Introduction

Tubes involved in the deformation processes are generally manufactured by rolling, extrusion, and drawing. It would lead to the mechanical properties of such tubes to exhibit anisotropic characteristics. If a constitutive relation is used to simulate a tube deformation process, the experimental data involving the anisotropic properties of the formed tube must be considered to meet the prediction accuracy. But to date, only the experimental data along a tube’s axial and circumferential directions can be obtained, including the simplest approach of the uniaxial tension test using a cambered specimen, the ring hoop tension test, and tube bulging experiments [1,2,3,4,5]. However, all the experimental characteristics are based on normal stress, and the principal axis is along the axial and circumferential direction, which belong to the same principal coordinate.

Similar to characterizing the anisotropic plastic deformation characteristics of rolled sheet metals, i.e., the experimental data of a uniaxial tension along 45° direction to the rolling (RD) is usually needed to identify the constitutive parameters, it is essential to find the second principal coordinate to build the anisotropic constitutive relation for simulations of formed tubes.

Trying to apply shearing experimental data of the formed tubes involving the principal stress directions along a non-axial and circumferential direction of a tube to define a constitutive relation should be another effective method. Free-end torsion and combined axial-torsion loading are often used to achieve the shear loading of thin-walled tubes [6,7,8,9]. However, these experiments cannot provide sufficient data for characterizing a constitutive relation for numerical simulations because these tests cannot maintain the direction of principal stress steady along the desired direction in the loading process with large deformation. It should be pointed out that the pure shear state is discussed in some research work on the calibration of constitutive functions using tubes loaded with axial force and internal pressure[5, 10, 11]. This kind of pure shear stress state is caused by the axial compressive stress and circumference tensile stress, which have the same absolute stress value. However, it is not a real shear deformation as the axial stress and circumferential stress are both normal stress.

A new pure shear test for thin-walled tubes that overcomes these shortcomings will be beneficial for describing the deformation behavior of anisotropic tubes. Mechanical property testing methods for thin-walled tubes are always carried out and referred to as those for sheet metals. There are three representative shear tests for sheet, e.g., a simple shear test proposed by Miyauchi [12], the well-known test setup based on the American Society for Testing and Materials standard ASTM B831 [13], and the twin bridge torsion shear test reported by Brosius et al. [14]. The three shear tests are commonly conducted using a translational movement, and the specific regions will be deformed by shear deformation.

A limitation of these sheet shear tests is that the shear deformation often accompanies the normal stress. To avoid instability and maintain deformation in the designed deformation zone, the shear test device has to contain two clamping structures to make the material move parallel to each other. Because of the influence of the constraint of the clamping areas, normal stress occurs inevitably. Previous analyses of the in-plane shear test results commonly neglected the occurrence of normal stress components [15,16,17,18]. However, some numerical studies have reported that the normal stress components can reach magnitudes similar to shear stress [19]. If there aren’t clamping structures, the material in the shear zone tends to rotate with large deformation, which means the direction of principal stress is not steady along the desired direction.

Following up on the principle of translational movement, we introduced a novel shear test for thin-walled tubes in this paper. This test can create a steady pure shear stress state in the center zone of a tubular specimen. Specimen shapes were analyzed to reduce the heterogeneity of stress and strain distribution. Differences in the stress-strain relationships between the one obtained from the uniaxial tension test and the one obtained from the shear test proposed herein are compared and analyzed. This test makes the experimental data of tubes for the second principal axes available and enables the construction of a realistic anisotropic constitutive relation for thin-walled tubes.

Principle of Pure Shear Loading of Tubes and Design of Experimental Device

The working principle of the novel shear device of thin-walled tubes is illustrated in Fig. 1. The device used in these tests comprises a specially designed tubular specimen and two semi-circle mandrels. The testing specimen is cut from a tube. Half of the tube is removed at two opposite ends while the center zone of the specimen remains intact. The mandrel is a scalar semi-circle cylinder. The diameter of the smaller semi-circle cylinder is the same as the inner diameter of the tube. In comparison, the diameter of the larger semi-circle cylinder is larger than the outer diameter of the tube. The two semi-circle mandrels are placed inside the tubular specimen, and a tensile force parallel to the tube axis is applied to the ends with the smaller diameter for both mandrels. One mandrel is loaded at the upper half-tube, and the other is loaded at the lower half-tube. Because the tensile force is applied separately to each half-tube of the specimen, the specimen will tend to rotate. With the support of the two semi-circle mandrels, the tubular specimen can only move along the axial direction. Thus, only a shear deformation occurs in the center zone of the tubular specimen, i.e., the connection zone of the two half-tubes.

Fig. 1
figure 1

Working principle of the shear test

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The stress state of the center zone of the shear zone is also shown in Fig. 1, where the first principal stress is equal to the maximum shear stress, and the angle between the first principal stress and the axis of the tube is 45°. The second principal stress is equal to zero. The magnitude of the third principal stress is equal to the maximum shear stress, but this stress is negative. The angle between the third principal stress and the axis of the tube is -45°.

  1. 1.

    Shear stress calculation.

    The acting region of the shear stress is the interface plane of the two half-tubes through the axis of the tube. The tensile force F is measured using the force sensor of the testing machine. Shear stress can be calculated according to equation (1).

    $$\uptau =\frac{F}{2\times L\times t}$$
    (1)

    where \(\tau\) is the real-time average shear stress, F is the force required to deform the two shear zones, L is the real-time length of the shear zone, and it decreases with increasing deformation, D is the outer diameter of the tube, and t is the real-time tube thickness. Strain is measured using a three-dimensional digital image correlation (DIC) system.

    As the corresponding semi-circle mandrel and the half-tube move together, there is no friction force between them. The friction force only exists between the interface plane of the two semi-circle mandrels, so the friction force should be counted when calculating the shear stress.

  2. 2.

    Measures to prevent rotation of the mandrels.

    There exists a rotation tendency of two mandrels as two parallel forces are added to two mandrels. To ensure the movement of the two mandrels strictly along the tube axis, two concentric cylinders are added to the ends of the two semi-circle mandrels on the same side as the guide pin used in the stamping die, as shown in Fig. 2. Deformation is more concentrated in the pre-set shear zone with the guiding mandrel structure, the stability of shear deformation is improved [20].

    Fig. 2
    figure 2

    Guide pin structure for the two mandrels to improve the stability of the shear test

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Stress and Strain Distribution Homogeneity with FEM Simulation

Simulation Model

To achieve a pure shear stress state, finite element simulations are conducted to obtain the optimized specimen shape. These factors below are considered to obtain the proper specimen dimensions. As the two half-tubes are under the same loading state, only half-tube A is discussed below.

  1. 1.

    Axial Stress

    For this method, the translational movement of the two half-tubes is caused by the compressive force Fz acting on the end plane (perpendicular to the tube axis) of each half-tube along the axial direction, as shown in Fig. 3(a). Increasing the shear zone length means greater compressive force on the end plane of the half-tube, which will make the axial normal stress bigger. If the value of the normal stress on the end plane of the half-tube is too high, the pure shear stress state will be strongly perturbed. So, the length of the shear zone should not be too long.

    Fig. 3
    figure 3

    Force diagram of the specimen

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    Long and thin slots are set to the two ends of the specimen’s shear zone to ensure a certain distance between the end plane and the shear zone. Thus, the axial normal stress in the shear zone is much smaller than that on the end plane. Meanwhile, if the length is too small, strain measurement error will increase, disturbing the results.

  2. 2.

    Circumferential stress

    The two parallel compressive forces act on the specimen, which causes a torque T to make the specimen rotate, as shown in Fig. 3(b). As the rotation of the specimen was hindered by the semi-circle mandrels, there will be a counterforce from the part of the semi-circle mandrel with a small diameter to the half-tubes of the specimen along the normal direction of the interface plane of the two half-tubes.

The semi-circle mandrels are named upper semi-circle mandrel and lower semi-circle mandrel, while the two half-tubes are named half-tube A and half-tube B. The counterforce from the upper mandrel to half-tube A is called Fupper-A, and the counterforce from the lower mandrel to half-tube B is called Flower-B, the two forces balance the torque together. The direction of the counterforce Fupper-A is the normal direction of the interface plane of the two half-tubes, thus it causes circumferential stress in the shear zone. Suppose the length of the half-tube is much longer than the length of the shear zone. In that case, the action area of the counterforce is larger, and the distance between the action area and the shear zone is larger, so the circumferential stress can be reduced to a minimum, which helps maintain the pure shear stress state.

Considering the above factors comprehensively, the dimension of the specimen is shown in Fig. 4. The length of the specimen is much larger than the length of the shear zone. The shear zone length L and the axial slot length Ls are to be decided from the simulation. The width of the axial slot is set as 0.3 mm, which is the minimum length that the wire-electrode cutting operation can machine.

Fig. 4
figure 4

Dimension of the specimen (unit: mm)

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As the shear deformation is under the plane stress and the plane strain state, the shear specimen is meshed with shell elements (S3 and S4R) in ABAQUS/Standard. To save the calculation time, the element edge length in the shear zone and the shell element thickness are set as 0.1 mm. The element size is a little bit large for the 0.15 mm radius of the axial slot but is enough for the deformation zone. Isotropic elastic-plastic material behavior, according to the Mises criterion, is used. The flow curve used for the input of the numerical simulation is extrapolated from the result of a uniaxial tension test along the axial direction of the tube. In the simulation, one-half of the tube is loaded with a specific axial displacement. At the same time, the other half is fixed to realize the axial shear deformation of the tube specimen. The two semi-circle mandrels are set to move along the axis of the tube, and there is no constraint for the specimen. The gap between the inner surface of the tube and the mandrel is set to 0.05 mm, the same as in the experiment. Simulation is carried out according to Table 1. The friction coefficient between the two semi-circle mandrels is set as 0.2.

Table 1 Simulation scheme
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As the elements at the two ends of the shear zone are severely distorted at larger strain and fracture is not considered in the used material model, the maximum shear strain is set as 0.3. The simulation can no longer describe the specimen behavior at this deformation stage.

Influence of the Length of the Axial Slot Along the Axial Direction on Stress Homogeneity

The length of the shear zone L is four times the thickness in this section. The nephogram of shear stress distribution is shown in Fig. 5 when the Ls/t ratio is set as 1. It is shown that the deformation of the specimen is concentrated at the interface of the two half-tubes, i.e., the pre-set shear zone. Shear stress at the two ends of the deformation zone increases fast initially, then the stress distribution in the interface plane becomes even.

Fig. 5
figure 5

Nephogram of shear stress distribution with different shear strain of the center point (Ls/t = 1, L/t = 4)

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The nephogram of shear stress distribution with different Ls/t ratios is shown in Fig. 6 when the shear strain of the center point equals 0.12. For the Ls/t ratio equals 1 and 2.5 (Fig. 6(a) and (b)), the shear strain at the two ends of the shear zone is bigger than the shear strain of the center point, and the shear strain is relative even distributed along the interface of the two half-tubes. For the Ls/t ratio equals 4 (Fig. 6(c)), the shear strain at the two ends of the shear zone is much bigger than the shear strain of the center point, and the uniformity of the stress distribution is lowered.

Fig. 6
figure 6

Nephogram of shear stress distribution with different Ls/t ratios (shear strain of the center point equals 0.12)

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Seven representative points, as shown in Fig. 7, are chosen to investigate the stress state. The seven points are located in the interface of the two half-tubes. Point 4 is the center point, while point 1 is at one end of the shear zone.

Fig. 7
figure 7

Location of the representative points

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The ratio of the axial/circumferential stress to shear stress along the interface of the shear zone, when the shear strain of the center point equals 0.05, 0.10, and 0.15, is shown in Fig. 8. Ratio of axial stress to shear stress is about 0.1–0.25 for all Ls/t values. This ratio of Ls/t equals 2.5 or 4 is smaller than that of Ls/t = 1. The ratio of circumferential stress to shear stress is bigger than that of the axial stress to shear stress. For Ls/t equals 2.5 or 4, the ratio is also smaller than Ls/t equals 1. The value is more even for Ls/t equals 2.5, so this value is chosen to conduct the simulation in the following section.

Fig. 8
figure 8

Ratio of axial/circumferential stress to shear stress along the interface of the shear zone

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Influence of the Length of the Shear Zone L

As shear stress can only be calculated according to the load and the real-time length of the shear zone, it is an average value. In this section, the influence of the length of the shear zone on stress homogeneity is analyzed. The Ls/t value is set as 2.5.

The nephogram of shear stress for different shear zone lengths when the shear strain of the center point equals 0.3 is shown in Fig. 9. It can be seen that shear deformation is concentrated in the pre-set zone, i.e., the interface of the two half-tubes.

Fig. 9
figure 9

Nephogram of shear stress of the shear zone for different L/t values

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The ratio of the shear stress of the representative points to that of the center point is used to evaluate the stress homogeneity. The relationship between the shear stress ratio and the shear strain of the center point(point 4) is shown in Fig. 10. With the increase of the L/t value, the shear stress ratio for points 1, 2, and 3 increased, which means the shear stress homogeneity decreased. For the L/t value equals 2.5, the shear stress ratio is closer to 1. This L/t value is chosen in the following simulation.

Fig. 10
figure 10

Relationship of shear stress ratio between representative points and center point with shear strain of the center point

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Influence of Torque-balanced Zone

The proper axial slot length and shear zone length were obtained through the simulation results carried on the original specimen shape shown in Fig. 4. However, it can be seen from Fig. 8(b) that the circumferential stress is relatively high, as it is about 0.3–0.5 times the shear stress.

It has been discussed that the counterforce from the upper semi-circle mandrel to half tube A caused the circumferential stress in the shear zone, as shown in Fig. 3(b). In fact, for the original shape specimen, there is also a counterforce from the lower semi-circle mandrel to the half tube A, which is named Flower-A, as shown in Fig. 11(a). Further, as the action zone of Flower-A is adjunct to the shear zone, the influence of the counterforce on the stress state, especially on the circumferential stress, cannot be ignored. To eliminate the counterforce Flower-A, the specimen shape is modified to make the half-tube A uncontacted with the lower semi-circle mandrel, as shown in Fig. 11(b). Simulation is carried out to investigate the influence of the shape modification on the stress states.

Fig.11
figure 11

The counterforce from the lower semi-circle mandrel to the half-tube A

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To accurately obtain the stress state of the endpoint of the deformation zone, the element size is set as 0.02 mm in the successive simulation. Meanwhile, because the stress value difference between the endpoint(point 1) and the neighboring point (point 2) is large, several representative points between the two points are chosen to investigate the stress state of this area.

The axial and circumferential stress distribution of representative points along the axial direction with different shear strains at the center point is shown in Fig. 12. Both axial stress and circumferential stress of the modified specimen are smaller than that of the original specimen, especially for circumferential stress at the end of the shear zone.

Fig. 12
figure 12

Normal stress distribution along the axial direction for the original specimen and modified specimen

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Mises equivalent stress is calculated using shear stress only or both normal and shear stress. The ratio of the two kinds of equivalent stress is shown in Fig. 13. The influence of the shear stress in deformation is more significant when the ratio is closer to 1. The ratio at the endpoint for the original shape specimen is only 69%. In contrast, it is about 82% for the modified shape specimen, which means that the influence of normal stress at the endpoint is still considerable. In other areas of the deformation zone, the ratio is 98% for the original shape specimen and nearly 100% for the modified specimen. It is indicated that the stress state is very close to the pure shear stress state for the modified specimen.

Fig. 13
figure 13

Ratio of Mises equivalent stress between using shear stress only and using normal stress and shear stress

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Similar to the calculation of Mises stress, Mises strain is calculated using shear strain only or both normal and shear strain for the original and modified specimen. The ratio of Mises strain between using shear strain only and using both normal and shear strain for different points is shown in Fig. 14.

Fig. 14
figure 14

Ratio between Mises strain using shear strain only and using both normal strain and shear strain

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The strain calculation results are also similar to that of the Mises stress. The proportion of the shear strain is smaller at the endpoint of the deformation zone for the original and modified specimen. The proportion of the shear strain is higher for the modified specimen than for the original specimen. It is indicated that the deformation for the modified shape specimen can be regarded as pure shear deformation.

Shear Experiment of a Thin-walled Aluminum Tube

Material and Dimension of the Specimen

An annealed thin-walled 5052 aluminum tube with an outer diameter of 50 mm and a nominal thickness of 1.2 mm was used in the experiment. The tubular specimen is shown in Fig. 15. According to the optimized result, the ratio of axial slot length to thickness Ls/t and the ratio of shear zone length to thickness L/t are both set as 2.5.

Fig. 15
figure 15

Tubular specimen for the pure shear test (L/t = 2.5)

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The length of the shear zone and the thickness of the specimen were measured for each specimen. There are two shear planes for one specimen; the areas of the two shear planes are calculated using the length and the thickness, and the shear zone with a smaller area was measured by the DIC system. The diameter of the combined two semi-circular mandrels was measured to calculate the gap between the inner surface of the tubular specimen and the circular mandrel. Two specimens were tested. The dimensions of the two specimens are shown in Table 2.

Table 2 Dimensions of the specimens (in mm, unless specified otherwise)
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Experimental Setup

The shear experiments were carried out using a WDW-T200 tension machine at HIT (Harbin Institute of Technology). A schematic illustration of the assembly testing device is shown in Fig. 16(a), and the testing device and its enlarged view are shown in Fig. 16(b). The tensile speed was set to 1 mm/min, which leads to a shear strain rate of 0.001 s−1. A lubricant was used between the contact surfaces of the two semi-circle mandrels to decrease the friction force between the two mandrels. The friction force between the two semi-circle mandrels was tested without the specimen, and the force was less than 10N, so it would not disturb the calculation of the shear stress.

Fig. 16
figure 16

Experimental device used for the pure shear test

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A three-dimensional digital image correlation (DIC) system developed by Xi’an Jiaotong University was used for the strain measurement.

The cracked specimen is shown in Fig. 17. The shear fracture occurred along the axial direction, and the two shear zones cracked at the same time for both specimens.

Fig. 17
figure 17

Cracked specimen (L/t = 2.5)

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Results and Discussion

Load-Displacement Relationship and Experimental Strain Distributions

The relationship between the axial tensile force and displacement is shown in Fig. 18. The endpoint of each curve means the time of the crack. The difference between the loading relationship of two specimens with the same shear zone length is minimal.

Fig. 18
figure 18

Relationship between the loading force and displacement

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The Mises equivalent strain distribution is shown in Fig. 19. It can be seen that deformation begins with the two ends of the shear zone, and then deformation extends to the center zone. The width of the shear zone with a higher equivalent strain is about 1 mm, which means deformation is highly concentrated in the interface plane of the two half-tubes.

Fig. 19
figure 19

Distribution of Mises equivalent strain (L/t = 2.5)

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The shear strain distribution along the interface plane is shown in Fig. 20. The horizontal ordinate is the relative position. The minimum shear strain is located in the center point. It should be noted that the shear strain distribution is symmetrical, which indicates nearly no rotation occurs for the two semi-circle mandrels. As the shear stress is an average value and only the shear strain of one point in the shear zone can be used in the shear stress-strain relationship, the specimen with a more even shear strain distribution will be closer to the actual shear stress-strain relationship. Though the shear strain in the two ends is bigger than that of the center point, the shear strain distribution for specimens with L/t = 2.5 is relatively even.

Fig. 20
figure 20

Distribution of the shear strain in the interface plane along the tube axis (the shear strain of the center point equals 5%, 10%, and 15%)

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Shear strain distribution in the surface of the shear zone is shown in Fig. 21. The X and Y coordinates are the axial and circumferential distance separately, and the Z coordinate is the shear strain value. The zero point for the circumferential direction is the interface of the two half-tubes. Furthermore, the zero point for the axial direction is the center of the shear zone. It can be seen that the shear strain is symmetrically distributed about the interface of the two half-tubes. The maximum shear strain value along the circumferential direction lies in the interface. The shear strain is decreased with increased distance from the interface. This means the shear deformation is highly concentrated in the pre-set shear plane, i.e., the interface plane. Thus, the shear stress value calculated using the load and the area of the interface is closer to the actual value, and the precision of the shear stress is improved.

Fig. 21
figure 21

Distribution of the shear strain (the shear strain of the center point equals 5%, 10%, and 15%)

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Prediction of Constitutive Relation

Generally, simulation models used in engineering applications do not consider the shear experimental properties of materials. That is to say, the shear properties of materials are generally equivalently represented by the tensile experimental properties of materials. However, due to the influence of anisotropic properties of materials, it is necessary to understand whether the shear properties of anisotropic materials can be approximately described by tensile experimental properties.

For a plane-stress anisotropic model, typical calibration experiments involve tensile tests in three orientations relative to the rolling direction of the sheet and an equal-biaxial test with each test providing both stress and R-value. The parameters are \({\sigma }_{0}\), \({\sigma }_{45}\), \({\sigma }_{90}\), \({\sigma }_{b}\), R0, R45, R90.

For the thin-walled tubes, \({\sigma }_{z}\) denotes the axial normal stress, \({\sigma }_{\theta }\) denotes the circumferential normal stress, \({\tau }_{\theta z}\) denotes the shear stress along the axial direction. The yield stress is set as the Mises equivalent strain equals 0.2%. \({\sigma }_{z0}\) and \({\sigma }_{\theta 0}\) are the yield stress in the axial and circumferential directions, where \({R}_{z}\) and \({R}_{\theta }\) are R-values in the axial and circumferential directions, respectively. \({\tau }_{\theta z0}\) is the shear yield stress. The experimental results are shown in Table 3.

Table 3 Experimental characteristics of thin-walled 5052 aluminum alloy tube
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The \({\sigma }_{45}\) value is not available for tubes, and the equal-biaxial tensile stress \({\sigma }_{b}\) cannot obtained using a general tension machine, so the two stress values are estimated using the axial and circumferential properties using equations (2) and (3).

$${\sigma }_{45}=\frac{{\sigma }_{\text{z}0}+{\sigma }_{\uptheta 0}}{2}$$
(2)
$${\sigma }_{\text{b}}=\frac{{\sigma }_{\text{z}0}+{\sigma }_{\uptheta 0}}{2}$$
(3)

The constitutive models are calibrated using the results of the uniaxial tension tests above. The uniaxial tension stress is converted to shear stress using the constitutive functions. Then, the shear strain is calculated according to the criterion of equal first-order incremental work, as shown in equation (4).

$$\overline{\upsigma }\Delta \overline{\upvarepsilon }=\tau \Delta \gamma$$
(4)

Mises, Tresca, Hill48, and Barlat-lian constitutive function are used to carry out the equivalent conversion, and the stress-strain relationship obtained from the uniaxial tension test using the cambered specimen along the axial direction is used in the conversion.

The Barlat-Lian constitutive function is shown as equation (5), where M = 8 for FCC materials. The equivalent stress is shown as equation (6), while the detailed description can be found in reference [21]. The coefficients used in the calculation of the equivalent stress of the constitutive function are shown in Table 4.

Table 4 Coefficients used in calculation of the equivalent stress of Barlat-Lian
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$$\text{f}=\text{a}{|{\text{k}}_{1}+{\text{k}}_{2}|}^{\text{M}}+\text{a}{|{\text{k}}_{1}-{\text{k}}_{2}|}^{\text{M}}+\text{c}{|2{\text{k}}_{2}|}^{\text{M}}=2{\overline{\upsigma } }^{\text{M}}$$
(5)
$$-{\sigma }=\sqrt[M]{\frac{2a+{2}^{M}c}{2}}q{\tau }_{\theta z}$$
(6)

The relationship between the shear stress and equivalent stress is shown in equation (7)

$$\overline{\upsigma }=1.653{\tau }_{\theta z}$$
(7)

The Hill48 constitutive function and the equivalent stress equation are shown in equations (8) and (9), respectively [22]. Hill’s coefficients can be calculated directly from the anisotropy coefficients.\({\sigma }_{b}\) is yield stress in biaxial tension and is calculated according to equation (3).

$$f={\sigma }_{z}^{2}-\left(\frac{{\sigma }_{z0}^{2}}{{\sigma }_{\theta 0}^{2}}-\frac{{\sigma }_{z0}^{2}}{{\sigma }_{b}^{2}}+1\right){\sigma }_{z}{\sigma }_{\theta }+\frac{{\sigma }_{z0}^{2}}{{\sigma }_{\theta 0}^{2}}{\sigma }_{\theta }^{2}+F{\tau }_{\theta z}^{2}={\overline{\sigma }}^{2}$$
(8)
$$\overline{\upsigma }=\sqrt{\rho {\sigma }_{z}^{2}+\xi {\sigma }_{\theta }^{2}+\lambda {({\sigma }_{z}-{\sigma }_{\theta })}^{2}+2\delta {\tau }_{\theta z}^{2}}$$
(9)

The coefficients of the Hill48 constitutive function are shown in Table 5, and the relationship between the shear stress and equivalent stress is shown in equation (10)

Table 5 Coefficients used in the calculation of the equivalent stress of Hill48
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$$\overline{\upsigma }=1.824{\tau }_{\theta z}$$
(10)

The stress-strain relationships predicted using the Mises, Tresca, Hill48, and Barlat-lian constitutive functions are shown in Fig. 22(a), and the relative error is shown in Fig. 22(b). The Barlat-Lian and Mises functions underestimate the shear stress at the initial stage, and both overestimate the shear stress at the subsequent stage. For the Tresca and Hill48 functions, the shear stress is always underestimated. It can be seen that the precision of all of the predictions is not good.

Fig. 22
figure 22

Predicted shear stress-strain relationships by different constitutive functions

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As the equal-biaxial tensile stress \({\sigma }_{b}\) is not obtained from the experiment, the \({\sigma }_{b}\) value is enlarged or lessened to investigate its influence on prediction precision. When the equal-biaxial tensile stress \({\sigma }_{b}\) equals 0.9 or 1.1 times the calculated average value of \({\sigma }_{z0}\) and \({\sigma }_{\theta 0}\), the equation is shown as equations (11) and (12).

When \({\sigma }_{\text{b}}=0.9*(\frac{{\sigma }_{\text{z}0}+{\sigma }_{\uptheta 0}}{2}\)),

$$\overline{\upsigma }=1.750{\tau }_{\theta z}$$
(11)

When \({\sigma }_{\text{b}}=1.1*(\frac{{\sigma }_{\text{z}0}+{\sigma }_{\uptheta 0}}{2}\)),

$$\overline{\upsigma }=1.878{\tau }_{\theta z}$$
(12)

The predicted results are shown in Fig. 23. The predicted shear stress decreased with the increased value of the equal-biaxial tensile stress. However, the prediction precision is not improved with the change of the equal-biaxial tensile stress.

Fig. 23
figure 23

Influence of the equal-biaxial tensile stress on prediction of shear stress-strain relationship of Hill48 constitutive function

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It is indicated that predictions of the models calibrated with the conventional method are inconsistent with the mechanics of shear deformation described in the shear experiment. In other words, the work hardening of the materials under a shear stress state does not appear to be the same as for uniaxial tension.

Summary and Conclusions

A novel testing method and associated device are proposed to achieve pure shear deformation along the axial direction of a thin-walled tube. Optimizations of the specimen shape and experimental device were carried out by using FE simulation. Moreover, it was verified by a shear experiment of a thin-walled aluminum tube. Based on the tests, the pure shear stress state can be maintained to a large deformation extent. The experimental shear stress-strain relationship was compared with the converted stress-strain relationship based on the uniaxial tension tests using the Mises, Tresca, Hill48, and Barlat-lian constitutive functions. A significant difference between the stress-strain relationships is found. This indicates that the uniaxial tension stress-strain relationship cannot accurately predict shear deformation.

The shearing characteristics obtained by this novel experimental method can be applied to the construction of anisotropic constitutive relations for simulations of tube deformation processes. This testing method can provide necessary empirical data with the principal stress directions along the direction at an angle of 45° to the tube axis. The advantage of this pure shear loading method is that the shearing deformation occurs in a fixed two-dimensional shearing plane rather than in a three-dimensional shear zone, as previous studies of shear experiments considered.