Mechanics of Materials 2010 Part 8 ppt

Draft Part III FRACTURE MECHANICS Draft Chapter 10 ELASTICITY BASED SOLUTIONS FOR CRACK PROBLEMS 1 Following the solution for the stress concentration around a circular hole, and as a transition from elasticity to fracture mechanics, we now examine the stress field around a sharp crack. This problem, first addressed by Westergaard, is only one in a long series of similar ones, Table 10.1. Problem Coordinate System Real/Complex Solution Date Circular Hole Polar Real Kirsh 1898 Elliptical Hole Curvilinear Complex Inglis 1913 Crack Cartesian Complex Westergaard 1939 V Notch Polar Complex Willimas 1952 Dissimilar Materials Polar Complex Williams 1959 Anisotropic Materials Cartesian Complex Sih 1965 Table 10.1: Summary of Elasticity Based Problems Analysed 2 But first, we need to briefly review complex variables, and the formulation of the Airy stress functions in the complex space. 10.1 †Complex Variables 3 In the preceding chapter, we have used the Airy stress function with real variables to determine the stress field around a circular hole, however we need to extend Airy stress functions to complex variables in order to analyze stresses at the tip of a crack. 4 First we define the complex number z as: z = x 1 + ix 2 = re iθ (10.1) where i = √ −1, x 1 and x 2 are the cartesian coordinates, and r and θ are the polar coordinates. 5 We further define an analytic function, f (z) one which derivatives depend only on z. Applying the chain rule ∂ ∂x 1 f(z)= ∂ ∂z f(z) ∂z ∂x 1 = f  (z) ∂z ∂x 1 = f  (z) (10.2-a) Draft 2 ELASTICITY BASED SOLUTIONS FOR CRACK PROBLEMS ∂ ∂x 2 f(z)= ∂ ∂z f(z) ∂z ∂x 2 = f  (z) ∂z ∂x 2 = if  (z) (10.2-b) 6 If f(z)=α + iβ where α and β are real functions of x 1 and x 2 ,andf(z) is analytic, then from Eq. 10.2-a and 10.2-b we have: ∂f(z) ∂x 1 = ∂α ∂x 1 + i ∂β ∂x 1 = f  (z) ∂f(z) ∂x 2 = ∂α ∂x 2 + i ∂β ∂x 2 = if  (z)  i  ∂α ∂x 1 + i ∂β ∂x 1  = ∂α ∂x 2 + i ∂β ∂x 2 (10.3) 7 Equating the real and imaginary parts yields the Cauchy-Riemann equations: ∂α ∂x 1 = ∂β ∂x 2 ; ∂α ∂x 2 = − ∂β ∂x 1 (10.4) 8 If we differentiate those two equation, first with respect to x 1 , then with respect to x 2 , and then add them up we obtain ∂ 2 α ∂x 2 1 + ∂ 2 α ∂x 2 2 =0 or ∇ 2 (α)=0 (10.5) which is Laplace’s equation. 9 Similarly we can have: ∇ 2 (β) = 0 (10.6) Hence both the real (α) and the immaginary part (β) of an analytic function will separately provide solution to Laplace’s equation, and α and β are conjugate harmonic functions. 10.2 †Complex Airy Stress Functions 10 It can be shown that any stress function can be expressed as Φ=Re[(x 1 − ix 2 )ψ(z)+χ(z)] (10.7) provided that both ψ(z) (psi) and χ(z) (chi) are harmonic (i.e ∇ 2 (ψ)=∇ 2 (χ) = 0) analytic functions of x 1 and x 2 . ψ and χ are often refered to as the Kolonov-Muskhelishvili complex potentials. 11 If f(z)=α + iβ and both α and β are real, then its conjugate function is defined as: ¯ f(¯z)=α − iβ (10.8) 12 Note that conjugate functions should not be confused with the conjugate harmonic functions. Hence we can rewrite Eq. 10.7 as: Φ=Re[¯zψ(z)+χ(z)] (10.9) 13 Substituting Eq. 10.9 into Eq. 9.19, we can determine the stresses σ 11 + σ 22 =4Reψ  (z) (10.10) σ 22 − σ 11 +2iσ 12 =2[¯zψ  (z)+χ  (z)] (10.11) and by separation of real and imaginary parts we can then solve for σ 22 − σ 11 & σ 12 . 14 Displacements can be similarly obtained. Victor Saouma Mechanics of Materials II Draft 10.3 Crack in an Infinite Plate, (Westergaard, 1939) 3 σ σ σ σ x x 0 0 0 0 1 2 2a Figure 10.1: Crack in an Infinite Plate 10.3 Crack in an Infinite Plate, (Westergaard, 1939) 15 Just as both Kolosoff (1910) and Inglis (1913) independently solved the problem of an elliptical hole, there are two classical solutions for the crack problem. The first one was proposed by Westergaard, and the later by Williams. Whereas the first one is simpler to follow, the second has the advantage of being extended to cracks at the interface of two different homogeneous isotropic materials and be applicable for V notches. 16 Let us consider an infinite plate subjected to uniform biaxial stress σ 0 with a central crack of length 2a, Fig. 10.1. From Inglis solution, we know that there would be a theoretically infinite stress at the tip of the crack, however neither the nature of the singularity nor the stress field can be derived from it. 17 Westergaard’s solution, (Westergaard 1939) starts by assuming Φ(z) as a harmonic function (thus satisfying Laplace’s equation ∇ 2 (Φ) = 0). Denoting by φ  (z)andφ  (z) the first and second derivatives respectively, and ¯ φ(z)and ¯ ¯ φ(z) its first and second integrals respectively of the function φ(z). 18 Westergaard has postulated that Φ=Re ¯ ¯ φ(z)+x 2 Im ¯ φ(z) (10.12) is a solution to the crack problem 1 . 19 †Let us verify that Φ satisfies the biharmonic equation. Taking the first derivatives, and recalling from from Eq. 10.2-a that ∂ ∂x 1 f(z)=f  (z), we have ∂Φ ∂x 1 = ∂ ∂x 1  Re ¯ ¯ φ  +     x 2 ∂ ∂x 1 Im ¯ φ(z)+Im ¯ φ(z) ∂x 2 ∂x 1  0     (10.13-a) =Re ¯ φ(z)+x 2 Imφ(z) (10.13-b) σ 22 = ∂ 2 Φ ∂x 2 1 = ∂ ∂x 1  Re ¯ φ(z)  +  x 2 ∂ ∂x 1 Imφ(z)+Imφ(z) ∂x 2 ∂x 1  (10.13-c) =Reφ(z)+x 2 Imφ  (z) (10.13-d) 1 Note that we should not confuse the Airy stress function Φ with the complex function φ(z). Victor Saouma Mechanics of Materials II Draft 4 ELASTICITY BASED SOLUTIONS FOR CRACK PROBLEMS 20 †Similarly, differentiating with respect to x 2 , and recalling from Eq. 10.2-b that ∂ ∂x 2 f(z)=if  (z), we obtain ∂Φ ∂x 2 = ∂ ∂x 2 (Re ¯ ¯ φ(z)+[x 2 ∂ ∂x 2  Im ¯ φ(z)  + ∂x 2 ∂x 2 Im ¯ φ(z)] (10.14-a) = −Im ¯ φ(z)+x 2 Reφ(z)+Im ¯ φ(z) (10.14-b) σ 11 = ∂ 2 Φ ∂x 2 2 = x 2 ∂ ∂x 2 Reφ(z)+Reφ(z) ∂x 2 ∂x 2 (10.14-c) = −x 2 Imφ  (z)+Reφ(z) (10.14-d) 21 †Similarly, it can be shown that σ 12 = − ∂ 2 Φ ∂x 1 ∂x 2 = −x 2 Reφ  (z) (10.15) 22 †Having derived expressions for the stresses and the second partial derivatives of Φ, substituting into Eq. ??, it can be shown that the biharmonic equation is satisfied, thus Φ is a valid solution. 23 †If we want to convince ourselves that the stresses indeed satisfy both the equilibrium and compati- bility equations (which they do by virtue of Φ satisfiying the bi-harmonic equation), we have from Eq. 6.18 in 2D: 1. Equilibrium: ∂σ 11 ∂x 1 + ∂σ 12 ∂x 2 = 0 (10.16-a) ∂σ 22 ∂x 2 + ∂σ 12 ∂x 1 = 0 (10.16-b) Let us consider the first equation ∂σ 11 ∂x 1 = ∂ ∂x 1  ∂ 2 Φ ∂x 2 2  = ∂ ∂x 1 [Reφ(z) − x 2 Imφ  (z)] (10.17-a) =Reφ  (z) − x 2 Imφ  (z) (10.17-b) ∂σ 12 ∂x 2 = ∂ ∂x 2  ∂ 2 Φ ∂x 1 ∂x 2  = ∂ ∂x 2 [−x 2 Reφ  (z)] (10.17-c) = −Reφ  (z)+x 2 ∂ ∂x 1 Imφ  (z) (10.17-d) = −Reφ  (z)+x 2 Imφ  (z) (10.17-e) If we substitute those two equations into Eq. 10.16-a then we do obtain zero. Similarly, it can be shown that Eq. 10.16-b is satisfied. 2. Compatibility: In plane strain, displacements are given by 2µu 1 =(1− 2ν)Re ¯ φ(z) − x 2 Imφ(z) (10.18-a) 2µu 2 = 2(1 −ν)Im ¯ φ(z) − x 2 Re φ(z) (10.18-b) and are obtained by integration of the strains which were in turn obtained from the stresses. As a check we compute 2µε 11 = ∂u 1 ∂x 1 (10.19-a) =(1− 2ν)Reφ(z) − x 2 Imφ  (z) (10.19-b) =(1− ν)[Reφ(z) − x 2 Imφ  (z)]    σ 11 −ν [Reφ(z)+x 2 Imφ  (z)]    σ 22 (10.19-c) =(1− ν)σ 11 − νσ 22 (10.19-d) Victor Saouma Mechanics of Materials II Draft 10.3 Crack in an Infinite Plate, (Westergaard, 1939) 5 Recalling that µ = E 2(1 + ν) (10.20) then Eε 11 =(1− ν 2 )σ 11 − ν(1 + ν)σ 22 (10.21) this shows that Eε 11 = σ 11 −ν(σ 22 −σ 33 ), and for plane strain, ε 33 =0⇒ σ 33 = ν(σ 11 + σ 22 )and Eε 11 =(1− ν 2 )σ 11 − ν(1 + ν)σ 22 24 So far Φ was defined independently of the problem, and we simply determined the stresses in terms of it, and verified that the bi-harmonic equation was satisfied. 25 Next, we must determine φ such that the boundary conditions are satisfied. For reasons which will become apparent later, we generalize our problem to one in which we have a biaxial state of stress applied on the plate. Hence: 1. Along the crack: at x 2 =0and−a<x 1 <awe have σ 22 = 0 (traction free crack). 2. At infinity: at x 2 = ±∞, σ 22 = σ 0 We note from Eq. 10.13-d that at x 2 =0σ 22 reduces to (σ 22 ) x 2 =0 =Reφ(z) (10.22) 26 Furthermore, we expect σ 22 → σ 0 as x 1 →∞,andσ 22 to be greater than σ 0 when | x 1 − a |>(due to anticipated singularity predicted by Inglis), thus a possible choice for σ 22 would be σ 22 = σ 0 1− a x 1 ,for symmetry, this is extended to σ 22 = σ 0  1− a 2 x 2 1  . However, we also need to have σ 22 = 0 when x 2 =0and −a<x 1 <a, thus the function φ(z) should become imaginary along the crack, and σ 22 =Re   σ 0  1 − a 2 x 2 1   (10.23) 27 Thus from Eq. 10.22 we have (note the transition from x 1 to z). φ(z)= σ 0  1 − a 2 z 2 (10.24) 28 If we perform a change of variable and define η = z − a = re iθ and assuming η a  1, and recalling that e iθ =cosθ + i sin θ, then the first term of Eq. 10.13-d can be rewritten as Reφ(z)=Re σ 0  η 2 +2aη η 2 +a 2 +2aη ≈ Re σ 0  2aη a 2 ≈ Reσ 0  a 2η ≈ Reσ 0  a 2re iθ ≈ σ 0  a 2r e −i θ 2 ≈ σ 0  a 2r cos θ 2 (10.25-a) 29 Recalling that sin 2θ =2sinθ cosθ and that e −iθ =cosθ −i sin θ, we substituting x 2 = r sin θ into the second term x 2 Imφ  = r sin θIm σ 0 2  a 2(re iθ ) 3 = σ 0  a 2r sin θ 2 cos θ 2 sin 3θ 2 (10.26) 30 Combining the above equations, with Eq. 10.13-d, 10.14-d, and 10.15 we obtain Victor Saouma Mechanics of Materials II Draft 6 ELASTICITY BASED SOLUTIONS FOR CRACK PROBLEMS σ 22 = σ 0  a 2r cos θ 2  1+sin θ 2 sin 3θ 2  + ··· (10.27) σ 11 = σ 0  a 2r cos θ 2  1 − sin θ 2 sin 3θ 2  + ··· (10.28) σ 12 = σ 0  a 2r sin θ 2 cos θ 2 cos 3θ 2 + ··· (10.29) 31 Recall that this was the biaxial case, the uniaxial case may be reproduced by superimposing a pressure in the x 1 direction equal to −σ 0 , however this should not affect the stress field close to the crack tip. 32 Using a similar approach, we can derive expressions for the stress field around a crack tip in a plate subjected to far field shear stresses (mode II as defined later) using the following expression of φ Φ II (z)=−x 2 Re ¯ φ II (z) ⇒ φ II = τ  1 − a 2 z 2 (10.30) and for the same crack but subjected to antiplane shear stresses (mode III) Φ  III (z)= σ 13  1 − a 2 z 2 (10.31) 10.4 Stress Intensity Factors (Irwin) 33 Irwin 2 (Irwin 1957) introduced the concept of stress intensity factor defined as:    K I K II K III    = lim r→0,θ=0 √ 2πr    σ 22 σ 12 σ 23    (10.32) where σ ij are the near crack tip stresses, and K i are associated with three independent kinematic movements of the upper and lower crack surfaces with respect to each other, as shown in Fig. 10.2: • Opening Mode, I: The two crack surfaces are pulled apart in the y direction, but the deformations are symmetric about the x −z and x − y planes. • Shearing Mode, II: The two crack surfaces slide over each other in the x-direction, but the defor- mations are symmetric about the x − y plane and skew symmetric about the x − z plane. • Tearing Mode, III: The crack surfaces slide over each other in the z-direction, but the deformations are skew symmetric about the x − y and x − z planes. 34 From Eq. 10.27, 10.28 and 10.29 with θ =0,wehave K I = √ 2πrσ 22 = √ 2πrσ 0  a 2r = σ 0 √ πa (10.33-a) where r is the length of a small vector extending directly forward from the crack tip. 2 Irwin was asked by the Office of Naval Research (ONR) to investigate the Liberty ships failure during World War II, just as thirty years earlier Inglis was investigating the failure of British ships. Victor Saouma Mechanics of Materials II Draft 10.5 Near Crack Tip Stresses and Displacements in Isotropic Cracked Solids 7 Figure 10.2: Independent Modes of Crack Displacements 35 Thus stresses and displacements can all be rewritten in terms of the SIF    σ 22 σ 12 σ 23    = 1 √ 2πr   f I 11 (θ) f II 11 (θ) f III 11 (θ) f I 22 (θ) f II 22 (θ) f III 22 (θ) f I 12 (θ) f II 12 (θ) f III 12 (θ)      K I K II K III    (10.34-a) i.e. σ 12 = K II √ 2πr sin θ 2 cos θ 2 cos 3θ 2    f II 22 (10.35) 1. Since higher order terms in r were neglected, previous equations are exact in the limit as r → 0 2. Distribution of elastic stress field at tip can be described by K I ,K II and K III . Note that this polar distribution is identical for all cases. As we shall see later, for anisotropic cases, the spatial distribution is a function of elastic constants. 3. SIF are additives, i.e. 4. The SIF is the measure of the strength of the singularity (analogous to SCF) 5. K = f(g)σ √ πa where f(g) is a parameter 3 that depends on the specimen, crack geometry, and loading. 6. Tada “Stress Analysis of Cracks”, (Tada, Paris and Irwin 1973); and Cartwright & Rooke, “Com- pendium of Stress Intensity Factors” (Rooke and Cartwright 1976). 7. One of the underlying principles of FM is that unstable fracture occurs when the SIF reaches a critical value K Ic . K Ic or fracture toughness represents the inherent ability of a material to withstand a given stress field intensity at the tip of a crack and to resist progressive tensile crack extensions. 10.5 Near Crack Tip Stresses and Displacements in Isotropic Cracked Solids 36 Using Irwin’s concept of the stress intensity factors, which characterize the strength of the singularity at a crack tip, the near crack tip (r  a) stresses and displacements are always expressed as: 3 Note that in certain literature, (specially the one of Lehigh University), instead of K = f(g)σ √ πa, k = f(g)σ √ a is used. Victor Saouma Mechanics of Materials II Draft 8 ELASTICITY BASED SOLUTIONS FOR CRACK PROBLEMS Pure mode I loading: σ xx = K I (2πr) 1 2 cos θ 2  1 − sin θ 2 sin 3θ 2  (10.36-a) σ yy = K I (2πr) 1 2 cos θ 2  1+sin θ 2 sin 3θ 2  (10.36-b) τ xy = K I (2πr) 1 2 sin θ 2 cos θ 2 cos 3θ 2 (10.36-c) σ zz = ν(σ x + σ y )τ xz = τ yz = 0 (10.36-d) u = K I 2µ  r 2π  1 2 cos θ 2  κ − 1+2sin 2 θ 2  (10.36-e) v = K I 2µ  r 2π  1 2 sin θ 2  κ +1− 2cos 2 θ 2  (10.36-f) w = 0 (10.36-g) Pure mode II loading: σ xx = − K II (2πr) 1 2 sin θ 2  2 + cos θ 2 cos 3θ 2  (10.37-a) σ yy = K II (2πr) 1 2 sin θ 2 cos θ 2 cos 3θ 2 (10.37-b) τ xy = K II (2πr) 1 2 cos θ 2  1 − sin θ 2 sin 3θ 2  (10.37-c) σ zz = ν(σ x + σ y ) (10.37-d) τ xz = τ yz = 0 (10.37-e) u = K II 2µ  r 2π  1 2 sin θ 2  κ +1+2cos 2 θ 2  (10.37-f) v = − K II 2µ  r 2π  1 2 cos θ 2  κ − 1 − 2sin 2 θ 2  (10.37-g) w = 0 (10.37-h) Pure mode III loading: τ xz = − K III (2πr) 1 2 sin θ 2 (10.38-a) τ yz = K III (2πr) 1 2 cos θ 2 (10.38-b) σ xx = σ y = σ z = τ xy = 0 (10.38-c) w = K III µ  2r π  1 2 sin θ 2 (10.38-d) u = v = 0 (10.38-e) where κ =3− 4ν for plane strain, and κ = 3−ν 1+ν for plane stress. 37 Using Eq. ??, ??,and?? we can write the stresses in polar coordinates Victor Saouma Mechanics of Materials II Draft 10.5 Near Crack Tip Stresses and Displacements in Isotropic Cracked Solids 9 Pure mode I loading: σ r = K I √ 2πr cos θ 2  1+sin 2 θ 2  (10.39-a) σ θ = K I √ 2πr cos θ 2  1 − sin 2 θ 2  (10.39-b) τ rθ = K I √ 2πr sin θ 2 cos 2 θ 2 (10.39-c) Pure mode II loading: σ r = K II √ 2πr  − 5 4 sin θ 2 + 3 4 sin 3θ 2  (10.40-a) σ θ = K II √ 2πr  − 3 4 sin θ 2 − 3 4 sin 3θ 2  (10.40-b) τ rθ = K II √ 2πr  1 4 cos θ 2 + 3 4 cos 3θ 2  (10.40-c) Victor Saouma Mechanics of Materials II [...]... 1.22 08 985 1 7494 1.2405 1.01 68 7929 1.2457 1.03 58 8259 1.2350 1.0536 87 23 1.2134 1.0 582 9029 1. 189 9 1.0571 9242 1.1476 1.0495 9513 1.1149 1.0409 9670 1.0904 1.0336 97 68 1.0649 1.0252 985 5 1.0395 1.0161 99267 1.01 78 1.0077 9976 β Pressurized Hole λ=1 λ=0 2 188 1725 30 58 2319 4 183 3334 49 58 3979 5551 4 485 6025 489 7 689 8 5 688 7494 6262 7929 6701 82 59 7053 87 23 7 585 9029 7971 9242 82 64 9513 86 77 9670 89 57... 11 .8 √ KI = βσ πa (11.11) where, using Newman’s solution β is given in Table 11.1 Victor Saouma Mechanics of Materials II Draft 6 LEFM DESIGN EXAMPLES a R 1.01 1.02 1.04 1.06 1. 08 1.10 1.15 1.20 1.25 1.30 1.40 1.50 1.60 1 .80 2.00 2.20 2.50 3.00 4.00 β Biaxial Stress λ = −1 λ = 1 λ=0 0.4325 0.3256 0.2 188 5971 4514 30 58 7 981 6 082 4 183 9250 7104 49 58 1.0135 784 3 5551 1.0775 84 00 6025 1.1746 9322 689 8 1.22 08. .. Mechanics of Materials II Draft 11.2 Stress Intensity Factors 7 Figure 11.9: Pressurized Hole with Radiating Cracks Figure 11.10: Two Opposite Point Loads acting on the Surface of an Embedded Crack Figure 11.11: Two Opposite Point Loads acting on the Surface of an Edge Crack Victor Saouma Mechanics of Materials II Draft 8 LEFM DESIGN EXAMPLES x a C 1 1.01 1.03 1.07 1.11 < 0.6 0.6-0.7 0.7-0 .8 0 .8- 0.9 >... solutions to some simple commonly used test geometries, followed by a tabulation of fracture toughness of commonly used engineering materials Finally, this chapter will conclude with some simple design/analysis examples 2 11.1 Design Philosophy Based on Linear Elastic Fracture Mechanics One of the underlying principles of fracture mechanics is that unstable fracture occurs when the stress intensity factor... b b and t ≤ 0 .8 For a approximately equal to 0.25, K is roughly independent of θ For shallow cracks b 1, Equation 11.24 reduces to t √ K = 1.13σ πb 1 − 08 For very long cracks b a b a 1 1.65 − 2 (11. 28) 1, Equation 11.24 reduces to √ K = 1.13σ πb 1 + 3.46 11.3 1 + 1.464 b a b t 2 + 11.5 b t 4 (11.29) Fracture Properties of Materials Whereas fracture toughness testing will be the object of a separate... not yet a standard for fracture toughness of concrete, Hillerborg (Anon 1 985 ) has proposed a standard procedure for determining GF Furthermore, a subcommittee of ASTM E399 is currently looking into a proposed testing procedure for concrete Rock: Ouchterlony has a comprehensive review of fracture toughnesses of numerous rocks in an appendix of (Ouchterlony 1 986 ), and a proposed fracture toughness testing... proposed fracture toughness testing procedure can be found in (Ouchterlony 1 982 ) Table 11.3 provides an indication of the fracture toughness of common engineering materials Note that √ stress intensity factors in metric units are commonly expressed in Mpa m, and that √ √ 1ksi in = 1.099Mpa m (11.30) Victor Saouma Mechanics of Materials II ... b a −1 + 14 1 − (11.26) b a 24 (11.27) Mechanics of Materials II Draft 10 LEFM DESIGN EXAMPLES Material KIc √ ksi in 49 190 20 20-30 70 100 18 0.7 16 0.5 1 3 Steel, Medium Carbon Steel, Pressure Vessel Hardened Steel Aluminum Titanium Copper Lead Glass Westerly Granite Cement Paste Concrete Nylon Table 11.3: Approximate Fracture Toughness of Common Engineering Materials b Newman and Raju report that... field uniform stress σ, we know that there is a stress concentration factor of 3 for a crack radiating from this hole, we consider two cases a Short Crack: D → 0, and thus we have an approximate far field stress of 3σ, and for an edge crack β = 1.12, Fig 11.6 thus √ KI = 1.12(3σ) πa √ = 3.36σ πa (11 .8) Victor Saouma Mechanics of Materials II Draft 4 LEFM DESIGN EXAMPLES Figure 11.4: Three Point Bend Beam... Metallic Alloys: 10 Testing procedures for fracture toughness of metallic alloys are standardized by various codes (see (399 n.d.) and (British Standards Institution, BS 5447, London 1977)) An exhaustive tabulation of fracture toughnesses of numerous alloys can be found in (Hudson and Seward 19 78) and (Hudson and Seward 1 982 ) Concrete: Fracture mechanics evolved primarily from mechanical and metallurgical . 0.2 188 .2 188 .1725 1.02 .5971 .4514 .30 58 .30 58 .2319 1.04 .7 981 .6 082 .4 183 .4 183 .3334 1.06 .9250 .7104 .49 58 .49 58 .3979 1. 08 1.0135 . 784 3 .5551 .5551 .4 485 1.10 1.0775 .84 00 .6025 .6025 . 489 7 1.15. .9322 . 689 8 . 689 8 .5 688 1.20 1.22 08 . 985 1 .7494 .7494 .6262 1.25 1.2405 1.01 68 .7929 .7929 .6701 1.30 1.2457 1.03 58 .82 59 .82 59 .7053 1.40 1.2350 1.0536 .87 23 .87 23 .7 585 1.50 1.2134 1.0 582 .9029. .7971 1.60 1. 189 9 1.0571 .9242 .9242 .82 64 1 .80 1.1476 1.0495 .9513 .9513 .86 77 2.00 1.1149 1.0409 .9670 .9670 .89 57 2.20 1.0904 1.0336 .97 68 .97 68 .9154 2.50 1.0649 1.0252 . 985 5 . 985 5 .93 58 3.00 1.0395