71
Chap 3. The Theory of PLAIN Concrete
3.1 Statical condition
= equilibrium cond. + boundary cond.
y
y
dy
y σ + ∂ σ
∂
Taylor’s series' '
( ) ( )
2( ) ( )
1! 2!
f a f a
f a + h = f a + h + h
3.2 Geometrical condition
-> displacement – strain relationship
=compatibility condition
σ
1σ
2σ
1σ
2z x
y
σ
xd
xd
yd
z xx x
x
σ d σ + ∂
∂
z x
y
σ
xd
xd
yd
z xx x
x
σ d σ + ∂
∂
z
z z
z
σ d σ + ∂
∂
x
x x
x
σ d σ + ∂
∂
y
y y
y
σ d σ + ∂
∂
z
z z
z
σ d σ + ∂
∂
x
x x
x
σ d σ + ∂
∂
y
y y
y
σ d σ + ∂
∂
72
3.3 Virtual Work
The equation in an identity if the statical conditions are fulfilled and is then valid for any displacement field satisfying the geometrical condition
if on the contrary, the equation is know to be valid for any displacement field, the statical conditions are fulfilled.
3.4 Constitutive equations (by normality rule)
i
i
q λ f θ
= ∂
∂
3.5 Plastic strains in coulomb material
1 3 c
2
k σ σ − = f = c k
Noteσ
3≤ σ
2≤ σ
1Where
2 2
1 sin 1 sin
( ) ( ) tan ( )
cos 1 sin 4 2
k ϕ ϕ π ϕ
ϕ ϕ
+ +
= = = +
−
Six yield conditions1 3 3 2 1
1. k σ σ − = f
c: σ ≤ σ ≤ σ
3 1 1 2 3
2. k σ σ − = f
c: σ σ ≤ ≤ σ
1 2 2 3 1
3. k σ σ − = f
c: σ ≤ σ ≤ σ
2 1 1 3 2
4. k σ σ − = f
c: σ σ ≤ ≤ σ
2 3 3 1 2
5. k σ σ − = f
c: σ ≤ σ σ ≤
3 2 2 1 3
6. k σ σ − = f
c: σ ≤ σ σ ≤
θ
2q
iθ
2q
i73 By normality rule
i
i
q f λ ∂ Q
= ∂
1 2 3
1
1. f , 0 ,
ε λ λ k ε ε λ
σ
= ∂ = = = −
∂
1 2 3
2. ε = − λ , ε = 0 , ε = λ k
1 2 3
3. ε = λ k , ε = − λ , ε = 0
1 2 3
4. ε = − λ , ε = λ k , ε = 0
1 2 3
5. ε = 0 , ε = λ k , ε = − λ
1 2 3
6. ε = 0 , ε = − λ , ε = λ k
1 2 3
( k 1)
ε ε ε λ
∴ + + = −
Along the edges, plastic strain vectors
1/5 linear combination of plastic strain on planes 1 & 5
1
k
ε λ =
ε
2= + 0 λ
2k ε
3= − − λ λ
1 24/5 edge :
ε
1= − λ
1ε
2= ( λ λ
1+
2) ε
3= − λ
22/4 edge :
ε
1= − ( λ λ
1+
2) ε
2= λ
2k ε
3= λ
1k
2/6 edge :
ε
1= − λ
1ε
2= − λ
2ε
3= ( λ λ
1+
2)k
3/6 edge :
ε
1= − λ
1k ε
2= − ( λ λ
1+
2) ε
3= − λ
2k
2/6 edge :
ε
1= ( λ λ
1+
2)
ε
2= − λ
2ε
3= − λ
1At the apex
1 2 3
1 f
cσ = σ = σ = k
−
From (3.4.4)
6
1 1 3 2 4 1
1
( ) k ( )
ε = λ λ + − λ λ + = ε
∑
6
2 4 5 3 6 2
1
( ) k ( )
ε = λ λ + − λ λ + = ε
∑
6
3 2 6 1 5 3
1
( ) k ( )
ε = λ λ + − λ λ + = ε
∑
Except for the apex, on planes
74
ε λ
+= k
∑
,∑ ε
−= λ k
On edges
1 2
( )k
ε
+= λ λ +
∑
,∑ ε
−= + λ λ
1 2∴ k
ε ε
+
−
∑ =
∑
On apex
k ε ε
+
−
∑ ≥
∑
Volume change
ε ε ε
1+ + =
2 3?
On planes
ε ε ε
1+ + =
2 3λ ( k − 1)
On edges
ε ε ε
1+ + =
2 3k ( λ λ
1+
2) ( − λ λ
1+
2)
Since
1 sin
1 sin k
ε ϕ
ε ϕ
+
−
= = +
−
∑
∑
Plastic strains of coulomb’s material
k ε ε
+
−
∑ =
∑
except for apex1 sin
1 sin
k ϕ
ϕ
= +
−
-①Volume change
1 2 3
( k 1)
ε ε ε + + = λ −
on plane -②1 2 3
k (
1 2) (
1 2)
ε ε ε + + = λ λ + − λ λ +
on edge -③From ①
1 sin 1 sin k
ε ϕ
ε ϕ
+
−
= = +
−
∑
∑
75 Eq ② is rewritten as
1 2 3
2 sin ( 1)
1 sin
k ϕ
ε ε ε ε ε
ϕ
− −
+ + = − =
∑ ∑ −
1 2 3
2 sin 1 1 sin k
k
ε ε ε ϕ ε
ϕ + + =
−− ∑
=
2 sin 1 sin 1 sin 1 sin
ϕ ϕ ε ε
ϕ ϕ ε
+
−
−
−
− + ∑ ∑ ∑
=
2 sin 1 sin
ϕ ε ϕ
+
+ ∑
1 2 3
( ε ε + + ε )(1 sin ) − ϕ = ∑ ε
−2sin ϕ
-④1 2 3
( ε ε + + ε )(1 sin ) + ϕ = ∑ ε
+2sin ϕ
-⑤④ + ⑤ Æ
1 2 3
2( ε ε + + ε ) = 2sin ( ϕ ∑ ε
−+ ∑ ε
+)
1 2 3
1 2 3
sin ϕ ε ε ε
ε ε ε + +
= + +
3.4.2 Dissipation formula for coulomb material along planes
- along the planes
1 1 2 2 3 3
W = σ ε σ ε σ ε + +
1
( k ) 0
3( )
σ λ σ λ
= + + −
1 3
( k ) f
cλ σ σ λ
= − =
-along the edges
1 1 2 2 3 3
W = σ ε σ ε σ ε + +
1 1 2 2 3
(
1 2)
W = σ λ k + σ λ k − σ λ λ +
1 1 2 2 1 1 2
( ) (
c)( )
k σ λ σ λ k σ f λ λ
= + − − +
2
(
1 2)
c(
1 2)
k λ σ σ f λ λ
= − + +
Since
σ σ
1=
21 2
( )
W = f
cλ λ +
76 -at apex
1 1 2 2 3 3
W = σ ε σ ε σ ε + +
Since 1 2 3
cot
1 f
ck c
σ = σ = σ = = ψ
−
1 2 3
( )
1 f
cW = k ε ε + + ε
−
1 2 3 4 5 6
( )
1 f
ck λ λ λ λ λ λ
= + + + + +
−
=summary
k
ε ε
+
−
= ∑
∑
,ε λ k
+
=
∑
,∑ ε
−= λ
c c c
W f f f
k ε
λ ε
+
= = ∑ =
−∑
for plane1 2
( )
c c cW f f f
k ε
λ λ ε
+
= + = ∑ =
−∑
for edge6 yield surfaces
1 3
1. k σ σ − − = f
c0
3 3
2. k σ σ − − = f
c0
1 2
3. k σ σ − − = f
c0
2 1
4. k σ σ − − = f
c0
2 3
5. k σ σ − − = f
c0
3 2
6. k σ σ − − = f
c0
Plane strain Æ
ε
3=01 2 3
ε ε ε
1 ( λ k , 0, − λ ) 2 ( − λ , 0, λ k ) 3 ( λ k , − λ , 0)
(−
λ λ
, k, 0)(−λ λ, ,o k)
(0,−λ λ, k)
(λk,−λ, 0)
( ,0, λ
k− λ )
(0,λ
k,−λ
)⑥
②
⑤
④
③
①
(−
λ λ
, k, 0)(−λ λ, ,o k)
(0,−λ λ, k)
(λk,−λ, 0)
( ,0, λ
k− λ )
(0,λ
k,−λ
)⑥
②
⑤
④
③
①
f
ck
f
ck
f
cf
c( , )
1 1
c c
f f
k− k−
f
ck
f
ck
f
cf
c( , )
1 1
c c
f f
k− k−
77
4 ( − λ , λ k , − λ ) 5 (0, λ k , − λ ) 6 (0, − λ , λ k )
3.4.3 & 3.4.4 Plastic strains & dissipation formulas for modified coulomb material
Plastic strains by normality rule
plane 1 2 3 4 5 6 7 8 9
ε
1λ k − λ λ k − λ
0 0λ
0 0ε
2 0 0− λ λ k λ k − λ
0λ
0ε
3− λ λ k
0 0− λ λ k
0 0λ
9 yield surfaces
1 3
1. k σ σ − − = f
c0
3 3
2. k σ σ − − = f
c0
1 2
3. k σ σ − − = f
c0
2 1
4. k σ σ − − = f
c0
2 3
5. k σ σ − − = f
c0
3 2
6. k σ σ − − = f
c0 7. σ
1− = f
t0 8. σ
2− = f
t0 9. σ
3− = f
t0
⑥
②
⑤
④
③
①
⑨
⑦ ⑧
σ
1σ
2σ
3⑥
②
⑤
④
③
①
⑨
⑦ ⑧
σ
1σ
2σ
378
Edge 1/5 4/5 2/4 2/6 3/6
ε
1λ
1k − λ
1− ( λ λ
1+
2) − λ
1λ
1k ε
2λ
2k ( λ λ
1+
2)k λ
2k λ
2− ( λ λ
1+
2) ε
3− ( λ λ
1+
2) − λ
2− λ
2( λ λ
1+
2)k λ
2k
Edge 1/3 1/7 3/7 6/9 2/7
ε
1( λ λ
1+
2)k λ
1k + λ
2λ
1k + λ
2 0− λ
1ε
2− λ
2 0− λ
1− λ
1 0ε
3− λ
1− λ
1 0λ
1k λ
1k + λ
2Edge 4/8 5/8 8/7 7/9 8/9
ε
1− λ
1 0λ
1λ
1 0ε
2λ
1k λ
1k + λ
2λ
2 0λ
1ε
3 0− λ
1 0λ
2λ
2W along edges
1/7 edge
1
(
1 2)
3(
1)
1(
1 3)
2 1W = σ λ k + λ + σ − λ = λ σ k − σ + λ σ = λ
1f
c+ λ
2f
t8/7 edge
1 1 2 2
(
1 2)
tW = σ λ σ λ + = λ λ + f
Apex
ε
1ε
2ε
34/5/8
− λ
1k ( λ λ
1+
2) + λ
3− λ
27/8/5/1
λ
1k + λ
3λ
2k + λ
4− ( λ λ
1+
2)
1/3/7
k ( λ λ
1+
2) + λ
3− λ
2− λ
13/7/6/9
λ
1k + λ
2− ( λ λ
1+
2) λ
2k + λ
42/6/9
− λ
1− λ
2k ( λ λ
1+
2) + λ
38/9/2/4
− ( λ λ
1+
2) λ
2k + λ
3λ
1k + λ
47/8/9
λ
1λ
2λ
3W at apex 4/5/8 apex
79
1 1 1 2 3 2 3 2
1 1 2 2 3 2 2 3
1 2 3
[ ( ) ] ( )
( ) ( ) ( )
( )
c t
W k
k k
f f
σ λ λ λ λ σ σ λ λ σ σ λ σ σ σ λ
λ λ λ
= − + + + + −
= − − − − +
= + +
7/8/5/1 apex
1 1 3 2 4 2 3 2 1
1 1 3 2 2 3 1 3 2 4
1 2 3 4
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )
c t
W k k
k k
f f
σ λ λ λ λ σ σ λ λ λ σ σ λ σ σ σ λ σ λ
λ λ λ λ
= − + + + + − −
= − + − + +
= + + +
1/3/9 apex
1 2 3
( )
c t
W = f λ λ + + f λ
3/7/6/9 apex
1 2 3 4
( ) ( )
c t
W = f λ λ + + f λ λ +
( )
c t
W = f ∑ ε
−+ f ∑ ε
+− k ∑ ε
− for plane stress In the apex, we havek ε ε
+
−
∑ ≤
∑
(Proof)
k 0 ε
ε
+
−
−≥
∑ ∑
At apex for coulomb material
1 2 3
1 3 2
2 3 1
1 2 3
2 3 1
3 1 2
, 0, 0
, 0, 0
, 0, 0
0, , 0
0, , 0
0, , 0
ε ε ε ε ε ε ε ε ε ε ε ε ε ε ε ε ε ε
≥ ≤
≥ ≤
≥ ≤
≤ ≥
≤ ≥
≤ ≥
80 Î 1 2 3
k ε ε + + ε
2 4 5 6
1 3 4 5
(
1 5) (
3 6)k
k λ λ λ λ
λ λ λ λ + + + λ λ λ λ
= + + + − − + + +
3 4 2 6
1 1
( )(1 ) ( )( k ) 1
k k
λ λ λ λ
= + − + + − ≥
Dissipation formula
1 1 2 2 3 3
W = σ ε σ ε σ ε + +
For coulomb material On plane and edges
c c
W f f
ε
−k ε
+= ∑ = ∑
At apex
1 2 3
cot ( )
W = c ϕ ε ε ε + +
For modified coulomb material
( )
c t
W = f ∑ ε
−+ f ∑ ε
+− k ∑ ε
−3.4.5 Planes and lines of discontinuity = yield line = failure line
I II
I II
△t
Con’c dissipation
Work?
thin layer
I II
I II
△t
Con’c dissipation
Work?
thin layer
t
before
after
movement
I
II
n α
u
δ
t
before
after
movement
I
II
n α
u
δ
81
n
sin
u = u α
t
cos u = u α
n
sin
n
u u
ε α
δ δ
= =
cos 0
t
u α
ε = =
∞ cos
nt
r u α
= δ
@A (0,
cos 2
u α
δ
)@B (
u sin α δ
,cos 2
u α
− δ
)1 n
R
ε = ε +
2 n
R
ε = − ε
Where 1 2
2 2
n t
m
ε ε ε = ε ε + = +
2 2
( ) 2
nt m
R = ε + r
2 2 2 2
2 2
sin cos
4 4 2
u α u α u
δ δ δ
= + =
1
sin
2 2
u α u
ε = δ + δ
2
sin
2 2
u α u
ε = δ − δ
Æ
(sin 1) 2
u α
δ ±
ε
2ε
12Θ
A
( , ) 2
nt t
ε γ
ε 2
γ
( , )
2
nt n
ε −γ
B
82
cos
tan 2 2 cot tan( )
sin 2
2 u u
α π
θ δ α α α
δ
= = = −
If the 1st principal strain coincides with the 1st principal stress
n 2
n 2
83 Plane strain
1) Coulomb material
k ε ε
+
−
∑ =
∑
1 2
(sin 1) 2
(sin 1) 2
u u k ε δ α
ε α
δ
= + =
− −
1 sin 1 sin 1 sin 1 sin
k α ϕ
α ϕ
+ +
= =
− −
tan α tan ϕ
∴ = ±
For
α ϕ =
,α π ϕ = −
W = f
c∑ ε
−(1 sin )
c
2
f u ϕ
= δ −
Dissipation per unit length
W
l= Wb
δ(1 sin )
c
2
f ub ϕ
= −
(1 sin )
c
2 f ub
k ϕ
= +
σ2 ε2
σ1 ε1 σ2 ε2
σ1 ε1
ϕ ϕ
at the apex
ϕ
ϕ
at the apex
84 Since
f
c= 2 c k
2 (1 sin )
l
2
W = c k ub − ϕ
For
ϕ α π ϕ ≤ ≤ −
1 1 1
cot ( )
W = c ϕ ε ε ε + +
1 1 1
cot ( )
W = c ϕ ε ε ε + + cot sin W
l= cub ϕ α α π ϕ = −
Plane strain
2) Modified coulomb material
( )
c c
W = f ∑ ε
−+ f ∑ ε
+− ∑ ε
−(1 sin ) [1 sin (1 sin )]
2 2
c c
u u
f α f α k α
δ δ
= − + + − −
(1 sin ) [ ( 1) ( 1) sin )]
2 2
c c
u u
f α f k k α
δ δ
= − + − − + +
Where
1 sin
1 sin 4
k α
α
= + =
−
for concrete
2 sin
1 1 sin
k α
− = α
−
,1 2
1 sin k + = − α
Plane stress
1) Coulomb material
(1 sin ) 2
ε u α
δ
−
= −
∑
For
α ϕ ≤
,α π ϕ ≤ −
σ2 ε2σ1 ε1
α π ϕ = −
α π =
ϕ α π ϕ ≤ ≤ −
σ2 ε2σ1 ε1
α π ϕ = −
α π =
ϕ α π ϕ ≤ ≤ −
85
W = f
c∑ ε
−1 (1 sin )
l
2
cW = f ub − α
For
α π ϕ ≤ −
1 [1 sin {( 1) ( 1) sin }]
2
t
l c
c
W f ub f k k
α f α
= − + − + +
σ2 ε2
σ1 ε1 α π ϕ= −
α π= ϕ α π ϕ≤ ≤ −
II: sliding II: sliding III : separation
I: crushing 0≤ ≤α ϕ
after
III II
ϕ ϕ σ2 ε2
σ1 ε1 α π ϕ= −
α π= ϕ α π ϕ≤ ≤ −
II: sliding II: sliding III : separation
I: crushing 0≤ ≤α ϕ
after
III II
ϕ ϕ
