Adhesive Joints Theory (and use of innovative joints) ERIK SERRANO STRUCTURAL MECHANICS, LUND UNIVERSITY

Transcription

1 Adhesive Joints Theory (and use of innovative joints) ERIK SERRANO STRUCTURAL MECHANICS, LUND UNIVERSITY

2 Wood and Timber Why I m intrigued

3 From this to this! via this Fibre deviation close to knots and this Deviation in fibre direction

4 Outline Wood adhesive bonds Basic behaviour (equations) Volkersen theory Fracture mechanics approaches Equivalent elastic layer approach A few words on testing Conclusions

5 An adhesive bond

6 0 m Thin or thick bond lines? m

7 Thin adhesive bond lines A thin layer where the most important stress components do not vary in the thickness direction

8 Stress Stress Material models Commonly used models Strength Stiffness Deformation Deformation

9 Stress Material models Less commonly used models Softening behaviour Deformation

10 Stress Softening behaviour Strength Fracture energy Shape Stiffness Deformation

11 Basics Luftfahrtforschung, Vol. 5, 938 Simplest possible analysis D-shear lag (Volkersen)» Constant shear across adhesive layer thickness (t 3 )» Pure axial action in adherends and pure shear in adhesive» Linear elastic behaviour of materials P P 2 E, t, b P 3 P 4 G 3, t 3, b 3 E 2, t 2, b 2

12 Basic variables N(0) N2(0) x L N(L) N2(L) b A t3 A2 dx Cross section N N+dN t3 N2 N2+dN2

13 b dx dn b dx dn t t t t A b dx dn A A b dx dn A t t A b A b, 2 Axial stress in adherends Horizontal equilibrium

14 / ) ( u u t u u Assumes constant shear strain in adhesive layer t E E G Assumes linear elastic materials 2a 3 Axial displacement of adherends 2b 2c

15 Derivation twice of and using and then 2a 2b 2c we obtain, using 3 and : 4 t 2 3 t 3 0 with the definition of G b Stiffness ratio shear/axial t3 A E A2 E2

16 Solution of governing equation The solution of t 2 3 t 3 0 is given by t C sinh( x) 6 cosh( x) C2 where constants C and C 2 are determined by the boundary conditions

17 Load case pull-pull E, t, b E 2, t 2, b 2 G 3, t 3, b 3 P P ) sinh( ) tanh( A E t PG C L A E L A E t PG C 7

18 Example Assume wood-wood joint Thicknesses (mm) E -E 2 -G (MPa) L: 0, 20, 50 and 00 mm, t 3 : 0. or.0 mm 0 mm 00 mm

19 Example Results Symmetric stress distribution Influence of L and stiffness ratio (G 3 /t 3 ) / (EA) Relative shear stress in bond line L=00 Different y-axis scales!!!

20 Example 2 Assume glass-wood-adhesive Thicknesses (mm) E -E 2 -G (MPa) L: varying 0, 20, 50 and 00 mm

21 Non-uniform stress distribution Relative shear stress in bond line

22 Conclusions so far Stress distribution depends on stiffness ratios expressed through joint parameter If L G b t3 L 2 3 << (small) => uniform stress distribution >> (large) => non-uniforms stress distribution A E A 2 E 2

23 Conclusions so far Parameter L << (small) small overlap length low bond line stiffness in relation to axial stiffness of adherends >> (large) large overlap length high bond line stiffness in relation to axial stiffness of adherends

24 Joint load-bearing capacity Assume joint capacity is reached when max shear stress equals bond line shear strength, then from 6 7 P max t f t3 G3 A E tanh( L) A 2 E 2 sinh( L) Where t f is the bond line shear strength A E (Assumes is chosen such that for which case A E max stress occurs at x=0) A A E E

25 Example: Influence of overlap length

26 Conclusions so far Joint capacity (N) depends on adhesive shear strength and parameter P max thus depends on geometry, strength AND stiffness parameters max ) sinh( ) tanh( L E A L A E G t P f t L E A A E t G b

27 Stress/strength analyses (using FEM) Conventional strength analysis: Linear elastic material response Mostly stressed point governs failure Sharp corners can give high stress ( ) For brittle/stiff joints (i.e. traditional wood adhesive joints)» Depicts the stress distribution at low load levels (?)» Difficult (impossible) to use for prediction of joint capacity?

28 Stress/strength analyses using FEM Elasto-plastic analysis Elasto-plastic material Mostly stressed point governs failure Sharp corners can give high strains ( ) Depicts the stress distribution in ductile joints at low load levels Can be used for capacity prediction of ductile/flexible joints

29 Fracture mechanics-based analysis Linear elastic fracture mechanics (LEFM) Assumes a brittle joint/bond line Assumes stress singularities (sharp corner/ existing crack) Can be used for capacity prediction of brittle joints

30 Linear elastic fracture mechanics Stress intensity approach Assume a small crack exists Calculate the stress intensity (stress concentration factor) Crack propagation approach (compliance method) Assume an existing crack propagates Calculate the change of compliance (flexibility) of the joint as the crack propagates J-integral Calculate the value of a path-independent integral of stress close to the crack tip

31 Example LEFM (compliance method) Finger joints Aicher & Radovic (999)

32 Non-linear fracture mechanics-based analyses Assume a non-linear material (bond line) behaviour including softening (Non-linear fracture mechanics=nlfm) Can be used for Any brittleness of the joint Any geometry Can be used for prediction of joint capacity in all cases from brittle to ductile joints

33 Stress Softening behaviour Strength Fracture energy Shape Stiffness Deformation

34 Bonded-in rods Influence of rod length Initial softening

35 Stress distributions Linear elastic At max load Shear Normal stress (peel)

36 NLFM Approach Softening Bonds Shear Tension perp. Serrano, E. Adhesive Joints in Timber Engineering Modelling and Testing of Fracture Properties. PhD thesis, Report TVSM-02, Lund University 2000.

37 Result presentation NLFM Joint brittleness ratio is given by Material» Bond line fracture energy and strength» Adherend material stiffness» Shape of softening curve Geometry» Joint shape and absolute size of joint Normalised strength at failure is given by: some stress measure / material strength Example: for a beam in bending: M max 6 bh 2 /f m

38 Normalised ultimate strength Influence of joint brittleness ratio Perfectly plastic NLFM LEFM Equivalent Elastic layer Joint brittleness ratio E. Serrano. Glued-in Rods for Timber Structures. A 3D Model and Finite Element Parameter Studies. International Journal of Adhesion and Adhesives. 2(2) (200) pp.5-27.

39 Comparison with tests Pull-out FEM Test E. Serrano and P. J. Gustafsson. Fracture mechanics in timber engineering Strength analyses of components and joints. Materials and Structures (2006) 40:87 96.

40 A generalised method Equivalent elastic fracture layer method Assume simplest possible stress-deformation behaviour Adapt stiffness (reduce it) in order to take into account fracture energy Perform linear elastic analysis Failure criterion: maximum stress in one point Stress(MPa) Material strength, f t G f Displacement (mm)

41 A generalised method How come this works? t (MPa) x (mm) Linear elastic solution (standard elastic stiffness used)

42 A generalised method How come this works? t (MPa) x (mm) Linear elastic solution (standard elastic stiffness used) NLFM solution

43 A generalised method How come this works? t (MPa) x (mm) Linear elastic solution (standard elastic stiffness used) NLFM solution Equivalent elastic layer (adapted stiffness)

44 Bonded-in rods Calculation Results

45 Vessby, J., Serrano, E., Enquist, B. Materials and Structures (200) 43:

46 Analyses by LEFM and NLFM Influence of lamination thickness on beam behaviour Serrano, E., Larsen, H.J. ASCE Journal of Structural Engineering (999) 25:

47 Bending strength (MPa) Results NLFM and LEFM Compliance method (analytical) Compliance method (analytical) Lamination thickness (mm)

48 Obstacles (at least some of them ) Material behaviour Time (creep) Moisture (hygroscopic materials) nonlinear (plasticity, damage, cracking) Experiments?

49 Testing Parameters needed Bond line strength Local strength of the bond line at a material point level (not joint strength ) Stiffness Failure strain Fracture energy Shape of response curve, e.g. softening behaviour

50 Testing for local strength How to test for local strength? Small specimen»uniform stress distribution»small amount of energy released at failure Large specimen»non-uniform stress distribution»large amount of energy released at failure

51 Example Stress distributions Stiff/brittle Soft/ductile

52 Standard test specimen (not very useful) Shear stress Normal stress (peel stress)

53 Linear elastic At max load Brittle adhesive Shear stress Peel stress Semi ductile adhesive Ductile adhesive

54 Brittle adhesive Semi-Ductile adhesive Ductile adhesive DIC-measurements Serrano, E., Enquist, B. Holzforschung, Vol. 59, pp , 2005 Brittle adhesive Ductile adhesive

55 Fracture mechanics tests (softening) Capture the complete response, including softening Deformation controlled testing Stiff test arrangement Testing machine Load cell Specimen and grips Fast response of control system

56 Stress Softening behaviour Deformation

57 Fracture mechanics tests (softening) The energy released during softening (diminishing load at increasing deformation) must be dissipated by the failure process in the bond line Test arrangements with high stiffness release small amounts of energy stable test performance can be achieved Small specimens required (3 5 mm bond line length in shear)

58 Serrano, E. Adhesive Joints in Timber Engineering Modelling and Testing of Fracture Properties. PhD thesis, Report TVSM-02, Lund University 2000.

59 Serrano, E. Adhesive Joints in Timber Engineering Modelling and Testing of Fracture Properties. PhD thesis, Report TVSM-02, Lund University 2000.

60 Conclusions Adhesive joint capacity (in N or MPa) is determined by Local strength of the bond line Material stiffness(es) Fracture energy of the bond line Shape of the softening curve of the bond line The geometry of the joint The absolute size of the joint Large adhesive joints need soft/ductile bond lines to be efficient NLFM can be considered a general theory for brittle to ductile joints

61 A few other factors affecting... (Marra,992)

62 Thanks for the attention questions?