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

All12 Chapters Covered
n n n




SOLUTIONS

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




CHAPTER 1 n




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




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




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




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




n




50 MPa n = 100 MPa n n




ac = 0.0796 m = 79.6 mm
n n n n n n




Total crack length = 2ac = 159 mmn n n n n n n




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




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




Ans:

(50 MPa m)
2 n



n n n
n




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



E 207,000 MPa
n


n n




=12.1 kJ/m2
n n




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




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




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




your undergraduate physics course. You decide to infer a relationship for time of flight of a
n n n n n n n n n n n n n n n n




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




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




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




by the following function:
n n n n




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




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




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




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




@
@SSeeisismmicicisisoolalatitoionn

,Solutions Manual n 3

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




sides of the above equation by
n n : n n n




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


n




h g h g n n




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




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




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




h
t= n n



g

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




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




reliable estimate of this constant. Note that =
n accordingto Newton's laws of n n n n n n n n n n n




motion.
n




CHAPTER 2 n




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




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




n necessary?


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




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




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




nConsequently, the material resistance to crack extension = 2 wf. n n n n n n n n n




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




into Eqs. (2.27) and (2.28), respectively.
n n n n n n




Ans:
(a) Load control.
P dCP
n




P d
G = 2B da  P dC
= 2B  da  = 2B da
n nn n
n n n nn n n n n n
n
n n
n n
n n n n




 P  P




@
@SSeeisismmicicisisoolalatitoionn

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




(b) Displacement control. n




 dP
G= −
n n




2Bda  
n

n n

n

n n




 dP 
nn n
d 1C ( )
n
n
n
n  dC n




  = n
=− n




 da  C2 da
n n



da n n




G = ( C ) dC = P
 2

n
n dC 2
n n


n




2B da 2B da n




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




n infinite plate when the stress is fixed. Suppose that a remote displacement (rather than load)
n n n n n n n n n n n n n n




n were fixed in this configuration. Would the driving force curves be altered? Explain. (Hint: see
n n n n n n n n n n n n n n




n Section 2.5.3). n




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




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




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




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




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




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




displacement are related as follows:
n n n n n




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




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




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




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




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




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




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




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




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




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




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




@
@SSeeisismmicicisisoolalatitoionn