# Fracture Mechanics_2nd Ed_Solution Manual

#### Description

Chapter 1 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

CHAPTER 1 THEORY OF ELASTICITY 1.1 A thin sheet made of an aluminum alloy having E = 67 GPa, G = 256 GPa and also v = 1/3 was used for two dimensional surface strain measurements. The measurements provided  xx  10.5 x10 5 ,  yy  20 x10 5 , and  xy  240 x10 5. Determine the corresponding stresses. Solution:  xx  E  xx  v yy   2.89 MPa       2   yy  (1  v )  yy  v xx   12.44 MPa 

 xy 

E xy

2(1  v)

 60.45 MPa

1.2 Determine (a) the principal stresses and strains and (b) the maximum shear stress for the case described in Problem 1.1. Solution: (a)  1, 2 

 xx   yy 2

  xx   yy   2 

2

    xy2  4.28 MPa  60.60 MPa 

 1  56.32 MPa  2  64.88 MPa The principal strains are  1, 2 

 xx   yy 2

  xx   yy   2 

   xy        2  2

2

 4.75 x10 5  1.20 x10 3

 1  1.15 x10 3  2  1.25 x10 3 (b) The maximum

 max

  xx   yy   2 

2

    xy2  60.60 MPa 

1

Chapter 1 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

1.3 Calculate the diameter of a 1-m long wire that supports a weight of 200 Newton. If the wire stretches 2 mm, calculate the strain and the stress induced by the weight. Let E = 207 GPa. Solution:



2 mm l  3  2 x10 3  0.20% l o 10 mm

4P  0.78 mm    E  414 MPa P  A  ( / 4)d 2  200 N d

1.4 Derive an expression for the local uniform strain across the neck of a round bar being loaded in tension. Then, determine its magnitude if the original diameter is reduced 80%. Solution: Volume constancy: AL  Ao Lo

 zz 

L

dL 2  lnL / Lo   ln d o / d  L Lo

 zz  2 ln d o / d 

If d  0.80d o , then  zz  2 ln 1 / 0.80  0.4463  zz  44.63%

1.5 The torsion of a bar containing a longitudinal sharp groove may be characterized by a warping function of the type [after F.A McClintock, Proc. Inter. Conf. on Fracture of Metals, r

Inst. of Mechanical Eng., London, (1956) 538] w   z    ( ydx  xdy ) . The displacements are 0

 x  0 and  y  rz , where  and r are the angle of twist per unit length and the crack tip radius, respectively. The polar coordinates have the origin at the tip of the groove, which has a radius (R). Determine w, the shear strains  rz and   z . In addition, predict the maximum of the shear strain   z .

2

Chapter 1 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

Solution:

 y  1   z   z  r   R 2  1  r    R 2   r  r r

Let x  R cos and y  R sin  so that

 z 

dx   R sin d and dy  R cos d

 z

r

0

0

 z    ( ydx  xdy)    R 2 d

If r  0 then   z   max  

 rz  0 1.6 A cantilever beam having a cross-sectional area of 1.5 cm2 is fixed at the left-hand side and loaded with a 100 Newton downward vertical force at the extreme end as shown in the figure shown below. Determine the strain in the strain gage located at 8 cm from the fixed end of the shown steel cantilever beam. The steel modulus of elasticity of is E = 207 GPa. -

+

Strain gage

P = 100 N h = 0.5 cm b = 3 cm

8 cm X = 20 cm

Solution:  M   0

M

FBD:

( Moments )

P X  X 1   M  0 M  P X  X 1 

X – X1

Thus, 6P X  X 1    P( X  X13)( h / 2 )   2  nh bh 12

Using Hooke’s law yields the elastic strain 



 E 6P X  X

Ebh 2

1



(6)(100 N )(20 m  8 m) x10  2 (207 x10 6 N / m)(3 x10  2 m)(0.50 x10  2 m) 2 9

3

 4.64 x10  4

  MC I Ch 2 3 bh I 12

Chapter 1 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

1.7 The stress-strain behavior of an annealed low-carbon steel (σys = 200 MPa and E = 207 GPa) obeys the Hollomon equation with k = 530 MPa and n =0.25. (a) Plot the true and engineering stress-strain curves. Calculate (b) the tensile strength (  ts ) and (c) the strain energy density up to the instability point. Solution: (a) Plots

 t  1     t  ln 1   

 t  k  t n

1     k ln1   n



k ln 1    1   

n

 t  n  0.25 @ Instability point

1000 True

 t  1200 t 

0.25

Stress 750

Engineering



500

1200ln 1    1  0.25

0.25

Yield Point

250



ys ,  ys   0.0043, 305.81 MPa 

 ts  0.25

00 00

0.05 0.05

0.10 0.1

0.15 0.15

0.20 0.2

0.25 0.25

0.30 0.3

Strain

(b) The engineering yield strain and the true strain at the instability point on the true stress-strain curve are, respectively

 ys 900 MPa   4.3478 × 10- 3 3 E 207 x10 MPa  t  n  0.25 @ Instability point  ys 

At the instability or ultimate tensile strength point (maximum),

 max  k tn  1,200 MPa 0.250.25  848.53 MPa  t  ln 1      1  exp( t )  1  exp(0.25)  0.28403

4

Chapter 1 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

 max  1    ts  ts   max / 1     ts  374.77 MPa  / 1  0.28403  291.87 MPa (c) The strain energy density (W) is just the area under the stress-strain curve. In general, the true and engineering strain energy densities are

   ts    ys   Wt   d    d  0      Elastic   ys  Plastic  ys

 ts

E ys2 k  ts 1 n   ys 1 n Wt   Ed   k d   2 1 n 0  ys n

Wt

2  207,000 4.3478 × 10- 3   

1,200 0.284031 0.25  4.3478 × 10-3 1 0.25  335.50 MPa 1  0.25

2 MN m Wt  335.50 2  335.50 MJ /m3 m m and    ts    ys   W    d     d  0      Elastic   ys  Plastic

n E ys2 k ln 1    k n 1 W   E d     d   ln  ts  1   ys  1 1    2 n 1 0  ys

 ys

 ts

2  207,000 4.3478 × 10- 3   W 

1,200  - 3 0.25 1   ln 0.28403 - 4.3478 × 10   0.25  1 

2 W  0.99932  0.95718i

Therefore, the engineering strain energy density can not be determined analytically using the modified Hollomon equation. Instead, the solution can be achieved numerically. Thus, 1200ln 1    0 207000 d  0.0043478  1   

0.0043478

W 

0.28403

0.25

d  170.60 MPa

W  170.60 MJ /m 3 Wt  2W

5

Chapter 1 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

1.8 The figure below shows a schematic cross-sectional view of a pressure vessel (hollow cylinder) subjected to internal and external pressures. Determine the stresses at a point P r,   in polar coordinates when (a) Pi  0 and Pi  0, (b) Po  0, (c) Pi  0 and (d) a  0 so that the hollow cylinder becomes a solid cylinder. (e) Plot  rr  f (r ) and  rr  f (r ) . Let a  450 mm, b  800 mm, Pi  0 and P  40 MPa. The valid radius range must be 0.45 m  r  0.80 m . Use the following Airy’s stress function   c1  c2 ln r  c3 r 2 Along with the boundary conditions  rr   Pi  r  0 @ r  a

 rr   Po

 r  0

Solution: (a) Derivatives of   c1  c2 ln r  c3 r 2  c2  0   2 c3 r  2 0 r r  r 2 2 c   0   22  2c3 2 r r  2 From eq. (1.58), 1  1  2 c2    2 c3 r r r 2  2 r 2  2 c    2   22  2c3 r r 1  1  2  r  2  0 r  r r

 rr 

6

@r b

Chapter 1 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

Using the boundary conditions yields c c  rr   Pi  22  2c3  Pi  22  2c3 r r c c  rr   Po  22  2c3  Po  22  2c3 r r Replace r for the proper radius and solve eq. (a) for c2 and c3

(a)

c2  2 c3 r2 c  Po  22  2c3 r from which c  Pi  22  2c3 a c 2c3   Pi  22 a and c  Po  22  2c3 b c 2c3   Po  22 b  Pi 

Then, c c Pi  22  Po  22 a b c2 c 2   Po  Pi a 2 b2 o  Pi

c2 1 a 2b2 a 2b2   Pi  Po    P  P  P   P  o o i o b2 b2 b2  a 2 b2  a 2  a2  a2P a2P a2P & 2c3  2 i 2  2 i 2  Po  2 i 2  Po  2  1 2 b a b a b a b a 

a 2b2 Po  Pi  c2  2 b  a2 a 2b2 Po  Pi  & c2  2 b  a2

2c3   Po 

2 c3 

c3 

a 2 Pi b 2 Po  b2  a 2 b2  a 2

a 2 Pi  b 2 Po 2b 2  a 2 

Thus,

a 2 Pi  b 2 Po a 2 b 2 Po  Pi   rr   b 2  a 2 r 2 b2  a 2

a 2 Pi  b 2 Po a 2 b 2 Po  Pi   b 2  a 2 r 2 b2  a 2 0

    r

7

Chapter 1 Fracture Mechanics, 2nd ed. (2015)

(b) If Po  0, then

Solution Manual

(c) If Pi  0, then

a P  rr  2 i 2 b a

 b2  1  2  r  

b 2 Po  rr  2 b  a2

 a2   2  1 r 

a 2 Pi b2  a 2

 b2  1  2  r  

  

b 2 Po b2  a 2

 a2   2  1 r 

2

  

 r  0

 r  0

(d) If a  0, then  rr   Po

    Po  r  0 (e) The plot

100 75

 (MPa)

 

50 25 0

 rr

-25 -50 0.4

0.5

0.6

8

0.7

0.8

Chapter 1 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

1.9 Consider an infinite plate with a central hole subjected to a remote uniform stress as shown in Example 1.5. The boundary conditions for this loaded plate are (1)  x   x  S and σy = τxy = 0 at r   and (2)  r   r  0 at r  a. Use the following complex potentials [12]

  z 

S 4

1

2a 2 z2

and   z   S2 1 

a2 z2

3a 4 z4

to determine σr, σθ and τrθ (in polar coordinates). Solution: The infinite plate with a central hole subjected to a remote uniform stress is

Use the following stress expressions

 r     2  z  2  z  4 Re   z     r  i2 r  2z  z    ze i2 Recall that the Euler's formula for z and z² are

z  re i  rcos   i sin  z  re i  rcos   i sin  z m  r m e im  r 3 cos m  i sin m and their real and imaginary parts are

Re z  r cos  & Re z m  r m cos m Im z  r sin  & Im z m  r m sin m Apply the Euler's formula to the above complex potentials so that 2 2   z  S 1  2a2  S 1  2a 2 i2 4 4 z r e 2 S 2a   z  1  2 e i2 4 r 2 4 Re   z  S 1  2a2 cos 2 r

(a) The first stress equation: (a)

 r     4 Re  z  S 1  

2a 2 r2

cos 2

Then, 9

Chapter 1 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

2  z  S  a 2 4 2z 2 2 2 2  z   2Sa3  Sa3  Sa  Sa3 e i3 3 i3 2z z r e r 2 Sa   z  3 cos 3 (Real part) r

Thus,

z  z  re i 

Sa 2 r 3 e i3

Sa 2 r2

e i4

(b) The second potential:

  z   S2 1 

a2 z2

3a 4 z4

  S2 1 

a2 r 2 e i2

3a 4 r 4 e i4

Hence,

    r  i2 r  2z  z    ze i2 2 2 i2 4 i2     r  i2 r  2 Sa2 e i4 e i2  S e i2  Sa2 e i2  3Sa4 ei4 2 r 2r e 2r e 2 2 4     r  i2 r  S 2a2 e i2  e i2  a2  3a4 e i2 r r r Real parts: 2 2 4     r  S 2a2 cos 2  cos 2  a2  3a4 cos 2

r

r

(b)

(c)

r

Add eqs. (a) and (c) and solve for σθ 2 4    S2 1  a2  S2 1  3a4 cos 2

(d1)

Use eq. (a) to solve for σr 2  r  S2 1  a2  S2 1 

(d2)

r

r

r

4a 2 r2

3a 4 r4

cos 2

Using the imaginary part of eq. (b) yields shear stress τrθ as 2 4 i2 r  S i 2a2 sin 2  i sin 2  i 3a4 sin 2 r r 2 4  r   S 1  2a2  3a4 sin 2 2 r r

(d3)

At r = a r  0

(f1)

   S  2S cos 2  S1  2 cos 2

(f2)

 r  0

(f3)

10

Chapter 1 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

1.10 Use the Cauchy-Riemann condition to show that (a) f(z) = 1/z is analytic and (b) its derivative is f(z)= -1/z². Solution: (a) If z = x + iy, then

1 z

1 z

1 xiy xiy xiy

1 xiy x x 2 y 2

xiy

x 2 y 2

iy x 2 y 2

Let

u

v 

x x 2 y 2

y x 2 y 2

Then x 2 y 2 x  x x 2 y 2

u x

u y



x 2 y 2

x 2 y 2 2x 2

2

2

x 2 y 2

y 2 x 2 x 2 y 2

2xy x 2 y 2

2

and 2xy v  2 2 x x  y 2  v  x 2  x 2 2 y x 2  y 2 

Thus, u  y

 v x

Therefore, f(z) = 1/z is analytic because ∂u/∂y = ∂v/∂x. (b) The derivative of f(z) = 1/z is

f  z  f  z 

d dz

 1z  

y 2 x 2 x 2 y 2

2

 

z2 zz 2

2

 i v x 2xy

i

iiyx  2 i x 2 y 2

u x

x 2 y 2

x 2 y 2

xiy 2



2

yix  2

x 2 y 2

2

2



1 z2 11

z2 xiyxiy 2

2

Chapter 1 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

1.12 Evaluate the Cauchy integral formula given below for the complex function when zo = π.

fz o  

1 2i

cos z z 2 1

dz

Solution: The integral has to be modified as

1 2i fz o   1 2i fz o  

fz o   1

z dz  zcos 2 1

cos z dz  z  1z  1

dz  zfz 1

dz  2ifz o   zfz 1 where

fz 

cos z z1

Then, dz  2ifz o   zfz 1 dz  2i  zfz 1

cos z o zo  1

dz  2i  zfz 1

cos  1

z o 

 1. 52i

12

Chapter 2 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

CHAPTER 2 INTRODUCTION TO FRACTURE MECHANICS 2.1 Show that the applied stress   0 as the crack tip radius   0. Explain Solution: From eq. (2.28), a  max  2 

 

and

1   max 2 a

Therefore,   0 as   0. This means that the applied stress   0 since the material has fractured or separated, at least into two pieces, and there is not a crack present so that   0. On the other hand,   0 means that the existing crack is very sharp as in the case of a fatigue crack.

2.2 For a Griffith crack case, crack propagation takes place if the strain energy satisfies the inequality  U ( a )  U ( a   a ) 2  a , where a  crack extension and  is the surface energy. Show that the crack driving force or the strain energy release rate at instability is G 

U(a) a

Solution: Use the Taylor’s series to expand the left side term of the inequality. Thus, 2 3 1  U (a ) 2 1  U ( a ) 3 1 U ( a ) U (a   a )  U ( a )  a  a   a  .......... .... 1! a 2! a 2 3! a 3

(2.1)

Using the first two terms yields U (a   a )  U (a ) 

U ( a )

U (a )  U (a   a )  

a U ( a )

a

(2.2)

a

a

(2.2)

1

.

Chapter 2 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

Then,  

U ( a ) a U ( a ) a

 a  2 a

(2.3)

 2

(2.4)

 G

Therefore,

G

U ( a ) a

(2.5)

2.3 A 1mmx15mmx100mm steel strap has a 3-mm long central crack is loaded to failure. Assume that the steel is brittle and has E  207,000 MPa,  ys  1500 MPa, and

K IC  70 MPa m . Determine the critical stress ( c ) and the critical strain energy release rate. Solution: a = 1.5 mm and a / w  0.10 The geometry correction factor:

  f (a / w) 

w a  tan    1.07 a  w 

The stress intensity factor K IC    f a

c 

2 K IC

a

 1020 MPa

The strain energy release rate 2 K IC E GIC  23.70 kPa m

GIC 

GIC  23.70 kN / m 2

2.4 Suppose that a structure made of plates has one cracked plate. If the crack reaches a critical size, will that plate fracture or the entire structure collapse? Explain. Answer: If the structure does not have crack stoppers, the entire structure will collapse since the cracked plate will be the source of structural instability. 2

Chapter 2 Fracture Mechanics, 2nd ed. (2015)

2.5

Solution Manual

What is crack instability based on according to Griffith criterion?

Answer: It is based on the energy consumption creating new surfaces or in developing crack extension. This energy is the surface energy 2.

2.6 Can the Griffith Theory be applied for a quenched steel containing 1.2%C, if a pennyshaped crack is detected? Answer: Despite the quenched steel is brittle, the Griffith Theory applies only if a plastic zone does not form at the crack tip.

2.7 Will the Irwin Theory (or modified Griffith Theory) be valid for a changing plastic zone size as the crack advances? Answer: No. It is valid if the plastic zone remains constant as it the case in many materials under plane strain conditions.

2.8 What are the major roles of the surface energy and the stored elastic energies in a crack growth situation? Answer: The surface energy acts as a retarding force for crack growth and the stored elastic energy acts to extend the crack. Thus, concurrent with crack growth is the recovery of elastic strain energy by the relaxation of atomic bonds above and below the fracture plane.

2.9

What does happen to the elastic strain energy when crack growth occur?

Answer: It is released as the crack driving force for crack growth. 2.10 What does U/a = 0 mean? Answer: Crack growth occurs at    f , as shown in eq. (2.23), since the magnitude of crack growth and the crack resistance force become equal. That is, 2s 

a 2 E

2 If 2s  a , then fracture occurs and the fracture toughness without any plastic zone

E

deformation is G  2γ at    f , as shown in eq. (2.31) with  p  0. Thus, G  2γ is related c

c

s

to an irreversible process of fracture. 3

s

Chapter 2 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

2.11 Derive eq. (2.28) starting with eq. (2.20). Solution: From eq. (2.20),

Kt 

 max a  2  

1  max  2

a  

Multiplying this expression by eq. (2.28) 1  max   2

 yields the stress intensity factor as per

a     a 

K I   a

4

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

CHAPTER 3 LINEAR ELASTIC FRACTURE MECHANICS 3.1 A steel strap 1-mm thick and 20-mm wide with a through-the-thickness central crack 4-mm long is loaded to failure. (a) Determine the critical load if KIC = 80 MPa m for the strap material. (b) Use the available correction factor,   f ( a / w) , for this crack configuration. Now, do the following three comparisons to have a better understanding: Feddersen:  c  ( fraction)  , Irwin:  c  ( fraction)  and Koiter-Benthem:  c  ( fraction)  . 2a

P

w

P

B=1 mm

Solution: B = 1mm, w = 20mm, 2a = 4 mm, a/w = 0.10, KIC = 80 MPa m , K I   c a f a / w , (a) Finite plate: From eqs. (3.31) and (3.32) along with a / w  0.10 rad .  (180o )( 0.10) /   5.7296o .   seca w  1025 w a  Irwin [13]: f  a w  tan   1017 . a  w  1 1  a  1304  . a Koiter-Benthem [14]: f  a w    w .  w   1021 a 1 2 f aw 

Feddersen [12]:

2

 c  984.64 MPa

 c  992.38 MPa  c  988.49 MPa

w

(b) Infinite plate approach [a/w  0 since f(a/w)  1]. K 80MPa m  c  IC   1009.25MPa a  2  10 3 m

Feddersen: Irwin: Koiter:

 c  97.56%   c (infinite)  c  98.33%   c (infinite)  c  97.94%   c (infinite)

Therefore, the calculated  c values do not differ much since the plate dimensions are large enough and the initial crack size is relatively small.

1

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

3.2 A steel tension bar 8-mm thick and 50-mm wide with an initial single-edge crack of 10-mm long is subjected to a uniaxial stress  = 140 MPa. (a) Determine the stress intensity factor KI. If KIC = 60 MPa m , is the crack stable? (b) Determine the critical crack size, and (c) determine the critical load. a w

P

P

B

Solution:

 = P/A= 140 MPa, KIC = 60 MPa m , B = 8 mm,

w = 50 mm,

a = 10 mm,

a  K I   a f   w  x = a/w = 0.20, A = Bw

From Table 3.1, K I    a and

  f  x   1.12  0.231x   10.55 x   21.71 x   30.38x    f  x  0.20  1.37 2

3

4

(a) The applied stress intensity factor KI = 34 MPa m KIC = 60 MPa m Therefore, the crack is stable because KI < KIC. (b) K IC   a c . f  x  2

1  K IC  ac    0.0311 m  31.10 mm   f x     (c) The critical load P   A A  Bw  8 mm 50 mm   400 mm 2  4  10 4 m 2  MN  P  A  4  10  4 m 2 140 2  0.056MN  56 kN (Applied) m    MN   4  10 4 60 2 m  AK IC  m   0.0988 MN  98.84 kN (Critical) Pc   3 f  x  a 1.37  10  10 m 

2

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

3.3 A very sharp penny-shaped crack with a diameter of 22-mm is completely embedded in a highly brittle solid. Assume that catastrophic fracture occurs when a stress of 600 MPa is applied. a) What is the fracture toughness for this solid? (Assume that this fracture toughness is for plane strain conditions). b) If a sheet 5-mm thick of this solid is prepared for fracturetoughness testing. Would the fracture-toughness value [(calculated in part a)] be an acceptable number according to the ASTM E399 standard? Use  ys = 1342 MPa. c) What thickness would be required for the fracture-toughness test to be valid? Solution:

2 2  a  600 MPa   11  10 3 m  71 MPa m  

a) K IC 

K b) B  2.5 IC   ys c) B  7 mm

2

2

     2.571 MPa m   7 mm > 5 mm (Not valid) 1342 MPa     

3.4 (a) One guitar steel string has a miniature circumferential crack of 0.009 mm deep. This implies that the radius ratio is almost unity, d / D  1 . b) Another string has a localized miniature surface crack (single-edge crack like) of 0.009 mm deep. Assume that both strings are identical with an outer diameter of 0.28 mm. If a load of 49 N is  applied to the string when being tuned, will it break? Given properties: K IC  15MPa m ,  ys  795MPa.

 Solution: (a) Circumferential crack Dd If a   0.009 mm , then d  0.262 mm 2 4P (4)(49 N )     908.87 MPa 2 d  (0.262 x10 3 m)

  f (d / D) 

1 2

2 3 D  1 D 3 d  d  d      0 . 36  0 . 73         d  2 d 8 D  D  D  

For d / D  0.9357,   1.14

3

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

K I    a

Solution Manual

K I  (1.14)908.87 MPa   0.009  10 3 m

K I  5.51 MPa m

(b) Surface edge crack

  1.12 K I    a

K I  (1.12)908.87 MPa   0.009  10 3 m

K I  5.41 MPa m Therefore, the strings will not break since K I  K IC in both cases. When d / D  1 and the crack size is very small, either approach gives similar results.

3.5 A 7075-T6 aluminum alloy is loaded in tension. Initially the 10mmx100mmx500mm plate has a 4-mm single-edge through–the-thickness crack. (a) Is this test valid? (b) Calculate the maximum allowable tension stress this plate can support, (c) Is it necessary to correct KI due to crack-tip plasticity? Why? or Why not? (d) Calculate the design stress and stress intensity factor if the safety factor is 1.5. Data:  ys = 586 MPa and KIC = 33 MPa m . Solution: 2

  (a) B  2.5K IC   7.93 mm and a w  0.0375  ys   Plane strain condition holds because the actual thickness is greater than the required value. (b) K I    a ; a = 4 mm and w = 100 mm

  f a   w

  f a

 

2

3

f 0.04  1.13

Thus,

 

K IC

4

a  a  a  a   1.12  0.231  10.55   21.71   30.38  w w  w  w  w 

a

 261 MPa

4

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

(c) Plastic zone correction: 2

1 K IC  r     0.17mm ys  6  K eff    a eff with a eff  a  r = 4.17 mm

K eff  1.13261 MPa   4.17 x10 3 m  K eff  33.76 MPa m

Therefore, there is no need for crack tip plasticity since K I ,eff  K IC .

 261 MPa   174 MPa SF 1.5 K 33 MPa m K Id  IC   22 MPa m SF 1.5

(d)  d 

3.6 A steel ship deck (30mm thick, 12m wide, and 20m long) is stressed in the manner shown below. It is operated below its ductile-to-brittle transition temperature (with KIc = 28.3 MPa m ). If a 65-mm long through–the-thickness central crack is present, calculate the tensile stress for catastrophic failure. Compare this stress with the yield strength of 240 MPa for this steel. Solution: 2a = 65mm w = 12 m L = 20m

  f a / w  1 since a/w  0 P

P

2a

KI

Transition Range.

(MPa m ) or CVN Energy (Joules) T

5

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

K IC    a

 

K IC

a

Solution Manual

K IC  28.3 MPa m and  ys  240 MPa 28.3 MPa m

65     10 3 m  2 

 88.57 MPa

Therefore,    ys and to assure structural integrity, the safety factor (SF) that can be used to avoid fracture or crack propagation is in the order of SF 

 ys 

240 MPa  2.71 88.57 MPa

3.7 Show that

KI  0 for crack instability in a large plate under a remote tensile external stress. da

Solution: K I   a

dK I 1     0 since   0 and a  0 da 2 a

3.8 The plate below has an internal crack subjected to a pressure P on the crack surface. The stress intensity factors at points A and B are

KA  KB 

P

P

a a

ax dx ax ax dx ax

2a P A B

Use the principle of superposition to show that the total stress intensity factor is of the form K I  P a Solution: According to the principle of superposition, the total stress intensity factor is KI  KA  KB 

P

 ax ax a dx dx  2 P    ax  a2  x2  ax

 a 

6

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

Furthermore,

ax  ax

ax  ax

ax

ax dx x  ar c sin 2 2 a a x

ax ax

(a  x)(a  x) ( a  x )( a  x )

2a a2  x2

and

a x arcsin  a

K I  2P

If x = a on the crack surface, then arcsin 1   / 2  90 o and

a x 2P  ar sin  a   P a  a  2

K I  2P

3.9 A pressure vessel is to be designed using the leak-before-break criterion based on the circumferential wall stress and plane strain fracture toughness. The design stress is restricted by the yield strength  ys and a safety factor (SF). Derive expressions for (a) the critical crack size and (b) the maximum allowable pressure when the crack size is equals to the vessel thickness. Solution: (a) The critical

crack size

K IC    ac

 ys SF

d  Thus,

K IC 

  ys ac SF

1 S  K ac   F   IC      ys 2

2

   

Select some alloy having known K IC and  ys values. The alloy with the highest

K IC    ys

2

  values should be used for constructing the pressure vessel.  

7

2 K IC and  ys

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

3.10 A stock of steel plates with GIC = 130 kJ/m², σys = 2,200 MPa, E = 207 GPa, v = 1/3 are used to fabricate a cylindrical pressure vessel (di = 5 m and B = 25.4 mm). The vessel fractured at a pressure of 20 MPa. Subsequent failure analysis revealed an internal semi-elliptical surface crack of a = 2.5 mm and 2c = 10 mm. (a) Use a fracture mechanics approach to predict the critical crack length this steel would tolerate. (b) Based on this catastrophic failure, another vessel was constructed with do = 5.5 m and di = 5 m. Will this new vessel fracture at a pressure of 20 MPa if there is an internal semi-elliptical surface crack having the same dimensions as in part (a)? Solution: (a) The vessel is based on the thin-wall theory since the thickness is d 5 x103 mm B i   250 mm (Requirement) 20 20 B  25.4 mm (Given) is less than the required thickness. So use the thin-wall theory. Thus, the fracture hoop stress is h  Pi di  2B

20 MPa

5x10 3 mm

2 25. 4 mm

 h  1, 968. 50 MPa

Using eq. (2.34) yields the critical crack length ac  ac 

EGIC  1  v 2 2h 207x10 3 MPa

130x10 3 MPa. m

 1  1/3 2  1, 968. 50 MPa

2

a c  2. 49 mm  2. 5 mm (b) For the new vessel, 5. 6 m  5 m B  do  di   0. 30 m 2 2 d o  5. 6  1. 12 5 di

Also, h 

d o /d i  2  1 d o /d i  2  1

P i  8. 86Pi

 h  8. 8620 MPa  177. 23 MPa    h  P i  9. 86P i  197. 23 MPa 8

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

Then, a/2c  2. 5/10  0. 25 /ys  197. 23/2, 200  0. 09

From eq. (3.44),  2 Q  2 Q  1. 63 Q

2

2

3  a 2 4 2c 3  0. 25 2 4

2

 7 ys 33 2  7 0. 09 2 33

2

Thus, the applied stress intensity factor is KI    a Q KI  197. 23 MPa

 2. 5x10 3  1. 63

KI  13. 69 MPa m

From eq. (2.33) at fracture, EGIC 1  v2

KIC 

207x10 3 MPa

KIC 

130x10 3 MPa. m

1  1/3 2

KIC  174 MPa m

Therefore, the new vessel will not fracture because KI < KIC. 3.11 A cylindrical pressure vessel with B = 25.4 mm and di = 800 mm is subjected to an internal pressure Pi. The material has K IC  31 MPa m and σys = 600 MPa. (a) Use a safety factor to determine the actual pressure Pi. (b) Assume there exists a semi-elliptical surface crack with a = 5 mm and 2c = 25 mm and that a pressure surge occurs causing fracture of the vessel. Calculate the fracture internal pressure Pf. (c) Calculate the critical crack length. Solution: (a) Let's figure out which pressure vessel wall theory should be used. B

di 20

800 mm 20

 40 mm

9

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

Therefore, the thin-wall theory should be used. The actual or applied pressure is  ys h  P i d i  2B SF 2 25. 4 mm 600 MPa 2B ys Pi   SFdi 2. 5 800 mm P i  15. 24 MPa (b) Using the principle of superposition yields   h  Pi  Pi di  Pi 2B d 800  1 P i i   1 Pi  2B 2x25. 4   16. 75P i  16. 75 15. 24 MPa   255. 24 MPa (Actua l)

Also, a  5  0. 20 2c 25   255. 24  0. 43  ys 600

a  5  0. 20 B 25. 4 M  Mk  1 [ See eq.(3.46) & Figure 3.6]

From eq. (3.44),

 2 Q  2 Q  1. 50 Q

2

2

3  a 4 2c 3  0. 2 2 4

2

2

 7 ys 33 2  7 0. 43 2 33

2

From eq. (3.41), K IC  MM k 

a Q

 16. 75P f

a Q

Now, the surge pressure is Pf 

K IC 16. 75

31 MPa m Q a  16. 75

1. 50  5x10 3 m

Pf  18. 09 MPa

10

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

(c) The critical crack length is KIC    a  16. 75P i Q Q ac  

KIC 16. 75Pi

2

 ac Q

 1.5

31 MPa m

2

16. 75  15. 24 MPa

a c  7. 04 mm

3.12 This is a problem that involves strength of materials and fracture mechanics. An AISI 4340 steel is used to design a cylindrical pressure having an inside diameter and an outside diameter of 6.35 cm and 12.07 cm, respectively. The hoop stress is not to exceed 80% of the yield strength of the material. (a) Is the structure a thin-wall vessel or a thick-wall pipe? (b) What is the internal pressure? (c) Assume that an internal semi-elliptical surface crack exist with a = 2 mm and 2c = 6 mm. Will the vessel fail? (d) Will you recommend steel for the pressure vessel? Why? or Why not? (e) What is the maximum crack length the AISI 4340 steel can tolerate? Explain. Solution:

Internal Crack

d i  6.35 m

External Surface

d O  12.07 m

a 2c a  2 mm and 2c  6 mm

From Table 3.2, σys = 1,476 MPa and KIC = 81 MPa for AISI 4340 steel. (a) Using the diameters yields d o  12. 07  1. 09 6. 35 di 12. 07 cm  6. 35 cm B  do  di  2 2 B  2. 86 cm

Therefore, the pressure vessel is based on the thick-wall theory because d o  1. 1 di B  d i  0. 32 20 (b) From eq. (3.43e), h 

d o /d i  2  1 d o /d i  2  1

P i  0. 8 ys

Pi  0. 8 ys

d o /d i  2  1 d o /d i  2  1 11

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

Thus, the internal pressure is P i  0. 8 1, 476 MPa

1. 09 2  1 1. 09 2  1

P i  101. 51 MPa

(c) Fracture mechanics:    h  P i  0. 8 ys  P i   0. 81, 476 MPa  101. 51 MPa   1, 282. 30 MPa Thus, a/2c = 2/6 = 0.33, σ/σys = 0.87 and the shape factor becomes  2 Q  2 Q  1. 66 Q

3  a 2 4 2c 3  0. 33 2 4

2

2

2

 7 ys 33 2  7 0. 87 2 33

2

Then, K I    a  1, 282. 30 MPa Q

 2x10 3 m 1. 66

K I  78. 89 MPa m

Therefore, the pressure vessel will not fail because KI < KIC. (d) Denote that KI is slightly below the KIC value and safety precautions should be taken because a minor increase in pressure due to a pressure surge will cause fracture. The margin of safety is very small and therefore, a new steel should be used having a higher KIC value. (e) The critical crack length is KIC    a c Q Q ac  

KIC 

2

 1.66

81 MPa m 1, 282. 30 MPa

a c  2. 11 mm

This result implies that the pressure vessel may fracture because the existing crack of 2 mm in length is approximately 5% smaller than its critical value. Therefore, a new steel should be used to design the pressure vessel having the same dimensions and being subjected to the design conditions. 12

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

3.13 A simple supported beam made of soda glass (E =71 MPa and GIC = 12 J/m²) is subjected to a uniformly distributed bending load as shown in the figure below. Assume an initial crack length of 0.1 mm due to either stress corrosion cracking (SCC) or a mechanical defect introduced during fabrication or handling. The design stress (working stress) and the crack velocity equation .83 are 0.10 MPa and v  K Im  6.24 m/s K 15 , respectively. Use this information to calculate the I lifetime of the beam if the maximum bending stress is given by σ = (MB/2)/I and M = σdL²/8, where the second moment of area is I = wB³/12. P B  15 mm w  200 mm L 2m

a

Solution: GI  12 J / m 2  12 Pa.m, E  71 GPa B  20 mm, w  200 mm, L  2 m

Given data:

The applied stress

 

3 d L2 30.10 MPa 2,000 mm    6.67 MPa 4 wB 2 4200 mm15 mm2 2

Using Griffith expression, eq. (2.34), gives the critical crack length ac 



EGIC 71x103 MPa 12 x10 6 MPa.m   6.10 mm  2  6.67 MPa 2

This is a large crack length for a brittle material such as soda glass. The applied and the critical stress intensity factors, eq. (3.29), are

K I    a  1.126.67 MPa   10 4 m  0.13 MPa m

K IC    ac  1.126.67 MPa   6.10 x10 3 m  1.03 MPa m

The crack velocity equation can be rearranged using the Chain Rule. Thus,

v

da da dK I  dt dK I dt

13

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

But, K I    a

a

K I2  2 2

Then, da 2K I  dK I  2 2

and v

da 2 K I dK I & v  K Im  6.24 m/s K I15.83  dt  2 2 dt

K Im 

2 K I dK I  2 2 dt

Rearranging this equation and integrating with respect to time yields 2K I dt  dK I 2 2   K Im tf

2 o dt   2 2  K IC

where

K IC

K

1 m I

1 m K I dK I  

KI

dK I

KI

1 K 2m m2

K IC KI

 

1 2 m K IC  K I2 m m2

Then,

tf  

2 1 2m K IC  K I2m 2    m  2 2

  1 2  215.83 2 15.83 t f    0.13 1.03  2 2   1.12  6.67 MPa  6.24 m/s  15.83  2  t f  2.3733x108 seconds  7.52 years Therefore, the beam will last 7.53 years prior to fracture.

14

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

3.14 Plot the given data for a hypothetical solid SE(T) specimen and determine the plane strain fracture toughness KIC. Here,   f ( a / w)  is the modified geometry correction factor. a 1/2 m 1/2  6 9.8 15 18 22  MPa

25

35

38

30 50 70 90 116 122 175 190

Solution: The plot

200

  150 MPa  100

Slope  5.0144 MPa m

50

0 0

10

20

a 1/ 2 m1/ 2  Linear curve fitting gives

   5.0144a -1/2 - 0.42832 Slope 

     5.0144 MPa m  a 1 / 2 

K IC    a  5.0144 MPa m

15

30

40

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

3.15 A pressure vessel (L >> B = 15 mm, di = 2 m) is to be made out of a weldable steel alloy having σys = 1,200 MPa and KIC = 85 MPa. If an embedded elliptical crack (2a = 5 mm and 2c = 16 mm) as shown below is perpendicular to the hoop stress, due to welding defects, the given data correspond to the operating room temperature and the operating pressure is 8 MPa, then calculate the applied stress intensity factor. Will the pressure vessel explode?

2a

B

2c

Solution: d o  d i  2B  2 m  2  15  10 3 m  2. 03 m d o  2. 03  1. 02 2 di 2  10 3 mm B  di   100 mm 20 20

The pressure vessel is of thin-wall type because d o /d i  1. 1 and B  d i /20. The hoop stress and the stress intensity factor for the embedded elliptical crack are, respectively

where

Then,

The pressure vessel will not explode because KI < KIC.

16

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

3.16 A steel pressure vessel subjected to a constant internal pressure of 20 MPa contains an internal semi-elliptical surface crack with dimensions shown below. Calculate K I when (a) Q and σ/σ ys in eq. (3.44) and (b) Q  1  1.464a/c 

1.65

as per reference [31]. (c) Compare results

and explain. (d) If fracture does not occur, calculate the safe factors S F( a ) and S F(b ) . Explain the results. Data: K IC  92 MPa m ,  ys  900 MPa , v  0.3 and B  30 mm.

2c = 8 mm

di = 1 m do = 1.04 m

B

a a = 2 mm

Solution: For d o /d i  1.04 and B  d o  d i / 2  1.04 m  1 m/ 2  0.02 m  20 mm Pi  20 MPa , a / 2c  2 / 8  1 / 4 Pi d i 1m   d     Pi  1  i Pi   1  20 MPa   520 MPa  σ ys 2B  2 B   2 x 0.02 m   520 / 900  0.57778

(a) σ  σ h  Pi 

 /  ys

2

2

2 2 7   π   3 a   Q           2   4 2c   33   ys  2

2 2 7 π   3 1   2 Q          0.57778  1.5581 2   4 4   33

a  2 x10 3  K I  1.12  1.12520 MPa   36.98 MPa m Q 1.5581 Validity as per ASTM E399 2

K IC      2.592 MPa m   2.6123x10 2 m  26.12 mm BASTM  2.5  900 MPa  σ     ys  B  30 mm  BASTM OK! 2

(b) For Q  1  1.464a / c 

1.65

K I  1.12

 1  1.4642 / 4 

1.65

 1.4665 ,

a  2 x10  3   1.12520 MPa   38.12 MPa m Q 1.4665

17

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

(c) Comparing these results yields K I( a )  0.97 K I( b ) and the percent error is

% Error 

38.12  36.98 x100  3% 38.12

This indicates that the Q equations give dissimilar results despite the different mathematical approaches being used. Nonetheless, both calculated K I values are below K IC and therefore, fracture does not occur. (d) Since K I  K IC the safety factor can be calculated. Thus, S F( a )  K IC / K I( a )  92 / 36.98  2.49 S F( b )  K IC / K I( b )  92 / 38.12  2.41 These are reasonably high values and therefore, the pressure vessel can be kept in service, but a nondestructive technique must be used to monitor any crack growth that may lead to a  a c and K I  K IC .

3.17 An aluminum-alloy plate has a plane strain fracture toughness of 30 MPa m . Two identical single-edge cracked specimens are subjected to tension loading. (a) One specimen having a 2-mm crack fractures at a stress level of 330 MPa. Calculate the geometry correction factor   f a / w . (b) Will the second specimen having a 1-mm crack fracture when loaded at 430 MPa? Solution: (a) The geometry correction factor is K IC 30 MPa m   f a / w     1.1469  a 330 MPa   2 x10 3 m (b) The stress intensity factor for the second specimen is K I    a  1.1469430 MPa   1x10 3 m  27.64 MPa m

Therefore, fracture will not occur because K I  K IC .

18

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

EXTRA PROBLEM NOT INCLUDED IN THE BOOK A thin-walled cylindrical pressure vessel contains gas at a pressure of 100 MPa. During the initial pressurization of the vessel, the axial and tangential strains were measured on the inside surface as xx = 500x10-6 and yy = 750x10-6, respectively. Calculate a) the axial and hoop stresses associated with these strains. Assume that an undetected internal semi-elliptical surface crack (a = 1 mm deep and 4 mm long) grows 0.5 mm/year and that the pressure remains constant. b) Deduce whether or not the vessel will fracture by calculating the applied stress intensity factor using the constant hoop stress (tangential stress) and c) if the vessel will not fracture, then calculate the time it takes for such unpleasant occurrence.

Note: Only the shape factor should be used to correct the stress intensity factor KI. Given data:  ys  600 MPa K IC  30 MPa m E  207 GPa v  0. 3 B  2 cm d  0. 50 m (Diameter)

19

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

Solution: a) From eq. (1.6) along with  zz  P, the inner strains are

xx

yy

 xx  v yy  P

1 E

(1.6)

 yy  v xx  P

Solving these two equations simultaneously yields

 xx 

E 1v

yy 

 xx  yy 1v

 yy  Eyy  v xx  P

vP 1v

E xx 1v 2

 vyy  

vP 1v

(a) (b)

Given data:

xx  500  10 6  ys  600 MPa

yy  750  10 6 K IC  30 MPA m

E  207, 000 MPa

Thus,

 xx 

207,000 MPa 10.3 2

500  10 6  0. 3 750  10 6

0.3 10.3

100 MPa

 xx  122. 06 MPa (Axial Stress)  yy  207, 000 MPa 750  10 6  0. 3 122. 06 MPa  100 MPa  yy  161. 87 MPa

(Hoop Stress or Tangential Stress)

b) ASTM E399 thickness requirement, eq. (3.30): 2 30 a, BASTM  2. 5  10 3 600  6. 25 mm B  BASTM , but a is not met for use LEFM. The valid average tangential or hoop stress regardless of the wall thickness is  av   yy,av 

Pd 2B

1000.510 2 cm 22 cm

 1, 250 MPa

The applied stress intensity at initial pressurization hoop stress is E  207 GPa K IC  30 MPa m v  0. 3    yy  161. 87 MPa  ys  600 MPa / ys  161. 87/600  0. 269 78 a  1 mm 2c  4 mm a/2c  0. 25 Q

 2

2

3 4

a 2c

2

2

7 33

K I   a/Q  161. 87 MPa

  ys

2

 1. 6134

 110 3 m 1.6134

20

 7. 14 MPa m

Chapter 3 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

 The the initial pressurization process of the vessel will not cause fracture because K I  K IC at    yy  161. 87 MPa.  If a  B  20 mm (leak-before-fracturecondition), then  2010 3 m

K I   yy a/Q  161. 87 MPa

1.6134

 31. 94 MPa m

Therefore, fracture will occur because K I  31. 94 MPa m  K IC  When  yy   av fracture will occur because K I   av a/Q  1, 250 MPa

 110 3 m 1.6134

 55. 16 MPa m

K I  K IC  30 MPa m c) If the tangential stress,  yy  161. 87 MPa, is assumed constant thoughtout the test, then the critical crack length becomes K IC   a c /Q Q

2

a c   K IC  yy a c  B  20 mm

1.6134 

10 3

30 161.87

2

 17. 64 mm

Thus, the life of the vessel is approximately t  a c  a/da/dt  17. 64 mm  1 mm / 0. 5 mm/year t  33. 28 years This is an unrealistic result because  yy   av after the initial pressurization process is completed. Consequently, the vessel fractures immediately and its lifetime is basically t  0.

21

Chapter 4 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

CHAPTER 4 ELASTIC STRESS FIELD EQUATIONS

4.1 (a) Calculate KI using the singularity and non-singularity stresses for a single-edge cracked plate having a = 4 mm, x = 0.1, and L/w ≥ 1.5, and subjected to 300 MPa in tension. (b) Plot the stress ratio Tx/ and the biaxiality ratio  as functions of x = a/w. Solution:

2) Single-edge Crack (SET) K I   a

L / w  1.5 and x  a/w

w

L

  1.12 - 0.23a/w   10.55 a / w  21. 71a / w 2

a

 30.38 a / w

3

4

Given data:   300 MPa, x  0.1, a  4 x10 3 m The stress ratio Tx 1   0.526  0.641x  0.2049 x 2  0.755 x 3  0.7974 x 4  0.1966 x 5 2  1  x 

Tx  0.56688  Tx   0.56688   0.56688300 MPa  Tx  170.06 MPa

The biaxiality ratio 1   0.469  0.1414 x  1.433x 2  0.0777 x 3  1.6195 x 4  0.859 x 5 1/ 2 1  x    0.46444

1

Chapter 4 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

Plots:

Parameter Parameter



Tx





x Stress ratio approach: T   170.06 MPa  3 K I( 2)  x a     4 x10 m    0.46444 

K I( 2)  41.05 MPa m Classical approach: The geometry correction factor   1.12  0.23a/w  10.55a/w2  21.71a/w3  30.38a/w4

  1.12  0.230.1  10.550.1  21.710.1  30.380.1   1.1838 2

The stress intensity factor

3

4

K I(1)   a  1.1838300 MPa   4 x10 3 m K

( 2) I

 39.81 MPa m

Therefore, the Tx stress approach gives slight higher values for K I .

2

Chapter 4 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

4.2 Using the information given in problem 4.1, calculate the stress intensity factor K I when the stress ratio is Tx  0. Solution: 2) Single-edge Crack (SET)  w

   1.12 - 0.23a/w   10.55 a / w  21. 71a / w 2

a

a  4 x10 3 m

 30.38 a / w

3

4

  300 MPa

x  a / w  0.1

  1.12  0.23x  10.55x 2  21.71x 3  30.38 x 4   1.12  0.230.1  10.550.12  21.710.13  30.380.14   1.1838

K I   a  1.1838300 MPa   4 x10 3 m  K I  39.81 MPa m

4.3 Assume that a single-edge crack in a plate is loaded in tension. Derive the dominant near crack-tip stresses in Cartesian coordinates using the Westergaard complex function Z z  

KI where K I   a 2z

Solution: Polar coordinates:

x  r cos , y  r sin  , r  x 2  y 2 , 1   sin   cos sin 2 2 2 From Euler’s formulas, z  reiθ / 2  r cos  i sin      z  r 1/ 2 e iθ / 2  r 1/ 2  cos  i sin  2 2  3 3   z  r 3 / 2 e i 3θ / 2  r 3 / 2  cos  i sin  2 2  

3

Chapter 4 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

Find the derivative of the Westergaard complex function KI KI KI KI Z z   & Z ' z     z 3 / 2   r 3 / 2 e i 3 / 2 3/ 2 2z 2 2 z 2 2 2 2 Using the Euler’s formula gives the Westergaard complex function in terms of trigonometric functions KI K K    Z z    I r 1 / 2 e iθ / 2  I r 1/ 2  cos  i sin  iθ / 2 2 2 2π 2π  2πre Z ' z   

KI KI 3 3   r 3 / 2 e i 3 / 2   r 3 / 2  cos  i sin  2 2  2 2 2 2 

The real and imaginary parts of the trigonometric functions are K K   Re Z  z   I r 1 / 2 cos Im Z  z    I r 1/ 2 sin 2 2 2π 2π K K 3 3 Re Z '  z    I r 3 / 2 cos Im Z '  z    I r 3 / 2 sin 2 2 2π 2π From eq. 4.11), the stress in the x-direction becomes K  3   K I 3 / 2  x  Re Z  z   y Im Z ' z   I r 1/ 2 cos  r sin   r sin  2 2  2π  2 2π K  1 3  K    3     x  I r 1/ 2  cos  sin  sin   I r 1 / 2  cos  sin cos sin  2 2 2  2 2 2 2  2π 2π   KI   3  x  cos 1  sin sin  2 2 2  2πr Similarly,

 y  Re Z  z   y Im Z '  z  

KI 2π

r 1/ 2 cos

 3   K I 3 / 2  r sin   r sin  2 2   2 2π

K I 1/ 2   1 3  K    3   r  cos  sin  sin   I r 1 / 2  cos  sin cos sin  2 2 2  2 2 2 2  2π 2π   KI   3  y  cos 1  sin sin  2 2 2  2πr

y 

and K I 3 / 2 3  K 1 3   xy   y Re Z '  z   r sin    r cos   I r 1 / 2 sin  cos 2  2 2 2π  2 2π KI   3  xy  cos sin cos 2 2 2 2πr

4

Chapter 4 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

4.4 Consider a unit circle with its center at the origin and let function F  z   Pz   iQz  be holomorphic inside the contour C. Also let the function F   take definite boundary values ______

where   e i gives the points on C. If the boundary condition is F    F    f   , then determine the Cauchy Integral formulae and the integral for F(z). Solution: Multiply the equation for the boundary condition by 1 d 2i z and integrate around C to get the Cauchy Integral formulae as 1 2i

C

Fd z

C

1 2i

Fd z

1 2i

C

fd z

The first integral is

Fz 

1 2i

C

Fd z

The second integral becomes

F0  a o  ia 1 which are unknown so far. Then,

Fz 

1 2i

C

fd z

 a o  ia 1

Let z = 0 so that

ao 

1 2i

C

Fd 

1 2

2

 o fd

The remaining work to completely solve this problem can be found in Muskhelishvili's book, 1977 published edition, page 309-311.

4.5 Show that the fundamental combination of stresses can be defined in terms of Cauchy Integral formulae.

hd  1 i   z 2

C

hd  1 i   z 2

hd  1 3 i  z  hd  12  iz C  2

C

hd  12 2 iz  z 

x  y  1 i  y   x  2i xy  1 i

C C

5

C

hd 2

C

hd   z 2

Chapter 4 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

Solution: z 

1 2i

C

h d  a z 1 z

  z h a d  z  z1 z h a1  a1  1  d 2i C  2 a 2  1  h3 d 2i C  z 

C

(4.124)

The resultant expressions are  x   y  2   z    z

(4.95a)

 1 i

C

h d  2a 1  1 i   z 2

 1 i

C

hd  1 2 i  z 

C

C

h d  2a 1   z 2

hd  1 2 i  z 

C

hd 2

 y   x  2i xy  2z  z    z  1 i

C

hd  1 i   z 3

C

2  z hd   2a21 2 2 z z   z

 1 i

C

hd  1 i   z 3

C

hd  12 iz   z 2

C

hd 2a 2a  21  21 z z   z 2

hd  1 i   z 3 hd  12  iz C  2

C

hd  12 iz   z 2

C

hd   z 2

 1 i

C

4.6 Suppose that a plate containing a single unstressed crack (Figure 4,15) is deformed by the unknown stresses at infinity and assume that the complex potential, f ' z     i xy , represents the stress distributions across the crack contour C. (a) Determine the stress distribution on y = 0 outside the crack and (b) the upper and lower boundary functions for x  a. Solution: (a) Use eq. (4.124d) for z = x + iy = x since y = 0.

 yy  i xy  

   y i xy  2

1

z z 2 a 2

In order to cover the stress distribution on the upper and lower planes the boundary function χ(z) has to be rewritten as   z    x2  a 2 for x > a and x < a, respectively. Therefore, |x| ≥ a and the stress distribution becomes

 yy  i xy 

   y i xy 

2

6

|x | x 2 a 2

1

Chapter 4 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

(b) The upper and lower boundary functions for |x| ≤ a are

  z    z  i a 2  x 2 since z – a = reiθ and z + a = reiθ so that for a unit disk with z = x + iy = x   z 

x  ax  a 

x 2  a2  i a2  x2

  z  i a 2  x 2

and   z  i a 2  x 2    z  i a 2  x 2

4.7 Consider an infinite plate subjected to a tensile remote stress (S) normal to the direction of a central crack. If the complex potentials for this type of crack are z  S 2 z 2  a 2  z 4   z  z  S 2

z

a2 z2  a2

then determine the crack tip stresses using the Westergaard stress function Z  z   2 '  z  when x  a. Solution:   z  S 4

2z 1 z  a2 2

Zz  2  z  S 2 Z  z  Z   S 2

z 1 z  a2 2

1 z2  3 z2  a2 z 2  a 2  2

The Westergaard stress functions, eq. (3.20) or (4.11), are

 x  Re Z  y ImZ   y  Re Z  y ImZ   xy  y Re Z 

7

Chapter 4 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

At y = 0 and z = x + iy = x > a, these stress functions become

 x  Re Z  S 2

z 1 z  a2

 y  Re Z  S 2

x 1 x  a2

2

 S 2

x 1 x  a2 2

#

2

 xy  0 For large z = x + iy = x since y = 0, the elastic stresses far from the crack tip vanish when x >> a along the x-axis. Therefore,

x  0 y  0  xy  0

4.8 Consider the elliptical crack shown below and assume that the crack is in the z-plane where z    a and p z   p . Derive the stress equations using the given crack geometry and the Westergaard complex method.

p z   p  

iy z ζ θ

x

a

Solution: Let us move the origin to crack tip. Then the coordinates of P can be expressed in complex variable z    a . Use the Westergaard complex equation so that Zz 

  1  a/z 2

z z  a2

Z 

  a    2a

a 1 

8

2

2a 1 

 a  2a

Chapter 4 Fracture Mechanics, 2nd ed. (2015)

Solution Manual

For |ζ|