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

All12ChaptersCovered
r r r




SOLUTIONS

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




CHAPTER 1 r




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




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




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




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




r




50 MPa r = 100 MPa r r




ac = 0.0796 m = 79.6 mm
r r r r r r




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




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




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




Ans:

(50 MPa m)
2 r



r r r
r




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



E 207,000 MPa
r


r r




=12.1kJ/m2
r r




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




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




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




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




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




the ground, h, and the weight of the object, mg, where m is the mass and g is the gravitational
r r r r r r r r r r r r r r r r r r r r




acceleration. Therefore, neglecting aerodynamic drag, the time of flight is given by the
r r r r r r r r r r r r r




following function:
r r




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




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




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




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




@
@SSeeisismmicicisisoolalatitoionn

,Solutions Manual r 3

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




sides of the above equation by
r : r r r r




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


r




h g h g r r




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




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




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




h
t= r r



g

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




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




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




motion.
r




CHAPTER 2 r




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




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




r necessary?


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




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




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




rConsequently, the material resistance to crack extension = 2 wf. r r r r r r r r r




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




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




Ans:
(a) Load control.
P dCP
r




P d
G = 2B da  P dC
= 2B  da  = 2B da
r rr rr r r rr r r r r r
r
r r
r r
r r r r




 P  P




@
@SSeeisismmicicisisoolalatitoionn

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




(b) Displacement control. r




 dP
G= −
r r




2Bda  
r

r r

r

r r




 dP 
rr r
d 1C ( )
r
r
r
r  dC r




  = r
=− r




 da  C2 da
r r



da r r




G = ( C ) dC = P
 2

r
r dC 2
r r


r




2B da 2B da r




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




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




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




2.5.3).
r




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




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




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




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




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




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




displacement are related as follows:
r r r r r




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




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




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




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




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




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




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




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




terms of the plate dimensions and elastic modulus, E. The stress in Eq. (2.24) is thenominal value;
r r r r r r r r r r r r r r r r r r




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




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




@
@SSeeisismmicicisisoolalatitoionn