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CHARACTERIZATION OF THE SHEAR BEHAVIOUR OF WOOD USING THE IOSIPESCU TEST J. C. Xavier Master of Science Thesis LMPF seminary - 11/12/2003 University of Trás-os-Montes e Alto Douro Vila Real, Portugal

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Plan Introduction The Iosipescu test Numerical simulation of the Variable Span Method Numerical simulation of the Iosipescu test Experimental work Presentation and discussion of the experimental results General conclusions and future work

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Introduction Wood modelling at the macroscopic level : Shear properties: ─ Shear moduli : G LR, G LT, G RT. ─ Shear strengths : S LR, S LT, S RT. L R T RT LT LR L R T RT LT LR f 11 f 12 f 13 f 21 f 22 f 23 f 31 f 32 f 33 f 44 f 55 f 66

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[5] prEN 408. European Committee for Standardization, [6] ASTM D American Society for Testing and Materials, [7] ASTM D American Society for Testing and Materials, [8] NP 623. Portuguese Standard, Drawbacks of the standardized tests for the identification of the shear properties of wood [3,4] : (i) Give only the shear properties parallel to the fibres (shear moduli : G LR, G LT and shear strengths : S LR e S LT ). (ii) The variable span method [5,6] proposed for the determination of E L and G LR (or G LT ), is not a fundamental test. (iii) The failure of the specimen of the shear block test [7,8] proposed for the identification of S LR and S LT, occurs under stress concentrations. [3] Yoshihara et all.. Journal of Wood Science, 44:15-20, [4] Rammer D.R. e L.A. Soltis. Res. Pap. FPL-RP-527, FPL, L [0,0025; 0,035] Nominal distribution of the shear stress Real distribution of the shear stress Increase of the shear stress

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Aim of this work: Investigation of the applicability of the Iosipescu shear test for characterizing the shear behaviour of wood Pinus Pinaster Ait. (i) simultaneous identification of the shear modulus and the shear strength, in a particular symmetry plane. Justifications for the choice of the Iosipescu test: (ii) possible application of this test method for all the symmetry planes of wood thanks to the small size of the specimen. Among different shear tests for orthotropic materials : Iosipescu test (standard test for synthetic composite materials [9] ). [9] ASTM D American Society for Testing and Materials, 1993.

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The Iosipescu test Iosipescu specimen [9] : [9] ASTM D American Society for Testing and Materials, 1993.

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General view of the Iosipescu fixture [9] : [9] ASTM D American Society for Testing and Materials, Wedge adjusting screw Specimen Stationary part of fixture Movable part of the fixture Fixture linear guide rod Base Adjustable wedges to tighten the specimen Attachment to the test machine

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Data processing [9] : Experimental information: +45º, –45º, P Engineering shear strain: Nominal shear stress: 6=P/A6=P/A Apparent shear modulus: Apparent shear strength: G 12 = 6 / 6 a S 12 = P / A a ult [9] ASTM D American Society for Testing and Materials, 6 = +45º – – 45º

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For an orthotropic material the distributions of 6 and 6 are not homogeneous [10,11]. The correction factors C e S are calculated through finite element analyses. G 12 = CSG 12 a [10] Pierron F. e A. Vautrin. Composite Science and Technology, 5:61-72, [11] Pierron F. Journal of Composite Materials, 32(22): , where C = 6 / (P/A) and S = 6 / 6 o o ros G 12 a Aspects about the identification of G 12 :

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[10] Pierron F. e A. Vautrin. Composite Science and Technology, 5:61-72, [11] Pierron F. Journal of Composite Materials, 32(22): , The distribution of 6 through the thickness of the specimen can be heterogeneous due to geometrical imperfections of its loading surfaces [10,11]. This effect is eliminated by considering 6 as the average of the shear strains measured over both lateral faces of the specimen [10,11]. Grandes deformações Pequenos módulos Pequenas deformações Grandes módulos Face frontal do provete Indeformado Deformado P Large deformations Small modulus Small deformations Large modulus Front face of the specimen Undeformed Deformed

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Aspects about the identification of S 12 : [12] Pierron F. e A. Vautrin. Composite Science and Technology, 57(12): , [13] Pierron F. e A. Vautrin. Journal of Composite Materials, 31(9): , [14] Odegard G. e M. Kumosa. Journal of Composite Materials, 33(21): , The failure of the specimens occurs under a homogeneous stress state although both 6 and 2 components exist [12-14]. S 12 should be determined through a failure criterion. S 12 = P / A represents an overestimated value. a ult 2 component should be calculated from finite element analysis, introducing in the model an suitable shear constitutive law.

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Numerical simulation of the variable span method Aim: Investigating the applicability of the variable span method [5,6] for validating the Iosipescu test. 3D models of the three-point-bending test developed in ABAQUS ®. Finite element models: Wood was modelled as: – continuous; – homogeneous; – orthotropic; – linear elastic. [5] prEN 408. European Committee for Standardization, [6] ASTM D American Society for Testing and Materials, 1994.

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L = 120, 135, 160, 200 and 400 mm Configuration of the specimens in the three-point-bending tests: Geometrical model used in the finite element analyses:

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Elastic properties used in the numerical models: E L (1) E R (1) E T (1) LR (1) LT (1) RT (1) G LR (2) G LT (2) G RT (2) (GPa) 15,131,911,010,470,050,591,111,100,18 (1) Pinus Pinaster Ait. [15]. (2) Pinus Tarda L. [16]. Calibration of the friction coefficient: Element C3D8 Mesh and boundary conditions of the model : [15] Pereira J.L. MSc Thesis, UTAD (in progress). [16] FPL. FPL-GTR-113, 1999.

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Euler-Bernoulli beam theory: E L = L 3 F 4h 4 f a f 1 = u y f 2 = u y f 3 = u y f 4 = u y – u y A B C C D D A B C x,L y,R A B C Numerical results:

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Timoshenko beam theory: f1f1 f2f2 f3f3 f4f4 E L (GPa) 16,63 (9,9%) 16,05 (6,1%) 16,01 (5,8%) 15,57 (2,9%) G LR (GPa) k = 1,2 0,74 (33,6%) 1,12 (0,6%) 1,22 (9,6%) 1,94 (74,6%) k = 1,5 0,92 (17,0%) 1,39 (25,8%) 1,52 (37,0%) 2,42 (118,3%) L h ELEL a 1 = ELEL G LR k

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This assumption is not verified at midspan ( AC ): Kinematical assumption of the Timoshenko beam theory : ─ The deflection is the same for each point belonging to the same vertical cross section, initially perpendicular to the neutral axis. A C B

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Numerical simulation of the Iosipescu test Aims: Determination of the stress and strain fields in the central region of the Iosipescu specimen of Pinus Pinaster Ait. Computation of the correction factors C and S. Finite element models: 2D models developed in ANSYS 7.0 ® and ABAQUS ®. The hypothesis and elastic properties are the same as the ones used in the numerical simulation of the variable span method.

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Nominal dimensions of the Iosipescu specimen: Mesh of the finite elements models: 5577 nodes and 1800 elements. Elements: PLANE82 (ANSYS) CPS8 (ABAQUS)

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Boundary conditions [17-19] : Flexão no plano i. base: ii. iterative (LR plane): iii. with contact: ANSYS ABAQUS [17] Pierron F. PhD, University of Lyon I, [18] Ho H. et al. Composite Science and Technology, 46: , [19] Ho H. et al. Composite Science and Technology, 50: , 1994.

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Average engineering shear strain ( ) Average shear stress (MPa) Base BC Iterative BC Contact BC Reference Average engineering shear strain ( ) Average shear stress (MPa) Base BC Iterative BC Contact BC Reference Comparaison and validation of the boundary conditions: Numerical results: (LR plane)

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Stress and strain fields for the LR specimen: LR /|P/A| ─ Stress field in the central region of the specimen: ─ Stress distribution along the vertical line between notches: RR /|P/A| LR /|P/A| RR /|P/A| R L

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─ Strain field over the strain gauge area : LR /| LR | RR /| LR |

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Stress and strain fields for the LT specimen: LT /|P/A| ─ Stress field in the central region of the specimen: ─ Stress distribution along the vertical line between notches: TT /|P/A| LT /|P/A| TT /|P/A| T L

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─ Strain field over the strain gauge area : LT /| LT | TT /| LT |

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Stress and strain fields for the RT specimen: RT /|P/A| ─ Stress field in the central region of the specimen: ─ Stress distribution along the vertical line between notches : TT /|P/A| RT /|P/A| TT /|P/A| T R RR /|P/A|

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─ Strain field over the strain gauge area : RT /| RT | TT /| RT |

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Calculation of the correction factors C and S : A C = 6 O F Y i=1 m S = +45º – –45º ) i=1 n i i n6n6 1 Symmetry planes Correction factors for woodCSCS LR0,970,990,95 (4,8%) LT0,920,990,91 (8,6%) RT1,040,971,01 (0,6%)

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Experimental work Preparation of the specimens: Material : wood of Pinus Pinaster Ait. (maritime pine), 74-year-old, from Viseu (Portugal).

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Iosipescu specimen: RT specimen LT specimen LR specimen ─ Moisture content: 9,5% – 12,1%; ─ Density: 0,537 – 0,623; ─ 0/90 strain gauge (CEA WT-350), bonded on both faces of the specimens, with the M-Bond AE-10 adhesive.

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EMSE fixture [20] : [20] Pierron F. Ecole des Mines de Saint-Etienne, No , Tightness of the wedges with a dynamometrical key : 1 Nm Experimental procedure:

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INSTRON 1125 Universal test machine with the capacity of 100 kN Data acquisition system : HBM SPIDER 8 Temperature of 23ºC (1ºC) and relative humidity of 45% (5%) Controlled displacement rate of 1 mm/min Experimental equipment : 5 kN Load cell

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Presentation and discussion of the experimental results LR specimens: Typical experimental data measured for the LR specimens: (A) (B) Linear deformation measured with strain gauges ( ) Load (N) Front face (A) Back face (B)

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Apparent average LR – LR curves : ─ The response of the specimens contain some variability. ─ The curves are nonlinear; the source of such nonlinearity can be attributed to [19,21] : (1) the nonlinear behaviour of the material; (2) the geometric nonlinearity; (3) the nonlinearity due to the contact conditions specimen/fixture. [21] Kumosa M. e Y. Han. Composite Science and Technology, 59: , Average engineering shear strain ( ) Average shear stress (MPa)

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Dispersion of the shear moduli values ( G LR, G LR and G LR ): a, A a, Ba G LR Mean (GPa)1,41 ± 0,15 1 1,54 ± 0,18 1 1,48 ± 0,12 1 C.V. 2 (%)14,115,010,3 a, A a, Ba ─ Reduction of the dispersion of G LR when the average of 6, between the measurements on both faces of the specimen, is considered. a (1) Confidence intervals at 95% confidence level; (2) Coefficient of variation (C.V.). Specimens Shear moduli (GPa)

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Moisture content ( u ), density ( d ) and shear moduli ( G LR, G LR ) : c Specimensu (%)dG LR (GPa) 111,90,5611,331,27 211,80,6071,591,52 312,10,6121,571,50 411,80,6051,621,54 512,10,6151,481,42 610,30,5381,321,23 710,00,5371,221,16 810,40,6091,531,46 99,40,6141,651,58 Mean11,10,5891,48 ± 0,12 1 1,41 ± 0,11 1 C.V. 2 (%)9,15,610,3 c a a ─ Applying the t test for equality of means between two samples it is concluded that G LR and G LR belongs to the same population, at a 95% confidence level. a c (1) Confidence intervals at 95% confidence level; (2) Coefficient of variation (C.V.).

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G LR shear modulus identified by the Iosipescu and off–axis [22] tests: Test method IosipescuOff-axis dG LR (GPa)d Mean0,5891,41 ± 0,11 1 0,5821,11 ± 0,04 1 C.V. 2 (%)5,610,34,07,0 ─ The dispersion of the G LR values are of the same order of magnitude. ─ Applying the t test of equality of means, it is concluded that the G LR values from both tests lead to different proprieties, at 95% confidence level. [22] Garrido N. MSc thesis, UTAD (in progress). ─ The average of G LR is great than G LR in 26%. IosipescuOff-axis (1) Confidence intervals at 95% confidence level; (2) Coefficient of variation (C.V.).

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LR – time curves : Initial cracks Geometric nonlinearity due to the rotation of the fibres Nonlinearity due to the specimen/fixture contact Crushing of the loading surfaces of the specimen Time (s) Shear stress (MPa)

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Shear stresses identified in the LR specimens : Specimens LR 114,414,9 212,616,3 317,218,6 416,217,9 519,1 613,213,8 714,915,0 815,916,8 919,5 Mean (MPa)15,9 ± 1,9 1 16,9 ± 1,6 1 C.V. 2 (%)15,212,1 1f ult ─ It is not possible to identify S LR using a suitable failure criterion, since the nonlinear shear constitutive law of wood Pinus Pinaster Ait. is not known. (1) Confidence intervals at 95% confidence level; (2) Coefficient of variation (C.V.).

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Shear stresses identified by the Iosipescu and off–axis [22] tests : Test method IosipescuOff-axis LR S LR 1 Mean (MPa)15,9 ± 1,9 1 16,9 ± 1,6 1 14,1 ± 0,9 1 16,5 ± 1,5 1 C.V. 3 (%)15,212,1 16,7 1f ult (1) Shear strength determined using the Tsai – Hill failure criterion; (2) Confidence intervals at 95% confidence level; (3) Coefficient of variation (C.V.). ─ The Iosipescu test gives a good estimation of S LR for wood Pinus Pinaster Ait.: LR < s LR < LR 1fult [22] Garrido N. MSc thesis, UTAD (in progress).

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LT specimen: Typical experimental data measured for the LT specimens: Linear deformation measured with strain gauges ( ) Load (N) Front face (A) Back face (B)

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Apparent average LT – LT curves : ─ The response of the specimens contain same variability. ─ The curves are nonlinear; the source of the nonlinearity can be attributed to [19,21] : (1) the nonlinear behaviour of the material; (2) the geometric nonlinearity; (3) the nonlinearity due to the contact conditions specimen/fixture. Average engineering shear strain ( ) Average shear stress (MPa)

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Dispersion of the shear moduli values ( G LT, G LT and G LT ): a, A a, Ba G LT Mean (GPa)1,33 ± 0,12 1 1,34 ± 0,10 1 1,34 ± 0,08 1 C.V. 2 (%)12,210,38,5 a, A a, Ba ─ Reduction of the dispersion of G LT when the average of 6, between the measurements on both faces of the specimen, is considered. a (1) Confidence intervals at 95% confidence level; (2) Coefficient of variation (C.V.). Specimens Shear moduli (GPa)

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─ Applying the t test for equality of means between two samples it is concluded that G LT e G LT belongs to the same population, at a 99% confidence level. Moisture content ( u ), density ( d ) and shear moduli ( G LT, G LT ) : c Provetesu (%)dG LT (GPa) 111,70,6031,381,26 211,70,5951,411,29 311,70,5901,431,31 411,50,5991,551,42 511,40,5921,291,17 610,80,5811,191,09 710,60,6061,381,26 811,30,5561,271,16 910,80,5741,211, ,50,5931,251,15 Média11,20,5891,34 ± 0,08 1 1,22 ± 0,07 1 C.V. 2 (%)4,52,68,5 c a a a c (1) Confidence intervals at 95% confidence level; (2) Coefficient of variation (C.V.).

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G LT shear modulus identified in the Iosipescu and off–axis [22] tests : Test method IosipescuOff-axis dG LT (GPa)d Mean0,5891,22 ± 0,07 1 0,5381,04 ± 0,05 1 C.V. 2 (%)2,68,54,08,1 (1) Confidence intervals at 95% confidence level; (2) Coefficient of variation (C.V.). ─ The dispersion of the G LT values are of the same order of magnitude. ─ Applying the t test of equality of means, it is concluded that the G LT values from both tests lead to different proprieties, at 95% confidence level. ─ The average of G LT is great than G LT in 17%. IosipescuOff-axis [22] Garrido N. MSc thesis, UTAD (in progress).

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LT – time curves : Initial cracks Geometric nonlinearity due to the rotation of the fibres Nonlinearity due to the specimen/fixture contact Crushing of the loading surfaces of the specimen Time (s) Shear stress (MPa)

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Shear stresses identified in the LT specimens : (1) These values does not follow a Normal distribution (Shapiro-Wilk test); (2) Confidence intervals at 95% confidence level; (3) Coefficient of variation (C.V.). Specimens LT 115,416,6 214,719,0 315,517,3 414,518,6 516,117,3 616,518,1 719,120,6 816,017,5 914,718,1 1016,118,1 Média (MPa)15,9 1 18,1 ± 0,8 2 C.V. 3 (%)8,46,1 1f ult ─ It is not possible to identify S LT using a suitable failure criterion, since the nonlinear shear constitutive law of wood Pinus Pinaster Ait. is not known.

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Shear stresses identified in the Iosipescu and off–axis [22] tests : Test method IosipescuOff-axis LT S LT 1 Mean (MPa)15,918,1 ± 0,8 1 14,0 ± 0,8 1 16,6 ± 1,0 1 C.V. 3 (%)8,46,19,510,9 [22] Garrido N. MSc Thesis, UTAD (in progress). 1f ult (1) Shear strength determined using the Tsai – Hill failure criterion; (2) Confidence intervals at 95% confidence level; (3) Coefficient of variation (C.V.). LT < s LT < LT 1fult ─ The Iosipescu test gives a good estimation of S LT for wood Pinus Pinaster Ait.:

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RT specimen: Typical experimental data measured for the RT specimens: Linear deformation measured with strain gauges ( ) Load (N) Front face (A) Back face (B)

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─ The response of the specimens contains some variability. ─ The curves are nonlinear. Apparent average RT – RT curves : Average engineering shear strain ( ) Average shear stress (MPa)

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Dispersion of the shear moduli values ( G RT, G RT and G RT ): a, A a, Ba G RT Mean (GPa)0,278 ± 0, ,286 ± 0, ,282 ± 0,038 1 C.V. 2 (%)27,212,416,2 a, A a, Ba (1) Confidence intervals at 95% confidence level; (2) Coefficient of variation (C.V.). ─ Reduction of the dispersion of G RT when the average of 6, between the measurements on both faces of the specimen, is considered. a Specimens Shear moduli (GPa)

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─ Applying the t test for equality of means between two samples it is concluded that G RT e G RT belongs to the same population, at a 95% confidence level. Moisture content ( u ), density ( d ) and shear moduli ( G RT, G RT ) : c Specimensu (%)dG RT (GPa) 111,30,5420,2210, ,60,5510,2540, ,70,5590,3410, ,70,5560,2710, ,10,5480,2490, ,20,6220,3380,345 79,80,6220,3110, ,40,6230,2800,285 Média11,20,5780,282 ± 0, ,288 ± 0,039 1 C.V. 2 (%)7,26,516,2 c a a a c (1) Confidence intervals at 95% confidence level; (2) Coefficient of variation (C.V.).

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G RT shear modulus identified by the Iosipescu and Arcan [23] tests: Ensaio de corte IosipescuArcan dG RT (GPa)d Média0,5780,288 ± 0, ,6500,229 ± 0,035 1 C.V. 2 (%)6,516,25,924,0 [23] Oliveira M. MSc thesis, UTAD (in progress). (1) Confidence intervals at 95% confidence level; (2) Coefficient of variation (C.V.). ─ The dispersion of the G RT values is slightly greater in the Arcan tests. ─ Applying the t test of equality of means, it is concluded that the G RT values from both tests lead to different proprieties, at 95% confidence level. ─ The average of G RT is great than G RT in 20%. IosipescuArcan

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RT – time curves : Cracks Time (s) Shear stress (MPa)

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Shear stresses identified in the RT specimens : Specimens RT 12,383,29 22,763,88 30,974,16 42,863,36 54,65 61,014,62 71,165,63 83,275,18 Mean (MPa)2,38 ± 1,08 2 4,35 ± 0,70 2 C.V. 2 (%)54,319,2 1f ult (1) Confidence intervals at 95% confidence level; (2) Coefficient of variation (C.V.).

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Shear stresses identified by the Iosipescu and Arcan [23] tests : Test method IosipescuArcan RT Mean (MPa)4,35 ± 0,70 1 4,54 ± 0,31 1 C.V. 2 (%)19,212,1 ult [23] Oliveira M. MSc Thesis, UTAD (in progress). ─ It was found a good agreement between RT values identified in both tests. However, as the failure of the Iosipescu RT specimens does not correspond to shear, it is not possible to say that the Iosipescu test gives a good estimation for s RT to wood Pinus Pinaster Ait. ult (1) Confidence intervals at 95% confidence level; (2) Coefficient of variation (C.V.).

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Comparison between the LR and LT specimens: ─ Applying the t test of equality of means between two samples it is concluded, for a 95% confidence level, that : (1) G LR and G LT are different properties, with G LR > G LT. (2) LR and LT are equal properties, suggesting that for wood Pinus Pinaster Ait.: S LR = S LT. ult

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General conclusions and future work The G LR, G LT and G RT shear moduli identified by the Iosipescu test are greater than the ones obtained by the off-axis and Arcan tests by 26%, 17% e 20%, respectively, leading to different properties at a 95% confidence level. Although it is not possible to directly identify the shear strengths S LR, S LT and S RT using the Iosipescu test, it was proved that this test gives a good estimation of these properties, at least for the LR and LT planes.

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Perpectives : ─ The use of identification technics, based on optical measurements and heterogenous fields, in order to identify several mechanical properties from only one test method. ─ The use of a micro/macro approach that allows the estimation of the macroscopic behaviour of wood through the characterization of its micro-structure.

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