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Validation of pure shear test device using finite element method and experimental methods
Engineering Fracture Mechanics 71 (2004) 233–243 www.elsevier.com/locate/engfracmech
Validation of pure shear test device using finite element method and experimental methods Jai-Sug Hawong a b
a,*
, Dong-Chul Shin a, Un-Cheol Baek
b
School of Mechanical Engineering, Yeung-nam University, Gyeong-San City, Gyeong-Buk 712-749, South Korea School of Mechanical and Aerospace Engineering, Seoul National University, San 56-1 Shinlim-dong, Kwanak-ku, Seoul 151-742, South Korea Received 29 July 2002; received in revised form 14 February 2003; accepted 16 February 2003
Abstract In this paper, we will validate the Iosipescu shear test (IST) and the Hawong Iosipescu shear test (HIST) used for the pure shear test device; the HIST is modified from the IST. This validation uses finite element method and experimental method. In the IST the loads are applied on the outside edges of the specimen, while in the HIST the loads are applied on the neutral surface of the specimen. It has been certified that HIST is more effective than IST because of the simplicity of the loading device, loading method and the distribution of pure shear stress, etc. We know that the specimen with a 110° V-notched angle is more effective than the specimen with a 90° notched angle for a pure shear test device. In specimens with a 90° notched angle, the maximum shear stress occurs at the end of V-notch. While in specimens with a 110° notched angle, it is produced at the center of specimen. In both HIST and IST the most ideal ratio of a=b is 0.3 although the ratios of 0.3, 0.4, 0.5 of a=b are also useful, except for 0.2. When HIST or IST is under mode fracture, HIST or IST with b ¼ 90° is more effective than those with b ¼ 0°, as a pure shear test device. Ó 2003 Elsevier Ltd. All rights reserved.
1. Introduction Composite materials have been developed in various types and characteristics, as they are used in several industrial fields. When used within the structures of an industrial field, the mechanical properties and behavior of the composite materials should be known. In the mechanical properties the shear modulus GLT and the shear stress fringe value fLT of L–T plane (L: the fiber direction, T : the direction transversally to the fiber direction), are very important in the stress analysis and the photoelasticity analysis of orthotropic materials respectively. It is very difficult to measure shear modulus and shear stress fringe value of orthotropic material. Therefore, various methods and devices of measuring shear modulus have been suggested.
*
Corresponding author. Tel.: +82-53-810-2445; fax: +82-53-813-3103. E-mail address: [email protected] (J.-S. Hawong).
0013-7944/$ - see front matter Ó 2003 Elsevier Ltd. All rights reserved. doi:10.1016/S0013-7944(03)00084-5
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There are several pure shear test devices including the off-axis tension test [1–3], Arcan circular disk (ACD: Arcan fixture) [4–7] and Iosipescu shear test (IST) [8–18], Among these devices, the pure shear device of IST is considerably precise, and the pure shear force is almost uniform in the central cross-section of the Iosipescu specimen. The pure shear device of IST has been uncountably analyzed by experimental method [8–16] and finite element method [14–18]. In order to achieve the goal that shear stress be uniformly distributed and stress concentration not occur in the specimen, notches are made in the specimen for IST. Most notch angles are 90° and 110° [16,17]. In the traditional IST device an auxiliary device is used to load the specimen. The loading device of the IST is complicated, and it is also difficult to apply a load to the specimen and control the loading point. Therefore, Hawong Iosipescu shear test (HIST) has been suggested by the first author and his associates [13]. HIST is simpler than the IST and easier than IST in applying the load to the specimen. In this research, comparing HIST with IST by experimental and finite element method, the validity of HIST as a pure shear test device is estimated.
2. Shear stress of HIST and IST Fig. 1 shows the shear force diagram (SFD), the bending moment diagram (BMD) and the configurations of IST and HIST. In the IST device compressive loads are applied on the outer sides of the specimen. Therefore, it is difficult to precisely apply the load to the specimen. In the HIST device, the loads are applied on the points along the neutral surface. It is simple and easy to apply the loads to the specimen precisely because loads are transferred to the specimen through the pin joints as shown in Fig. 1(c). In this case, the diameter of the pin is less than 2 mm and the applied load is
Fig. 1. HIST and IST: (a) stress component, shear force and bending moment diagram, (b) IST device, (c) HIST device.
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Table 1 Load (Pi ), shear force (Vi ) and bending moment (M) for IST and HIST Load device
P1
P2
V1
HIST & IST
Pb ða þ bÞ
Pa ða þ bÞ
Pa ða þ bÞ
V2 P ðb aÞ ða þ bÞ
M P ðb aÞ ða þ bÞ
P ¼ P1 þ P2 , Pi : Load (i ¼ 1, 2), Vi : Shear force (i ¼ 1, 2), M: Moment.
tension. Specimen configurations, applying directions of load, stress components in the plane, BMD, SFD and so on are shown in Fig. 1(a) and their force magnitudes are revealed in Table 1. On the x-axis and y-axis of IST and HIST, only shear stresses are produced. The average stress and bending moment produced along the line A–B in Fig. 1 are given in Eqs. (1) and (2) respectively. sAB ¼
M¼
P ðb aÞ wtða þ bÞ
Paðb aÞ aþb
ð1Þ
ð2Þ
where P is the external load (¼ P1 þ P2 ), a and b are the distance in the direction of the x-axis from y-axis. w is width between points A and B and t is the thickness of the specimen.
3. Finite element model ANSYS 5.3 is the finite element program used in this research. We used several finite element models. Among them, the best finite element model was selected in this research. The finite element model uses a two-dimensional element with eight nodal points. To compare the results of HIST with those of IST,
Fig. 2. Finite element model (t ¼ 6 mm): (a) supports and loading of IST, (b) supports and loading of HIST, (c) grid, (d) detail drawing of A.
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Table 2 Loading conditions a=b
P1
P2
V2
0.2 0.3 0.4 0.5
1125 1285.7 1500 1800
225 385.7 600 900
900
identical geometrical and loading conditions for HIST and IST are employed. In order to have a shear force between V-notch equal to 900 N, P1 and P2 are controlled. Angles ( ¼ a) of V-notch are changed from 90° and 110°. Geometrical conditions and boundary conditions of finite element analyses are shown in Fig. 2. Loading conditions used in this research are given in Table 2.
4. Experiment and experimental method Specimen configuration used in this research is shown in Fig. 1(a). When HIST and IST are analyzed by fracture mechanics, a central slanted crack is created. The slanted angle is b and the crack length is 2a. Strain gages are used to measure strain distribution between points A and B. Strain gages are mounted at the center of the specimen and are 45° with respect to the AB-line. Strain distributions are measured with the variations of load. HIST and IST are investigated for the locations with pure shear strain distributions. To compare the actual isochromatic fringe patterns obtained from the photoelastic experiment with isochromatic fringe patterns obtained from finite element method, the photoelastic experiment is applied to the HIST and IST. To evaluate HIST and IST as a pure shear test device, caustics is applied to the HIST and IST. To validate HIST and IST as a pure shear test device in high stress conditions, the photoelastic experiment is applied to the specimen with cracks for HIST and IST devices. To obtain the optimal loading point position for HIST and IST as pure shear test devices, loading points are variously changed. Stress intensity factors are obtained from a least-squares method based on photoelastic experiment [19]. Material properties of epoxy resin used in photoelasticity are shown in Table 3. Figs. 3–5 and Fig. 6, respectively, show the stress contour plots (¼ rx ; ry ; sxy ), graphic and experimental isochromatic fringe patterns of HIST and IST when a=b ¼ 0:3 and a ¼ 90° and a ¼ 110°. Stress contours are obtained from the finite element method. Graphic isochromatic fringe patterns and experimental isochromatic fringe patterns are obtained respectively from the finite element method and the photoelastic experiment. For the normal stress components (rx ; ry ), A, B, C, . . ., M etc., respectively, indicate )7.5, )6.0, )4.5, . . ., 0, . . ., 10.5 MPa. In the shear stress ðsxy Þ, A, B, C, . . ., M etc., respectively, indicate 1.5, 2.25, 3.0, . . ., 10.5 MPa. As shown in Figs. 3 and 4, when a is 90°, the stress distributions of rx and ry obtained from HIST is more symmetric than those from IST. But their magnitudes are very similar to each other. They are almost zero in the neighbor region of AB line as shown in Fig. 3 and 4. Maximums of sxy occur at points A and B (See Fig. 1) in HIST and IST devices. In the shear stress distributions, HIST is more symmetric than IST. In the distribution of isochromatic fringe patterns, experimental isochromatic fringe patterns obtained from HIST are more symmetric than those from IST. Isochromatic fringe patterns of HIST obtained from the Table 3 Material properties of epoxy resin used in this research E [GPa]
G [GPa]
m
fr [kN/m]
3.09
1.10
0.396
11.15
E: YoungÕs modulus, G: Shear elastic modulus, m: PoissonÕs ratio, fr : Stress fringe value.
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Fig. 3. Stress contour plots and isochromatics of HIST (a=b ¼ 0:3, a ¼ 90): (a) rx , (b) ry , (c) sxy , (d) isochromatic fringe patterns (d.1) FEM, (d.2) experiment.
Fig. 4. Stress contour plots and isochromatics of IST (a=b ¼ 0:3, a ¼ 90°): (a) rx , (b) ry , (c) sxy , (d) isochromatic fringe patterns (d.1) FEM, (d.2) experiment.
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Fig. 5. Stress contour plots and isochromatics of HIST (a=b ¼ 0:3, a ¼ 110°): (a) rx , (b) ry , (c) sxy , (d) isochromatic fringe patterns (d.1) FEM, (d.2) experiment.
finite element method are more symmetric than those patterns of IST from finite element method. High shear stresses occur around the A–B line of HIST or IST device. But normal stresses (rx and ry ) are zero around the A–B line of the HIST or IST device. These results match with the elementary strength of materials theory. The results outlined above illustrate that the HIST device is more effective than the IST device. As shown in Figs. 5 and 6, similar results as those shown in Figs. 3 and 4 are obtained. From Figs. 3–6, we can know that the HIST device with a ¼ 110° and the HIST device with a ¼ 90° are, respectively, the best and the second best pure shear test devices. The IST device with a ¼ 90° is similar to the IST device with a ¼ 110° as a pure shear test device. Fig. 7 shows the normalized shear stress distributions on the y-axis (x ¼ 0 mm) obtained from the finite element method. The normalized shear stresses are normalized by the average shear stress sAB occurring along the notch line A–B, h is half of the AB-length. As shown in Fig. 7, rx =sAB of HIST is zero with the entire region y=h when a ¼ 90° or 110°. Maximum values of sxy =sAB occur respectively at the end of the notch (at points A and B) and at the center of specimen (jy=hj ¼ 0) when a (notch angle) is 90° or 110°. When jy=hj is less than 0.7 and a is 110°, maximum value of sxy =sAB is greater than when a is 90°. All values are very similar to each other in the region, except for HIST in which a=b is 0.2. When a is 110°, distributions of sxy =sAB are more equal to 1 than those when a is 90°. This means that the notch angle of 110° is more effective to produce pure shear than that of 90°. Therefore, it is known that when a is 90°, IST is better than the HIST in the shear stress distributions whereas when a is 110° and a=b is 0.3, HIST is better than IST. It is known that HIST is better than IST in the normal stress (rx ; ry ) distributions when a is 90° or 110°. It is certified that a HIST in which a=b is 0.3 and a is 110° is most effectively used as pure shear test device.
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Fig. 6. Stress contour plots and isochromatics of IST (a=b ¼ 0:3, a ¼ 110°): (a) rx , (b) ry , (c) sxy , (d) isochromatic fringe patterns (d.1) FEM, (d.2) experiment.
(a)
(b)
Fig. 7. Normalized shear stress distributions on the y-axis (x ¼ 0 mm) obtained from the finite element method: (a) V-notch angle: 90°, (b) V-notch angle: 110°.
Fig. 8 reveals normalized shear stress distributions along the x-axis (y ¼ 0 mm) from the finite element method. Distributions of sxy =sAB on the x-axis (y ¼ 0 mm) are different from those on the y-axis (x ¼ 0 mm). Distributions of sxy =sAB with a ¼ 110° are closer to one than those with a ¼ 90°. It is known from Figs. 3–8 that HIST and IST with a ¼ 110° are more effective than HIST and IST with a ¼ 90° for pure shear testing
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(a)
(b)
Fig. 8. Normalized shear stress distributions on the x-axis (y ¼ 0 mm) obtained from the finite element method: (a) V-notch angle: 90°, (b) V-notch angle: 110°.
and that HIST is more effective than IST for a ¼ 90° or a ¼ 110°. When a=b ¼ 0:2, IST or HIST are not good for pure shear testing because of the effect of the loading points. As far as HIST is concerned, HIST with an arbitrary a=b can be used as pure shear test except for when a=b ¼ 0:2. When a=b is large, in order to produce the corresponding stress, the large load should be applied to the HIST. As far as IST is concerned, when jy=hj is less than 0.7, a=b ¼ 0:5 gives almost zero for rx and a=b ¼ 0:3 gives almost zero for ry . Their magnitudes are very small compared with magnitudes of sxy . Therefore, HIST or IST with a=b ¼ 0:3 can be used as a pure shear test device. But HIST is more effective than IST as a pure shear test device with respect to the precision of shear stress production, load applying method and so on. Fig. 9 shows strain distributions of HIST and IST. Here, ACD represents Arcan circular disk [20] used as a specimen of the pure shear test. Two strain gages normal to each other are used in this research. The strain gage is mounted at the center of the specimen. The adhesive direction of the strain gage corresponds
Fig. 9. Strain distributions of HIST and IST.
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Fig. 10. Isochromatics of HIST and IST with a crack (b ¼ 100 mm, w ¼ 40 mm, t ¼ 6 mm): (a) HIST with V-notch (b ¼ 0°, a=b ¼ 0:3, 2a=w ¼ 0:6, V2 ¼ 627 kN), (b) IST with V-notch (b ¼ 0°, a=b ¼ 0:3, 2a=w ¼ 0:5, V2 ¼ 627 kN).
with the notch direction when a is 90°. Vf is the fiber volume ratio. Principal stress directions of IST and HIST specimens are 45° and 135° with AB direction. Therefore, the absolute values of principal strains (e11 and e22 ) obtained from principal stresses are theoretically equal to each other for isotropic or orthotropic materials. Therefore, if e11 ¼ e22 , the slope of the graph should be 1. As shown in Fig. 9, the slope of strain distributions obtained from HIST is about 1, but that from IST is less than 1. In this case, HIST has a better linearity than IST. Therefore, it can be concluded that HIST is better for pure shear testing. Fig. 10 shows isochromatic fringe patterns of HIST and IST with a central crack [21]. As shown in Fig. 10, HIST is under less influence of the loading point on the stress distributions and more symmetric in the distributions of isochromatic fringe patterns than IST. Therefore, it is known that HIST is more effective than IST for pure shear testing. Fig. 11 shows stress intensity factors ( ¼ KI =K0 ) of the HIST and IST with the load applying point ratio (a=b) when 2a=w ¼ 0:5. Stress intensity factors are obtained from the linear least squares method of photoelastic experiment [19]. When 0:2 6 a=b < 0:3 and b ¼ 90°, KII =K0 of HIST is varied with a=b in the arbitrary crack length. When 0:3 6 a=b 6 0:5 and b ¼ 90 °, KII =K0 of HIST and IST is almost constant
pffiffiffiffiffiffi Fig. 11. Stress intensity factors of the HIST and IST with V-notch versus a=b ð2a=w ¼ 0:5; K0 ¼ 2 pa; s ¼ V2 =AB tÞ.
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Fig. 12. Caustics pattern of IST and HIST (b ¼ 45°): (a) IST with V-notch (V2 ¼ 1016:8 N), (b) HIST with V-notch (V2 ¼ 747 N).
with a=b. When b is 0°, stress intensity factor KII =K0 Õs are almost constant with the loading point ratio ( ¼ a=b) and KII =K0 Õs of HIST and IST are less than that when b ¼ 90°. KI =K0 Õs are almost zero with arbitrary crack length( ¼ 2a=w) and loading point ratio ( ¼ a=b). When 2a=w ¼ 0:3 or 0.6, KI =K0 Õs of HIST are almost equal to the KI =K0 Õs of IST. When 2a=w ¼ 0:4 or 0.5, KI =K0 Õs of HIST and IST are almost zero. As shown in Fig. 11, it is known that HIST or IST are effective as pure shear test devices. When 0:3 6 a=b 6 0:5, stress intensity factors of HIST are respectively similar to those of IST when b ¼ 0° or 90°. But when b ¼ 90°, stress intensity factors are greater than those when b ¼ 0°. In this case, the load applying point ratio had better be within 0:3 6 a=b 6 0:5. When 0:3 6 a=b 6 0:5, HIST or IST with b ¼ 90° is better than those with b ¼ 0° as a pure shear test device. It is known that the HIST or IST device is not under the influence of a load applying point when 0:3 6 a=b 6 0:5. In viewing the load applying method, the symmetry of the isochromatic fringe pattern and the precision of stress intensity factor, it is known that HIST device is better than IST device as pure shear test device and that HIST with b ¼ 90° is best as a pure shear test device. Fig. 12 shows caustics patterns obtained from IST and HIST with V-notch when b is 45° [22]. They are perfect circles. The perfect circle in caustics patterns means that the specimen with a crack is under the perfect mode I load. This indicates that HIST and IST perform their roles well as pure shear test devices. This is because the normal stresses are only produced on the surface of a 45° direction in the pure shear stress plane. Therefore, it is known through Fig. 12 that HIST or IST are valid as pure shear test devices. With the view of stress distributions obtained from the finite element method, isochromatic fringe patterns, caustics patterns, strain distributions, stress intensity factors and so on, it is known that HIST is more effective than IST as a pure shear test device. Additionally, HIST can be effectively used as a pure shear test device when a=b ¼ 0:3 and a ¼ 110° and IST can be used as a pure shear test device when a=b ¼ 0:3 and a ¼ 90°.
5. Conclusions The following conclusions are drawn from previous discussions. (1) HIST is more effective than IST as a pure shear test device and can be effectively used as a pure shear test device in the arbitrary stress analysis. It is easier and more precise to apply the loads to the specimen of HIST whose load is then applied to the neutral specimen. The specimen of ISTÕs load is applied to the outside edge of the specimenÕs surface. (2) The maximum values of sxy are produced at the tips of the notches of HIST or IST when the angle of notch a is 90°. When a is 110°, they originate at the center of specimen. When jx=hj or jy=hj are less than
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0.7, the maximum value of sxy of a ¼ 110° is greater than that of a ¼ 90° and sxy =sAB of the former is closer to 1 than that of the latter. Therefore, HIST or IST with a ¼ 110° is more effective than those with a ¼ 90°. (3) When the effective load applying ratio (a=b) is between 0.3 and 0.5, the smaller jx=hj or jy=hj produces a better HIST or IST. When jx=hj or jy=hj are less than 0.7, HIST or IST can be effectively used as a pure shear test device. (4) When a ¼ 90°, the shear stress distribution of IST with a=b ¼ 0:3 is the most uniform case among those investigated. When a is 110°, the shear stress distribution of HIST with a=b ¼ 0:3 is the most uniform. (5) When HIST and IST are under the mode II load, HIST and IST with b ¼ 90° are more effective than those with b ¼ 0° for pure shear testing.
References [1] Edward M, Rodney LT. Off-axis test of a composite. J Compos Mater 1968;2(4):523–6. [2] Chamics CC, Sinclair JH, Ten-deg, off-axis test for shear properties in fiber composite. NASA TECH NOTE, no. NASA TND8251, 1976 (Exp Mech 17 (9), 339–46). [3] Sun CT, Berreth SP. A new end tab design for off-axis tension test of composite materials. J Compos Mater 1988;22:766–79. [4] Goldenberg N, Arcan M, Nicolau E. On the most suitable specimen shape for testing shear strength of plastics. In: Int. Symp. Plastic Test and Stand., Philadelphia, ASTM STP 247, 1958. p. 115–21. [5] Arcan M, Hashin Z, Voloshin A. A method to produce uniform plane stress state with applications to fiber-reinforced materials. Exp Mech 1978:141–6. [6] Banks-Sills L, Arcan M, Bui HD. Toward a pure shear specimen for KII c determination. Int J Fract 1983;22(1):R9–R14. [7] Yen SC, Craddock JN, Tech KT. Evaluation of modified Arcan fixture for the in-plane shear test of materials. Exp Tech 1988: 22–5. [8] Iosipescu N. New accurate procedure for single shear testing of Metals. J Mater 1967;2(3):537–66. [9] Walrath DE, Adams DF. The Iosipescu shear as applied to composite materials. Exp Mech 1983:105–10. [10] Mircea A. The Iosipescu shear as applied to composite materials. Exp Mech 1984:66–7. [11] Sullivan JL, Kao BG, Van Oene H. Shear properties and a stress analysis obtained from vinyl-ester Iosipescu specimens. Exp Mech 1984:223–32. [12] Bergner HW, Davis JG, Herakvoich CT. Analysis of shear test Methods for Composite Laminates, VPI-E-77-14. Virginia Polytechnic Institute and State University, Blacksburg, Va; NASA CR152704, 1977. [13] Hawong JS, Shimamoto A, Takahashi S, Choi SH. A method of measuring physical properties from stress fringe values. JSME Int J Ser 1990;33(4):497–505. [14] Balakrishnan MV, Bansal B, Kumosa M. Biaxial testing of unidirectional carbon-epoxy using biaxial Iosipescu test fixture. J Compos Mater 1997;31(5):486–508. [15] Bansal A, Kumosa M. Experimental and analytical studies of failure modes in Iosipescu specimens under biaxial loadings. J Compos Mater 1995;29(3):334–58. [16] Pindera MA, Choksi G, Hidde JS, Herakovich CT. A methodology for accurate shear characterization of unidirectional composites. J Compos Mater 1987;21:1164–85. [17] Adams DF, Walrath DE. Current status of the Iosipescu shear test method. J Compos Mater 1987;21:494–507. [18] Gascoigne HE, Abdallah MG, Adams DS. Influence of composite layup on shear strain distribution in the Iosipescu shear test specimen, SEM Spring Conf. Exp. Mech., 1986. p. 314–28. [19] Sanford RJ, Dally JW. A general method for determining mixed mode stress intensity factor from isochromatic fringe patterns. Engng Fract Mech 1979;11:621–33. [20] Arcan M, Hashin Z, Voloshin A. A method to produce uniform plane stress state with applications to fiber-reinforced materials. Exp Mech 1978:141–6. [21] Hawong JS, Baek UC, Shin DC, Lee HJ. Evaluation of pure shear test device by the transparent photoelastic experiment. Trans Kor Soc Mech Engrs (Ser A) 1999;23(4):660–8. [22] Hawong JS. A study on the evaluation of pure shear test device by the caustic method. J Inst Indus Technol, Yeungnam Univ 1998;26(1):27–42.
Validation of pure shear test device using finite element method and experimental methods Jai-Sug Hawong a b
a,*
, Dong-Chul Shin a, Un-Cheol Baek
b
School of Mechanical Engineering, Yeung-nam University, Gyeong-San City, Gyeong-Buk 712-749, South Korea School of Mechanical and Aerospace Engineering, Seoul National University, San 56-1 Shinlim-dong, Kwanak-ku, Seoul 151-742, South Korea Received 29 July 2002; received in revised form 14 February 2003; accepted 16 February 2003
Abstract In this paper, we will validate the Iosipescu shear test (IST) and the Hawong Iosipescu shear test (HIST) used for the pure shear test device; the HIST is modified from the IST. This validation uses finite element method and experimental method. In the IST the loads are applied on the outside edges of the specimen, while in the HIST the loads are applied on the neutral surface of the specimen. It has been certified that HIST is more effective than IST because of the simplicity of the loading device, loading method and the distribution of pure shear stress, etc. We know that the specimen with a 110° V-notched angle is more effective than the specimen with a 90° notched angle for a pure shear test device. In specimens with a 90° notched angle, the maximum shear stress occurs at the end of V-notch. While in specimens with a 110° notched angle, it is produced at the center of specimen. In both HIST and IST the most ideal ratio of a=b is 0.3 although the ratios of 0.3, 0.4, 0.5 of a=b are also useful, except for 0.2. When HIST or IST is under mode fracture, HIST or IST with b ¼ 90° is more effective than those with b ¼ 0°, as a pure shear test device. Ó 2003 Elsevier Ltd. All rights reserved.
1. Introduction Composite materials have been developed in various types and characteristics, as they are used in several industrial fields. When used within the structures of an industrial field, the mechanical properties and behavior of the composite materials should be known. In the mechanical properties the shear modulus GLT and the shear stress fringe value fLT of L–T plane (L: the fiber direction, T : the direction transversally to the fiber direction), are very important in the stress analysis and the photoelasticity analysis of orthotropic materials respectively. It is very difficult to measure shear modulus and shear stress fringe value of orthotropic material. Therefore, various methods and devices of measuring shear modulus have been suggested.
*
Corresponding author. Tel.: +82-53-810-2445; fax: +82-53-813-3103. E-mail address: [email protected] (J.-S. Hawong).
0013-7944/$ - see front matter Ó 2003 Elsevier Ltd. All rights reserved. doi:10.1016/S0013-7944(03)00084-5
234
J.-S. Hawong et al. / Engineering Fracture Mechanics 71 (2004) 233–243
There are several pure shear test devices including the off-axis tension test [1–3], Arcan circular disk (ACD: Arcan fixture) [4–7] and Iosipescu shear test (IST) [8–18], Among these devices, the pure shear device of IST is considerably precise, and the pure shear force is almost uniform in the central cross-section of the Iosipescu specimen. The pure shear device of IST has been uncountably analyzed by experimental method [8–16] and finite element method [14–18]. In order to achieve the goal that shear stress be uniformly distributed and stress concentration not occur in the specimen, notches are made in the specimen for IST. Most notch angles are 90° and 110° [16,17]. In the traditional IST device an auxiliary device is used to load the specimen. The loading device of the IST is complicated, and it is also difficult to apply a load to the specimen and control the loading point. Therefore, Hawong Iosipescu shear test (HIST) has been suggested by the first author and his associates [13]. HIST is simpler than the IST and easier than IST in applying the load to the specimen. In this research, comparing HIST with IST by experimental and finite element method, the validity of HIST as a pure shear test device is estimated.
2. Shear stress of HIST and IST Fig. 1 shows the shear force diagram (SFD), the bending moment diagram (BMD) and the configurations of IST and HIST. In the IST device compressive loads are applied on the outer sides of the specimen. Therefore, it is difficult to precisely apply the load to the specimen. In the HIST device, the loads are applied on the points along the neutral surface. It is simple and easy to apply the loads to the specimen precisely because loads are transferred to the specimen through the pin joints as shown in Fig. 1(c). In this case, the diameter of the pin is less than 2 mm and the applied load is
Fig. 1. HIST and IST: (a) stress component, shear force and bending moment diagram, (b) IST device, (c) HIST device.
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Table 1 Load (Pi ), shear force (Vi ) and bending moment (M) for IST and HIST Load device
P1
P2
V1
HIST & IST
Pb ða þ bÞ
Pa ða þ bÞ
Pa ða þ bÞ
V2 P ðb aÞ ða þ bÞ
M P ðb aÞ ða þ bÞ
P ¼ P1 þ P2 , Pi : Load (i ¼ 1, 2), Vi : Shear force (i ¼ 1, 2), M: Moment.
tension. Specimen configurations, applying directions of load, stress components in the plane, BMD, SFD and so on are shown in Fig. 1(a) and their force magnitudes are revealed in Table 1. On the x-axis and y-axis of IST and HIST, only shear stresses are produced. The average stress and bending moment produced along the line A–B in Fig. 1 are given in Eqs. (1) and (2) respectively. sAB ¼
M¼
P ðb aÞ wtða þ bÞ
Paðb aÞ aþb
ð1Þ
ð2Þ
where P is the external load (¼ P1 þ P2 ), a and b are the distance in the direction of the x-axis from y-axis. w is width between points A and B and t is the thickness of the specimen.
3. Finite element model ANSYS 5.3 is the finite element program used in this research. We used several finite element models. Among them, the best finite element model was selected in this research. The finite element model uses a two-dimensional element with eight nodal points. To compare the results of HIST with those of IST,
Fig. 2. Finite element model (t ¼ 6 mm): (a) supports and loading of IST, (b) supports and loading of HIST, (c) grid, (d) detail drawing of A.
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Table 2 Loading conditions a=b
P1
P2
V2
0.2 0.3 0.4 0.5
1125 1285.7 1500 1800
225 385.7 600 900
900
identical geometrical and loading conditions for HIST and IST are employed. In order to have a shear force between V-notch equal to 900 N, P1 and P2 are controlled. Angles ( ¼ a) of V-notch are changed from 90° and 110°. Geometrical conditions and boundary conditions of finite element analyses are shown in Fig. 2. Loading conditions used in this research are given in Table 2.
4. Experiment and experimental method Specimen configuration used in this research is shown in Fig. 1(a). When HIST and IST are analyzed by fracture mechanics, a central slanted crack is created. The slanted angle is b and the crack length is 2a. Strain gages are used to measure strain distribution between points A and B. Strain gages are mounted at the center of the specimen and are 45° with respect to the AB-line. Strain distributions are measured with the variations of load. HIST and IST are investigated for the locations with pure shear strain distributions. To compare the actual isochromatic fringe patterns obtained from the photoelastic experiment with isochromatic fringe patterns obtained from finite element method, the photoelastic experiment is applied to the HIST and IST. To evaluate HIST and IST as a pure shear test device, caustics is applied to the HIST and IST. To validate HIST and IST as a pure shear test device in high stress conditions, the photoelastic experiment is applied to the specimen with cracks for HIST and IST devices. To obtain the optimal loading point position for HIST and IST as pure shear test devices, loading points are variously changed. Stress intensity factors are obtained from a least-squares method based on photoelastic experiment [19]. Material properties of epoxy resin used in photoelasticity are shown in Table 3. Figs. 3–5 and Fig. 6, respectively, show the stress contour plots (¼ rx ; ry ; sxy ), graphic and experimental isochromatic fringe patterns of HIST and IST when a=b ¼ 0:3 and a ¼ 90° and a ¼ 110°. Stress contours are obtained from the finite element method. Graphic isochromatic fringe patterns and experimental isochromatic fringe patterns are obtained respectively from the finite element method and the photoelastic experiment. For the normal stress components (rx ; ry ), A, B, C, . . ., M etc., respectively, indicate )7.5, )6.0, )4.5, . . ., 0, . . ., 10.5 MPa. In the shear stress ðsxy Þ, A, B, C, . . ., M etc., respectively, indicate 1.5, 2.25, 3.0, . . ., 10.5 MPa. As shown in Figs. 3 and 4, when a is 90°, the stress distributions of rx and ry obtained from HIST is more symmetric than those from IST. But their magnitudes are very similar to each other. They are almost zero in the neighbor region of AB line as shown in Fig. 3 and 4. Maximums of sxy occur at points A and B (See Fig. 1) in HIST and IST devices. In the shear stress distributions, HIST is more symmetric than IST. In the distribution of isochromatic fringe patterns, experimental isochromatic fringe patterns obtained from HIST are more symmetric than those from IST. Isochromatic fringe patterns of HIST obtained from the Table 3 Material properties of epoxy resin used in this research E [GPa]
G [GPa]
m
fr [kN/m]
3.09
1.10
0.396
11.15
E: YoungÕs modulus, G: Shear elastic modulus, m: PoissonÕs ratio, fr : Stress fringe value.
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Fig. 3. Stress contour plots and isochromatics of HIST (a=b ¼ 0:3, a ¼ 90): (a) rx , (b) ry , (c) sxy , (d) isochromatic fringe patterns (d.1) FEM, (d.2) experiment.
Fig. 4. Stress contour plots and isochromatics of IST (a=b ¼ 0:3, a ¼ 90°): (a) rx , (b) ry , (c) sxy , (d) isochromatic fringe patterns (d.1) FEM, (d.2) experiment.
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Fig. 5. Stress contour plots and isochromatics of HIST (a=b ¼ 0:3, a ¼ 110°): (a) rx , (b) ry , (c) sxy , (d) isochromatic fringe patterns (d.1) FEM, (d.2) experiment.
finite element method are more symmetric than those patterns of IST from finite element method. High shear stresses occur around the A–B line of HIST or IST device. But normal stresses (rx and ry ) are zero around the A–B line of the HIST or IST device. These results match with the elementary strength of materials theory. The results outlined above illustrate that the HIST device is more effective than the IST device. As shown in Figs. 5 and 6, similar results as those shown in Figs. 3 and 4 are obtained. From Figs. 3–6, we can know that the HIST device with a ¼ 110° and the HIST device with a ¼ 90° are, respectively, the best and the second best pure shear test devices. The IST device with a ¼ 90° is similar to the IST device with a ¼ 110° as a pure shear test device. Fig. 7 shows the normalized shear stress distributions on the y-axis (x ¼ 0 mm) obtained from the finite element method. The normalized shear stresses are normalized by the average shear stress sAB occurring along the notch line A–B, h is half of the AB-length. As shown in Fig. 7, rx =sAB of HIST is zero with the entire region y=h when a ¼ 90° or 110°. Maximum values of sxy =sAB occur respectively at the end of the notch (at points A and B) and at the center of specimen (jy=hj ¼ 0) when a (notch angle) is 90° or 110°. When jy=hj is less than 0.7 and a is 110°, maximum value of sxy =sAB is greater than when a is 90°. All values are very similar to each other in the region, except for HIST in which a=b is 0.2. When a is 110°, distributions of sxy =sAB are more equal to 1 than those when a is 90°. This means that the notch angle of 110° is more effective to produce pure shear than that of 90°. Therefore, it is known that when a is 90°, IST is better than the HIST in the shear stress distributions whereas when a is 110° and a=b is 0.3, HIST is better than IST. It is known that HIST is better than IST in the normal stress (rx ; ry ) distributions when a is 90° or 110°. It is certified that a HIST in which a=b is 0.3 and a is 110° is most effectively used as pure shear test device.
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Fig. 6. Stress contour plots and isochromatics of IST (a=b ¼ 0:3, a ¼ 110°): (a) rx , (b) ry , (c) sxy , (d) isochromatic fringe patterns (d.1) FEM, (d.2) experiment.
(a)
(b)
Fig. 7. Normalized shear stress distributions on the y-axis (x ¼ 0 mm) obtained from the finite element method: (a) V-notch angle: 90°, (b) V-notch angle: 110°.
Fig. 8 reveals normalized shear stress distributions along the x-axis (y ¼ 0 mm) from the finite element method. Distributions of sxy =sAB on the x-axis (y ¼ 0 mm) are different from those on the y-axis (x ¼ 0 mm). Distributions of sxy =sAB with a ¼ 110° are closer to one than those with a ¼ 90°. It is known from Figs. 3–8 that HIST and IST with a ¼ 110° are more effective than HIST and IST with a ¼ 90° for pure shear testing
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(a)
(b)
Fig. 8. Normalized shear stress distributions on the x-axis (y ¼ 0 mm) obtained from the finite element method: (a) V-notch angle: 90°, (b) V-notch angle: 110°.
and that HIST is more effective than IST for a ¼ 90° or a ¼ 110°. When a=b ¼ 0:2, IST or HIST are not good for pure shear testing because of the effect of the loading points. As far as HIST is concerned, HIST with an arbitrary a=b can be used as pure shear test except for when a=b ¼ 0:2. When a=b is large, in order to produce the corresponding stress, the large load should be applied to the HIST. As far as IST is concerned, when jy=hj is less than 0.7, a=b ¼ 0:5 gives almost zero for rx and a=b ¼ 0:3 gives almost zero for ry . Their magnitudes are very small compared with magnitudes of sxy . Therefore, HIST or IST with a=b ¼ 0:3 can be used as a pure shear test device. But HIST is more effective than IST as a pure shear test device with respect to the precision of shear stress production, load applying method and so on. Fig. 9 shows strain distributions of HIST and IST. Here, ACD represents Arcan circular disk [20] used as a specimen of the pure shear test. Two strain gages normal to each other are used in this research. The strain gage is mounted at the center of the specimen. The adhesive direction of the strain gage corresponds
Fig. 9. Strain distributions of HIST and IST.
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Fig. 10. Isochromatics of HIST and IST with a crack (b ¼ 100 mm, w ¼ 40 mm, t ¼ 6 mm): (a) HIST with V-notch (b ¼ 0°, a=b ¼ 0:3, 2a=w ¼ 0:6, V2 ¼ 627 kN), (b) IST with V-notch (b ¼ 0°, a=b ¼ 0:3, 2a=w ¼ 0:5, V2 ¼ 627 kN).
with the notch direction when a is 90°. Vf is the fiber volume ratio. Principal stress directions of IST and HIST specimens are 45° and 135° with AB direction. Therefore, the absolute values of principal strains (e11 and e22 ) obtained from principal stresses are theoretically equal to each other for isotropic or orthotropic materials. Therefore, if e11 ¼ e22 , the slope of the graph should be 1. As shown in Fig. 9, the slope of strain distributions obtained from HIST is about 1, but that from IST is less than 1. In this case, HIST has a better linearity than IST. Therefore, it can be concluded that HIST is better for pure shear testing. Fig. 10 shows isochromatic fringe patterns of HIST and IST with a central crack [21]. As shown in Fig. 10, HIST is under less influence of the loading point on the stress distributions and more symmetric in the distributions of isochromatic fringe patterns than IST. Therefore, it is known that HIST is more effective than IST for pure shear testing. Fig. 11 shows stress intensity factors ( ¼ KI =K0 ) of the HIST and IST with the load applying point ratio (a=b) when 2a=w ¼ 0:5. Stress intensity factors are obtained from the linear least squares method of photoelastic experiment [19]. When 0:2 6 a=b < 0:3 and b ¼ 90°, KII =K0 of HIST is varied with a=b in the arbitrary crack length. When 0:3 6 a=b 6 0:5 and b ¼ 90 °, KII =K0 of HIST and IST is almost constant
pffiffiffiffiffiffi Fig. 11. Stress intensity factors of the HIST and IST with V-notch versus a=b ð2a=w ¼ 0:5; K0 ¼ 2 pa; s ¼ V2 =AB tÞ.
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Fig. 12. Caustics pattern of IST and HIST (b ¼ 45°): (a) IST with V-notch (V2 ¼ 1016:8 N), (b) HIST with V-notch (V2 ¼ 747 N).
with a=b. When b is 0°, stress intensity factor KII =K0 Õs are almost constant with the loading point ratio ( ¼ a=b) and KII =K0 Õs of HIST and IST are less than that when b ¼ 90°. KI =K0 Õs are almost zero with arbitrary crack length( ¼ 2a=w) and loading point ratio ( ¼ a=b). When 2a=w ¼ 0:3 or 0.6, KI =K0 Õs of HIST are almost equal to the KI =K0 Õs of IST. When 2a=w ¼ 0:4 or 0.5, KI =K0 Õs of HIST and IST are almost zero. As shown in Fig. 11, it is known that HIST or IST are effective as pure shear test devices. When 0:3 6 a=b 6 0:5, stress intensity factors of HIST are respectively similar to those of IST when b ¼ 0° or 90°. But when b ¼ 90°, stress intensity factors are greater than those when b ¼ 0°. In this case, the load applying point ratio had better be within 0:3 6 a=b 6 0:5. When 0:3 6 a=b 6 0:5, HIST or IST with b ¼ 90° is better than those with b ¼ 0° as a pure shear test device. It is known that the HIST or IST device is not under the influence of a load applying point when 0:3 6 a=b 6 0:5. In viewing the load applying method, the symmetry of the isochromatic fringe pattern and the precision of stress intensity factor, it is known that HIST device is better than IST device as pure shear test device and that HIST with b ¼ 90° is best as a pure shear test device. Fig. 12 shows caustics patterns obtained from IST and HIST with V-notch when b is 45° [22]. They are perfect circles. The perfect circle in caustics patterns means that the specimen with a crack is under the perfect mode I load. This indicates that HIST and IST perform their roles well as pure shear test devices. This is because the normal stresses are only produced on the surface of a 45° direction in the pure shear stress plane. Therefore, it is known through Fig. 12 that HIST or IST are valid as pure shear test devices. With the view of stress distributions obtained from the finite element method, isochromatic fringe patterns, caustics patterns, strain distributions, stress intensity factors and so on, it is known that HIST is more effective than IST as a pure shear test device. Additionally, HIST can be effectively used as a pure shear test device when a=b ¼ 0:3 and a ¼ 110° and IST can be used as a pure shear test device when a=b ¼ 0:3 and a ¼ 90°.
5. Conclusions The following conclusions are drawn from previous discussions. (1) HIST is more effective than IST as a pure shear test device and can be effectively used as a pure shear test device in the arbitrary stress analysis. It is easier and more precise to apply the loads to the specimen of HIST whose load is then applied to the neutral specimen. The specimen of ISTÕs load is applied to the outside edge of the specimenÕs surface. (2) The maximum values of sxy are produced at the tips of the notches of HIST or IST when the angle of notch a is 90°. When a is 110°, they originate at the center of specimen. When jx=hj or jy=hj are less than
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0.7, the maximum value of sxy of a ¼ 110° is greater than that of a ¼ 90° and sxy =sAB of the former is closer to 1 than that of the latter. Therefore, HIST or IST with a ¼ 110° is more effective than those with a ¼ 90°. (3) When the effective load applying ratio (a=b) is between 0.3 and 0.5, the smaller jx=hj or jy=hj produces a better HIST or IST. When jx=hj or jy=hj are less than 0.7, HIST or IST can be effectively used as a pure shear test device. (4) When a ¼ 90°, the shear stress distribution of IST with a=b ¼ 0:3 is the most uniform case among those investigated. When a is 110°, the shear stress distribution of HIST with a=b ¼ 0:3 is the most uniform. (5) When HIST and IST are under the mode II load, HIST and IST with b ¼ 90° are more effective than those with b ¼ 0° for pure shear testing.
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