Solutions Manual for Fracture Mechanics: Fundamentals and Applications (4th Edition, 2018) - Anderson

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dr dr dr




SOLUTIONS

,2 Fracture Mechanics: Fundamentals and dr dr dr


Applications
dr




CHAPTER 1 dr




1.2 d r d rA flat plate with a through-thickness crack (Fig. 1.8) is subject to a 100 MPa
dr dr dr dr dr dr dr dr dr dr dr dr dr dr


dr (14.5 ksi) tensile stress and has a fracture toughness (KIc) of 50.0 MPa
dr dr m (45. dr dr dr dr dr dr dr dr dr dr d r d r


dr ksi in ). Determine the critical crack length for this plate, assuming the material
d r dr d r dr dr dr dr dr dr dr dr dr dr


is linear elastic.
Ans:
At KIc  KI d r
dr . d r Therefore,
fracture,
dr

d r
dr




50 MPa dr = d r 100 MPa dr




ac = dr d r 0.0796 m dr d r = d r 79.6 mm dr




Total crack length dr dr d r = d r 2ac d r = d r 159 mm dr




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


E = 207,000 MPa (30,000 ksi)..
dr dr dr dr dr dr




Ans:

50 MPa 
2 d r

d r dr d r m dr


G 
  0.0121 MPa mm  12.1 kPa m
d r
KIc d r c dr dr dr dr dr dr dr


E
dr
dr 207,000 MPa dr




12.1 kJ/m2
dr dr




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




1.4 d r Suppose that you plan to drop a bomb out of an airplane and that you are
d r dr dr dr dr dr dr dr dr dr dr dr dr dr dr dr

interested in the time of flight before it hits the ground, but you cannot remember
dr dr dr dr dr dr dr dr dr dr dr dr dr dr dr

the appropriate equation from your undergraduate physics course. You decide to infer a
dr dr dr dr dr dr dr dr dr dr dr dr dr


relationship for time of flight of a falling object by experimentation. You reason that
dr dr dr dr dr dr dr dr dr dr dr d r dr dr

the time of flight, t, must depend on the height above the ground, h, and the weight
dr dr dr dr dr dr dr dr dr dr dr dr dr dr dr dr dr

of the object, mg, where m is the mass and g is the gravitational acceleration.
dr dr dr dr dr dr dr dr dr dr dr dr dr dr dr

d rTherefore, neglecting aerodynamic drag, the time of flight is given by the following
dr dr dr dr dr dr dr dr dr dr dr dr

function:
dr




t  f (h, m, g)
dr dr dr dr dr




Apply dimensional analysis to this equation and determine how many experiments
dr dr dr dr dr dr dr dr dr dr

would be required to determine the function f to a reasonable approximation,
dr dr dr dr dr dr dr dr dr dr dr dr


assuming you know the numerical value of g. Does the time of flight depend on
dr dr dr dr dr dr dr dr d r dr dr dr dr dr dr

the mass of the object?
dr dr dr dr dr




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, Solutions Manual dr 3

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


divide both
dr dr



sides of the above equation by
dr dr : dr dr dr




t f h,m, g

dr dr dr dr dr

dr

h h
 g g
The left side of this equation is now dimensionless. Therefore, the right side
dr dr dr dr dr dr dr dr d r dr dr dr


must also be dimensionless, which implies that the time of flight cannot
dr dr dr dr dr dr dr dr dr dr dr dr


depend on the mass of the object. Thus dimensional analysis implies the
dr dr dr dr dr dr dr d r dr dr dr dr


following functional relationship:
dr dr dr




h
t  dr dr

g

where is a dimensionless constant. Only one experiment would be required to
dr dr dr dr dr d r dr dr dr dr dr dr


estimate
dr , but several trials at various heights might be advisable to
dr dr dr dr dr dr dr dr dr dr dr


drobtain a d r



reliable estimate of this constant. Note that
dr  according to Newton's laws of dr dr dr d r dr dr dr dr dr dr dr


motion.
dr




CHAPTER 2 dr




2.1 d r According to Eq. (2.25), the energy required to increase the crack area a unit amount
d r dr dr dr dr dr dr dr dr dr dr dr dr dr dr

dr is equal to twice the fracture work per unit surface area, wf. Why is the factor of 2
dr dr dr dr dr dr dr dr dr dr dr d r dr dr dr dr dr


dr in this equation necessary?
dr dr dr




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


area. The former is defined as the projected area of the crack. The surface
dr dr dr dr dr dr dr dr dr dr dr dr d r dr


area is twice the crack area because the formation of a crack results in the
dr dr dr dr dr dr dr dr dr dr dr dr dr dr dr


creation of two surfaces. Consequently, the material resistance to crack
dr dr dr dr d r dr dr dr dr dr


extension = 2 wf.
dr dr dr dr




2.2 Derive Eq. (2.30) for both load control and displacement control by substituting Eq.
dr dr dr dr dr dr dr dr dr dr dr dr


(2.29) into Eqs. (2.27) and (2.28), respectively.
dr dr dr dr dr dr dr




Ans:  
(a) Load control.  
 d CP
dr

P  d P P dC
G   

dr d r dr drdr dr dr dr dr dr dr dr d r


 dr
 dr dr
dr
dr dr

2B da 2B da 2B da dr dr dr d r


 P  P
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, 4 Fracture Mechanics: Fundamentals and dr dr dr


Applications
dr




(b) Displacement control. dr


  dP 
G 
d r dr dr

  dr d r

2B  da 
dr
dr dr




 dP 
drdr dr
d 1C  dC dr   dr
dr dr d r



  dr dr
  dr  dr


 da  da C2 da dr d r




 
2

G  C dC
2
P dC

dr d r d r
dr
d r

2B da 2B da d r




2.3 d r d rFigure 2.10 illustrates that the driving force is linear for a through-thickness crack
dr dr dr dr dr dr dr dr dr dr dr dr

dr in an infinite plate when the stress is fixed. Suppose that a remote displacement (rather
dr dr dr dr dr dr dr dr dr dr dr dr dr dr

dr than load) were fixed in this configuration. Would the driving force curves be altered?
dr dr dr dr dr dr dr dr dr dr dr dr dr


dr Explain. (Hint: see Section 2.5.3). dr dr dr dr




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


remote displacement would not effect the load, since the crack comprises a
dr dr dr dr dr dr dr dr dr dr dr dr


negligible portion of the cross section. Thus a fixed remote displacement implies
dr dr dr dr dr dr d r dr dr dr dr dr


a fixed load, and load control and displacement control are equivalent in this
dr dr dr dr dr dr dr dr dr dr dr dr dr


case. The driving force curves would not be altered if remote displacement,
dr d r dr dr dr dr dr dr dr dr dr dr


rather than stress, were specified.
dr dr dr dr dr


Consider the spring in series analog in Fig. 2.12. The load and remote dr dr dr dr dr dr dr dr d r dr dr dr


displacement are related as follows:
dr dr dr dr dr




T d r = d r (C + C m) P T dr dr dr dr
d r
 C  C m P
dr dr dr
dr
dr




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


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


is the additional compliance that results from the presence of the crack.
dr dr dr dr dr dr dr dr dr dr dr dr


When the crack is small compared to the plate dimensions, Cm >> C. If the
d r dr dr dr dr dr dr dr dr dr dr dr dr d r dr



crack were to grow at a fixed T, only C would change; thus load would also
dr dr dr dr dr dr dr dr dr dr dr dr dr dr dr dr


remain fixed.
dr dr




2.4 A plate 2W wide contains a centrally located crack 2a long and is subject to a
dr dr dr dr dr dr dr dr dr dr dr dr dr dr dr

tensile load,
dr dr

P. Beginning with Eq. (2.24), derive an expression for the elastic compliance, C (=
d r dr dr dr dr dr dr dr dr dr dr dr dr


/P) in terms of the plate dimensions and elastic modulus, E. The stress in Eq. (2.24)
dr dr dr dr dr dr dr dr dr dr dr d r dr dr dr dr


is the nominal value; i.e.,
dr dr = P/2BW in this problem. (Note: Eq. (2.24) only
dr dr dr dr dr d r dr dr dr d r d r dr dr

applies when a << W; the expression you derive is only approximate for a finite
dr dr dr dr dr dr dr dr dr dr dr dr dr dr dr


width plate.)
dr dr




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