Fracture Mechanics_2nd Ed_Solution Manual

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


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


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


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


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


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


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


(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

Radius r (m)

8

0.7

0.8


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


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


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


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


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

.


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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




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


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


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


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


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


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


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


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




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


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


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


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


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


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


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


Solution Manual

For |ζ| << a, a   a 2a 2

Z  Z    

 a 2 3/2

But ζ = reiθ and Zr,    Zr,   Z  r,   Z  r,  

 a i 3/2

2re 

 a 2r  a



 a 2r 3/2

e i/2

cos   i sin  2 2

i 3/2

2re   a

i 3/2

2re 

 

 a 2r 3/2  a 2r 3/2

e i3/2 cos 3  i sin 3 2 2

Then,

 y  Re Z  y Im Z   a y  cos  1  sin  sin 3 2 2 2 2r and

 xy

 a

cos  1  sin  sin 3 2 2 2 2r  a  cos  sin  sin 3 2 2 2 2r

x 

4.9 Assume that an infinite plate contains a through-central crack along the x-axis. If the plate is subjected to a remote or infinite stress loading condition,  y   y  S ,  x   xy  0, then use the Sanford [54] modified Irwin-Sih complex potential

2  z  Zz    z Here, γ′(z) may be defined by a Cauchy integral, Z(z) is the Westergaard complex function, ψ∗(z) is a complex polynomial and z is the complex variable define as z = x + iy. Based on this information, determine a function for the stress intensity factor KI and expand ψ∗(z) when n = 0 and 1, and z = a, half the crack length. 9


Solution Manual

Solution: The Westergaard complex function and the complex polynomial are KI 2z

Zz 

m

  z 

 b n zn/2 n0

The given complex function yields KI 2z

 2  z   n0 b n z n/2 m

For n = 0 and 1 and z = a, KI 

2a 2  z  b o  b 1 z 1/2 

10


Solution Manual

CHAPTER 5 CRACK TIP PLASTICITY

5.1 Use the inequality K IC  K I as a criterion for crack instability where K I is defined by Irwin’s plastic zone corrected expression for a finite size, to determine if a steel pressure vessel is susceptible to explode under   200 MPa hoop stress. The vessel contains an internal circular crack perpendicular to the hoop stress. If K IC  60 MPa m ,  ys  700 MPa, and the crack size is 20 mm, (a) determine the ASTM E399 thickness requirement and the minimum thickness to be used to prevent explosion, (b) Will crack propagation occur at 200 MPa? (c) Plot B (thickness) vs.  /  ys for a = 10, 20, and 30 mm, and (d) Will the pressure vessel explode when the crack size is 30 mm? Why? or Why not? and e) When will the pressure vessel explode?

Solution: (a) From eq. (5.12) along with  



K I   a 1  0.5( /  ys ) 2



2 , 

Let K IC  K I so that  K IC    ys

2

     KI     ys

K 2.5 IC   ys

   

2

2

 K   2.5 I    ASTM  ys

2

    Actual

Thus, 2

B ASTM

K  2.5 I   ys

    Actual

B ASTM

  a 1  0.5( /  ) 2 ys  2.5  YS  



 

2

   2   2 . 5  a       ys 

1

   

2

 1  1    2   ys 

   

2

    


Solution Manual

If the equality is used, then B ASTM  Bmin is the minimum thickness required by the ASTM E399 standard testing method. Substituting values, such as  /  ys  0.285714, a = 20 mm, and

  2 /  in the derived thickness expression, yields B ASTM  Bmin  5.41 mm .



(b) K I   a 1  0.5( /  ys 

2

  32.56 MPa

m . Therefore, crack propagation will not

occur because K I  K IC . In this case, the crack is stable. . (c) Plot



 

2

B ASTM

  a 1  0.5( /  ys ) 2  2.5   YS 

B ASTM

2 2     1       2     1    2.5a     2       ys    ys   

 

160 140 120

B ASTM

a  30 mm

100

20

80 60

10

40 20 0

0.2

0.4

0.6

0.8

1

 /  ys



(d) It will not explode since K I   a c 1  0.5( /  ys 



(e) Using K IC   a c 1  0.5( /  ys 2



2

  40 MPa

m and K I  K IC .

yields the critical or maximum crack size as

a c  67.91 mm . Therefore, the pressure vessel will explode when a  a c  67.91 mm.

2


Solution Manual

5.2 A project was carried out to measure the elastic-strain energy release rate as a function of normalized stress  / ys of large plates made out a hypothetical brittle solid. All specimens had a single-edge crack of 3-mm long. Plot the given data and do regression analysis on this data set. Determine (a) the maximum allowable  /  ys ratio for G IC  30 kPa.m and (b) K IC in MPa m . Given data: v  0.3 ,  ys  900 MPa and E  207 GPa .

 / ys

G IC kPa.m 

0

0.10 0.20 0.30 0.40 0.50

0.60

0.70

0.80

0.90

0

0.40 1.70 1.90 7.00 12.00 19.00 26.00 36.00 48.00

Solution: The required regression equation and the plot are given below and the 2

         53.4440    14.9570   G IC  0.2322  7.3168        ys   ys   ys 

3

50

G IC

kPa.m 

40 30 20 10

0

0.2

0.4

0.6

0.8

1

 / ys (a) Combining eqs. (3.5) and (5.12) yields





 2 1  v 2 a ys  /  ys 2 1  0.5 /  ys 2 E 2 2  1.12  1  0.3 3 x10  3 m 900 MPa x 2 1  0.5 x 2  G IC   42.0980 x 2  21.0940 x 4 207,000 MPa where x   /  ys  0.75 at G IC  30 kPa.m. G IC 

(b) The fracture stress and the plane strain fracture toughness

  0.75900 MPa   675 MPa EG IC 207,000 MPa 0.03 MPa.m   82.61 MPa m  2 1 v 1  0.32 --------------------------------------------------------------------------------------------------------------------K IC 

3


Solution Manual

5.3 Determine the critical crack length of problem 5.2. Solution:

 / ys

G IC kPa.m 

0

0.10 0.20 0.30 0.40 0.50

0.60

0.70

0.80

0.90

0

0.40 1.70 1.90 7.00 12.00 19.00 26.00 36.00 48.00

50

G IC

kPa.m 

40 30 20 10

0

0.2

0.4

0.6

0.8

1

 / ys Combining eqs. (3.5) and (5.12) yields





 2 1  v 2 a ys  /  ys 2 1  0.5 /  ys 2 G IC  E 2 2  1.12  1  0.3 3x10 3 m 900 MPa x 2 1  0.5 x 2  G IC   42.098 x 2  21.094 x 4 207,000 MPa where x   /  ys  0.75 at G IC  30 kPa.m. The fracture stress and the plane strain fracture toughness

  0.75900 MPa   675 MPa K IC 

EG IC  1  v2

207,000 MPa 0.03 MPa.m   82.61 MPa 1  0.32

m

Thus, K IC

2          a c 1  0.5     ys     1

2 2 2     1  K IC   1  82.61 MPa m  2 1      ac    3 mm  1  0.5  1  0.50.75        ys     1.12 675 MPa     ---------------------------------------------------------------------------------------------------------------------



4




Solution Manual

5.4 A large brittle plate containing a central crack 4-mm long is subjected to a tensile stress of 800 MPa. The material has KIC = 80 MPa m , ys = 1200 MPa and v = 0.30. Calculate (a) the applied KI, (b) the plastic zone size using the Von Mises yield criterion and prove that r = rmax when    o . Consider all calculations for plane stress and plane strain conditions, and (c) draw the entire plastic zone contour where the crack tip is the origin of the coordinates. Solution: (a) K I   a  (800 MPa)  2  10 3 m   63.41 MPa m with   f (a / w)  1 The crack is stable since KI < KIC. (b) From eq. (5.52), 2 KI 3 2  r sin   h1  cos   ; 2  4 ys  2 

For plane stress   1 h  2 1  2v  For plane strain 

If  = 0, then r

hK I 2 2 ys

2

0.44 mm plane stress  and r    0.07 mm plane strain 

Using eq. (5.52) yields 2

r 1  K I   sin  3 cos  h    2 ys  For maximum KI and critical plastic zone size r = rc, let

r  0 so that 

sin  o 3 cos o  h   0 h 3 70.53o o   o 86.94 cos o 

For plane stress   For plane strain 

Thus, 2

rmax 

KI 2 4 ys

3 2  0.59 mm   sin   h 1  cos  o o  2    0.37 mm 

Furthermore, KI r  sin  3 cos   h  4 ys 2

5

For plane stress   For plane strain 


Solution Manual







KI KI  2r  3 cos  3 sin 2   h cos  3  h  cos  3 sin 2  2 2 2  4 ys 4 ys 2

 2r  2

o

2



 0.44 mm plane stress     0.63 mm plane strain 

Therefore, r = rmax at o since

 2r  0 in both cases.  2

(c) The contours in polar coordinates are as per r 

KI

2

4 ys

2

3 2   2 sin   h1  cos    

r mm  0.4

Crack Tip

0.2

-0.1

0

0.1

0.2

0.3

0.4

 Radians 

-0.2 -0.4

Plane Strain

Plane stress

---------------------------------------------------------------------------------------------------------------------

5.5 (a) Use the data given in Example 3.3 for a pressure vessel containing a semi-elliptical crack (Figure 3.6) to calculate Irwin’s and Dugdale’s (a) plastic zones, (b) K I using Kabayashi’s finite size correction factor ( ) and plasticity correction factor. (c) Compare results and determine the percent error against each case. (d) Is it necessary to include a plastic correction factor? Explain.

Solution: Data from Example 3.3: a = 3 mm, 2c = 10 mm, B = 6 mm,   420 MPa,  ys  700 MPa, a/(2c) = 0.30, a/B = 0.50,

K IC  60 MPa m ,  /  ys  0.60, M = 1, Mk = 1.12, and Q = 1.70 . The correction factor and the stress intensity factor are, respectively

  1.12 M k M / Q  0.127 K I   a  5.18 MPa m 6


Solution Manual

Comparison: (a) Irwin’s Approach:   1.12 M k M / Q  0.127 a 2 r   /  ys   0.54 mm Eq. (5.13) 2

Dugdale’s Approach:   1.12 M k M / Q  0.127 a 2 r   / 2 ys   1.33 mm 2

(b) Using a finite correction factor K I    (a  r )



2



2

K I   a 1  0.5 /  ys 

K I   a 1  0.5 /  ys 



Eq. (5.24)

K I    (a  r )

Eq. (5.12)



K I ( Irwin' s )  5.63 MPa m

   K I   a 1  0.5  2   ys 

   

2

   K I   a 1  0.5  2   ys 

   

2

   

Eq.(5.25)

   

K I ( Dugdale' s )  6.22 MPa m

Comparisons: K I ( eq. 3.29)  5.18 MPa m < K I ( Irwin' s ) < K I ( Dugdale' s )  6.22 MPa m Dugdale’s - Irwin’s Error  100%(6.22  5.63) / 5.63  10.48% Irwin’s - eq. (3.29) Error  100%(5.63  5.18) / 5.18  8.69% Dugdale’s - eq. (3.29) Error  100%(6.22  5.18) / 5.18  20.08% Observe that eq. (3.29) does not include plasticity correction and it yields a smaller value than both Dugdales’s and Irwin’s expressions. The latter expression gives an error of 26.5%, which is slightly large. On the other hand, 8.3% error seems to be more acceptable for comparison purposes. This implies that a plastic zone correction may not be necessary. (c) Using Irwin’s approach yields a plastic zone of 0.54 mm for one side of the semi-ellipse and as a result, r << a and the plastic zone size could have been excluded. On the other hand, Dugdale’s approach gives a plastic zone of 1.33 mm, which is large enough to be excluded; therefore, a plasticity correction is needed to obtain more a accurate KI- result. ---------------------------------------------------------------------------------------------------------------------

7


Solution Manual

5.6 A 50- mm thick pressure vessel is to support a hoop stress of 300 MPa at room temperature under no action of corrosive agents. Assume that a semi-elliptical crack (Figure 3.6) is likely to develop on the inner surface with the major axis 2c = 40 mm and semi-minor axis a = 10 mm. A 300-M steel, which is normally used for airplane landing gear, is to be considered. Will crack propagation occur at 300 MPa hoop stress? Make sure you include the Irwin’s plastic zone correction in your calculations and explain if it is necessary to do so. Use the data below and select the suitable tempered steel.  ys

300-M Steel

K IC

(MPa )

o

650 Temper 300o Temper

( MPa m)

1070 1740

152 65

Solution: a = 10 mm, 2c = 40 mm, B = 50 mm, a/2c = 0.25, a/B = 0.20 Mk = 1.02 (From Figure 3.6b)   1.12M k / Q since M < 0.5 [lower limit according to eq. (3.46)] From eq. (3.42), 2 Q   2  0.212 /  ys   /2

 /2  c2  a2  2   1  sin   d   1  0.75 sin 2   d 0   c2    0   0.13169 (From a Table of Elliptic Integral of the Second Kind)



Thus,



K I   a 1  0.50 /  ys 

2

a  /  ys 2 2 300-M Steel (650o C Temper)  /  ys  0.28 Q  1.72   0.87 K I  47.16 MPa m  K IC r = 0.39 mm r << a = 10 mm



r

300-M Steel (300o C Temper)  /  ys  0.17 Q  1.67   0.88 K I  47.13 MPa m  K IC r = 0.15 mm r << a = 10 mm

These results are similar and crack propagation will not occur since both steels have KI < KIC. Select 300-M Steel (650o C Temper) since it has a larger K IC value. The plastic zone correction was not necessary. Thus, K I   a  46.26 MPa m  K IC . Both steels can be used for the sought application. In fact, the calculated K I values are similar for both steels. This is purely accidental. Anyway, the plastically corrected KI values are approximately 2% higher. 8


Solution Manual

5.7 If localized plasticity is to be considered, explain the physical meaning of the following a 2 inequality   t  ys . E Solution: If the applied stress (  ) is unchanged, but causes crack growth, then  t  ys is constant since

 ys  constant and  t   c . In this case, the crack length is the only changing variable that increases; therefore, the inequality holds. --------------------------------------------------------------------------------------------------------------------5.8 Show that r   t /( ys ), where r is the plastic zone due to dislocation networks within the plastic zone area ahead of the crack tip. Solution: K  ys K I2 If  t  and E  , then,  t   ys  I   ys E  ys  ys

2

 2 ys    2 

 KI    ys

2

   2 ys  r  

t . 2 ys --------------------------------------------------------------------------------------------------------------------Thus, r 

5.9 Show that  t /  ys  K I /  ys  and give a reasonable interpretation of this equality. 2

Solution: Apparently,  t /  ys is related to the crack size [  t /  ys  a] since K I /  ys  is 2

associated with the plastic zone [r  K I /  ys  ]. Therefore, r   t /  ys . Here  means “proportional.” --------------------------------------------------------------------------------------------------------------------2

5.10

A large plate containing a through-the-thickness central crack of 2 a c  20 mm has

E  207 GPa,  ys  1,275 MPa, and  c  9.47 m at service temperature. Determine (a) the plane strain fracture toughness, (b) the design stress intensity factor for a safety factor (SF) of 2, (c) the critical fracture stress, and (d) the design service stress.

Solution: 2 K IC (a)  c   ys E

K IC  50 MPa m K (b) K ID  IC  25 MPa m SF

(c)  c  K IC / a c  282 MPa (d)  d   c / SF  141 MPa

9


Solution Manual

4 a2  x2 E Solution: Glass is a brittle material and there is no need for plasticity correction; therefore,  t  0 @ x  a. ---------------------------------------------------------------------------------------------------------------------

5.11 Predict  t for glass using  

5.12 Develop a  t  expression for a von Mises material. Compare it with  t for a Tresca material under plane strain condition. Assume that crack growth occurs along the crack plane. Solution: von Mises  t  expression: From eq. (5.53), hK I2 K r r and  t  I 2 2 ys E 2

 t von Mises  

h1 / 2 K I2 2E ys

Tresca  t  expression: From eq. (5.59) with   0 , K I2 K r r and  t  I 2 2 ys E 2

 t Tresca  

Combining  t Mises  and  t Tresca  yields

K I2 2E ys

 t Mises   h1 / 2 t Tresca   t Mises   1  2v  t Tresca 

For Poisson’s ratio v  1,  t Mises    t Tresca  because r Mises  r Tresca --------------------------------------------------------------------------------------------------------------------5.13 A material has E  70 GPa,  ys  500 MPa and v  1 / 3. It has to be used as a plate in a large structure. Non-destructive evaluation detects a central crack of 50 mm long. If the displacement at fracture is 0.007 mm and the plate width is three times the thickness, calculate (a) the crack tip opening displacement, (b) the plane strain fracture toughness, (c) the plane strain energy release rate, (d) the plate thickness and (e) What’s the safety factor being indirectly included in this elastic-plastic fracture mechanics approach? Assume plane strain conditions as per eq. (5.49) and a fracture load of 200 kN. Solution: Given data: a = 25 mm,  f  0.007 mm ,   3 and v  1 / 3.  ys  500 MPa, P = 200 kN and E  70 GPa . (a)  tc  2 f  0.014 mm  1.40 x10 5 m (b) From eq. (5.49) with   3  1.7321 , which is defined below eq. (5.5), the plane strain fracture toughness ( K IC ), as per Irwin’s model, is

10


K IC 

G IC

Solution Manual

 3 1.40 x10 m70 x10 MPa 500 MPa   25.82 MPa 1  1 / 925.82 MPa m   8.46 kJ / m 

1 1  tc E ys   2 2

5

3

2

2 (1  v 2 ) K IC  E

2

3

70 x10 MPa

(c)  c  K IC / a  92.13 MPa m and  c  B

Pc  3 c

m

200 kN 3 92.13x10 3 kN / m 2





Pc P  c2 wB 3B

 26.90 mm and w  3B  80.70 mm (Width)

(d) SF   ys /  c  500 / 92.13  5.43 --------------------------------------------------------------------------------------------------------------------5.14 Repeat problem 5.13 using eq. (5.41). Compare results. Solution: Data: a = 25 mm,  f  0.007 mm ,   3 and v  1 / 3.

 ys  500 MPa, P = 200 kN, E  70 GPa and  c  2  f  0.014 mm  1.4 x10 5 m   3  4v  5 / 3 For plane strain (a) c =

2 2 (  1)(1  v) K IC 8 K IC  4E ys 9E ys

K IC 

(b) G IC

9 c E ys 8



9 1.4 x10 5 m 70 x10 3 8



2 (1  v 2 ) K IC 1  1 / 9 41.61 MPa m   E 70 x10 3 MPa

(c)  c 

K IC

a





MPa 500 MPa 



 41.61 MPa m

2

 0.022 MJ / m 2  22 kJ / m 2

41.61 MPa m   148.48 MPa  25 x10 3 m 

Pc P  c2 wB 3B Pc 200 kN B  3 c 3 148.48 x10 3 kN / m 2

c 



(d) SF 

 ys

c Problem 5.13 Problem 5.14



 21.19 mm and w  3B  63.57 mm (Width)

 3.37

K IC  41.61 MPa m

 c  148.48 MPa

B  21.19 mm

SF  5.43

K IC  25.82 MPa m

 c  92.13 MPa

B  26.90 mm

SF  3.37

11


Solution Manual

5.15 A hypothetical large metallic plate containing a 10-mm central crack is 30-mm wide and 5mm thick and mechanically loaded in tension. This plate has E  69 GPa,  ys  500 MPa , v  1 / 3, and   0.3% (plane stress strain). Determine (a)  t , (b) G , and c)  as per Irwin, Dugdale, Burdekin, Rice, and Average equations, and compare the results.

Solution: Data: w  30 mm, B  5 mm, v  1 / 3, a  5 mm ,   0.3%,  ys  500 MPa, E  69 GPa Calculations:

  B  5 mm 0.003  0.015 mm and K I   a G I   ys  1.5 x10 5 m 500 MPa   0.0075 m.MPa  7.5  kJ / m 2 (a) In general,

t 

 t E ys K I2 a c2  and   E ys E ys a

Irwin–eq. (5.31):

 E   t ys  4a

Dugdale–eq. (5.32):

 

Burdekin–eq. (5.40):   Rice–eq. (5.41):

 

 t E ys 2a

 t E ys a  t E ys a

  

0.015 mm 69 x103 MPa 500 MPa   161 MPa 45 mm  0.015 mm69 x10 3 MPa 500 MPa   128 MPa 2 5 mm 0.015 mm 69 x10 3 MPa 500 MPa   182 MPa  5 mm  0.015 mm 69 x10 3 MPa 500 MPa   322 MPa 5 mm 

Rice equation yields  Rice   2 Irwin  .

12


Solution Manual

Determine (a) the critical crack tip opening displacement (  c ), (b) the plastic zone size

5.16

(r) and (c) the fracture stress ( c ) for a large aluminum alloy plate containing a central crack of 5-mm long. Use the data given below and assume that plane strain conditions exist. Data: K IC  25 MPa m ,  ys  500 MPa, and E  70 GPa. Solution: (a) From eq. (5.40),





2

K2 25 MPa m  c  IC   0.018 mm E ys 70 x103 500 MPa  (b) For plane strain condition,

1 r 6

 K IC    ys

2

2

     1  25 MPa m   0.13 mm   6  500 MPa  

From eq. (3.24),

c 

K IC

a

25 MPa m



 2.5 x10 3 m

 282 MPa

--------------------------------------------------------------------------------------------------------------------2

 E  2 5.17 Show that    t 1     t for plane stress conditions, where  is the crack  8a  opening displacement (COD) and  t is the crack tip opening displacement (CTOD).

Schematically, plot   f  t  for various  and fixed a values. Solution: From eq. (5.24), 4 t  2ar E 2  E t  2ar     4 

(5.30) (a)

 

1  E t  r   2a  4 

(b)

If a  x, then

(c)

 

r2 

1 2a 2

From eq. (5.23a), 2 4    a2  r  x2 E 4 E

a 2  2ar  r 2  x 2

(5.29) (d)

2

 E t     4 

4

13

4 E

2ar  r 2

(e) (f)


Solution Manual

Inserting eqs. (b) and (c) into (f) yields 4  E

 E  2  1  E  4   t    t 2   4   2a   4  2

4

 1  E  4     t2    t 2   2a   4  2

2

 E  2   t 1    t  8a 

The plot is



 a  a o  fixed

Increasing

t --------------------------------------------------------------------------------------------------------------------5.18 If the plane strain fracture toughness (KIC) and the yield strength (ys) of a 12-mm thick C(T) steel specimen are 71 MPa.m1/2 and 1,896 MPa, respectively, determine (a) the validity of the fracture mechanics tension test as per ASTM E399 for the plate containing a single-edge crack of 10-mm long at fracture, (b) the fracture stress if the plate is 20-mm wide, (c) the critical crack tip opening displacement, (d) the plastic zone size, and (e) interpret the results with regard to plane strain condition. Use a Poisson's ratio of 1/3 and assume that the elastic modulus of the steel 207 GPa. Solution:

14


Solution Manual

(a) Using eq. (3.5) yields the critical stress intensity factor

The minimum size requirements can be computed using eq. (3.30). Thus,

Therefore, the test is valid because a, B10, 12   amin , Bmin 3.51 mm and a/w  0.5 is within the 0.2  a/w  1 valid range. So proceed with the required calculations. (b) The fracture stress is determined from eq. (3.29) along with α=9.6591 since a/w=0.5 (Consult Table 3.1).

Thus,

(c) The crack-tip opening displacement is calculated using Rice’s equation, eq. (5.41), with   3  4v  5 / 3 and   11  v   32 / 9.

15


Solution Manual

(d) The plastic zone can be calculated using eq. (5.53) along with the constant h = (1 - 2v)2 = 1/9

(e) The above results suggests that the plate met the ASTM E399 size requirements because a, B  amin , Bmin and 0.2  a/w  1 . The C(T) specimen breaks in a brittle manner because both the plastic zone size and the crack-tip opening displacement are very small. ---------------------------------------------------------------------------------------------------------------------

5.19 Assume that isotropic solid material having a single-edge crack is subjected to a remote tensile stress at room temperature. Let the properties of the material be  ys  500 MPa, Poisson’s ratio v  1 / 3 and E  72 MPa. Let the applied stress intensity factor for mode I loading be K I  20 MPa m . Excluding microstructural details and microscale defects, use the Tresca yielding criterion to derive (a) an expression for the critical plastic zone angle (  c ) and its magnitude when minimum principal stresses are equal (  2   3 ), (b) determine when  min   2 and  min   3 by knowing the value of  c , (c) the plastic zone size at  c and K I  20 MPa m . The Tresca yielding criterion is based on the maximum shear stress reaching a critical or failure level. Hence, the definition of the maximum shear stress for this criterion is

 max  0.5 max   min   0.5 ys Here  max and  min are principal stresses and  ys is the monotonic tensile yield strength of a solid material. Let  max   1 and  min   2 or  min   3 . Solution: Needed equations for solving the problem. The principal stresses are

Hence, 16


Solution Manual

(a) The expression for the critical plastic zone angle (  c ). For plane strain and  c   conditions,

If v = 1/3, then

and the minimum principal stresses become

(b) The minimum stresses. Let's calculate the values of  2 and  3 at    c and    c . For    c  0.67967 rad  38.942o , say   35o so that

For   20o , 17


Solution Manual

Therefore, the calculated values of the principal stress  2 and  3 at    c and fixed

r

indicate that  min   3 . For    c  0.67967 rad  38.942o , say   40o so that

For   60o ,

Therefore, the calculated values of the principal stress  2 and  3 at    c and fixed indicate that  min   2 .

18

r


Solution Manual

(c) The plastic zone sizes for plane stress and plane strain at  c  38.942o are

As expected, r for plane stress is greater than that for plane strain.

19


Solution Manual

5.20 Consider a ductile steel plate containing a 50-mm through-thickness central crack subjected to a remote tensile stress of 40 MPa. If the yield strength of the steel is 300 MPa, then calculate K I as per LEFM, Irwin’s approximation and the Dugdale’s strip yield criterion. Compare results. Repeat all calculations for 290 MPa remote stress. Explain. Solution: Given data:   40 MPa and  ys  300 MPa LEFM:

Irwin’s:





 50  K I   a  40 MPa     10 3 m  11.21 MPa m  2 KI 

Dugdale’s: K I 

40 MPa 

 a

1  0.5 /  ys 

2



 a 8     ln sec   2   2 ys 

 50    10 3 m   2

1  0.540 / 300

2

 11.26 MPa m

 50    10 3 m   2  11.25 MPa m 8   x 40  ln sec   2   2 x300 

40 MPa  

These results are similar. However, if   290 MPa, then K I LEFM:

K I  81.27 MPa m

Irwin’s:

K I  115.18 MPa m

Dugdale’s: K I  130.01 MPa m Therefore, K I values are dubious. See Figure 5.4 for comparing Irwin’s and Dugdale’s approximation schemes.

20


Solution Manual

5.21 A single-edge SE(B) specimen with B = 20 mm, w = 40 mm is used to determine the critical CTOD  tc and J IC . See Example 5.4 for the specimen configuration and the load-displacement plot. Assume plane strain conditions and let Pmax  38 kN ,  ys  800 MPa, E  207 GPa, v  0.3, K IC  60 MPa m B  20 mm; w  40 mm;  p  1.00 mm; S  150 mm; z  1.5 mm; a  12 mm

Solution: From Example 5.4,

The geometry correction factor  : x  a / w  12 / 40  0.3 3x1 / 2 1.99  x1  x  2.15  3.93 x  2.7 x 2  3/ 2 21  2 x 1  x 







30.3

1/ 2



1.99  0.31  0.32.15  3.930.3  2.70.3   1.5212 2

21  2 * 0.31  0.3

3/ 2

The applied stress intensity factor as per LEFM is P S 1.521238000 N  150 x103 m K I  max   54.19 MPa m 3/ 2 Bw3 / 2 20 x103 m 40 x10 3 m





Then, CTOD becomes











2

K I2 54.19 MPa m  te    8.866 x103 m  0.009 mm m ys E 2800 MPa 207000 MPa 

 tp 

0.44w  a  p



0.4440 mm  12 mm1 mm  0.47715 mm 0.4440 mm  12 mm  12 mm  1.5 mm

0.44w  a   a  z  tc   te   tp  0.008866  0.47715  4.78 mm

Thus, the J-integral is J IC  m ys tc  2 800 MPa  4.78 x103 m  7.65 MPa.m  7.65MN/m 2





21


Solution Manual

22

Chapter 6 Fracture Mechanics, 2nd Ed. (2015)

Solution Manual

CHAPTER 6 THE ENERGY PRINCIPLE 6.1 Use Dugdale’s model for a fully developed plane stress yielding confined to a narrow plastic zone. Yielding is localized to a narrow size roughly equal to the sheet thickness (B). This is a fully elastic case, in which the plastic strain may be defines as ε  δt /B, where t is the crack tip opening displacement. If the J-integral is defined by dJ  d , then show that a 2 2 t  E ys Solution: If the strain energy density and the J-integral at yielding are defined by 



0

0

W   d    ys d 

J  BW  B   ys d   t  ys o

Thus, 

t

o

o

J  B   ys d    ys d Solving the J-integral yields t

J    ys d   t  ys o

Then,

K I2 K I2 J and  t ys  E E 2 2 2 KI a  t   E ys E ys Here,  is a correction factor.

1


Solution Manual

6.2 The crack tip opening displacement (t) for perfectly plastic solution to the Dugdale model was derived by Rice in 1966 [21] as k  1a 0     t  log sec  G   2 0  where  o   ys and   Remote tensile stress (    o ), a  Crack size, and G = Shear modulus. Show that the path-independent J-integral is defined by

J

k  11   a 2 E

 f a,  

Solution:      yields sec 2  0    2 2 y 5 4 y sec y   1   y  ...  1  2 24 2

    and on log Using Taylor’s series on sec  2 0  y

 2 0

and

Let x  sec y  so that

 x  1 1  x  1 3  log x   2     ...   x  1 3  x  1   1   sec y   1  x 1 log x   2  2  2    sec y   1  x 1   1  

For x  0

    2   y2      1  y2    2 o   2  2   2 2 2 y2 4  y    1  4        2   2   o 

         and log sec   If  1, then 4   o 2  o  2 o     2

Thus,  t 

k  1a 0   2G

  2   o

1    2  2 o

  

2

2

From eq. (6.66), J   t 0 

k  1a o2   2G

 k  1a 2    2  8G  o 2

The elastic shear modulus of elasticity is G 

J

k  11  v a 2 4E

E . Thus, J becomes 21  v 

 f a,   2


6.3

Solution Manual

A bending test specimen made out of carbon steel showed a load-displacement behavior

A = Area =10 Joules E = 207 GPa ys = 1,500 MPa B = 25 mm b = 25 mm

P Area

 If the area under the P vs.  curve is 10 joules at the onset of crack growth, determine (a) the fracture toughness in terms of J IC as per ASTM E-813 Standard, (b) K IC and its validity as per ASTM E399 testing method, and (c)  tc [according to eq. (5.31) with   3 ] ,  f and r . Explain the meaning of the results. Solution: (a) J IC 

210 joules  2A   32 kJ/m 2  32 x10  3 m. MPa Bb 25  10  3 m25  10 3 m 

207 x10 MPa32 x10 m.MPa  81.39 MPa  2.5K /    7.36 mm  B and B B

(b) K IC  EJ IC  Validity: BASTM

3

3

m

2

IC

ys

min

Actual

ASTM

Therefore, K IC  81.39 MPa m is valid. (c) From eq. (5.31),





2 4 81.39 MPa m 4 K IC  tc    0.016 mm E ys   3 207 x103 MPa 1,500 MPa 

 

2



Using eqs. (5.44) and (5.13) yields the fracture strain and the plastic zone size, respectively

f 

 c 0.016 mm   6.4 x10 4 or 0.064% B 25 mm

The plastic zone size:

1 r 6

 K IC    ys

2

2

     1  81.39 MPa m   0.16 mm  a  6  1,500 MPa  

Therefore, the material behaved in a brittle manner implying that very little plastic deformation took place at the crack tip. These are very small values indicating that the material is brittle and undergoes very little plastic deformation. With regard to the critical crack tip displacement of 0.016 mm, it indicates that the crack tip displacement is very small; that is,  y   c / 2  0.008 mm .

3


Solution Manual

6.4 If J I   ys  t is used to determine the fracture toughness, will  t be a path-independent entity? Explain. Solution: Using Dugdale’s model, Figure 5.3 and eq. (6.33) yields y P

Plastic Zone

2 y

x

P 2a

JI  

ar

a

r

    ds  Wdy  T x  

According to Figure 5.3, dy  0 from a to a  r and T   y   ys . Thus,

J I  

ar

a

 ys



  y   y x

 dx 



t

o

 ys d  y   y dx  t  ys

Therefore, both J I and  t are path-independent. Only the initial and final values of  t are needed to solve the integral. In fact,  t is just an opening as well as a path-independent entity.

2 6.5 Assume that crack growth occurs when J I  J IC , where J IC  (1   2 ) K IC / E . If a welldeveloped plastic flow occurs, will these expressions be valid? Explain.

Solution: It is valid in the elastic regime where a  r.

4


Solution Manual

6.6 A double cantilever beam (DCB) is slowly loaded in tension up 10 MN as shown schematically below. Assume that there is no rotation at the end of the beam and that the beam is made of isotropic steel having the following properties: K IC  47 MPa m and E  207 GPa. Will fracture occur?

AokN  B h PP  10 M  Pa

M

B a = 20 mm B = 20 mm h = 10 mm

h h

  2

M

a P

Solution: From eq. (6.42) along with E '  E / 1  v 2  and I  Bh 3 / 12, P 2 a 2 121  v 2 P 2 a 2 121  0.32 10 x10 3 MN  20 x10 3 m  GI    BE ' I EB 2 h 3 207,000 MN / m 2 20 x103 m 2 10 x103 m 3 2

2

G I  5.28 x10 3 MPa.m EG I 207,000 MPa 5.28 x10 3 MPa.m   1  v2 1  0.32 K I  34.66 MPa m KI 

Therefore, fracture will not occur because K I  34.66 MPa m  K IC  47 MPa m

5


Solution Manual

CHAPTER 7 PLASTIC FRACTURE MECHANICS 7.1 Determine (a) the J-integral (J) and (b) the dJ/da for a hypothetical steel plate containing a central crack of 114 mm long loaded at 276 MPa and exposed to room temperature air. What does TJ < TR mean? Assume plane strain conditions and that the stainless steel obeys the Ramberg-Osgood relation with curve fitting parameters such as  ' 1.69 and n  5.42. Explain the results based on the strain hardening effect. Data: E  206,850 MPa ,  ys  207 MPa,

  0.3 and J IC  130 kJ/m 2 . Dimensions: w  4a  228 mm, L  2w



2w



2a 2L

Solution: (a) For a central crack, use Table 3.1 for  x  a/w  57 / 228  0.25

  1  0.5 x 2  20.46 x 4  81.72 x 6  1  0.50.25  20.460.25  81.720.25  1.0635 2

4

6

  sec 0.25  1.1892 K I   ae

The effective crack length needed for calculating K I  K I ae  is given below.  1 ry    

 n  1  K I    n  1   o

2

  1  n  1            n  1   o

2

  ae 

&  

  1  n  1   1 ae  a  w ry  a     2  1  P / Po     n  1   o With   sec 0.25  1.1892

1 2 1  P / Po 

2

  ae 

  1  5.42  1   1.1892276  1 ae  57 mm    2    ae 207 1  125.86 / 81.75  6  5.42  1   ae  57 mm  8.5597 x10 2 ae ae  62.336 mm 2

1

 6


Solution Manual

K I   ae  1.1892276 MPa   62.336 x10 3 m K I  145.25 MPa m

The elastic J-integral values that include the effects strain hardening is

K I2 (1   2 ) K I2 (1  0.32 )145.25    9.2815x10 2 MPa  m E' E 206,850 J e  92.82 kPa.m  92.82 kJ/m 2 2

Je 

Plastic Part (continued): Table 7.2 for a central crack εo  σ o /E  0.001 and   0.001

Po  4(w  a)σ o / 3  81.75 MN/m & P  2 wσ  2228 x103 m276 MPa   125.86 MN/m w  a  228  57  171 mm

The Plastic zone: 2  1  5.42  1  145.25  3 r  ry      10   17.855 mm  6  5.42  1  207  From eq. (7.25) and Table 7.1 with n = 5.42, I n  6.2511  0.29381n  1.1724 x10 2 n 2  5.0031

  J p   ' o o I n r   o  J p  87.17 kJ/m 2

n 1

 276   1.69 207 MPa 0.0015.003117.855 x10 m   207  3

Thus, eq. (7.36) gives

J  J e a e   J p a, n   92.82 kJ/m 2  87.17 kJ/m 2

J  J I  180 kJ/m2 Therefore, crack grows occurs since J  290.96 kJ/m 2  J IC  130 kJ/m 2 (b) Crack instability K I2    

2

Je 

K I2 (1   2 ) K I2  (1   2 ) a   E' E E

2

5.42 1

10  3


 1 r  ry    

 n  1  K I    n  1   o

  J p   ' o o I n r   o 

n 1

Solution Manual

2

  1  n  1            n  1   o

2

  a 

 1  n  1     '  o o I n       n  1   o

  

2

    o 

n 1

 1  n  1     (1   2 ) a J  Je  J p    ' o o I n 2     E    n  1   o  2  1  n  1    dJ  (1   2 )     ' o o I n 2     da E    n  1   o 

a n 3

a

n3

0

dJ  (1  0.32 )1.1892 x 276 MPa   da 206,850 MPa

2

 1  5.42  1  276   1.69207 MPa 0.0015.00311.0635      6  5.42  1  207  2

dJ  4.05 MPa  4.05 MJ/m3 da From eq. (6.70), the tearing modulus is

E dJ 206,850 MPa   4.05 MPa   19.55  o2 da 207 MPa 2 TJ  19.55 TJ 

Therefore, crack growth is stable if TJ  TR .

3

5.42  3


Solution Manual

7.2 Plot the given uniaxial stress-strain data for an ASTM A533B steel at 93ºC, and perform a regression analysis based on the Ramberg-Osgood equation. A single-edge-cracked plate made out of this steel is loaded in tension at 93ºC. Determine the total elastic-plastic J-integral when the applied load is   500 MPa . Will the crack grow in a stable manner? Why? or Why not? Assume plane strain conditions and necessary assumptions. Let a  100 mm, v = 0.30, E = 207 GPa, L = 3w, w = 400 mm, JIC = 1.20 MPa.m, h1  0.523 and B = 150 mm.

Strain (x10-3)

0

1.0 0

2.2 4

2.3 0

2.3 5

5.0 0

7.5 0

10

20

40

Stress (MPa)

0

381

414

415

416

450

469

483

519

557

Solution:

Single-edge cracked specimen L w





a

Stress v s. Stra in

Stress (M Pa )

800 600 400 200 0 0

0.05

0.1

0.15

0.2

0.25

0.3

Stra in

Data for plane strain condition:  ys   o  414 MPa (From the above Figure) v  0.30 E  207 GPa in  kips (From Ref. [8]) J IC  1.20 MPa  m  6.8 in 2   500 MPa (Applied) 4

a  100 mm w  400 mm

B  150 mm

L  3w  1,200 mm b  w  a  300 mm (Ligament)


Solution Manual

Elastic Part: n

  Nonlinear regression using     0   yields  '  1.12 and n  9.71 . From Table 7.2,  0  '

P  wσ  400 x103 m500 MPa   200 MN/m 1/ 2

1/ 2

  a 2    100 2  a 100 x  1    1   100  0.82137       w  a   400  100   400  a   w  a   P0  1.455x w  a  0  1.4550.82137300 x103 m 414 MPa   148.43 MN/m

1  w  a a / w   400  100100 / 400  75 mm Also,   6 and

 n  1  K I   n  1    0  1 w 2 1  P/P0   1 ry    

  1  n  1   ae       n  1    0    2

2

  ae  n  1              n  1     0 

2

The crack size can be calculated using ae  a  w ry 2

  1  n  1      1  9.71  1  500  1 1 ae  a       ae  a    ae  2  2   1  200 / 148.43  6  9.71  1  414  1  P/P0     n  1   0  ae  a  0.05814ae

ae  1.0617a

a  100 mm ae  1.0617100 mm  106.17 mm For a / w  100 / 400  0.25,

  1.12  0.23a / w   10.55a / w   21.71a / w   30.38a / w   1.5013 2

3

4

K I   ae  1.5013500 MPa   106.17 x10  3 m  469.10 MPa m



K 2 1  ν 2 K I2 1  0.32  469.10 MPa m J e  I'   E E 207 x103 MPa



5

2

 0.97 m.MPa  0.97 MJ/m 2


Solution Manual

Plastic part: 2

 a  n  1     106.17 mm  9.71  1  500  r  ry   e          21 mm 6  9.71  1  414     n  1   0   2

From eq. (7.51) and Table 7.2, h1  0.523 ,  '  1.12 and 1  75 mm  75 x103 m

P J p    0 01h1    P0  '

n 1

with ε0  σ 0 /E  0.002 1 9.71

 200  J p  1.12414 MPa 0.00275 x10 m0.523   148.43  J p  0.89 MJ/m 2 3

Thus, the total value of the J-integral is

J  J e  J p  0.97  0.89  1.86 MJ/m2 Therefore, crack will grow because J  J IC  1.20 m. MPa. Stable? Unstable? Let’s find out how the crack will grow. Let









K 2 1   2 K I2 ae 1   2   K I   ae ; J e  I'   E E E so that the total value of the J-integral can be computed by P a 1   2  2 J  Je  J p  e   ' 0 01h1   E  P0 

2

and

P J p    0 01h1    P0  '

n 1

If ae  1.0617a and 1  w  a a / w  , then





n 1   1   2  2  P  ' J    1.0617a     0 0 h1   w  a a / w  E   P0    

Thus, at constant load dJ/da is

n 1  1   2  2   '  P   w  a a  dJ  1.0617   1  26 MPa     0 0 h1     da E w   P0    w    E dJ 207 x103 MPa  TJ  2  26 MPa   31.40  o da 414 MPa 2

Therefore, the crack will grow in a stable manner TJ  31.40  TR 6

n 1


Solution Manual

7.3 Repeat Problem 7.2 using the Hollomon equation   K n for the plastic region. Curve fitting should be performed using this equation for obtaining K and n. Assume plane strain conditions and make the necessary assumptions. Compare the result from Problem 7.2. What can you conclude from these results? Assume a contour as shown in Figure 7.6 with y23  70 mm. Solution:

Single-edge cracked specimen

Stress v s. Stra in

Stress (M Pa )

800 600 400 200 0 0

0.05

0.1

0.15

0.2

0.25

0.3

Stra in

Data for plane strain condition:  ys   o  414 MPa (From the above Figure)

a  100 mm

E  207 GPa v  0.30

w  400 mm B  150 mm

J IC  1.20 MPa  m  6.8

  500 MPa (Applied)

in  kips (From Ref. [8]) in 2


Elastic Part: For a / w  100 / 400  0.25,

  1.12  0.23a / w   10.55a / w   21.71a / w   30.38a / w   1.5013 2

3

4

K I   ae  1.5013500 MPa   106.17 x10  3 m  469.10 MPa m






7

2

 0.97 m.MPa  0.97 MJ/m 2


Solution Manual

Plastic part: Nonlinear regression using   k n yields k  776 MPa and n  0.103. The plastic strain energy density can be defined using the Hollomon equation as 1/ n     k n and     k 



0

0

Wp     d   k n d

k  776 MPa and

k n 1  k    Wp      n 1  n  1  K 

n 1 n

n  0.103 10.7087

 776 MPa  500 MPa      1.103  776 MPa 

W p  6.35 MPa  6.35 MJ/m3

Assume a contour as shown in Figure 7.6 with y23  70 mm so that J p  2 y23W p as per eq. (7.43), which represents a near-field condition. Thus,





J p  2 y23W p  2  70 x103 m 6.35 MPa   0.89 m. MPa J I  J e  J p  0.97  0.89  1.86 MPa  m  1.86 MJ/m 2

Crack growth occurs since J  J IC . Note that if the value of y23 changes so will J p and J I . Anyway, both problems 7.2 and 7.3 (P7.2 and P7.3) give the same result as per chosen contour segment y23  70 mm; otherwise J p P7.2   J p P 7.3 or J I P7.2   J I P7.3 . Then, K I   a and K I2  a  

2

K I2  1   2   a Je  '  E E 2

K n 1  K    Wp      n 1  n  1  K 

n 1 n

K n 1  K    J p  2 y23W p    2 y23    n 1  n  1  K 

n 1 n

 1   2  2 a  K    JI  Je  J p   2 y23    E  n  1  K  2 dJ I  1   2  2 2  1  0.32 1.5013x500    da E 207 x103

n 1 n

dJ I  7.78 MPa  7.78 MJ/m3 da E dJ 207 x103 MPa  TJ  2  7.78 MPa   9.40  o da 414 MPa 2 Therefore, the crack will grow in a stable manner TJ  9.40  TR . 8


Solution Manual

7.4 A steel plate having a single-edge crack is loaded as shown below. Calculate (a) the loadline displacement (  ) and (b) the crack opening displacement ( ) that corresponds to a point on the resistance curve (not included). Assume plane strain conditions and use the following data: v  0.3 , h1  0.523, h2  1.93, h3  3.42, B  0.15 m, w  0.40 m, L  1.2 m and a  0.1 m. J IC  1.20 kJ/m 2 ,   500 MPa and E  206,850 MPa Solution: L w





a

(a) From Table 7.2, P J p   ' 0 0 w  a h1    P0  P  p   ' a 0 h3    P0 

n

n 1

1

 P   p n so that       P0   ' a 0 h31 

Eliminating P/Po yields  p   J p   ' 0 0 w  a h1    ' a 0 h3 

 h w  a     J p  0 1 1 / n  p   '  0   ah3 

 n  1 / n

n 1 / n

n

 J p  n 1   p  ah3  '  0 1 /( n 1)    01h1  Similarly, n

n

P  J p  n 1     ' a 0 h2    ah2  '  0 1 /( n 1)   P0    01h1  These are the equations to be used in this problem. However, we need the value of J p at

  500 MPa. Let’s get.

9


Solution Manual

Data for plane strain condition:  ys   o  414 MPa (From the above Figure)

E  207 GPa

a  100 mm w  400 mm

v  0.30

J IC  1.20 MPa  m  6.8

  500 MPa (Applied)

in  kips (From Ref. [8]) in 2

B  150 mm


Elastic Part: n

  Nonlinear regression using     0   yields  '  1.12 and n  9.71 . From Table 7.2,  0  P  wσ  400 x103 m500 MPa   200 MN/m '

1/ 2

1/ 2

  a 2  x  1       w  a  

  100 2  a 100   1   100  0.82137    w  a   400  100   400  a P0  1.455x w  a  0  1.4550.82137 300 x103 m 414 MPa   148.43 MN/m





1  w  a a / w   400  100100 / 400  75 mm Also,   6 and  1  n  1  K I r  ry        n  1   0 1 w 2 1  P/P0 

2

  1      

 n  1   ae    n  1   0

2

  ae  n  1              n  1     0 

2

The crack size can be calculated using ae  a  w ry 2

  1  n  1      1  9.71  1  500  1 1   ae  a     ae  a    ae  2  2    n  1  6 9 . 71  1 414     1  P/P 1  200 / 148 . 43              0 0   ae  a  0.05814ae ae  1.0617a

a  100 mm ae  1.0617100 mm  106.17 mm

10


Solution Manual

For a / w  100 / 400  0.25,

  1.12  0.23a / w   10.55a / w   21.71a / w   30.38a / w   1.5013 2

3

4

K I   ae  1.5013500 MPa   106.17 x10  3 m  469.10 MPa m






2

 0.97 m.MPa  0.97 MJ/m 2

J e  0.97 MJ/m2

From Problem 7.2, 2

 a  n  1     106.17 mm  9.71  1  500  r  ry   e          21 mm 6  9.71  1  414     n  1   0   2

From eq. (7.51) and Table 7.2, P  wσ  400 x103 m500 MPa   200 MN/m 1/ 2

  a 2  x  1       w  a  

1/ 2

  100 2  a   1     w  a   400  100  





100 100  0.82137 400  a



P0  1.455x w  a  0  1.4550.82137 300 x103 m 414 MPa   148.43 MN/m

1  w  a a / w   400  100100 / 400  75 mm h1  0.523 ,  '  1.12 and 1  75 mm  75 x103 m P J p    0 01h1    P0  '

n 1

with ε0  σ 0 /E  0.002 1 9.71

 200  J p  1.12414 MPa 0.00275x10  3 m0.523   148.43  J p  0.89 MJ/m2

J  J e  J p  0.97 MJ/m2  0.89 MJ/m2  1.86 MJ/m2

11


Solution Manual

Thus, n

 J p  n1   p  ah3  '  0 1/( n1)    01h1 

 p  0.1 m 3.421.120.002

0.0934

  0.89 m.MPa  414 MPa 0.3 m 0.523   

0.9066

 p  3.90 mm

(b) The crack opening displacement is n

n

P  J p  n 1     ' a 0 h2    ah2  '  0 1 /( n 1)   P0    01h1 

  0.1 m1.931.12 0.002 

0.0934

  2.20 mm

  0.89 m. MPa  414 MPa 0.3 m0.523   

0.9066

Therefore,  p  3.90 mm and   2.20 mm at the onset of crack growth since J IC  1.2 MN/m2 and crack blunting required these values.

7.5 Determine the strain hardening exponent (n) for a steel with  ys = 400 MPa, E = 207 GPa. Assume that it obeys the Hollomon equation  p  k n . Consider the maximum plastic stress in your calculations. Solution: and  max  k  700 MPa

 p  max @    p  1   max

n @ yielding,  ys  k ysn ,  max  k max . and  ys 

 ys k ysn    ysn  max k max   @  ys ln   max n  ln  ys 

   

 400  ln   700   0.0895 ln 1.93  10 3





12

 ys E

 1.93  10 3


Solution Manual

7.6 (a) Derive an expression for the J-integral J p /J e  f(σ(σo ) ratio as per Rice model. (b) Plot the resultant expression for a remote stress to yield stress ratio range 0   /  o  1 . Here,

 o   ys . Explain the resultant trend. Solution: (a) The plastic case: k  1a o     t  log  sec  G   2 o 

J p   o t 

k  1a o2 G

     log  sec 2  o   

The elastic case:   k  1K I2  k  1a 2   Je  8G 8G            2 log sec   Jp  2 o    f ( /  o )  J  2      e      2 o    From the plot, J p /J e   as σ/σ o  1 since secπ/ 2    and σ  σ o .

5

They significantly differ from each other due to a large plasticity. At σ/σ o  0.8 , JP dominates.

4

J p /J e 3 2 1

0

0.2

0.4

0.6

0.8

1

σ/σ o

13


Solution Manual

7.7 Calculate the total J-integral (J) for a 2024 Al-alloy plate containing a single-edge crack under plane stress conditions. Plot b) the   f   and J I  f  . Use the following data to carry out all calculations: a = 1.40 mm, w = 19 mm, B = 0.8 mm, L = 10 cm,  o  64 MPa, E  72,300 MPa, α′ = 0.35, n = 5 and F = 1.01 kN. Solution: Elastic part: a) The following calculations are self-contained. Thus, 64 o   o   8. 852  10 4 A  wB  15. 20x10 6 m 2 E 72, 300 1. 01x10 3 MN   F   66. 45 MPa   ys A 15. 20x10 6 m 2 For a single-edge crack, x  a/w  1. 4/19  0. 073684

  1. 12  0. 230. 073684  10. 550. 073684 2  21. 710. 073684 3  1. 1516   1. 12  0. 23x  10. 55x 2  21. 71x 3  1. 1516 Je 

2 2 1. 40x10 3 m1. 1516 2 66. 45 MPa K 2I  a   E E 72, 300 MPa

2

J e  3. 5623  10 4 MPa. m  356. 23 J/m 2

J e  356. 23 J/m 2 Plastic part: From Table 7.2 and eqs. (7.60),  1  w  aa/w  19  1. 41. 4/19  1. 2968 mm  

a 1   wa 2 

a wa



1

1.4 191.4

2



1.4 191.4

 0. 92361

P o  1. 072w  a o  1. 072  0. 92361  19  1. 4  10 3  64  1. 1153 MN/m P  w  19  10 3 m66. 45 MPa  1. 2626 MN/m h1 

h1 

3a n 4 1n  1

 o 

31.4 5 41 1n 1.2968

n1

Po P

66.45 64

n1

51

1.1153 1.2626

51

 3. 3853

From eq. (7.54a), J p    o o  1 h 1

P Po

n1

 0. 35  64  8. 852  10 4  1. 2968  10 3   3. 3853

J p  183. 23 J/m 2 J I  J e  J p  356. 23 J/m 2  183. 23 J/m 2  539. 46 J/m 2 14

1.2626 1.1153

51

10 6 


Solution Manual

(b) These results indicate that J p  0.34 J and J e  0.66 J . Therefore, J e  J p and contributes 66% to J. (c) The required stress-strain curve can be determined using eq. (7.30). Thus,

   ys

  ys

1/n

 322. 08 MPa 5  80 80 70

Stress  (MPa)

60 60 50

40 40

Load Point

30

20 20 10

0

0

0

0.0002 0.0003 10.0001 2 3 -4 Strain  Strain  x10





0.0004

4

The J-integral plot along with the load point is based on the following equation:

J I  2. 1247  10 2  2  1. 0638  10 9  6 200

150 JI 2 (J/m )

Loading Point

100

50

0 0

20

40

 MPa 

60

80

The trend of the J I  f   resembles the trend shown in Figure 7.5.

15


Solution Manual

7.8 The plastic J-integral (Jp) for some configurations can be defined by J  ηW/(Bb), where W  absorbed energy, Bb = cross sectional area, b = (w – a) = ligament and   constant. This integral can then be separated into elastic and plastic components. For pure tension,

J  Je  J p 

K I2  pWp  E' Bb

(1)

Consider a strain hardenable material and a specimen with unit thickness B. If the plastic load is defined by P  C pn , where C is the compliance and n is the strain hardening exponent, then the load-displacement and J-P profiles are schematically shown below. Jlimit J

P W



P

Po = Plimit

Derive an expression for  p . Solution: The area under the curve is the strain energy density define by Wp  

p

o

p

Pd p   C np d p  o

n n C np 1  P p since P  C pn n 1 n 1

(2)

Since W p is a maximum energy then P reaches Po (the load limit). Thus, Wp 

n Po  p n 1

(3) n

P From eq. (7.56),  p   ' a o h3   and  Po  P n Wp  Po ' a o h3   n 1  Po 

n

(4)

Substituting this expression into the J-integral equation above yields the plastic part as

 n P Jp  p Po ' a o h3   Bb n  1  Po 

n

(5)

16


Solution Manual

Equate eq. (7.54) and (5) and solve for  p

p n P P Po ' a o h3     '  o o1h1   b n 1  Po   Po  n

p 

n

n  1 b 2  o h1 n a Po h3

17


Solution Manual

CHAPTER 8 MIXED-MODE FRACTURE MECHANICS

8.1 A large plate (2024-0 Al-alloy) containing a central crack is subjected to a combined mode III loading. The internal stresses are  y = 138 MPa and xy = 103 MPa. Use the Maximum Principal Stress Criterion ( -criterion) to calculate the fracture toughness (KIC). Use the following data: crack length 2a = 76 mm,  = 1/3, and E = 72,300 MPa. Solution: a = 38 mm,  = 138 MPa,  = 103 MPa K I   y a  47.68 MPa m K II   xy a  35.60 MPa m

K I / K II  1.3393 Using eq. (8.29a) yields 1  x x  3 x2  8 0   ar cos 1  x2  9  3  



  48.60  

From eq. (8.33),    K IC  K I cos 3 0  3K II cos 2 0 sin 0 2 2 2 K IC  72.50 MPa m

o

From eq. (8.34), 3 K IIC  K IC  62.77 MPa m 4

1


8.2

Solution Manual

Repeat problem 8.1 using the Strain Energy Density Factor (S) Criterion (S-criterion).

Solution: K I   a  47.68 MPa m K I   xy a  35.60 MPa m

K II / K I  0.75 Use eq. (8.53) to solve for the fracture angle





sin  0 2 cos 0  k  1K I2  4 cos 2  0  k  1 cos  0  2 K I K II  sin  0 6 cos 0  k  1K II2  0

  48.10 o Using Equation (8.54) yields 4E 2 K IC  a K 2  2a12 K I K II  a 22 K II2 k  11    11 I





K IC  59.35MPa m

From eq. (8.57) and (8.51),

K IIC  0.9045K IC  53.68 MPa m S c  14.26 kPa m  14.26 kJ / m 2

8.3 Determine (a) the incline angle  and (b) the applied tension  in Example 8.1. Solution: (a) K I   a  sin 2 

(b) K I   a sin 2 

K II   a  sin  cos 

 

KI 47.68MPa m  tan   K II 35.60MPa m   53.25º

  214.95 MPa

KI

a  sin  2



a = 38 mm 47.68MPa m

 38  10 3 m sin 2 53.25º 

8.4 Calculate the critical stress (fracture stress) for the problem described in Problem 8.1 according to the   - criterion. Will crack propagation take place? Solution: From problem 8.1 & 8.2:

From Example 8.4:

  53.25º   215MPa (Applied)

a = 38 mm E = 72,300 MPa

2

o = -48.60º  = 1/3


Solution Manual

From eq. (8.35),

c 

K IC sin  a

b

11



1 / 2

 2b12 cos   b22 cos 2    0  492 MPa

where 1 3 b11  1  cos 0   0.57314 8 3 2 b12  sin  0 1  cos  0   0.77635 8 9 b22  sin 2  0 1  cos 0   1.05161 8 Crack propagation will not take place since    c . 8.5 Repeat problem 8.4 using the S-criterion with   215 MPa. Solution: K I   a  sin 2 

K I   2a  sin 4 

K II   a  sin  cos 

K II   2a  sin 2  cos 2 

K I K II   2a  sin 2  cos 

From eq. (8.54),

c 

K IC sin  a



(   1 )( 1   ) a11 sin 2   2a12 cos   a 22 cos 2  4E

where



1 / 2

  o

1 1  cos k  cos   1  v  1  cos  k  cos  3.84 x10 6 16G 8E 1 a12  sin   2 cos   k  1 1  v  sin 2 cos   k  1 1.15 x10 6 16G 8E  1  v a 22  k  11  cos    1  cos 3 cos  1 5.90 x10 6 8E a11 

Thus,

 c  204 MPa   215 MPa (Applied) Therefore, crack propagation will occur because    c .

3


Solution Manual

8.6 Show that the stress intensity factor is K I  Poisson’s ration is v  1 / 3 . Solution:





From eq. (8.13) with E '  E / 1  v 2 and K II  0, E ' (1   ) 2 2 K IC  K I2  K III with K I  2 K III E E (1   ) 2 (1   ) 2 2 K IC  K I2  K III  K I2  K III 2 1  v 1  v  E 1  v  2 K IC  K I2 

2 K III K I2  K I2   1  v  41  v 

 1  2 1  41  v  K I  

  2 11 2 1 2 K IC  1  K I   8 K I    41  1 / 3 Thus,

KI 

8 K IC 11

4

8 / 11K IC when K I  2 K III and that the


Solution Manual

8.7 Use the experimental mixed mode fracture data or Solid A and Solid B given below to compare the mixed mode fracture criteria discussed in this chapter. Determine which criterion is the most suited to predict the mixed mode fracture behavior of these solids. Solid A K II

Solid B

KI

K II

KI

(MPa m  (MPa m  (MPa m  (MPa m  0.49

0

0.42

0

0.43

0.52

0.41

0.10

0.33

1.02

0.43

0.21

0.31

1.10

0.38

0.40

0.28

1.20

0.40

0.50

0.26

1.25

0.37

0.60

0.16

1.40

0.39

0.65

0

1.50

0.37

0.80

0.31

1.20

0.27

1.26

0.20

1.55

0

1.65

Solution: Plot the given data and use all possible mixed mode criteria in order to determine the most suited criterion. 2

KI

MPa m 

1.5

Solid A 1

Solid B

0.5

0 0

0.125

0.25



K II MPa m

5



0.375

0.5


Solution Manual

From the given data, Solid A:

K IC  1.50 MPa m & K IIC  0.49 MPa m

Solid B:

K IC  1.65 MPa m & K IIC  0.42 MPa m

For Solid A, KI

2

KI

MPa m 

K I  K  IC

1.5

2

 K II      K IIC

2

    1 

1 2

K I K II  K IC  K IIC

0.5

    1 

0 0

0.125

0.25

0.375



K II MPa m



0.5 KII

For Solid B, KI

2

KI

MPa m 

1.5

2

K I  K  IC

 K II      K IIC



0.375

2

    1 

1


0.5

0 0

0.125

2

    1 



0.25

K II MPa m

K I Therefore, the criterion  K  IC

2

 K II      K IIC

2

0.5 KII

 K I K II    1 fits both data sets better than K   K IC   IIC

6

2

    1. 


Solution Manual

8.8 A large and wide plate has a small through-the-thickness central crack as shown in the figure below. Use the free-body diagram (FBD) to derive expressions for (a) the normal (σN) and shear (τ) stresses and (b) the intensity factors KI and KII as functions of β.

FBD

 

N

 A

2a

Ao cos 

B



P Ao

 



  Ao  Area

Solution: (a) Using the FBD yields the normal and shear stresses

F N N

y

F

  N A   Ao cos   0

x

  A   Ao cos   0

Ao   Ao cos   0 cos     cos  cos    sin  . cos 

Ao   Ao cos   0 cos    cos 2    sin 2 



(b) The stress intensity factor equations are K I   N a 

K II   a

K I   a sin 2  

K II   a sin  cos 

7


Solution Manual

8.9 Two identical cracked plates as shown in problem 8.8 are to be tested to determine K IC and K IIC . The observed critical tension loads and the incline angles were 120 MPa @   0 and 130 MPa @    / 4, respectively. Use the equations given below to determine the fracture toughness for mode I and II. K I K II  K IC  K IIC

2

    1 

K I  K  IC

and

2

 K II      K IIC

2

    1 

Solution: For case   120 MPa @   0 and    / 2  90o ,





K I   a sin 2    120 MPa   20 x10 3 m 1  30 MPa m K II   a sin   cos   0 MPa m 0


2

    1 

K IC  K I  30 MPa m

For case   130 MPa @    / 4 and    / 4  45o ,



K I   a sin    130 MPa   20 x10 2

3

2

 2  m  2   16.30 MPa m  



 2  2    K II   a sin   cos   130 MPa   20 x10 3 m   2  2  16.30 MPa m   






2

    1 

K IC  30 MPa m

K IIC





K IIC  16.30 MPa m 1  16.30 / 30

1 / 2

2

 K II      K IIC

K IIC

 K I  K II 1   K   IC

1 / 2

 KI   K II  1  K   IC  

2

K I  K  IC



    1; K IC  30 MPa m       2

1 / 2



K IIC  16.30 MPa m 1  16.30 / 30 

K IIC  24.12 MPa m



2 1 / 2

K IIC  19.42 MPa m

These results indicate that the care must be taken in selecting a particular fracture criterion. K I Nevertheless, the fracture criterion  K  IC K II K when compared to I   K IC  K IIC

2

 K II      K IIC

2

    1  8

2

    1 gives conservative values for K IIC 


Solution Manual

CHAPTER 9 FATIGUE CRACK GROWTH  E ys  da   t n / 2 , where  t is the 9.1 Show that Paris equation can take the form  A 2  dN 1 v  change of crack tip opening displacement, E is the modulus of elasticity, and A is a constant. n/2

Solution: From eq. (7.45) or (8.10) with KII = KIII = 0

J





K I2 1  v 2 K I2  E' E

KI 

EJ 1 v2

K 

EJ 1 v2

Substituting K into Paris expression, eq. (9.6) yields  EJ  da n  AK   A 2  dN  1 v 

n

From eq. (6.66), J   ys t and J   ys  t . Thus,  E ys  da   A 2  dN 1  v  

n/2

 x n / 2

da A n  and K e  K max 1  R  is  C K e  , where C  n( 1 ) dN 1  R  da da n n Walker’s effective stress intensity factor range. (b) Plot  C K  and  AK  for a dN dN 4340 steel having  ys  1,254 MPa,  ts  1,296 MPa, K IC  130 MPa m ,   0.42, n  3.24 ,

9.2

(a) Show that

R  0.7 , and A  5.11x10 11 . .

Solution (a) K e  K max (1  R )

R

K min K max  K max  K min K max  K K   1 K max K max K max K max

1


Solution Manual

K  K max (1  R ) K K max  1 R Then, K e 

K and (1  R)1 n

 K  da A  A( K e ) n  A  K n 1  n( 1 ) dN (1  R ) (1  R ) 

where C

A ( 1  R ) n( 1 )

For R  1

Thus, da n  C K  dN

If R  0 , then C  A.

(b) If  ys  1,254 MPa,  ts  1,296 MPa, K IC  130 MPa m ,   1 / 3, n  3, R  0.7 and

A  5.11x10 11

MN n  mm cycles

2n 2 2

, then n(1   )  31  1 / 3  2

Thus, C

A 5.11x10 11 MN 3  mm 4 / cycles   5.68 x1010 MN 3  mm 4 / cycles 2 n (1 ) (1  R) 1  0.7 

Comparison: Walker: Paris:

da  MN 3  mm 4  K 3   5.68  10 10 dN  cycles  da  MN 3  mm 4  K 3   5.11  10 11 dN  cycles 

2


Plot

Solution Manual

da vs . K for 20 MPa m  K  100 MPa m dN

2.5e-08

Walker 2e-08 1.5e-08

da mm / cycles  dN 1e-08

Paris

5e-09

0

20

40



60

K mpA M



80

100

9.3 Suppose that a single-edge crack in a plate grows from 2 mm to 10 mm at a constant loading frequency of 0.02 Hz. The applied stress ratio and the maximum stress are zero and 403 MPa, respectively. The material has a plane strain of 80 MPa.m1/2 and a crack growth behavior described by da 4  3.68 x10 12 K I  dN Here, da / dN is in m/cycle and K I is in MPa m . Determine the time it takes for rupture to occur. a  

Solution:

  1.12

K IC  80 MPa m

a o  2 mm

a f  10 mm

R0

A  3.68 x10 12

n4

 min  0

 max  403 MPa

Using K IC   max a c yields a c  a f  10 mm

3

f  20 Hz

  403 MPa




da n  AK I   A  a dN



af

2  a da  A  

t

4

Nf f



278.31 cycles  13,916 sec 0.020 cycles / sec

Nf

t  3.87 hours

0

a a Nf

4

  dN

ao 1 o



Solution Manual

1 f

 1.51N f

2 x10 

m   10 x10 3 m  1.51 m 1 .cycles 1 3

1

1

N f  278.31 cycles

9.4 If a large component is subjected to a cyclic loading under   300 MPa and R  0. The material behaves according to Paris Law da 2.45  2 x10 8 K I  dN where da / dN is in m/cycles and K I is in MPa MPa m . Determine the plane strain fracture toughness for the component to endure 37,627 cycles so that a single-edge crack grows from 2 mm to a c . Solution:



da 2.45  2 x10 8 K I   A 1.12  dN ac

Nf

ao

0



2.45

K IC  1.12 a c

a 2.45

K IC  1.12 300 MPa   7 x10 3 m 

 2.45  a da  0.12577  dN  4 ,732.30

K IC  50 MPa m

ao1.45  a c1.45  4,732.30 1.45 a c  a f  7 mm

9.5 Consider a part made of a polycrystalline metal that is stresses in the elastic stress range. If the metal contains inclusions, has an imperfectly smooth exterior surface, and natural dislocation, would the metal experience irreversible changes in a micro-scale? Explain. Answer: Many dislocations would move and contribute to a small irreversible process since the dislocation configurations would eventually change. This is one mechanism would lead to fatigue cracking due to dislocations.

4


9.6

Solution Manual

Why most service fatigue fractures are normally not clear?

Answer: Most service fatigue fractures experience varying stress amplitude, leading to irregularly spaced striations, if any.

9.7 What is the physical meaning of the slope n of the Stage II line according to the Paris model? da n  AK I  , the higher the value of the slope n , the dN n higher the crack growth rate da / dN . For instance, if K  0 and n   , then K I    and da / dN   since A is a constant.

Answer: According to Paris law,

9.8 Suppose that d ( 2a ) / dN  0.001 mm / cycle and n  4 in the Paris equation for 7075-T6 (FCC), 2024-T3 (FCC), Mo (BCC), and steel (BCC). Determine a) the constant A and its units, and b) which of these materials will have the higher crack growth, rate? Solution: From Figure 9.6, 10 3 mm / cycle d ( 2a ) 4  AK I  and A  dN K I 4 Material 7075-T6 (FCC) 2024-T3 (FCC) Mo (BCC) Steel (BCC)

K I 10 15 20 30

MPa

m)



A

MN

4

 m 5 / cycle



1.00 x10 10 1.98 x10 11 6.25 x10 12 1.23 x10 12

Therefore, 7075-T6 (FCC) alloy has the highest fatigue crack growth rate because it has the highest constant A.

5


Solution Manual

9.9 A Ti-6Al-4V large plate containing a 4-mm long central crack is subjected to a steady cyclic loading R = 0.10. The plate plane strain and the threshold fracture toughness are 70 MPa m and 14.7 MPa m . Determine (a) the minimum stress range, (b) the maximum applied stress range for a fatigue life of 3,000 cycles, and (c) the critical crack size for 3,000 cycles. Let the Paris equation be applicable so that n  4 and A  10 12 MN 4  m12 / cycles .



Solution:



2a

2a o  4 mm

N  10 3 cycles

K IC  70 MPa m

a o  2 mm

R  0.40

K th  14.7 MPa m

(a) Using eq. (9.2b) yields K th 14.7 MPa m  th    185.45 MPa  a o  2 x10 3 m





 min   th  185.45 MPa (b) From eq. (9.9) with N o  0 since ao exists,

C1    C2   n

n 2

 C3  0

( n  2)

C1  A N  N o n  2K IC   0.50 m n

C 2  ( 2 /  )( K IC /  ) 2  3.12 x10 3 MPa 2  m C 3  ( 2a 1o n / 2 )( 1 /  )n / 2 ( K IC /  )n  2.43 x10 9 MPa 2  m 0.50 4  3.12 x10 3  2  2.43x10 9  0 Thus,

C2 1  C 22  4C1C 3   3.12 x10 4 MPa 2  3.81x10 5 MPa 2 2C1 2C1   642 MPa   max (Positive real root)

 2  

 min and    max   min  max R max   min and  min   max   R max   max   642.31 MPa   max    713 MPa 1 R 1  0.1 Then, 2 1  K IC  ac    3.1 mm and 2a c  6.2 mm    max  (c) R 

6

 1


Solution Manual

9.10 Plot da / dN vs . K for 403 S.S. using the Paris, Forman and Broek/Schijve equations. Use the data given in Table 9.2 and a (20 mm)x(300 mm)x(900 mm) plate containing a single-edge crack of 2-mm long. Let 20 MPa m  K  80 MPa m .

Solution: a  2 x10 3 m a / w  0.007 and

B  20 mm w  300 mm   f ( a / w )  1.12

L  900 mm

Using data for AISI 403 stainless steel yields K IC  36 MPa m R   min /  max R max   min and

K th  1.37 MPa m  ys  1172 MPa R  0.1

and

and

   max   min

 min   max  

R max   max  

 max   / 1  R 

n  3.5

for R  1

A  5.98 x10 13

Need to calculate K max for each given K . For instance, using the upper limit in the range 20 MPa m  K  80 MPa m , the procedure is as follows: K max   max a 

 a and K   a  80 MPa m 1 R

Solve for the stress range,

  901.12 MPa (Maximum)    max   / 1  R   1001 MPa (upper limit)   K max   max a  89 MPa m (upper limit)

  a c 

1  K IC    max

2

   48.30 mm  

It is suggested that you use a computer program to handle all the calculations needed in order to plot smooth curves. Paris : Forman : Broek/Schijve :

da n  A K  dN n da AK  = dN 1  R K IC  K da 2  AK max  K n dN

(9.6)

Plane strain 

(9.10) (9.11)

7


Solution Manual

9.11 Plot the data given below and use eq. (9.6) as the model to draw a curve fitting line on loglog scales. Determine the constants in such an equation. No . 1 2 3 4 5 6 7

K

MPa m 

da / dN m / cycle

20 30 40 50 60 70 80

2.14x10-8 8.84x10-8 2.42x10-7 5.29x10-7 1.00x10-6 1.72x10-6 2.74x10-6

Solution: Since the Stage II data is for a linear behavior in a log-log scale, the constants in eq. (9.6) can be estimated as follows:

 da  n    AK 1   dN 1  da  n    dN 1  K 1     da   K 2     dN  2 8 lnda / dN 1 / da / dN 2  ln 2.14 x10 / 8.84 x10 8 n  lnK 1 / K 2  ln20 / 30 n  3.50



da / dN 1 A K1 n



 da  n    AK 2   dN  2

and



2.14 x10 8 m / cycles

20 MPa m 

3.50

A  5.98 x10 13 MPa 3.50  m 8 / cycles Then, the equation for curve fitting is of the following nature da 3.50  5.98 x10 13 K  dN Perhaps, the curve fitting procedure gives slightly different results. 4e-06

3e-06

da mm / cycles  dN

2e-06

1e-06

0 20

40

60



K MPa m

8

80



100


Solution Manual

9.12 A steel plate containing a single edge crack was subjected to a uniform stress range  at a stress ratio of zero. Fatigue fracture occurred when the total crack length was 0.03 m. Subsequent fatigue failure analysis revealed a striation spacing per unit cycle of 7.86x10-8 m. The hypothetical steel has a modulus of elasticity of 207 GPa. Predict (a) the maximum cyclic stress for a crack length of 0.01 m, (b) the striation spacing per unit cycle when the crack length is 0.02 m, (c) the Paris equation constants and (d) the plane strain fracture toughness and e) the fatigue crack growth rate assuming that the Paris equation is applicable nearly up to fracture. Solution:

a





(a) x  7.86 x10 8 m a  da   7.86 x10 8 m / cycles     dN  a N K a  E

x 7.86 x10 8 m  207,000 MPa   23.69 MPa m  K max 6 6

Thus,  

K 23.69 MPa m   133.66 MPa   max a  0.01 m 

(b) For a  0.02 m ,

K b  K max   a  133.69 MPa   0.02 m   33.50 MPa m Then, 2

 33.50 MPa m   K    1.57 x10 7 m x  6   6   E   207,000 MPa  a x 1.57 x10 7 m  da    1.57 x10 7 m / cycle    b  dN  N  N 1 cycle  b 2

(c) The exponent in the Paris equation is  da  n    AK a   dN  a  da  n    AK b   dN  b

 da / dN a   7.86 x10 8  ln  ln  da / dN b  1.57 x10  7    n  2  K a   23.69  ln  ln    33.50   K b 

9


Solution Manual

and the constant is A

da / dN a K a n



7.86 x10 8 m / cycle

23.69 MPa m 

2

 1.40 x10 10 MPa  2 / cycle

da 2  1.40 x10 10 K  dN (d) For ac = 0.03 m, K IC   max a c  133.66 MPa   0.03 m 

K IC  41 MPa m (e) Finally, A

da / dN a K a n



7.86 x10 8 m / cycle

23.69 MPa m 

2

 1.40 x10 10 MPa  2 / cycle

da 2 2  1.40 x10 10 K IC   1.40 x10 10 41  2.35 x10 7 m / cycle dN

9.13 A 2-cm thick pressure vessel made of a high strength steel welded plates burst at an unknown pressure. Fractographic work using a scanning electron microscope (SEM) revealed a semielliptical fatigue surface crack (a = 0.1 cm and 2c = 0.2 cm) located perpendicular to the hoop stress and nearly in the center of one of the welded plates. The last fatigue band exhibited three striations having an average length of 0.34 mm at 10,000 magnification. The vessel internal diameter was 10 cm. Calculate a) the pressure that caused fracture and b) the time it took for fracture to occur due to pressure fluctuations. Assume a pressure frequency of 0.1 cycles per minute (cpm). Given data: ys = 600 MPa, E = 207 GPa, K IC  75 MPa m and da/dN = 4.50x10-7(K)2. Solution: a 0.1   0.5 2c 0.2 a 0.1   0.05 B 2 da a 0.00034 m    1.13x10  4 m / cycle dN N 3 cycle

10


Solution Manual

Using the given crack growth rate equation, K becomes 1/ 2

 da / dN  K     A 

1/ 2

 1.13x10  4 m / cycle     2 4 . 50 x 10 MPa / cycle  

K  15.85 MPa m From eq. (9.23), x 0.00034 m K  E  207,000 MPa  610,000 6 K  15.58 MPa m which is similar to the previous K value. The average becomes K  15.72 MPa m . Assume the following stress ratio  /  ys  0.64 so that   384 MPa , which is verified below. Thus, the shape factor becomes

  Q  2

2

2

2

2

2 2  3  a 2  7        3  0.1   7 2        0.64   2.38         4  2c   33   ys   2   4  0.2   33

The hoop stress range along with the average value of K is  

K 15.72 MPa m   386.32 MPa  a / Q 1.12  0.001 m  / 2.38

Thus,

 386.32   0.64  ys 600 Therefore, the assumed stress ratio is reasonably correct and fortunately, a single iteration is sufficient. The pressure is estimated as P

2 B 2 2 cm 384 MPa    154.40 MPa d 10 cm

(b) The critical crack length is 2

2 1 K  1  75 MPa m    9.56 x10 3 m  0.956 cm ac   IC          1.12 x386.32 MPa 

11


Solution Manual

Integrating eq. (9.15) along with N o  0 yields c c da da 1 da ln a / a  dN    N a AK 2 a A  a 2 A 2 a a  Aco2 o o o o ln ac / ao  ln(0.956 / 0.1) N   8.53 cycles 2 A   4.50 x10 7 1.12 x386.322 Then,

ac

N

a

a







t



N 8.53 cycles   85.30 min cpm 0.1 cycle / min

9.14 Show that ac  aK IC / K max 

2

Solution: From,

K IC   max a c and K max   max a

 max a c K IC   K max  max a  K IC   K max

ac a

2

 a   c a 

Then,  K a c  a  IC  K max

  

2

12


Solution Manual

CHAPTER 10 FRACTURE TOUGHNESS CORRELATIONS

10.1 (a) Plot the given CVN–T data for a carbon steel. (b) Calculate KIC using the CVN values up to zero oC and Plot KIC-cal and KIC-exp. vs. Temperature. Is there a significant difference? If so, explain. T (oC) CVN (J) KIC-exp. (MPa.m1/2)

-125 12

-100 18

-60 20

-40 28

-12 40

0 78

40

50

80

88

50 210

10 98

25 110

40 125

Solution: (a) Plots 140 120 100

CVN J 

80 60 40 20 0

-100

-50

0

T

1

 C o

50

100

45 126

60 28

80 30


Solution Manual

K IC  205.05 + 4. 5721T + 5.0958x10 -2 T 2 + 2.0082x10 -4 T 3 240 220 200

K IC

MPa m 

180 160 140 120 100 80 60 40 20

-120

-100

-80

-60

T

-40

 C

-20

0

o

(b) Do nonlinear curve fitting on K IC and CVN data set as per eqs. (10.66) and (10.67). Then, use eq. (10.68) for calculating K IC values. Thus, plot KIC-cal and KIC-exp. vs. Temperature. Below is a linear curve fitting result for plane strain fracture toughness and Charpy impact energy.

250

K IC  17.963  2.603U K IC

MPa m 

200 150 100 50

0

10

20

30

40

U (J )

2

50

60

70

80


Solution Manual

100

K IC

MPa m 

80 60 40

Exp.

20

0

-120

-100

-80

-60

T

-40

-20

0

 C o

10.2 A mild steel plate has a through the thickness single-edge crack, a yield strength of 800 MPa and a static fracture strength is  f  3.2 ys . If the plate is loaded in tension and fractures at 600 MPa, calculate the plane strain fracture toughness of the steel plate and the critical crack length. Solution:

 f  3.2 ys 

f  ys

and

 3.2 and

 ys  800 MPa   20 m 1/ 2

From eq. (10.12), K IC 

  1 ys 3.2  1800 MPa    88 MPa 

20 m 1 / 2

m

If   600 MPa, then K IC  1.12 a c 2

2 1  K IC  1  88 MPa m  ac        5.46 mm   1.12    1.12 600 MPa 

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Solution Manual

10.3 A standard Charpy specimen with B = 5 cm, w = 5 cm and S/w = 4 was tested at a room temperature. The measured impact energy was 30 joules. The tested material has a modulus of elasticity of 70,000 MPa. Calculate the KIC for this hypothetical specimen having x = a/w = 0.2. Solution: Data: x = a/w = 0.2 E = 70,000 MPa

w = 5×10-2 m B = 5×10-2 m U = 30×10-6 MJ = 30×10-6 MN.m

From eqs. (10.57) and (10.43), respectively, 2 2    0.45473 7x 7 0.2  K IC  K I 

UE  Bw

30 x10

MN .m70,000 MPa  0.454735x10 2 m5x10 2 m 6

K IC  42.98 MPa m

10.4 Suppose that a design code calls for a Charpy impact energy of 22 Joules for building a large pressure vessel containing an inert gas. If A533B and A723 (  ys  1,100 MPa ) steel plates are available for such a purpose, then (a) select the steel that will tolerate the largest critical crack length (depth of a surface semi-elliptical crack) when the hoop stress is 500 MPa and (b) determine the minimum plate thickness as per ASTM E399 standard for the selected steel. Solution: For   500 MPa, U  22 J and  ys  1,100 MPa (A723),  196 )  K IC A533B  20+139 -1  U 

- 0 .54

 65.50 MPa m

K IC A723  0.644U ys  0.006 ys2  91.24 MPa m

eq. (10.71) eq. (10.74)

(a) The critical length 2

2 1K  1  65.50 MPa m    5.46 mm ac (A533B)   IC          500 MPa  2

2 1  K IC  1  91.24 MPa m    10.60 mm ac (A723)            500 MPa  The ASTM A723 steel allows a larger critical crack length; therefore, this material is selected for such an application.

(b) The plate thickness for the selected A723 steel is 2

2

K   91.24 MPa m    17.20 mm  0.6772 inches B  2.5 IC   2.5    1 , 100 MPa ys     4


Solution Manual

10.5 Plot the fracture toughness data for a hypothetical polymer and determine the brittle-ductile transition temperature. T (oC) K IC ( MPa m )

-160 4.10

-120 4.11

-90 -80 -75 -70 -60 -50 -30 4.00 4.05 4.10 4.50 5.50 6.60 11.00

Solution:

20 20

K IC

15 15

MPa m 

10 10  80o C

55 Brittle

Ductile

00 -200 -200

-150 -150

-100 -100

T

 C

-50 -50

00

O

10.6 For linear motion during Charpy impact tests, the energy lost by the striker ( U S ) and the kinetic energy ( mv 2 ) data for some polymers with different masses and spans to depth (S/w) ratios [47] are given below. Let m/M  1 and a) plot U S  f(mv²) and estimate the coefficient of restitution (e) for these data. What does 0  e  1 mean? Theoretically, determine b) the U S  f(mv²) ratio for the first bound which is transformed into specimen strain energy and c) the U S  f(mv²) ratio when there is no bouncing. mv 2 mJ 0 50 100 150 200 U S mJ

0 75 150 222 297

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Solution Manual

Solution: (a) The plot and the coefficient of restitution are

y  0. 6  1. 482x or U S  0. 6  1. 482mv 2 

Polynomial fit:

350 300 y

US

250

(mJ )

200 150 100 50 0 0

50

100

150

200 250 2 mv ( mJ ) x

From eq. (10.39) along with m/M << 1,

U S  1  e mv² Slope  1  e  15 e05 This result, e  0.5, means that a repeated impact of converging energy into strain energy occurs until the last contact is achieved. (b) The U S  f(mv²) ratio for the first bound which is transformed into specimen strain energy requires that the coefficient of restitution be e = 1. Thus, U S / mv²   1  e  2 c) The U S  f mv²  ratio when there is no bouncing means that e  0 . Thus, U S / mv²   1  e  1

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Solution Manual

10.7 A large and thick ASTM A533B-1 steel plate containing a 4-mm long through-the-thickness central crack fractures when it is subjected to a tensile stress of 7 MPa. Plot U, KIC, and GIC at  200o C  T  100o C . Which of the plots is more suitable for determining the transition temperature? Explain. Data: E = 207,000 MPa and v = 1/3. Solution: From eq. (10.72) and (10.73),

U

196 1  exp(0.0297T )

 196  K IC  20  139  1  U 

in Joules  0.54

in MPa m

200 K IC U ( JU ) K IC

MPa m 

U

150

100

To  65o C

50

0 -200

-150

-100

-50 T

 C o

0

50

100

T

Denote that the transition temperature ( To ) is easily determined using the U  f (T ) plot. This is shown with dash lines. GIC 

1  v  K 2

2 IC



10 1  1/ 3 20  95.50 exp0.016T  6

2

E 207,000 GIC  4.294120  95.50 exp0.016T 

7

in Joules


Solution Manual

4

GICG(J) IC

3

(J ) 2

T0  55o C 1

0 -200

-150

-100

-50 T C

 

0

50

100

o

T (C)

It is clearly shown that the U IC  f (T ) plot is more suitable for determining the transition temperature T0  65o C .

10.8 A large plate made of 18Ni-8Co-3Mo Grade 200 alloy is part of structure exposed to relatively high temperature. Charpy impact tests were carried out and the average impact energy is 60 J. Use this information to calculate (a) the plane strain fracture toughness, (b) the minimum thickness ASTM requirement. The plate width is at least twice the thickness. Is this thickness practical? (c) Assume that a single-edge through the thickness crack develops. What will the critical crack length be? Will its value be reasonable? Solution: (a) If  60 J and  ys  1,310 MPa (From Table 10.1), then eq. (10.76) yields

K IC  ys

2

 4. 69 Uys

 0. 20

K IC   ys 4. 69 Uys K IC  1, 310 MPa

 0. 20 4. 69

60 1, 310

 0. 20

K IC  159. 42 MPa m

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Solution Manual

(b) From eq. (3.30), the minimum plate thickness is IC B  2. 5 K  ys

2

 2. 5 159. 42 1310 B  0. 30424 m  304. 24 mm  12 inches

This is a very impractical thick for plate, but it complies with the ASTM plane strain conditions. (c) Now if a single-edge crack develops during service, the critical crack length under a stress half the yield strength will be

 ys a c 2 1 2K IC ac    ys

K IC 

2

3  10 

2  159. 42 1310

2

a c  19 mm This is a reasonable result for a plate so thick.

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Fracture Mechanics_2nd Ed_Solution Manual

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