><sup>α</sup>

It is found that the coefficient between transverse and axial strains is affected by the helical angle α and Poisson's ratio υ. When the twist pitch of the fifth stage cable is of 427, 450, and 476 mm, the corresponding helical angle of the petal equals to 80, 80.5, and 81°, respectively. Substituting these values into Eq. (50), one can see that the axial strain of the cable ε<sup>0</sup> has a linear relationship with the transverse strain

At the two ends of the petal per twist pitch, the rotation is zero, and the

<sup>ε</sup> <sup>¼</sup> <sup>h</sup> � <sup>h</sup>

<sup>h</sup> <sup>¼</sup> <sup>ξ</sup> <sup>þ</sup>

sin α cos α <sup>4</sup>ð Þ <sup>r</sup>=<sup>R</sup> <sup>2</sup>

The angle of twist per unit length τ of the petal can be defined by the expression

1 þ ξ tan <sup>α</sup> � <sup>Δ</sup><sup>α</sup> � �

cos <sup>2</sup>α <sup>4</sup>ð Þ <sup>r</sup>=<sup>R</sup> <sup>2</sup>

sin α cos α <sup>4</sup>ð Þ <sup>r</sup>=<sup>R</sup> <sup>2</sup>

cos <sup>2</sup>α <sup>4</sup>ð Þ <sup>r</sup>=<sup>R</sup> <sup>2</sup>

C2C<sup>3</sup> � C1C<sup>4</sup>

<sup>1</sup>�2 sin <sup>2</sup><sup>α</sup>

<sup>1</sup>þ<sup>υ</sup> <sup>þ</sup> 2 sin <sup>2</sup><sup>α</sup> � � � 1.

<sup>1</sup>þ<sup>υ</sup> <sup>þ</sup> 2 sin <sup>2</sup><sup>α</sup> � �,

is only related to the Δr and Δα can be written as

Δr r þ

ð Þ 1 þ υ

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

From Eqs. (44) and (45), one can get

Δr r þ

rτ<sup>c</sup> ¼ r

ð Þ <sup>1</sup> <sup>þ</sup> <sup>υ</sup> � cos <sup>α</sup>

" # <sup>Δ</sup><sup>r</sup>

ð Þ 1 þ υ

" # <sup>Δ</sup><sup>r</sup>

ð Þ <sup>1</sup> <sup>þ</sup> <sup>υ</sup> � cos <sup>α</sup>

Δr r þ

sin α

ð Þ <sup>1</sup>þ<sup>υ</sup> , <sup>C</sup><sup>2</sup> <sup>¼</sup> cos <sup>α</sup>

sin <sup>α</sup>, <sup>C</sup><sup>4</sup> <sup>¼</sup> cos <sup>2</sup><sup>α</sup>

<sup>4</sup>ð Þ <sup>r</sup>=<sup>R</sup> <sup>2</sup>

εTrans, which is displayed in Figure 14. We carried out an experiment by

<sup>ξ</sup> <sup>¼</sup> <sup>υ</sup> sin <sup>2</sup><sup>α</sup> cos <sup>2</sup><sup>α</sup> <sup>4</sup>ð Þ <sup>r</sup>=<sup>R</sup> <sup>2</sup>

<sup>ε</sup> <sup>¼</sup> <sup>υ</sup> sin <sup>2</sup><sup>α</sup> cos <sup>2</sup><sup>α</sup> <sup>4</sup>ð Þ <sup>r</sup>=<sup>R</sup> <sup>2</sup>

That is,

8 >>>>><

>>>>>:

97

expressed as

<sup>r</sup>τ<sup>c</sup> <sup>¼</sup> <sup>υ</sup> sin <sup>α</sup> cos <sup>2</sup><sup>α</sup> <sup>4</sup>ð Þ <sup>r</sup>=<sup>R</sup> <sup>2</sup>

�ε<sup>0</sup> <sup>¼</sup> <sup>υ</sup> sin <sup>2</sup><sup>α</sup> cos <sup>2</sup><sup>α</sup> <sup>4</sup>ð Þ <sup>r</sup>=<sup>R</sup> <sup>2</sup>

<sup>0</sup> <sup>¼</sup> <sup>υ</sup> sin <sup>α</sup> cos <sup>2</sup><sup>α</sup> <sup>4</sup>ð Þ <sup>r</sup>=<sup>R</sup> <sup>2</sup>

where <sup>C</sup><sup>1</sup> <sup>¼</sup> <sup>υ</sup> sin <sup>2</sup><sup>α</sup> cos <sup>2</sup><sup>α</sup> <sup>4</sup>ð Þ <sup>r</sup>=<sup>R</sup> <sup>2</sup>

ð Þ <sup>1</sup>þ<sup>υ</sup> � cos <sup>α</sup>

<sup>C</sup><sup>3</sup> <sup>¼</sup> <sup>υ</sup> sin <sup>α</sup> cos <sup>2</sup><sup>α</sup> <sup>4</sup>ð Þ <sup>r</sup>=<sup>R</sup> <sup>2</sup>

ð Þ 1 þ υ

Figure 13. (a) Uniform forces acting on the petal; (b) geometric relation of the petal centerline.

Assume that the petals were contact with each other in the original state. The curvature and the twist per unit length of the petal is k, k<sup>0</sup> , τ; then the changes can be written as Δk, Δk<sup>0</sup> , Δτ. They can be expressed as [19]

$$
\Delta \kappa' = \frac{\cos^2 \overline{a}}{\overline{r}} - \frac{\cos^2 a}{r} = \frac{-\cos^2 a}{r} \frac{\Delta r}{r} - 2 \frac{\sin a \cos a}{r} \Delta a \tag{38}
$$

$$
\Delta \tau = \frac{\cos \overline{a} \sin \overline{a}}{\overline{r}} - \frac{\cos a \sin a}{r} = \frac{-\sin a \cos a}{r} \frac{\Delta r}{r} + \frac{(1 - 2 \sin^2 a)}{r} \Delta a \tag{39}
$$

In Eqs. (38) and (39), α and α is the initial and final helical angle of the petal, Δα and Δr stand for the change of helical angle and radius of the petal, and r and r denote the original and final radius, respectively. The loads and moments can be deduced as.

$$\text{tr}\,G'=\frac{E\pi R^4}{4}\Delta\kappa';H=\frac{E\pi R^4}{4(1+\nu)}\Delta\tau;T=\pi ER^2\xi\tag{40}$$

Here, T, G<sup>0</sup> , H stand for the axial load, bending moment, and twist moment of the petal, respectively; E is of the axial stiffness of the petal, v is Poisson's ratio, and ξ stands for the strain in the axial direction of the petal. If the petal is free in the initial state and there is no contact force under the compression state, then the uniform load between the petals is equal to zero. According to the thin rod model presented by Costello [19], the following equation is satisfied:

$$X = N'\frac{\sin a \cos a}{r} - T\frac{\cos^2 a}{r} = 0\tag{41}$$

$$N' = H \frac{\cos^2 a}{r} - G' \frac{\sin a \cos a}{r} \tag{42}$$

In Eqs. (41) and (42), X stands for the resultant contact force per unit length of a petal; N<sup>0</sup> is of the shear force acting on the petal. From Eqs. (38)–(40) and Eq. (42), the shearing force N<sup>0</sup> acting on the petal can be written as

$$\frac{N'}{\left(\overline{E}\overline{R}^2\right)^2} = \frac{\pi\nu\sin a\cos^3 a}{4(r/R)^2(1+\nu)}\frac{\Delta r}{r} + \frac{\pi\cos^2 a}{4(r/R)^2} \left(\frac{1-2\sin^2 a}{1+\nu} + 2\sin^2 a\right)\Delta a \tag{43}$$

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

Substituting Eqs. (40) and (43) into Eq. (41), the axial strain of the petal ξ which is only related to the Δr and Δα can be written as

$$\xi = \frac{\nu \sin^2 a \cos^2 a}{4(r/R)^2 (1+\nu)} \frac{\Delta r}{r} + \frac{\sin a \cos a}{4(r/R)^2} \left(\frac{1-2\sin^2 a}{1+\nu} + 2\sin^2 a\right) \Delta a \tag{44}$$

The deformed configuration of the petal in Figure 13 yields

$$
\varepsilon = \frac{\overline{h} - h}{h} = \xi + \frac{\Delta a}{\tan a} \tag{45}
$$

From Eqs. (44) and (45), one can get

$$\varepsilon = \frac{\nu \sin^2 a \cos^2 a}{4(r/R)^2 (1+\nu)} \frac{\Delta r}{r} + \left[ \frac{\cos a}{\sin a} + \frac{\sin a \cos a}{4(r/R)^2} \left( \frac{1 - 2 \sin^2 a}{1+\nu} + 2 \sin^2 a \right) \right] \Delta a \tag{46}$$

The angle of twist per unit length τ of the petal can be defined by the expression

$$r\tau\_c = r\frac{\overline{\theta} - \theta}{h} = \frac{r}{\overline{r}} \left( \frac{1 + \xi}{\tan a} - \Delta a \right) - \frac{1}{\tan a} \tag{47}$$

That is,

Assume that the petals were contact with each other in the original state. The

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<sup>r</sup> <sup>¼</sup> � cos <sup>2</sup><sup>α</sup> r

<sup>r</sup> <sup>¼</sup> � sin <sup>α</sup> cos <sup>α</sup> r

and Δr stand for the change of helical angle and radius of the petal, and r and r denote the original and final radius, respectively. The loads and moments can be

> ; H <sup>¼</sup> <sup>E</sup>πR<sup>4</sup> 4 1ð Þ þ ν

the petal, respectively; E is of the axial stiffness of the petal, v is Poisson's ratio, and ξ stands for the strain in the axial direction of the petal. If the petal is free in the initial state and there is no contact force under the compression state, then the uniform load between the petals is equal to zero. According to the thin rod model

In Eqs. (38) and (39), α and α is the initial and final helical angle of the petal, Δα

Δr

, H stand for the axial load, bending moment, and twist moment of

<sup>r</sup> � <sup>T</sup> cos <sup>2</sup><sup>α</sup>

<sup>r</sup> � <sup>G</sup><sup>0</sup> sin <sup>α</sup> cos <sup>α</sup>

In Eqs. (41) and (42), X stands for the resultant contact force per unit length of a petal; N<sup>0</sup> is of the shear force acting on the petal. From Eqs. (38)–(40) and Eq. (42),

> π cos <sup>2</sup>α <sup>4</sup>ð Þ <sup>r</sup>=<sup>R</sup> <sup>2</sup>

r

<sup>1</sup> � 2 sin <sup>2</sup><sup>α</sup>

<sup>1</sup> <sup>þ</sup> <sup>υ</sup> <sup>þ</sup> 2 sin <sup>2</sup>

Δr r þ

<sup>r</sup> � <sup>2</sup> sin <sup>α</sup> cos <sup>α</sup> r

<sup>Δ</sup>τ; T <sup>¼</sup> <sup>π</sup>ER<sup>2</sup>

<sup>1</sup> � 2 sin <sup>2</sup> ð Þ <sup>α</sup> r

, τ; then the changes can be

Δα (38)

ξ (40)

(42)

Δα (43)

<sup>r</sup> <sup>¼</sup> <sup>0</sup> (41)

α

Δα (39)

curvature and the twist per unit length of the petal is k, k<sup>0</sup>

<sup>r</sup> � cos <sup>2</sup><sup>α</sup>

(a) Uniform forces acting on the petal; (b) geometric relation of the petal centerline.

<sup>Δ</sup>κ<sup>0</sup> <sup>¼</sup> cos <sup>2</sup><sup>α</sup>

<sup>r</sup> � cos <sup>α</sup> sin <sup>α</sup>

<sup>G</sup><sup>0</sup> <sup>¼</sup> <sup>E</sup>πR<sup>4</sup> 4

Δκ<sup>0</sup>

presented by Costello [19], the following equation is satisfied:

the shearing force N<sup>0</sup> acting on the petal can be written as

Δr r þ

ð Þ 1 þ υ

<sup>X</sup> <sup>¼</sup> <sup>N</sup><sup>0</sup> sin <sup>α</sup> cos <sup>α</sup>

<sup>N</sup><sup>0</sup> <sup>¼</sup> <sup>H</sup> cos <sup>2</sup><sup>α</sup>

, Δτ. They can be expressed as [19]

written as Δk, Δk<sup>0</sup>

Figure 13.

deduced as.

Here, T, G<sup>0</sup>

N0

96

ER<sup>2</sup> <sup>¼</sup> <sup>π</sup> <sup>v</sup> sin <sup>α</sup> cos <sup>3</sup><sup>α</sup> <sup>4</sup>ð Þ <sup>r</sup>=<sup>R</sup> <sup>2</sup>

<sup>Δ</sup><sup>τ</sup> <sup>¼</sup> cos <sup>α</sup> sin <sup>α</sup>

$$r\tau\_{\varepsilon} = \left[\frac{\nu\sin a\cos^2 a}{4(r/R)^2(1+\nu)} - \frac{\cos a}{\sin a}\right] \frac{\Delta r}{r} + \left[\frac{\cos^2 a}{4(r/R)^2} \left(\frac{1-2\sin^2 a}{1+\nu} + 2\sin^2 a\right) - 1\right] \Delta a \tag{48}$$

At the two ends of the petal per twist pitch, the rotation is zero, and the compression strain obeys: ε ¼ �ε0; then, Eq. (46) and Eq. (48) become

$$\begin{cases} -\varepsilon\_0 = \frac{\nu \sin^2 a \cos^2 a}{4(r/R)^2 (1+\nu)} \frac{\Delta r}{r} + \left[ \frac{\cos a}{\sin a} + \frac{\sin a \cos a}{4(r/R)^2} \left( \frac{1 - 2 \sin^2 a}{1+\nu} + 2 \sin^2 a \right) \right] \Delta a \\\ 0 = \left[ \frac{\nu \sin a \cos^2 a}{4(r/R)^2 (1+\nu)} - \frac{\cos a}{\sin a} \right] \frac{\Delta r}{r} + \left[ \frac{\cos^2 a}{4(r/R)^2} \left( \frac{1 - 2 \sin^2 a}{1+\nu} + 2 \sin^2 a \right) - 1 \right] \Delta a \end{cases} \tag{49}$$

The relation between axial strain <sup>ε</sup><sup>0</sup> and transverse strain <sup>ε</sup>Trans <sup>¼</sup> <sup>Δ</sup><sup>r</sup> <sup>r</sup> can be expressed as

$$
\varepsilon\_{T\text{trans}} = \frac{C\_4}{C\_2 C\_3 - C\_1 C\_4} \varepsilon\_0 \tag{50}
$$

where  $\mathbf{C}\_{1} = \frac{\nu \sin^{2} a \cos^{2} a}{4(r/R)^{2}(1+\nu)}$ ,  $\mathbf{C}\_{2} = \frac{\cos a}{\sin a} + \frac{\sin a \cos a}{4(r/R)^{2}} \left(\frac{1 - 2\sin^{2} a}{1+\nu} + 2\sin^{2} a\right),$   $\mathbf{C}\_{3} = \frac{\nu \sin a \cos^{2} a}{4(r/R)^{2}(1+\nu)} - \frac{\cos a}{\sin a},$   $\mathbf{C}\_{4} = \frac{\cos^{2} a}{4(r/R)^{2}} \left(\frac{1 - 2\sin^{2} a}{1+\nu} + 2\sin^{2} a\right) - \mathbf{1}.$   $\mathbf{1},$   $\mathbf{1},$ 

It is found that the coefficient between transverse and axial strains is affected by the helical angle α and Poisson's ratio υ. When the twist pitch of the fifth stage cable is of 427, 450, and 476 mm, the corresponding helical angle of the petal equals to 80, 80.5, and 81°, respectively. Substituting these values into Eq. (50), one can see that the axial strain of the cable ε<sup>0</sup> has a linear relationship with the transverse strain εTrans, which is displayed in Figure 14. We carried out an experiment by

STP (short twist pitch) shows an opposite phenomenon, the Tcs becomes a constant even have an enhanced. These experimental results presented in Ref. [30] have a contrast with analytical prediction for many years. Using the theoretical model proposed in this chapter, the Tcs enhancement and degradation behaviors can be predicted quantitatively. Figure 15 shows a comparison of calculated results and experimental results, where the lines show the theoretical results and the dots are the experimental results. It is found that during the primary stage of the EM cycle there was a quick increase of axial strain that will lead to the Tcs drop dramatically. With the same reason, samples with LTP and baseline show an obviously degradation; this same rule applies to TF conductors. However, several STP CS conductors have an increase of Tcs, which means that the initial axial compression strain has been released by the EM loads, squeezing the cables at the high-field zone into the

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

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

While it is easy to imagine when the expansions of the cable in the high-field zone get stacked, the strain will not be released and the Tcs will drop (see the black dash line in Figure 15). If the inner part of the conductor is smooth enough, the axial compression strain is released completely, the Tcs reaches its upper boundary which is shown by the orange dash line with circle symbol corresponding to the length of the high-field region is 400 mm. If the high-field region's length is 1000 mm, then the lower boundary of the enhancing Tcs is given with purple line with triangle symbol. We can find that the theoretical results agree with the exper-

In the past half-century, the structure of CICC conductor was under continuous optimization and improvement. The CICC conductors have so many advantages such as good self-support, high operational stability, high current carrying capacity, low AC loss, etc., and they are widely used in the superconducting magnets. The nuclear fusion device CFTER built by China has also chosen the CICC conductors. However, there still have some problems to be solved: (1) there is a necessity for theoretical model to explain the degradation of critical current caused by mechanical deformation for the Nb3Sn strand; (2) accurate description of the mechanical response of complex strand structures in the coupling fields remains a challenging problem; (3) the untwisting behavior during the cable penetration is still not suppressed completely. One needs a more effective model to optimize the

manufacturing process; and (4) the long-term stability and real-time monitoring of superconducting magnets are also a challenge for the engineers. Based on this, the equivalent mechanical parameters of CICC conductors and their mechanical behavior under coupled fields will be further studied. The authors hope that these

This work is supported by the Funds of the National Natural Science Foundation

of China (Nos. 11622217, 11802291, and 11872196), the National Key Project of Scientific Instrument and Equipment Development (11327802), and National Program for Special Support of Top-Notch Young Professionals. This work is also supported by the Fundamental Research Funds for the Central Universities

models can provide a valuable reference for the related researchers.

low-field region.

imental results very well.

Acknowledgements

99

(lzujbky-2017-ot18 and lzujbky-2017-k18).

6. Discussion

#### Figure 14.

Relationship between the axial strain ε<sup>0</sup> and the transverse strain εT. The colored lines are obtained by using the presented theoretical model. The green triangle symbols denote the experimental results.

compressing a CS cable which was fabricated in CASIPP to validate the analytical model. When the CS cable specimen is compressed along axial direction, it extends transversely with high resolution. The transverse extension can be measured by using a laser sensor. The experimental setup is schematically illustrated in the inset of Figure 14. One can find that the theoretical model shows perfect consistency with the experiment.

In Figure 14, we can get that the CS conductors with baseline and LTP (long twist pitch) shows a Tcs degradation after EM cycles, while for the samples with

Figure 15. A comparison of the experimental results and theoretical results based on the presented model [18].

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

STP (short twist pitch) shows an opposite phenomenon, the Tcs becomes a constant even have an enhanced. These experimental results presented in Ref. [30] have a contrast with analytical prediction for many years. Using the theoretical model proposed in this chapter, the Tcs enhancement and degradation behaviors can be predicted quantitatively. Figure 15 shows a comparison of calculated results and experimental results, where the lines show the theoretical results and the dots are the experimental results. It is found that during the primary stage of the EM cycle there was a quick increase of axial strain that will lead to the Tcs drop dramatically. With the same reason, samples with LTP and baseline show an obviously degradation; this same rule applies to TF conductors. However, several STP CS conductors have an increase of Tcs, which means that the initial axial compression strain has been released by the EM loads, squeezing the cables at the high-field zone into the low-field region.

While it is easy to imagine when the expansions of the cable in the high-field zone get stacked, the strain will not be released and the Tcs will drop (see the black dash line in Figure 15). If the inner part of the conductor is smooth enough, the axial compression strain is released completely, the Tcs reaches its upper boundary which is shown by the orange dash line with circle symbol corresponding to the length of the high-field region is 400 mm. If the high-field region's length is 1000 mm, then the lower boundary of the enhancing Tcs is given with purple line with triangle symbol. We can find that the theoretical results agree with the experimental results very well.
