Mechanics of Materials 2010 Part 9 ppsx

Thus, the design stress will be given by σ d= σ y 2 and from K Ic = σ d √ πa cr ⇒ a cr= π1 K I c σy 2 2 Thus, Yield Stress Design Stress Fracture Toughness Critical Crack Total Crack We

Ksi ksi in

Table 11.4: Fracture Toughness vs Yield Stress for 45C − N i − C r − M o Steel

11.4 Examples

Assume that a component in the shape of a large sheet is to be fabricated from 45C −N i −C r −M osteel, with a decreasing fracture toughness with increase in yield stress, Table 11.4 The smallest crack size

(2a) which can be detected is approximately 12 in The specified design stress is σ y

2 To save weight,

an increase of tensile strength from 220 ksi to 300 ksi is suggested Is this reasonable?

At 220 ksi K Ic = 60 ksi√

in, and at 300 ksi K Ic= 30 ksi√

in Thus, the design stress will be given

by σ d= σ y

2 and from K Ic = σ d √

πa cr ⇒ a cr= π1

K I c

σy

2

2 Thus,

Yield Stress Design Stress Fracture Toughness Critical Crack Total Crack

We observe that for the first case, the total crack length is larger than the smallest one which can be detected (which is O.K.); Alternatively, for the second case the total critical crack size is approximately five times smaller than the minimum flaw size required and approximately eight times smaller than the flaw size tolerated at the 220 ksi level

Hence, σ y should not be raised to 300 ksi

Finally, if we wanted to use the flaw size found with the 300 ksi alloy, we should have a decrease

in design stress (since K Ic and a cr are now set) K Ic = σ d √

πa vis ⇒ σ d = √ K πa Ic

vis = 30ksi

√

in

√

0.06π = 69 ksi, with a potential factor of safety of one against cracking (we can not be sure 100% that there is no crack of that size or smaller as we can not detect it) We observe that since the design stress level is approximately half of that of the weaker alloy, there will be a two fold increase in weight

A small beer barrel of diameter 15” and wall thickness of 126” made of aluminum alloy exploded when

a pressure reduction valve malfunctioned and the barrel experienced the 610 psi full pressure of the CO2 cylinder supplying it with gas Afterwards, cracks approximately 4.0 inch long by (probably) 07 inch deep were discovered on the inside of the salvaged pieces of the barrel (it was impossible to measure their depth) Independent tests gave 40 ksi√

in for K Ic of the aluminum alloy The question is whether the cracks were critical for the 610 psi pressure?

For a cylinder under internal pressure, the hoop stress is σ = pD 2t =610 lb

in2 2(.126)15 inin = 36, 310 psi =

36.3 ksi This can be used as the far field stress (neglecting curvature).

First we use the exact solution as given in Eq 11.24, with a = 2 in, b = 07 in, and t = 126 in.

upon substitution we obtain:

Draft12 LEFM DESIGN EXAMPLES

M1 = 1.13 − 0.09



.07

2.



= 1.127



0.2 +



.07

2.

−1

− 0.54

= 3.247



0.65 +



.07

2

−1 + 14



1 −



.07

2

24

= 4.994

Substituting

π.07

!

1.127 + 3.247



.07 126

2

+ 4.994



.07 126

4" !

1 + 1.464



.07

2

1.65"−1

!

.07

2.

2

0 + 1

"12

1 +

!

0.1 + 0.35



.07 126

2"

(1− 1)2

/

= 44.2ksi √

in This is about equal to the fracture toughness

Note that if we were to use the approximate equation, for long cracks we would have obtained:

K = (1.13)(36.3)

π(.07)

!

1 + 3.46



.07 126

2

+ 11.5



.07 126

4"

= 60.85

> K Ic

11.5 Additional Design Considerations

11.5.1 Leak Before Fail

10 As observed from the preceding example, many pressurized vessels are subject to crack growth if internal flaws are present Two scenarios may happen, Fig 11.14

Break-through: In this case critical crack configuration is reached before the crack has “daylighted”,

and there is a sudden and unstable crack growth

Leak Before Fail: In this case, crack growth occur, and the crack “pierces” through the thickness of

the vessel before unstable crack growth occurs This in turn will result in a sudden depressurization, and this will stop any further crack gowth

11 Hence, pressurized vessels should be designed to ensure a leak before fail failure scenario, as this would

usually be immediately noticed and corrrected (assuming that there is no leak of flamable gas!)

12 Finally, it should be noted that leak before break assessment should be made on the basis of a complete

residual strength diagram for both the part through and the through crack Various ratios should be

considered

2c t

2a=2t

Initial Elliptical Crack

Figure 11.14: Growth of Semielliptical surface Flaw into Semicircular Configuration

13 Fracture mechanics is not limited to determining the critical crack size, load, or stress combination It can also be applied to establish a fracture control plan, or damage tolerance analysis with the following objectives:

1 Determine the effect of cracks on strength This will result in a plot of crack size versus residual

strength, or Residual Strength Diagram

2 Determine crack growth with time, resulting in Crack Growth Curve.

Chapter 12

THEORETICAL STRENGTH of

SOLIDS; (Griffith I)

1 We recall that Griffith’s involvement with fracture mechanics started as he was exploring the disparity

in strength between glass rods of different sizes, (Griffith 1921) As such, he had postulated that this can be explained by the presence of internal flaws (idealized as elliptical) and then used Inglis solution

to explain this discrepancy

2 In this section, we shall develop an expression for the theoretical strength of perfect crystals (theoret-ically the strongest form of solid) This derivation, (Kelly 1974) is fundamentally different than the one

of Griffith as it starts at the atomic level

12.1 Derivation

3 We start by exploring the energy of interaction between two adjacent atoms at equilibrium separated

by a distance a0, Fig 12.1 The total energy which must be supplied to separate atom Cfrom C’ is

where γ is the surface energy1, and the factor of 2 is due to the fact that upon separation, we have two distinct surfaces

12.1.1 Tensile Strength

12.1.1.1 Ideal Strength in Terms of Physical Parameters

We shall first derive an expression for the ideal strength in terms of physical parameters, and in the next section the strength will be expressed in terms of engineering ones

Solution I: Force being the derivative of energy, we have F = dU

da , thus F = 0 at a = a0, Fig 12.2, and

is maximum at the inflection point of the U0− a curve Hence, the slope of the force displacement

curve is the stiffness of the atomic spring and should be related to E If we let x = a − a0, then

the strain would be equal to ε = a x0 Furthermore, if we define the stress as σ = a F2, then the σ − ε

1From watching raindrops and bubbles it is obvious that liquid water has surface tension When the surface of a liquid

is extended (soap bubble, insect walking on liquid) work is done against this tension, and energy is stored in the new surface When insects walk on water it sinks until the surface energy just balances the decrease in its potential energy For solids, the chemical bonds are stronger than for liquids, hence the surface energy is stronger The reason why we do not

notice it is that solids are too rigid to be distorted by it Surface energy γ is expressed in J/m2and the surface energies

of water, most solids, and diamonds are approximately 077, 1.0, and 5.14 respectively.

Figure 12.1: Uniformly Stressed Layer of Atoms Separated by a0

curve will be as shown in Fig 12.3 From this diagram, it would appear that the sine curve would

be an adequate approximation to this relationship Hence,

σ = σ theor max sin 2π x

and the maximum stress σ theor

max would occur at x = λ

4 The energy required to separate two atoms

is thus given by the area under the sine curve, and from Eq 12.1, we would have

2γ = U0 =

 λ

2 0

σ max theorsin

2π x

λ

2π σ

theor max [− cos ( 2πx

λ )]| λ2

2π σ

theor max [−

−1

   cos (2πλ

2λ ) +

1

  

σ theor max

(12.6)

For very small displacements (small x) sin x ≈ x, Eq 12.2 reduces to

σ ≈ σ theor max

2πx

elliminating x,

σ max theor ≈ E

a0

λ

Substituting for λ from Eq 12.6, we get

σ theor max ≈

1

Eγ

Draft12.1 Derivation 3

Interatomic Distance

a 0

a 0

Force

Displacement a Inflection Point

Figure 12.2: Energy and Force Binding Two Adjacent Atoms

Solution II: For two layers of atoms a0apart, the strain energy per unit area due to σ (for linear elastic

systems) is

U = 12σεa o

3

U = σ

2a o

If γ is the surface energy of the solid per unit area, then the total surface energy of two new fracture surfaces is 2γ.

For our theoretical strength, U = 2γ ⇒ (σ max theor)2a0

2E = 2γ or σ theor

max = 2

0

γE

a0

Note that here we have assumed that the material obeys Hooke’s Law up to failure, since this is seldom the case, we can simplify this approximation to:

σ theor max =

1

Eγ

which is the same as Equation 12.9

Example: As an example, let us consider steel which has the following properties: γ = 1 m J2; E =

2× 10 11 N

m2; and a0≈ 2 × 10 −10 m Thus from Eq 12.9 we would have:

σ max theor ≈

1 (2× 1011)(1)

≈ 3.16 × 1010N

Thus this would be the ideal theoretical strength of steel

Figure 12.3: Stress Strain Relation at the Atomic Level

12.1.1.2 Ideal Strength in Terms of Engineering Parameter

We note that the force to separate two atoms drops to zero when the distance between them is a0+ a where a0 corresponds to the origin and a to λ2 Thus, if we take a = λ2 or λ = 2a, combined with Eq.

12.8 would yield

σ max theor ≈ E

a0

a

4 Alternatively combining Eq 12.6 with λ = 2a gives

σ theor max

(12.16)

5 Combining those two equations

γ ≈ E

a0

a

π

2

(12.17)

6 However, since as a first order approximation a ≈ a0then the surface energy will be

γ ≈ Ea0

7 This equation, combined with Eq 12.9 will finally give

σ max theor ≈ √ E

which is an approximate expression for the theoretical maximum strength in terms of E.

8 Similar derivation can be done for shear What happen if we slide the top row over the bottom one Again, we can assume that the shear stress is

τ = τ max theor sin 2π x

Draft12.2 Griffith Theory 5

x

h

Figure 12.4: Influence of Atomic Misfit on Ideal Shear Strength

and from basic elasticity

and, Fig 12.4 γ xy = x/h.

9 Because we do have very small displacement, we can elliminate x from

τ max theor sin 2π x

λ = γG =

x

h G

⇒ τ theor max = Gλ

10 If we do also assume that λ = h, and G = E/2(1 + ν), then

τ max theor  E

12(1 + ν)  E

12.2 Griffith Theory

11 Around 1920, Griffith was exploring the theoretical strength of solids by performing a series of

exper-iments on glass rods of various diameters He observed that the tensile strength (σ t) of glass decreased

with an increase in diameter, and that for a diameter φ ≈ 1

10,000 in., σ t = 500, 000 psi; furthermore, by

extrapolation to “zero” diameter he obtained a theoretical maximum strength of approximately 1,600,000 psi, and on the other hand for very large diameters the asymptotic values was around 25,000 psi

12 Griffith had thus demonstrated that the theoretical strength could be experimentally approached, he now needed to show why the great majority of solids fell so far below it

12.2.1 Derivation

13 In his quest for an explanation, he came across Inglis’s paper, and his “strike of genius” was to assume that strength is reduced due to the presence of internal flaws Griffith postulated that the theoretical strength can only be reached at the point of highest stress concentration, and accordingly the far-field applied stress will be much smaller

14 Hence, assuming an elliptical imperfection, and from equation ??

σ theor max = σ cr act



1 + 2

1

a ρ



(12.24)

15 Asssuming ρ ≈ a0and since 20

a

a0 & 1, for an ideal plate under tension with only one single elliptical

flaw the strength may be obtained from

σ max theor

   micro

= 2σ cr act

1

a

a0

   macro

(12.25)

hence, equating with Eq 12.9, we obtain

σ theor max = 2σ act cr

1

a

a o

   Macro

=

1

Eγ

a0

   Micro

(12.26)

16 From this very important equation, we observe that

1 The left hand side is based on a linear elastic solution of a macroscopic problem solved by Inglis

2 The right hand side is based on the theoretical strength derived from the sinusoidal stress-strain assumption of the interatomic forces, and finds its roots in micro-physics

17 Finally, this equation would give (at fracture)

σ act cr =

1

Eγ

18 As an example, let us consider a flaw with a size of 2a = 5, 000a0

σ cr act =

0

Eγ

4a

10

/

σ act

0

E2

40a a o a

a0 = 2, 500

/

σ cr act =

0

E2

100,000 = E

100√

19 Thus if we set a flaw size of 2a = 5, 000a0 in γ ≈ Ea0

10 this is enough to lower the theoretical fracture strength from √ E

10 to a critical value of magnitude E

100√

10, or a factor of 100

20 Also

σ theor max = 2σ act

cr

0

a

a o

a o = 1˚A = ρ = 10 −10 m





σ max theor = 2σ cr act

1

10−6

10−10 = 200σ

act

21 Therefore at failure

σ act

cr = σ theor max

200

σ theor max = 10E

/

σ cr act ≈ E

which can be attained For instance for steel 2,000 E = 30,000 2,000 = 15 ksi

Chapter 13

ENERGY TRANSFER in CRACK GROWTH; (Griffith II)

1 In the preceding chapters, we have focused on the singular stress field around a crack tip On this basis, a criteria for crack propagation, based on the strength of the singularity was first developed and then used in practical problems

2 An alternative to this approach, is one based on energy transfer (or release), which occurs during crack propagation This dual approach will be developed in this chapter

3 Griffith’s main achievement, in providing a basis for the fracture strengths of bodies containing cracks, was his realization that it was possible to derive a thermodynamic criterion for fracture by considering the total change in energy of a cracked body as the crack length increases, (Griffith 1921)

4 Hence, Griffith showed that material fail not because of a maximum stress, but rather because a certain energy criteria was met

5 Thus, the Griffith model for elastic solids, and the subsequent one by Irwin and Orowan for elastic-plastic solids, show that crack propagation is caused by a transfer of energy transfer from external work and/or strain energy to surface energy

6 It should be noted that this is a global energy approach, which was developed prior to the one of

Westergaard which focused on the stress field surrounding a crack tip It will be shown later that for linear elastic solids the two approaches are identical

13.1 Thermodynamics of Crack Growth

13.1.1 General Derivation

7 If we consider a crack in a deformable continuum aubjected to arbitrary loading, then the first law of thermodynamics gives: The change in energy is proportional to the amount of work performed Since only the change of energy is involved, any datum can be used as a basis for measure of energy Hence energy is neither created nor consumed

8 The first law of thermodynamics states The time-rate of change of the total energy (i.e., sum of the

kinetic energy and the internal energy) is equal to the sum of the rate of work done by the external forces and the change of heat content per unit time:

d

energy, W the external work, and Q the heat input to the system.

9 Since all changes with respect to time are caused by changes in crack size, we can write

∂

∂t =

∂A

∂t

∂

and for an adiabatic system (no heat exchange) and if loads are applied in a quasi static manner (no

kinetic energy), then Q and K are equal to zero, and for a unit thickness we can replace A by a, then

we can rewrite the first law as

∂W



∂U e

∂U p

∂a

 +∂Γ

10 This equation represents the energy balance during crack growth It indicates that the work rate supplied to the continuum by the applied external loads is equal to the rate of strain energy (elastic and plastic) plus the energy dissipated during crack propagation

11 Thus

Π = U e − W (13.4)

− ∂Π

∂U p

∂Γ

∂a (13.5)

that is the rate of potential energy decrease during crack growth is equal to the rate of energy dissipated

in plastic deformation and crack growth

12 It is very important to observe that the energy scales with a2, whereas surface energy scales with a.

It will be shown later that this can have serious implication on the stability of cracks, and on size effects

13.1.2 Brittle Material, Griffith’s Model

13 For a perfectly brittle material, we can rewrite Eq 13.3 as

Gdef= − ∂Π

∂a =

∂W

∂a − ∂U e

∂Γ

the factor 2 appears because we have two material surfaces upon fracture The left hand side represents

the energy available for crack growth and is given the symbol G in honor of Griffith Because G is

derived from a potential function, it is often referred to as the crack driving force The right hand side represents the resistance of the material that must be overcome for crack growth, and is a material constant (related to the toughness)

14 This equation represents the fracture criterion for crack growth, two limiting cases can be considered

They will be examined in conjunction with Fig 13.1 in which we have a crack of length 2a located in

an infinite plate subjected to load P Griffith assumed that it was possible to produce a macroscopical load displacement (P − u) curve for two different crack lengths a and a + da.

Two different boundary conditions will be considered, and in each one the change in potential energy

as the crack extends from a to a + da will be determined:

Fixed Grip: (u2 = u1) loading, an increase in crack length from a to a + da results in a decrease in stored elastic strain energy, ∆U ,

∆U = 1

2P2u1−1

= 1

kinetic energy and the internal energy) is equal to the sum of the rate of work done...

11 Around 192 0, Griffith was exploring the theoretical strength of solids by performing a series of

exper-iments on glass rods of various diameters He observed that... law of thermodynamics gives: The change in energy is proportional to the amount of work performed Since only the change of energy is involved, any datum can be used as a basis for measure of energy