Mechanics.Of.Materials.Saouma Episode 4 pps
Mechanics of Materials 2010 Part 4 ppsx
...
J+J
c
)
(4. 73)
or:
E
11
=
∂u
1
∂X
1
(4. 74- a)
E
12
=
1
2
∂u
1
∂X
2
+
∂u
2
∂X
1
(4. 74- b)
··· = ··· (4. 74- c)
Note the similarity with Eq. 4. 7.
Victor Saouma Mechanics of Materials II
Draft
4. 3 Strain ... T
0
n
0
(4. 149 )
and for which
df = t
0
dA
0
= tdA ⇒ t
0
=
dA
ddA
0
t (4. 150)
using Eq. 4. 147 -b and 4. 149 the preceding equation becomes
T
0
n
0
=
dA
dA
0
Tn =...
Mechanics.Of.Materials.Saouma Episode 1 doc
... B
12
C
11
D
21
+ B
12
C
12
D
22
A
12
= B
11
C
11
D
11
+ B
11
C
12
D
12
+ B
12
C
11
D
21
+ B
12
C
12
D
22
A
21
= B
21
C
11
D
11
+ B
21
C
12
D
12
+ B
22
C
11
D
21
+ B
22
C
12
D
22
A
22
= B
21
C
21
D
11
+ ... Mechanics 1
11. 2 Stress Intensity Factors 2
11 .3 Fracture Properties of Materials 10
11 .4 Examples 11
11 .4 .1 Example...
Mechanics.Of.Materials.Saouma Episode 2 potx
... σ
(2)
and n
2
i
be the direction
cosines of this axis,
2n
2
1
+ n
2
2
+ n
2
3
=0
n
2
1
− n
2
2
+2n
2
3
=0
n
2
1
+2n
2
2
− n
2
3
=0
⇒ n
2
1
=
1
√
3
; n
2
2
= −
1
√
3
; n
2
3
= −
1
√
3
(2. 44)
Finally, ... (2. 27)
II
σ
= −(σ
11
σ
22
+ σ
22
σ
33
+ σ
33
σ
11
)+σ
2
23
+ σ
2
31
+ σ
2
12
(2. 28)
=
1
2
(σ
ij
σ
ij
− σ
ii
σ
jj
)=
1
2
σ
ij
σ
ij
−
1
2
I...
Mechanics.Of.Materials.Saouma Episode 3 docx
... e
2
]+x
2
1
x
2
[e
1
⊗ e
3
]
+x
2
2
x
3
[e
2
⊗ e
1
]+2x
1
x
2
x
3
[e
2
⊗ e
2
]+x
1
x
2
2
[e
2
⊗ e
3
] (3. 39-a)
+x
2
x
2
3
[e
3
⊗ e
1
]+x
1
x
2
3
[e
3
⊗ e
2
]+2x
1
x
2
x
3
[e
3
⊗ e
3
]
= x
1
x
2
x
3
2 x
1
/x
2
x
1
/x
3
x
2
/x
1
2 ... (3. 27-a)
=8i − j − 10k at (1, −2, −1) (3. 27-b)
n =
2i − j − 2k
(2)
2
+(−1)
2
+(−2)
2
=
2
3
i −
1
3
j −
2
3
k (3. 27-c)
∇φ...
Mechanics.Of.Materials.Saouma Episode 4 pps
... X
2
plane.
Solution:
[F]=
λ
1
00
00−λ
3
0 λ
2
0
(4. 43-a)
det F = λ
1
λ
2
λ
3
(4. 43-b)
∆V = λ
1
λ
2
λ
3
(4. 43-c)
∆A
0
= 1 (4. 43-d)
n
0
= −e
3
(4. 43-e)
∆An = (1)(det F)(F
−1
)
T
(4. 43-f)
= λ
1
λ
2
λ
3
1
λ
1
00
00−
1
λ
3
0
1
λ
2
0
0
0
−1
=
0
λ
1
λ
2
0
(4. 43-g)
∆An ... 4. 101 du =(u∇
X
)
P
dX in the direction of −X
2
or
{du} =
41 0
012
0...
Mechanics.Of.Materials.Saouma Episode 5 ppsx
... P
2
(5. 5-a)
C
A·dr = 0 along a closed contour line (5. 5-b)
5. 3 Integration by Parts
5 The integration by part formula is
b
a
u(x)v
(x)dx = u(x)v(x)|
b
a
−
b
a
v(x)u
(x)dx
(5. 6)
5. 4 Gauss; ... deformation to the corresponding spatial vector dx at x. Thus, if we let
d
˜
f =
˜
tdA
0
(4. 155 -a)
and
df = Fd
˜
f (4. 155 -b)
where d
˜
f is the pseudo differential force which tra...
Mechanics.Of.Materials.Saouma Episode 6 pps
... =
V
T:
˙
εdV +
V
v·(∇·T)dV (6. 35-b)
41 We now evaluate P
ext
in Eq. 6. 31
P
ext
=
S
t
i
v
i
dS +
V
ρv
i
b
i
dV (6. 36- a)
=
V
v·(ρb + ∇·T)dV +
V
T:
˙
εdV (6. 36- b)
Using Eq. 6. 17 (T
ij,j
+ ρb
i
= ... publish his law when first
discovered it in 166 0. Instead he published it in the form of an anagram “ceiinosssttuu” in 167 6 and
the solution was given in 167 8. Ut tensio sic vi...
Mechanics.Of.Materials.Saouma Episode 7 docx
... any deformation W>0thus
λ +
2
3
G ≡ K>0 (7. 47- a)
G>0 (7. 47- b)
ruling out K = G = 0, we are left with
E>0; −1 <ν<
1
2
(7. 48)
37 The isotropic strain energy function can be alternatively ... substituting into Eq. 7. 37 we
obtain:
σ
xx
σ
yy
τ
xy
=
1
1 − ν
2
1 ν 0
ν 10
00
1−ν
2
ε
xx
ε
yy
γ
xy
(7. 55-a)
ε
zz
= −
1
1 − ν
ν(ε
xx
+ ε
yy
) (...
Mechanics.Of.Materials.Saouma Episode 8 potx
... −
K
III
(2πr)
1
2
sin
θ
2
(10. 38- a)
τ
yz
=
K
III
(2πr)
1
2
cos
θ
2
(10. 38- b)
σ
xx
= σ
y
= σ
z
= τ
xy
= 0 (10. 38- c)
w =
K
III
à
2r
1
2
sin
2
(10. 38- d)
u = v = 0 (10. 38- e)
where =34 for plane strain, ... −
a
2
r
2
+
1+3
a
4
r
4
−
4a
2
r
2
1
2
σ
0
cos 2θ (9. 78- a)
σ
θθ
=
σ
0
2
1+
a
2
r
2
−
1+
3a
4
r
4
1
2
σ
0
cos 2θ (9. 78- b)
σ
rθ
= −
1 −
3a
4
r
4
+
2a
2
r
2
1
2...
Mechanics.Of.Materials.Saouma Episode 9 potx
... .90 29 . 797 1
1.60 1.1 899 1.0571 .92 42 .92 42 .8264
1.80 1.1476 1.0 495 .95 13 .95 13 .8677
2.00 1.11 49 1.04 09 .96 70 .96 70 . 895 7
2.20 1. 090 4 1.0336 .97 68 .97 68 .91 54
2.50 1.06 49 1.0252 .98 55 .98 55 .93 58
3.00 ... .6 898 .6 898 .5688
1.20 1.2208 .98 51 .7 494 .7 494 .6262
1.25 1.2405 1.0168 . 792 9 . 792 9 .6701
1.30 1.2457 1.0358 .82 59 .82 59 .7053
1.40 1.2350 1.0536 ....
Mechanics.Of.Materials.Saouma Episode 10 pptx
... 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
a
a
o
a =10
6
m =1àm
a
o
=1
A = =10
10
m
theor
max
=2
act
cr
10
6
10
10
= 200
act
cr
(12.29)
21 ... 000a
0
σ
act
cr
=
Eγ
4a
γ =
Ea
0
10
σ
act
cr
=
E
2
40
a
o
a
a
a
0
=2, 500
σ
act
cr
=
E
2
100 ,000
=
E
100
√
10
(12.28)
19 Thus if we set a flaw si...
Mechanics.Of.Materials.Saouma Episode 11 pdf
... and σ
min
= 50 MPa. The aircraft is made up of aluminum which has
R =15
kJ
m
2
,E=70GPa, C =5ì 10
11
m
cycle
,andn = 3. The smallest detectable aw is 4mm. How
long would it be before the crack will ... can be rewritten as :
∆N =
∆a
C [∆K(a)]
n
(15.3)
or
N =
dN =
a
f
a
i
da
C [∆K(a)]
n
(15.4)
11 Thus it is apparent that a small error in the SIF calculations would be magnified greatly as...
Mechanics.Of.Materials.Saouma Episode 12 ppt
... c
2233
E
22
E
33
+2c
2223
E
22
E
23
+2c
2231
E
22
E
31
+2c
2 212
E
22
E
12
+
1
2
c
3333
E
2
33
+2c
3323
E
33
E
23
+2c
3331
E
33
E
31
+2c
3 312
E
33
E
12
+2c
2323
E
2
23
+4c
2331
E
23
E
31
+4c
2 312
E
23
E
12
+2c
3131
E
2
31
+4c
3 112
E
31
E
12
+2c
121 2
E
2
12
(18.31)
we ... W and obtain for instance
T
12
=
∂W
∂E
12
=2c
6
+ c
1 112
E
11
+ c
2 212
E
22
+ c
3 312
E
33
+ c...
Mechanics.Of.Materials.Saouma Episode 13 pps
... (hardening)
uniformly (see figure 19 .13) .
s
1
s
2
s
3
n+1
n+1
(Hardening)
F = 0 for
p
n
F = 0 for
(Perfectly Plastic)
H = 0
(Softening)
F = 0 for
H < 0
p
p
H > 0
Figure 19 .13: Isotropic Hardening/Softening
1. ... φ
(19.63)
Victor Saouma Mechanics of Materials II
Draft
19.4 Plastic Yield Conditions (Classical Models) 13
c Cot
π Plane
3
2
1
σ
σ
ξ
ρ
σ
Φ
σ
1
’
ρ
ty
ρ
cy
σ
t
σ
t
σ
1...
Mechanics.Of.Materials.Saouma Episode 14 ppsx
... elastic stiffness E
o
to the secant stiffness E
s
due to material damage is defined by
E
s
=(1−D)E
o
(20 .14)
where the damage parameter D provides a measure of the reduction of the stiffness due to the
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