dy, f = f(x,y) = -\u00b1- \\f{x^e^da (14) Applying this transform to the governing equation, it can be expressed as an ordinary differential equation. ( ^ - 3 \u00ab V , _ + 3 a 4 ^ - a V ) - ^ , _ - 2 a 2 ^ - r - a V ) = 0 (15) Keeping only roots that vanish for x \u2014> \u00b0 o , the solution is ^ = ^ j ( A 1 ( a ) e \" W x + A 2 ( a ) x e > k + A 3 ( \u00ab ) e - f o ) e - ' ^ J a r , R2=a2+k2 (16) To simplify the notation a is assumed to be positive in the remainder of the development. To satisfy symmetry about the crack plane, the shear and pinching stress resultants must vanish. A^ (0 ,y ) = - ^ , ( 0 , y ) = 0 (17) #,(0,y) = 0 (18) 71 From (7), the pinching shear condition requires wx(0,y) = 0. Using equations (10) and (11), w may be expressed in terms of the average in-plane stress. w \u2022 1 V 2 J V - A ^ 7]2k2 'J This leads to an additional symmetry condition. V > , ; c ( 0 , y ) - ^ 2 V V , ; c ( 0 , y ) = 0 (19) (20) Using the two symmetry conditions (17, 20), two constants may be eliminated from (16), and the stress function takes the following form. 0 = 1_^[ + A-+\u2014e-^A (a) 0 2 - k 2 ^ a p2+Xj +Rfi26 l J 2 { a ) ' P ~ k rj2 (21) To form the line-spring model, the remaining constant must be expressed in terms of the displacement of the crack faces. From compatibility (5) and the constitutive equations (6a,b,d). 1 \u00a3xx'yy 2h N N Nr AT .1 Ez 'J (22) A n expression for may be found from equations (9) and (19). This relationship shows that the pressurization of the plate arises due to curvatures in the in-plane stress field. \" \u00ab = 2 2,2 V- V2N Equation (22) may now be expressed as follows. 2hEe^yy = 2(1 + v)N\u201eJt - i V v _ + vNajac + ^ V 2 ^ (23) ' xy,xy *\"yy,xx ' \"'xx.xx ' (24) Next we express the in-plane stresses in terms of 0, and integrate to find the displacement. As the displacement must vanish as [x,y]\u2014>\u00b0\u00b0, the integration constant must also vanish. 2hEu x Vy=-Z,x. (20) From the definition of y\/, and constitutive equations (9e,f), we find the following. V2Z = 2GzVy (21) A differential equation for % may now be found from equation (16). V4Z-kp2Z = 0 (22) Equations (18), (15), and (22) allow the analysis of transverse shear stresses and pressurization in plate bending. 4.4 Hyper-singular integral equations In this section, the governing equations are reduced to a hyper-singular integral relating the crack face deflection to the bending stresses along the crack plane. The authors apply the integral transform method used by Erdogan and Joseph [7,8] to analyse a Reissner plate. The Fourier transform of a function and its inverse are defined as follows. oo - oo \/ = f(x,a) = \\f(x,y)eia>dy , \/ = f(x,y) = \u2014 [f\\x,a)e^da (23) \u2014oo \u2014oo Applying this transform, the governing equations may be written as ordinary differential equations. (w,xax\u00ab - 3 \u00ab r 2 w , \u00ab \u00ab +3a4w,xx -a6w)-kl(w,xxxx-2a2w,xx+a*w)=0 (24) (^-\u00abV)-^> = 0 (25) 95 Keeping only the roots that vanish as x \u2014> \u00b0\u00b0 , one finds the following expressions for w and \\j\/. w = \u2014 [(A, (a)e~Hx + A2 {d)xe~w + A 3 (a)e~Kx) eiayda, Rw2 = a2+k2w (26) IK J \u00a5 = \u00b1- J(B, ( o r ) * \" v ) e-\"\u00bbda, R\u00a52 = a2+k2y (27) Symmetry requires vanishing shear stresses along the line of the crack. Accordingly: Mxy(0,y) = 0, Vx(0,y) = 0, Rx(0,y) = 0 (28a,b,c) From the constitutive equations, the deflections along the line of the crack are limited as follows: Ux,y + Uy,x = 0 , ux + hw.x = 0 , uzx = 0 (29a,b,c) One must express these in terms of w and y\/. The following arise from (29a): 1 ]_ 2 (30) The first symmetry condition arises from (29a) by substitution from (29b) and (30b). y\/+2hw,xy = 0 (31) To find the second symmetry condition, we require an expression for Mxxx(0,y) under the restrictions (29). The following may be obtained from constitutive equation (9a). r,x 2 1-V-2TJ 2 \\ (32) The second symmetry condition wi l l then arise from equilibrium equation (8a). W,y=0 (33) r \\ - v - 2 r , 2 ^ A general equation for the transverse displacement arises from equations 13 and 14. ,2 1 \/ i rs\u201e2 A 2 V * 1 z \/ i z 1 - v k2 v l - ^ 2 J (34) After taking the derivative, the final symmetry condition may be found from (29c). 96 f 1 + v ^ l - v - 2 \/ 7 2 (35) Substitution of (24) and (25) into symmetry conditions (31), (33), and (35) leads to the following restrictions on the integration constants. fl, + 2h^Axa2 - A2a +A3 Rwa\\ = 0 ia 2A2ai-KRwA3+ \u2014 2h \\-v-2r] 2 A IB, = 0 R..A, = l a 2 ( \u201e2 V , , , ^ r\\L [1 + v I-772 kl-v (36a) (36b) (36c) After some manipulation, we find that A, and A, may be expressed as follows. 1 + v ^ oA, =-l - v A2-RWA\u201e - L B i = - ^ A 2 2h l - v (37a,b) The symmetry conditions eliminate all but one integration constant, and the crack face rotation, = \u2014w,x (0, y), may now be written in terms of A2 (a) . 2n h - v 2 (38) The last integration constant arises from the boundary condition M^(0,y) =-M(y), but one must first find an expression for either uxx(0,y) or uyy(0,y). Transforming equation (22) and applying the symmetry conditions, the stress function x is given by: .(e^'-e-^e-^da 1 -r-iihG. Z~2x\u00b1kl(l-v) From constitutive equations (9e,f) we find expressions for uxx (0, y) and uy (0, y). (39) X-xy *\u2022 - Z'xy-hw U = \u2014-flW,xx, H , = x,x 2 G y,y 2 ( J yy (40) From (9a), we now find an expression for the boundary condition M ( y ) = -M(0, y) 4h2G M(y) = u x x + \u2014 0 ^ V 2 J (41) 97 A n integral equation arises by substituting for ux x and w. M(y) = - ^ - ^ \\f(.a)A2e-ia*da 7>7T 11 - V J J 1 + v Here, f(a) is given as follows. (42) f(cc) = 3 + v \\ Aa2{a-Rv) \\ 1 + v + -k2a + v) \u2022 + \u2022 2n\u00a3 1-772 , a a -4 A R (43) w J We may now form a hyper-singular equation for the crack face rotation. Inverting equation (38) we find an expression for integration constant A2. 2-\u00b1= \\ m 1 - v 3 e,cada (44) B y substitution, an integral arises relating the moment to the crack face rotation. T T , . h?E M(y)=\u2014jP(t) \\f(a)eia\u00b0-y)da dt (45) Note the following integral identities, where K0 and K2 are modified Bessel functions of the second kind. - 1 \\ a e ^ d a = o {t-y) o (t-y) ^ \\ R a 2 e ' a ^ d a = \u00b1(K2(k\\t- y\\)-K0{k\\t-y\\))+ * K2(k\\t- y\\) 1 rflr4 JaO-y) da 1 + -k2(t-y)2 K2(k\\t-y\\) (46a) (46b) (46c) (46d) Applying these identities, equation (45) may be written as a hyper-singular integral. The regular and weakly singular parts have been separated into the functions L and . M(y): -tiE 3^(1 -rf) [Pit) 1 , * > l - 7 ? 2 (t-y)2 2 1 + V L\u00a5(k^-y\\) + kl7i2UkJ\\t-y\\i 1 dt (47) 98 The function L\u00a5 describes the transverse shear deflections, and function is the same expression found by the authors [6] for generalized plane strain plate extension, and accounts for the effects of pressurization near the crack tip. L\u00a5 (z) = ~ + 4 + 4 ( * 2 (z) - K0 (z)) + ^ K2 (z) z z z L(z) = 2 1 3 V \/ \\ 1 2 1 1 + \u2014 | | ^ 2 U ) - \u2014 - \u2014 z 'J z z (48) (49) Equation (47) may be written in terms of the bending stress in the outer fibre as follows. (55) A stiffness matrix for the as-reinforced crack may now be developed by considering the crack and the patch as a set of springs acting in parallel. The result is an as-reinforced stiffness matrix applicable to repaired cracks of arbitrary length. m \\ab j k ~T~ k ^ mm m 1 ^bb + J f..T \\ Kmb \\Ub J (56) B y inverting the combined stiffness matrix of equation (56), a new compliance matrix may be generated for any particular crack length that accounts for the stiffness of the cracked pate: \\Ub j \"bm ^mbi .To -0 > _0 J (57) The as-reinforced crack face displacements can now be calculated directly, and the as-reinforced membrane and bending stress intensity may be calculated by the applying the new compliance matrix to the energy method described in Section 5.2.3. This is a new application of the energy method, which has in the past only been used to calculate long-crack limit solutions. This method of interpolation does not require any assumptions regarding the ratio of the membrane and bending components, overcoming a significant problem with the method of Wang, Rose and Calinan [19]. This new approach to the interpolation allows for the simplified analysis of linear coupled bending and extension of a repair and includes the effects of transverse normal and shear stresses and accompanying shear deflections. It is a valuable tool for patch design and damage tolerance. In subsequent sections, this new interpolation model wi l l be validated against a numerical line-spring model for generalized plane strain plates and against a three-dimensional finite element model of a bonded composite repair. 137 5.3 A line-spring model for crack-bridging In this section we combine the generalized plane strain crack-bridging models for extension and bending [6,25] to address the problem of a crack bridged by coupled springs. The results w i l l be used to validate the closed-form equations developed in Section 5.2 above. The springs bridging the crack are characterized by equation (5), i.e. the stiffness matrix relating the membrane and bending stresses to the crack face displacements. Using an approach similar to that of Wang and Rose [1], the line-spring models for extension and bending may be combined to include the effect of the coupling springs. The significant differences in the current approach are the use of the generalized plane strain plate models and the choice to express the bending moment and displacement in terms of the crack face displacement and the bending stress in the outer fibre of the plate. The generalized plane strain plate models lead to additional terms that account for pressurization and transverse shear about the crack tip. This leads to the following set of coupled Fredholm integral equations with hyper-singular integrands that are solvable for the crack face displacements. The left side of the equation is the net crack opening stress along the crack face, and the right side describes the reaction of the plate. -k kmmUJy) - 1 arEpu\u201e(t) 2K J 1-JJ2 -a ' (t-y)2 +T]2(Kr)2Lpres(Kr\\t-y\\) (58) __ '35 0) .c \u2022o < 20 40 60 80 100 Remote Applied Stress [MPa] Figure 7-23: Adhesive shear stress 120 140 140 Interpolation: 160 mm crack -A\u2014 Interpolation: 80 mm crack - \u2022 - Interpolation: 40 mm crack - \u2022 - Interpolation: 20 mm crack o FEM: 160 mm crack A FEM: 80 mm crack o FEM: 40 mm crack o FEM: 20 mm crack \u2014 2 D FEM 140 Remote Applied Stress [MPa] Figure 7-24: Adhesive peel stress B y a similar technique, it is possible to calculate the stresses in the composite patch. Here, it is possible to calculate the crack bridging component of the membrane and bending stresses in the patch directly from the resultant force and moment given by the 209 effective crack opening stresses in equation (15). The nominal stresses in the patch may be determined from the two-dimensional model presented in Section 7.3. 1 \\ \u00b0 b j 0 > fa0 ^ mc _ 0 \\ G b c J (17) 'J Figures 7-25 and 7-26 show the results of this calculation. It is again observed that the results follow the correct trends. It must be noted that the patch bending stresses are very high for single-sided repairs, and therefore must be carefully considered during design of the resulting hybrid structural repair. n Q. (0 (0 0) 600 500 400 300 1 1 4 ^ Joseph and Erdogan [3,4] and Wang and Rose [7] have used this method to form crack-bridging models. In subsequent sections, the method is applied to the six plate models 281 described above, showing the evolution toward the advanced models resented in chapters three and four. D . l Plane stress extension The plane stress solution follows by assuming vanishing transverse stresses through the thickness of the plate, and is considered valid for thin plates or thick plates with small stress gradients. The governing equations can be derived from an assumed set of displacements with identical results, but here the displacement method has been used to introduce concepts that wi l l be applied to more advanced models. For a plate of thickness 2h, the displacement field is approximated as shown. u x = u x ( x , y ) , u = u ( x , y ) , u = \u2014 w ( x , y ) (6a,b,c) h Integrating through the thickness, and observing the traction-free surfaces of the plate, equilibrium reduces to the following three conditions. For models in which the stresses vary through the thickness, the integrated stresses and equilibrium equations wi l l be expressed in terms of stress resultants, or line forces. \u00b0xx,x + ^ = 0 (7a>D) The compatibility equations reduce to a single condition, automatically satisfied by the assumed displacements. This is a general characteristic of displacement-based solutions. u + u =y (8) x,xyy y,xxy I xy,xy V \/ Given the constant rj = vz __

__ and y\/ are required throughout the solution as the equilibrium equations are not so easily satisfied. D . l . l A cracked plate in plane stress Using Fourier transforms, we may find the general solution to equation (16). Keeping only the roots that vanish as JC, y \u2014> \u00b0\u00b0 . O = \u2014 ] ( v ~ W * + *A2e-Wx) e-iayda (17) The frequency, a, is assumed to be the absolute value in the remainder of the development. Symmetry dictates that shear stresses vanish along the line of the crack. c V 0 , y ) = - O ^ ( 0 , y ) = 0 (18) The stress function reduces to the following. 0 = \u2014 \"\\(e-m+axe-m)Al e'iayda (19) 283 From equilibrium (7a), we note that \u00b0 o ; \u2014oA1 = \\u{t) eic\u00b0da (25) B y substitution, we find an integral relating u(t) to the applied stress. __

U x + h w . x = 0 - (101a,b) The following useful relationships also arise from (101). \"*,y=^> uyiX=~vr (102a,b) The first symmetry condition arises from (100a) by substitution from (101a). y\/ + 2hWxy=0 (103) From equilibrium equation (54a), we find that Mxxx(Q,y) = 0. Substituting into constitutive equation (90a), we find the second symmetry condition + { ( ! - \u00bb ' \u00a5 . , = 0 (104) Symmetry then leads to the following restrictions on the integration constants. Bx + 2h^A}a2 - A2a\\=Q, 2A2a+ * ( 1 ~ V ) \u00a3 , = 0 (105a,b) 2h After some manipulation, we find that A, andfi, may be expressed as follows. (106a,b) aA,=- | A 2 , \u00b1 . B ] = \u2014 A 2 2h 1 l - v 2 V l - V y The rotation of the crack face, fl - \u2014w x (0, y), may be expressed in a simple form. 296 2K J\\-V (107) The final integration constant arises from the boundary condition M H ( 0 , y) = - M ( y ) , however one must first find an expression for either ux x (0, y) or u (0, y). This requires consideration of the stress function %. From equation (98) we find: Z = -^ j(c,\u00abHa|x + C2e~R*x) e-iayda From (100b), we find that C, + C 2 = 0 and x m a v D e expressed as follows. X = \u2014 \\ c \\ e - H x - e R A e - i a y d a 2K i (108) (109) From equation (97), we find that. C, = -2(1 - v)DBx - -2DhiaA2 (110) From constitutive equations (90d,e) we find expressions for uxx(0, y) and uy y(0, y). X, 2G, -hwrr, u \u2014 y-y 2G \u2022 hw yy ( l l l a , b ) B y constitutive equation (90a), we define the boundary condition M ( y ) = M ( 0 , y) M(y) = -D J (112) Substituting for x, w. 2;r (3 + v ) a + \u2014 ) a 2 W,e-\"\u00bbda (113) We may now form a hyper singular equation for the crack face rotations. Inverting (107), we find an expression for integration constant A2. --?-A2 = )j3(t) e,a,dt (114) Substitution into (113) leads to an equation relating the rotation to the bending moment. 297 a+ \u2014(a-R\u00a5)a2 ,ia(t-y) da dt (115) Noting the integral identities (3a-c), this may be rewritten as a hyper-singular integral, resulting in the form found from the Riessner bending model [3,4]. Differences are due to the consideration of a transversely isotropic plate and changes to kv resulting from use of a displacement- rather than stress-based formulation. Denoting z = k\u00a5(t - y). M{y) = Dhk I ( l - v ) AK J>)| 2(3 + v) + 48 _ 4 ^ ( z ) _ ^ ( z ) ) _ 24 ^ ( z ) V ( 1 1 6 ) z z z J Separating the strongly singular parts from the Modified Bessel functions, one may define a function Lv(z) from the regular and weakly singular parts such that. 48 4 24 L\u00a5(z) = -\u2014 + \u2014 + 4{K2 (Z)-K0(z))+ \u2014 K2 (z) z z z The moment along the line x = 0 may now be expressed as follows. ' 2 ( l + v) (117) - -Dhk2v(\\-v) . , M{y) = T Mti An i 1 + L(z) dt (118) Expressed in terms of the bending stress, this may be written in the simpler form. \u2014, . Ghk2 f 2(1+ v) + LM) W 2K I v z- j (119) For a short crack, Lv(z) vanishes, and we find the expression for plane stress extension. Eh f P{t) -dt 2K [(t-yY D.6 Shear-deformable plane strain bending (120) Here we develop a plane strain formulation for a shear deformable plate, assuming a form for the displacements by which the transverse strains vanish. z z u\u201e=\u2014ux(x,y), u =-u(x,y), uz=w(x,y) h h (121a,b,c) The stress resultant and equilibrium equations are identical to equations (51) through (53), and the assumed displacement field satisfies compatibility. The constitutive equations (77) may be used to express the stress resultants in terms of the displacements. 298 Mr=D\\ u _ r + V + T] -U l-TJ2 y'yl] Myy=D\\ V + 7J2 ,1 U H \u2014U y-y 1 ~2 1 _ 2 x.x | 1-^ J %(1 + V)> + M l - v - 2 \/ 72 1 ~2 l - ^ V , = 2 G z l \u00ab , + * w . , J , v + \" , J The plate bending constant, D, takes the form for plane strain bending. ,2 A D = Ah2r \\-V-2T]2 (122a,b) (122c,d) (122e,f) (123) The remainder of the derivation is identical to that for shear deformable plane stress bending. The result is the same set of equations, which describe a 6 t h order system. V 4 w = 0. 3G^ h2G (124a) (124b) The stress function x takes the same form. V 4 ^ r - < V 2 ^ = 0 (125) D.6.1 A shear deformable cracked plate under plane strain conditions Differences between the plane stress and plane strain shear deformable plate models arise only from the definition of the constitutive equations (122) and the plate constant (123). Solutions to (124a,b) and (125) are: J _ In \\{\\eHx + A2xe^x) e~,ayda. y\/ = \u2014 j(fl,e~v) e~iayda, where R* = a2+k2. X = j^\u00b0\\{cxeMx+C2e-R*x) e^da (126a) (126b) (126c) Symmetry requires M (0, y) = 0 and Vx (0, y) = 0, leading to the following conditions. W+2hWxy=0, \\-V-lTJ1 (127a,b,c) 299 Observing these symmetry conditions, equations (126a,b,c) w i l l become: w 1 \u00b0\u00b0 ( = A f 2K \\ x \\ \u00b1 ^ L V e ^ ' e ^ d a . (128a) (128b) 1 - v - 2TJ2 a 1 0 0 1 2 \u2014 {4ih V ,aA2e~R*xe~ia>da 2K_{ \\-v-2rj2 % = \u2014[-2ihDaA2(eMx e'^da (128c) 2TZ_1 x ' The rotation of the crack face, \/? = - w ^ O , y), may now be expressed in a simple form. fly)=J-]2 i - , ^ 2A2e'ia>'da (129) 2\/r_i l - v - 2 T J The last boundary condition M x t ( 0 , y) = - M ( y ) may be arises from equation (122a). , 2 M(y) = -D l - V - 2 \/ 7 2 , v + 77 2 , 1 y \u2014 \/iw hw 2 G z ( l - ^ 2 ) \/ t \" x y 1 \" 2 (130) Substituting for %, w. M ( y ) -Dh 2K V 2 + 1 + v hr + \u2014 (a-Rw)a2 \\A2e~ia>'da (131) Inverting (129) and substituting for A2 results in an integral equation relating the crack face rotation to the bending moment. AK \\-v-2rf 2 + b + \u2014 (a-RJa2 Ya('-y)da dt (132) B y (3a-c), a hyper-singular integral arises. Here, z = k (t - y) and Zy(z) is unchanged. M ( y ) : -Dhk 2 \/ 4;r 1 -V-2T] 2 A JL 1 + V 1-\/7 2 (133) This may be expressed in terms of the stress in the outer fibre. \u2014, , -Ghk2 , f 1 + v 1-\/7 2 - + ' M z ) P (134) 300 For a short crack, the regular part vanishes and this reduces to the form for plane strain extension. D.7 References [1] Reissner E . The effect of transverse shear deformation on the bending of elastic plates. Journal of Applied Mechanics 12, 1945, A69-77 [2] Mind l in R D . Influence of rotary inertia and shear on flexural motions of isotropic, elastic plates. Journal of Applied Mechanics 18, 1951, 31-38 [3] Joseph P F and Erdogan F. Surface crack problems in plates. International Journal of Fracture 41, 1989, 105-131 [4] Joseph PF and Erdogan F. Plates and shells containing a surface crack under general loading conditions. N A S A CR178232, N A S A Langley Research Centre, 1987. [5] Muskhelishvil l i NI . Some Basic Problems in the Mathematical Theory of Elasticity. P. Noordhoff, 1953 [6] Westergaard H M . Bearing Pressures and Cracks. Journal of Applied Mechanics 6, 1939, 49-53 [7] Wang C H and Rose L R F . A crack bridging model for bonded plates subjected to tension and bending. International Journal of Solids and Structures 36, 1999, 1985-2014 [8] Irwin G R . Analysis of Stresses and Strains Near the End of a Crack Traversing a Plate. Journal of Applied Mechanics 24, 1957, 361-364 (135) 301 Appendix E: Material properties E . l Boron\/Epoxy 5521\/4 Laminate Following are the material properties for Textron Specialty Materials Boron 5521\/4 Thickness Resin content (wt.) Cure cycle Cure pressure Test temperature Longitudinal Tensile Modulus Transverse Tensile Modulus Longitudinal Compressive Modulus Longitudinal Poisson's Ratio Transverse Poisson's Ratio Ultimate Tensile Stress Ultimate Compressive Stress Ultimate Interlaminar Shear Stress Coef. O f Thermal Expansion Density Table E-l: Material 0.132 mm (0.0052 in) 33% 2.2-3.3\u00b0C (4-6\u00b0F) per minute to 120\u00b0C (250\u00b0F) 60 minutes at 120\u00b0C (250\u00b0F) 345-586 kPa (50-85 psi) Room temperature 210 GPa (30 msi) 25 GPa 210 GPa (30 msi) 0.21 0.019 1520 M P a (220 msi) 2930 M P a (425 msi) 97 M P a (14.1 msi) 4.5 PPM\/\u00b0C (2.5 PPM\/\u00b0F) 2 g\/cm 3 (0.072 lbm\/in 3) properties for 5521\/4 boron\/epoxy 302 E.2 Aluminum 2024-T3 (A-Basis) Mechanical properties for 2024-T3 aluminum in the rolling direction [98] are; Thickness Test temperature Tensile Modulus Compressive Modulus Shear Modulus Poisson's Ratio Tensile Y i e l d Stress Compressive Y i e l d Stress Shear Y ie ld Stress Ultimate Tensile Stress 3.175 mm (0.125 in) Room temperature 72.4 G P a (10.5 msi) 73.8 G P a (10.7 msi) 27.6 G P a 0.33 331 M P a (48 ksi) 276 M P a (40 ksi) 276 M P a (40 ksi) 448 M P a (65 ksi) Table E-2: Material properties for aluminum 2024-T3 (A-basis) 303 E.3 Adhesive F M 73M The properties for Cytec's F M 7 3 M follow. Note the properties were measured without pre- or post-bond environmental exposure [99]. Thickness 0.25 mm (0.010 in) Nominal weight 300 g\/m 2 (0.06 psf) Primer B R 127 Cure cycle 30 minutes to 120\u00b0C 60 minutes at 120\u00b0C Cure pressure 280 M P a (40 psi) Test temperature 24\u00b0C Shear Modulus 842 M P a (122 ksi) Elastic Shear Stress 17.3 M P a (2510 psi) Elastic Shear Strain 0.021 Ultimate Shear Stress 40.9 M P a (5.93 ksi) Ultimate Shear Strain 0.873 Table E-3: Material properties for adhesive FM-73 304 ","type":"literal","lang":"en"}],"http:\/\/www.europeana.eu\/schemas\/edm\/hasType":[{"value":"Thesis\/Dissertation","type":"literal","lang":"en"}],"http:\/\/www.europeana.eu\/schemas\/edm\/isShownAt":[{"value":"10.14288\/1.0080740","type":"literal","lang":"en"}],"http:\/\/purl.org\/dc\/terms\/language":[{"value":"eng","type":"literal","lang":"en"}],"https:\/\/open.library.ubc.ca\/terms#degreeDiscipline":[{"value":"Mechanical Engineering","type":"literal","lang":"en"}],"http:\/\/www.europeana.eu\/schemas\/edm\/provider":[{"value":"Vancouver : University of British Columbia Library","type":"literal","lang":"en"}],"http:\/\/purl.org\/dc\/terms\/publisher":[{"value":"University of British Columbia","type":"literal","lang":"en"}],"http:\/\/purl.org\/dc\/terms\/rights":[{"value":"For non-commercial purposes only, such as research, private study and education. Additional conditions apply, see Terms of Use https:\/\/open.library.ubc.ca\/terms_of_use.","type":"literal","lang":"en"}],"https:\/\/open.library.ubc.ca\/terms#scholarLevel":[{"value":"Graduate","type":"literal","lang":"en"}],"http:\/\/purl.org\/dc\/terms\/title":[{"value":"Damage tolerance of bonded composite aircraft repairs for metallic structures","type":"literal","lang":"en"}],"http:\/\/purl.org\/dc\/terms\/type":[{"value":"Text","type":"literal","lang":"en"}],"https:\/\/open.library.ubc.ca\/terms#identifierURI":[{"value":"http:\/\/hdl.handle.net\/2429\/31275","type":"literal","lang":"en"}]}}