Solutions Manual for Fracture Mechanics: Fundamentals and Applications, 4th Edition by Ted L. Anderson | Complete Worked Solutions and Explanations

All12 ChaptersCovered
c c c




SOLUTIONS

,2 Fracture Mechanics: Fundamentals and Applications c c c c




CHAPTER 1 c




1.2 c c A flat plate with a through-thickness crack (Fig. 1.8) is subject to a 100 MPa (14.5 ksi) tensile
c c c c c c c c c c c c c c c c c




stress and has a fracture toughness (KIc) of 50.0 MPa m (45. ksi in ). Determine the
c c c c c c c c c c c c c c c c c




critical crack length for this plate, assuming the material is linear elastic.
c c c c c c c c c c c c




Ans:
At fracture, KIc = KI =
c
c
c
c
. Therefore, c




c




50 MPa c = 100 MPa c c




ac = 0.0796 m = 79.6 mm
c c c c c c




Total crack length = 2ac = 159 mmc c c c c c c




1.3 Compute the critical energy release rate (Gc) of the material in the previous problem for E =
c c c c c c c c c c c c c c c c




207,000 MPa (30,000 ksi)..
c c c c




Ans:

(50 MPa m)
2 c



c c c
c




KIc
G c= c c = =0.0121 MPa mm =12.1kPa m c c c c c c c



E 207,000 MPa
c


c c




=12.1kJ/m2
c c




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




1.4 c c Suppose that you plan to drop a bomb out of an airplane and that you are interested in the time
c c c c c c c c c c c c c c c c c c c




of flight before it hits the ground, but you cannot remember the appropriate equation from your
c c c c c c c c c c c c c c c c




undergraduate physics course. You decide to infer a relationship for time of flight of a falling
c c c c c c c c c c c c c c c c




object by experimentation. You reason that the time of flight, t, must depend on the height
c c c c c c c c c c c c c c c c




above the ground, h, and the weight of the object, mg, where m is the mass and g is the
c c c c c c c c c c c c c c c c c c c c




gravitational acceleration. Therefore, neglecting aerodynamic drag, the time of flight is given
c c c c c c c c c c c c




by the following function:
c c c c




t = f (h,m,g)
c c c c c




Apply dimensional analysis to this equation and determine how many experiments would be
c c c c c c c c c c c c




required to determine the function f to a reasonable approximation, assuming you know the
c c c c c c c c c c c c c c




numerical value of g. Does the time of flight depend on the mass of the object?
c c c c c c c c c c c c c c c c




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,Solutions Manual c 3

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




sides of the above equation by
c :c c c c




t f (h,m,g)
=
c c c c c


c




h g h g c c




The left side of this equation is now dimensionless. Therefore, the right side must also
c c c c c c c c c c c c c c




be dimensionless, which implies that the time of flight cannot depend on the mass of
c c c c c c c c c c c c c c c




the object. Thus dimensional analysis implies the following functional relationship:
c c c c c c c c c c




h
t= c c



g

where is a dimensionless constant. Only one experiment would be required to
c c c c c c c c c c c c




estimate , but several trials at various heights might be advisable to obtain a
c c c c c c c c c c c c c c




reliable estimate of this constant. Note that =
c accordingtoNewton'slawsof c c c c c c c c c c c




motion.
c




CHAPTER 2 c




2.1 c c According to Eq. (2.25), the energy required to increase the crack area a unit amount is equal to
c c c c c c c c c c c c c c c c c




c twice the fracture work per unit surface area, wf. Why is the factor of 2 in this equation
c c c c c c c c c c c c c c c c c




c necessary?


Ans:
The factor of 2 stems from the difference between crack area and surface area. The
c c c c c c c c c c c c c c




former is defined as the projected area of the crack. The surface area is twice the crack
c c c c c c c c c c c c c c c c c




area because the formation of a crack results in the creation of two surfaces.
c c c c c c c c c c c c c c




cConsequently, the material resistance to crack extension = 2 wf. c c c c c c c c c




2.2 Derive Eq. (2.30) for both load control and displacement control by substituting Eq. (2.29) into
c c c c c c c c c c c c c c




Eqs. (2.27) and (2.28), respectively.
c c c c c




Ans:
(a) Load control.
P dCP
c




P d
G = 2B da  P dC
= 2B  da  = 2B da
c cc cc c c cc c c c c c
c
c c
c c
c c c c




 P  P




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, 4 Fracture Mechanics: Fundamentals and Applications c c c c




(b) Displacement control. c




 dP
G= −
c c




2Bda  
c

c c

c

c c




 dP 
cc c
d 1C ( )
c
c
c
c  dC c




  = c
=− c




 da  C2 da
c c



da c c




G = ( C ) dC = P
 2

c
c dC 2
c c


c




2B da 2B da c




2.3 c c Figure 2.10 illustrates that the driving force is linear for a through-thickness crack in an infinite
c c c c c c c c c c c c c c c




plate when the stress is fixed. Suppose that a remote displacement (rather than load) were fixed in
c c c c c c c c c c c c c c c c c




this configuration. Would the driving force curves be altered? Explain. (Hint: see Section
c c c c c c c c c c c c c




2.5.3).
c




Ans:
In a cracked plate where 2a << the plate width, crack extension at a fixed remote
c c c c c c c c c c c c c c c




displacement would not effect the load, since the crack comprises a negligible portion
c c c c c c c c c c c c c




of the cross section. Thus a fixed remote displacement implies a fixed load, and load
c c c c c c c c c c c c c c c




control and displacement control are equivalent in this case. The driving force curves
c c c c c c c c c c c c c




would not be altered if remote displacement, rather than stress, were specified.
c c c c c c c c c c c c




Consider the spring in series analog in Fig. 2.12. The load and remote c c c c c c c c c c c c




displacement are related as follows:
c c c c c




T = (C + Cm) PT =(C+Cm )P c
c c c c c
c
c c c
c
c




where C is the “local” compliance and Cm is the system compliance. For the present
c c c c c c c
c
c c c c c c




problem, assume that Cm represents the compliance of the uncracked plate and C is the
c c c c
c
c c c c c c c c c c




additional compliance that results from the presence of the crack. When the crack is
c c c c c c c c c c c c c c




small compared to the plate dimensions, Cm >> C. If the crack were to grow at a fixed
c c c c c c c
c
c c c c c c c c c c




T, only C would change; thus load would also remain fixed.
c c c c c c c c c c c




2.4 A plate 2W wide contains a centrally located crack 2a long and is subject to a tensile load,
c c c c c c c c c c c c c c c c c




P. Beginning with Eq. (2.24), derive an expression for the elastic compliance, C (= /P) in
c c c c c c c c c c c c c c c




terms of the plate dimensions and elastic modulus, E. The stress in Eq. (2.24) is the nominal
c c c c c c c c c c c c c c c c c




value; i.e.,
c = P/2BW in this problem. (Note: Eq. (2.24) only applies when a << W; the
c c c c c c c c c c c c c c c c c




expression you derive is only approximate for a finite width plate.)
c c c c c c c c c c c




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