...
xz
zx
zy
yz
yx
xy
zxx
z
yz
z
y
xyy
x
∂∂
γ∂
∂
ε∂
∂
ε∂
∂∂
γ∂
∂
ε∂
∂
ε∂
∂∂
γ∂
∂
ε∂
∂
ε∂
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
=+
=+
=+
(1 .2. 11.1b)
The compatibility relations imply that the displacements within the material
vary smoothly throughout the specimen. Solutions to problems ...
Equilbrium
(Stable)
U
g
= 4gc
8c
3
E
d
3
h
2
U
s
=
Energy (U)
Crack
length
c
d
c
P
h
(a)...
... of obtaining 0, 1, or 2 heads in two tosses of a coin.
f(X=x)
x
0 1 2
0.5
0 .25
F(X£ x)
x
0 1 2
1.0
0.5
0 .25
Delayed Fracture in Brittle Solids
58
()
2
1
22
1
2
1
2
−
−
⎟
⎠
⎞
⎜
⎝
⎛
−πσ=σ
n
n
C
n
afp
YnDKt
... normal to the
crack. The total stress intensity factor is given by integrating Eq. 2. 5.2c with
F
replaced by
dF = σ(b)db.
12
1
12
22
0
2( )σ
=
π...
...
109
P
dt
d
m −=
2
2
δ
(6.3.1)
Equating Eqs. 6 .2. 1f and 6.3.1,
6
multiplying both sides by velocity and inte-
grating, we obtain:
*21 25
2
*21 23
*21 23
*21 23
3
4
5
2
2
1
3
4
3
4
3
4
ERmv
dERdvmv
dt
d
ER
dt
dv
mv
ER
dt
dv
m
o
v
v
o
δ
δδ
δ
δ
δ
=
−=
−=
−=
∫∫
... distribution r < a
Sphere
1
2
3
21
2
2
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−−=
a
r
p
m
z
σ
Cylinder 2- D
1
2
21
2...
... full plasticity, then:
2. 2
23 .0
1
2
=
∴=
C
p
Y
m
(9.3.4.2a)
The theory appears to be inconsistent with their requirement that the pressure
distribution across the contact area be unchanged ... finite value of the contact radius at the initiation of
yield, we obtain:
() ()
()
2 2
2 2
21 * 1
1ln 4 12
32 61
paaaa
EY
YaR
aa
⎡ ⎤
⎛⎞
⎛⎞
⎡⎤
⎛⎞
⎢ ⎥
⎜⎟
=+ − +−
⎜⎟
⎢⎥
⎜⎟
⎜⎟
⎜⎟
′−...