Solutions Manual: Conv for Fracture Mechanics Fundamentals and Applications (4th Edition) - Graded A+ Comprehens...

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2 Fracture Mechanics: Fundamentals and
·




Applications
···




CHAPTER 1

1.2 A flat plate with a through-thickness crack (Fig. 1.8) is subject to a 100 MPa
(14.5 ksi) tensile stress and has a fracture toughness (KIc) of 50.0 MPa m (45.
ksi in ). Determine the critical crack length for this plate, assuming the
material is linear elastic.
Ans:
At fracture, K  K  
Ic I ac . Therefore,
···




·




50 MPa m = 100 MPa  ac

ac = 0.0796 m = 79.6 mm···




Total crack length = 2ac = 159 mm
··




1.3 Compute the critical energy release rate (Gc) of the material in the previous
problem for E = 207,000 MPa (30,000 ksi)..


Ans:

 
2
50 MPa m
KIc
G    0.0121 MPa mm  12.1 kPa
c
E m 207,000 MPa
 12.1 kJ/m2

Note that energy release rate has units of energy/area.

1.4 Suppose that you plan to drop a bomb out of an airplane and that you are
interested in the time of flight before it hits the ground, but you cannot
remember the appropriate equation from your undergraduate physics course.
You decide to infer a relationship for time of flight of a falling object by
experimentation. You reason that the time of flight, t, must depend on the
height above the ground, h, and the weight of the object, mg, where m is the
mass and g is the gravitational acceleration. Therefore, neglecting
aerodynamic drag, the time of flight is given by the following function:

t  f (h, m, g)

Apply dimensional analysis to this equation and determine how many
experiments would be required to determine the function f to a reasonable
approximation, assuming you know the numerical value of g. Does the time of




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Solutions 3
Manual
Ans:
Since h has units of length and g has units of (length)(time)-2, let us divide both
···




sides of the above equation by h :
··




g
t f h, m, g 

·




h g h
g
The left side of this equation is now dimensionless. Therefore, the right side must
·




also be dimensionless, which implies that the time of flight cannot depend on the
·




mass of the object. Thus dimensional analysis implies the following functional
··




relationship:

h
t
g

where  is a dimensionless constant. Only one experiment would be required to
estimate , but several trials at various heights might be advisable to obtain a
reliable estimate of this constant. Note that   2 according to Newton's laws
of motion.
CHAPTER 2

2.1 According to Eq. (2.25), the energy required to increase the crack area a unit
amount is equal to twice the fracture work per unit surface area, wf. Why is
the factor of 2 in this equation necessary?


Ans:
The factor of 2 stems from the difference between crack area and surface area.
The former is defined as the projected area of the crack. The surface area is twice
the crack area because the formation of a crack results in the creation of two
surfaces. Consequently, the material resistance to crack extension = 2 wf.

2.2 Derive Eq. (2.30) for both load control and displacement control by
substituting Eq. (2.29) into Eqs. (2.27) and (2.28), respectively.


Ans:
(a) Load control.
P  d P  d P dC
 
G
 
  CP   
2B da 2B da 2B da
 P  P


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4 Fracture Mechanics: Fundamentals and
Applications
(b) Displacement control.
  dP 
G  2 
B  da

 dP  d  1C   dC
   
 d  C 2 da
a da


 C  dC P
2
G
 dC
2B  da 2B da


2.3 Figure 2.10 illustrates that the driving force is linear for a through-thickness
crack in an infinite plate when the stress is fixed. Suppose that a remote
displacement (rather than load) were fixed in this configuration. Would the
driving force curves be altered? Explain. (Hint: see Section 2.5.3).


Ans:
In a cracked plate where 2a << the plate width, crack extension at a fixed remote
displacement would not effect the load, since the crack comprises a negligible
portion of the cross section. Thus a fixed remote displacement implies a fixed
load, and load control and displacement control are equivalent in this case. The
driving force curves would not be altered if remote displacement, rather than
stress, were specified.
Consider the spring in series analog in Fig. 2.12. The load and remote
displacement are related as follows:

T = (C + Cm) P T  C  Cm  P

where C is the “local” compliance and Cm is the system compliance. For the
present problem, assume that Cm represents the compliance of the uncracked plate
and C is the additional compliance that results from the presence of the crack.
When the crack is small compared to the plate dimensions, Cm >> C. If the crack
were to grow at a fixed T, only C would change; thus load would also remain
fixed.

2.4 A plate 2W wide contains a centrally located crack 2a long and is subject to a
tensile load,
P. Beginning with Eq. (2.24), derive an expression for the elastic compliance, C
(= /P) in terms of the plate dimensions and elastic modulus, E. The stress in
Eq. (2.24) is the nominal value; i.e.,  = P/2BW in this problem. (Note: Eq.
(2.24) only applies when a << W; the expression you derive is only approximate

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