本文主要内容包括:
- 1. Hill 应变度量 与 Seth 应变度量
- 2. Hill -Seth 应变度量的 Lagrange 描述
- 2.1. Green-Lagrange 应变张量
- 2.2. 物质 Biot 应变张量/工程应变
- 2.3. 右 Henkey 应变张量/Lagrange 型对数应变
- 2.4. Piola 应变张量
- 3. Hill -Seth 应变度量的 Euler 描述
- 3.1. Finger 应变张量
- 3.2. 左 Henkey 应变张量/Euler 型对数应变
- 3.3. 空间 Biot 应变张量
- 3.4. Almansi 应变张量
- 4. 应变协调方程
1. Hill 应变度量 与 Seth 应变度量
无论是立足于参考构型还是当前构型,某一代表性物质点领域内的变形(某方向的长度比、任意两方向夹角的变化、面元体元的改变)均可通过主长度比/主方向加以描述。因此,任何能够确定长度比/主方向的张量均可作为应变的度量,来描述代表性物质点邻域内的变形状态。基于上述观点, Rodney Hill 于 1968 年定义了 通类的 应变度量函数(应变张量的主轴表示):
- Lagrange 描述:
E H i l l = E = ∑ α = 1 3 f ( λ α ) L ⃗ α ⊗ L ⃗ α \bold{E}_{Hill}=\bold{E}=\sum_{\alpha =1}^3f(\lambda_\alpha)\vec{L}_\alpha\otimes\vec{L}_\alpha EHill=E=α=1∑3f(λα)Lα⊗Lα - Euler 描述:
e H i l l = e = ∑ α = 1 3 f ( λ α ) l ⃗ α ⊗ l ⃗ α \bold{e}_{Hill}=\bold{e}=\sum_{\alpha =1}^3f(\lambda_\alpha)\vec{l}_\alpha\otimes\vec{l}_\alpha eHill=e=α=1∑3f(λα)lα⊗lα
其中, f ( λ ) f(\lambda) f(λ) 为单调可微的标量函数,且满足:
(1) 主伸长比为1(无变形)时,应变为零:
f
(
1
)
=
0
f(1)=0
f(1)=0
(2) 应变是严格单调递增的函数(较大的主长度比对应更大的应变):
f
′
(
λ
)
>
0
f'(\lambda)>0
f′(λ)>0
(3) 小变形条件下,Hill 应变 可退化为已有的 柯西应变:
f
′
(
1
)
=
1
f'(1)=1
f′(1)=1
对条件(3)作如下说明:
1822 年 Cauchy 提出在小变形条件下,可用熟知的六个柯西应变分量来度量材料的线段变形与角度变形,又
λ
=
1
+
ε
\lambda=1+\varepsilon
λ=1+ε
其中,柯西主应变
ε
\varepsilon
ε 为一小量,那么
f
(
λ
)
=
f
(
1
+
ε
)
=
f
(
1
)
+
f
′
(
1
)
ε
+
o
(
ε
2
)
≈
f
′
(
1
)
ε
f(\lambda)=f(1+\varepsilon)=f(1)+f'(1)\varepsilon+o(\varepsilon^2)\approx f'(1)\varepsilon
f(λ)=f(1+ε)=f(1)+f′(1)ε+o(ε2)≈f′(1)ε
式中运用了无变形无应变的条件。可见当且仅当
f
′
(
1
)
=
1
f'(1)=1
f′(1)=1 时
f
(
λ
)
≈
ε
f(\lambda)\approx\varepsilon
f(λ)≈ε
从Hill应变度量的定义来看,
E
H
i
l
l
\bold{E}_{Hill}
EHill 或
e
H
i
l
l
\bold{e}_{Hill}
eHill 均为对称张量,且二者满足:
E
=
R
T
⋅
e
⋅
R
\bold E = \bold{R^T\cdot e\cdot R}
E=RT⋅e⋅R
其中,
R
\bold R
R 为转动张量。
印度力学家 Seth 提出如下形式的
f
f
f ,此时的应变称作 Seth 应变度量。
f
(
λ
)
=
{
1
2
n
(
λ
2
n
−
1
)
(
n
∈
R
,
n
≠
0
)
l
n
λ
(
n
=
0
)
f(\lambda)= \begin{cases} \dfrac{1}{2n}(\lambda^{2n}-1) &(n\in\mathbb{R},n\ne0) \\\\ ln\ \lambda&(n=0) \end{cases}
f(λ)=⎩
⎨
⎧2n1(λ2n−1)ln λ(n∈R,n=0)(n=0)
其中,
n
=
0
n=0
n=0 时的定义源自
n
≠
0
n\ne0
n=0 时的极限 (应用洛必达法则),即
lim
n
→
0
1
2
n
(
λ
2
n
−
1
)
=
lim
n
→
0
e
2
n
⋅
l
n
λ
−
1
2
n
=
lim
n
→
0
(
2
l
n
λ
)
e
2
n
⋅
l
n
λ
2
=
l
n
λ
\lim_{n\rightarrow0}\dfrac{1}{2n}(\lambda^{2n}-1) =\lim_{n\rightarrow0}\dfrac{e^{2n\cdot ln\lambda}-1}{2n} =\lim_{n\rightarrow0}\dfrac{(2ln\lambda)e^{2n\cdot ln\lambda}}{2} =ln\lambda
n→0lim2n1(λ2n−1)=n→0lim2ne2n⋅lnλ−1=n→0lim2(2lnλ)e2n⋅lnλ=lnλ
2. Hill -Seth 应变度量的 Lagrange 描述
Lagrange 描述下的 Seth 应变张量为:
E
=
{
∑
α
=
1
3
1
2
n
(
λ
α
2
n
−
1
)
L
⃗
α
⊗
L
⃗
α
=
1
2
n
(
C
n
−
I
)
=
1
2
n
(
U
2
n
−
I
)
(
n
∈
R
,
n
≠
0
)
∑
α
=
1
3
l
n
λ
α
L
⃗
α
⊗
L
⃗
α
≜
l
n
U
(
n
=
0
)
\bold E= \begin{cases} \sum\limits_{\alpha=1}^3\dfrac{1}{2n}(\lambda_\alpha^{2n}-1)\vec{L}_\alpha\otimes\vec{L}_\alpha =\dfrac{1}{2n}(\bold{C}^n-\bold I) =\dfrac{1}{2n}(\bold{U}^{2n}-\bold I)&(n\in\mathbb{R},n\ne0)\\\\ \sum\limits_{\alpha=1}^3ln\ \lambda_\alpha\ \vec{L}_\alpha\otimes\vec{L}_\alpha \triangleq ln\bold{U} &(n=0) \end{cases}
E=⎩
⎨
⎧α=1∑32n1(λα2n−1)Lα⊗Lα=2n1(Cn−I)=2n1(U2n−I)α=1∑3ln λα Lα⊗Lα≜lnU(n∈R,n=0)(n=0)
2.1. Green-Lagrange 应变张量
当
n
=
1
n=1
n=1 时,
E
(
1
)
=
1
2
(
C
−
I
)
\bold{E}^{(1)}=\dfrac{1}{2}(\bold{C-I})
E(1)=21(C−I)
对 Green-Lagrange 应变张量的几何意义 进行说明:
立足于参考构型,经历变形后,任意方向线元
L
⃗
\vec{L}
L 的长度比
λ
L
\lambda_{L}
λL 为:
(
d
s
d
s
0
)
2
=
L
⃗
⋅
C
⋅
L
⃗
\left(\dfrac{ds}{ds_0}\right)^2=\vec{L}\cdot\bold C\cdot\vec{L}
(ds0ds)2=L⋅C⋅L
则,
L
⃗
⋅
E
(
1
)
⋅
L
⃗
=
L
⃗
⋅
[
1
2
(
C
−
I
)
]
⋅
L
⃗
=
1
2
[
(
d
s
d
s
0
)
2
−
1
]
\vec{L}\cdot\bold E^{(1)}\cdot\vec{L} =\vec{L}\cdot\left[\dfrac{1}{2}(\bold{C-I})\right]\cdot\vec{L} =\dfrac{1}{2}\left[\left(\dfrac{ds}{ds_0}\right)^2-1\right]
L⋅E(1)⋅L=L⋅[21(C−I)]⋅L=21[(ds0ds)2−1]
在 物质坐标系 下,Green-Lagrange 应变张量可由位移进行表示:
E
(
1
)
=
1
2
(
F
T
⋅
F
−
I
)
=
1
2
[
(
I
+
∂
u
⃗
∂
X
A
⊗
G
⃗
A
)
T
⋅
(
I
+
∂
u
⃗
∂
X
B
⊗
G
⃗
B
)
−
I
]
=
1
2
(
▽
0
u
⃗
+
u
⃗
▽
0
+
▽
0
u
⃗
⋅
u
⃗
▽
0
)
\begin{aligned} & \bold{E}^{(1)} =\dfrac{1}{2}(\bold{F^T\cdot F-I}) \\\ \\ &\quad \ \ \ =\dfrac{1}{2}\left[\left(\bold I+\dfrac{\partial \vec{u}}{\partial X^A}\otimes\vec{G}^A\right)^T\cdot \left(\bold I+\dfrac{\partial \vec{u}}{\partial X^B}\otimes\vec{G}^B\right)-\bold I\right]\\\\ &\quad \ \ \ =\dfrac{1}{2}\left(\triangledown_0\vec{u}+\vec{u}\triangledown_0+\triangledown_0\vec{u}\cdot\vec{u}\triangledown_0\right) \end{aligned}
E(1)=21(FT⋅F−I) =21[(I+∂XA∂u⊗GA)T⋅(I+∂XB∂u⊗GB)−I] =21(▽0u+u▽0+▽0u⋅u▽0)
写成分量形式:
E
A
B
(
1
)
=
1
2
(
U
A
∣
B
+
U
B
∣
A
+
U
M
∣
A
U
M
∣
B
)
{E}^{(1)}_{AB}=\dfrac{1}{2}\left(U_A|_B+U_B|_A+U^M|_AU_M|_B\right)
EAB(1)=21(UA∣B+UB∣A+UM∣AUM∣B)
2.2. 物质 Biot 应变张量/工程应变
当
n
=
1
2
n=\dfrac{1}{2}
n=21 时,
E
(
1
2
)
=
U
−
I
\bold{E}^{(\frac{1}{2})}=\bold{U-I}
E(21)=U−I
2.3. 右 Henkey 应变张量/Lagrange 型对数应变
当
n
=
0
n=0
n=0 时,
E
(
0
)
=
l
n
U
=
∑
α
=
1
3
l
n
λ
α
L
⃗
α
⊗
L
⃗
α
\bold{E}^{(0)}=ln\bold U=\sum\limits_{\alpha=1}^3ln\ \lambda_\alpha\ \vec{L}_\alpha\otimes\vec{L}_\alpha
E(0)=lnU=α=1∑3ln λα Lα⊗Lα
2.4. Piola 应变张量
当
n
=
−
1
n=-1
n=−1 时,
E
(
−
1
)
=
1
2
(
I
−
C
−
1
)
\bold{E}^{(-1)}=\dfrac{1}{2}(\bold{I-\overset{-1}{C}})
E(−1)=21(I−C−1)
3. Hill -Seth 应变度量的 Euler 描述
Euler 描述下的 Seth 应变张量为:
e
=
{
∑
α
=
1
3
1
2
n
(
λ
α
2
n
−
1
)
l
⃗
α
⊗
l
⃗
α
=
1
2
n
(
B
n
−
I
)
=
1
2
n
(
V
2
n
−
I
)
(
n
∈
R
,
n
≠
0
)
∑
α
=
1
3
l
n
λ
α
l
⃗
α
⊗
l
⃗
α
≜
l
n
V
(
n
=
0
)
\bold e= \begin{cases} \sum\limits_{\alpha=1}^3\dfrac{1}{2n}(\lambda_\alpha^{2n}-1)\vec{l}_\alpha\otimes\vec{l}_\alpha =\dfrac{1}{2n}(\bold{B}^n-\bold I) =\dfrac{1}{2n}(\bold{V}^{2n}-\bold I)&(n\in\mathbb{R},n\ne0)\\\\ \sum\limits_{\alpha=1}^3ln\ \lambda_\alpha\ \vec{l}_\alpha\otimes\vec{l}_\alpha \triangleq ln\bold{V } &(n=0) \end{cases}
e=⎩
⎨
⎧α=1∑32n1(λα2n−1)lα⊗lα=2n1(Bn−I)=2n1(V2n−I)α=1∑3ln λα lα⊗lα≜lnV(n∈R,n=0)(n=0)
3.1. Finger 应变张量
当
n
=
1
n=1
n=1 时,
e
(
1
)
=
1
2
(
B
−
I
)
\bold{e}^{(1)}=\dfrac{1}{2}(\bold{B-I})
e(1)=21(B−I)
3.2. 左 Henkey 应变张量/Euler 型对数应变
当
n
=
0
n=0
n=0 时,
e
(
0
)
=
l
n
V
=
∑
α
=
1
3
l
n
λ
α
l
⃗
α
⊗
l
⃗
α
\bold{e}^{(0)}=ln\bold V=\sum\limits_{\alpha=1}^3ln\ \lambda_\alpha\ \vec{l}_\alpha\otimes\vec{l}_\alpha
e(0)=lnV=α=1∑3ln λα lα⊗lα
3.3. 空间 Biot 应变张量
当
n
=
−
1
2
n=-\dfrac{1}{2}
n=−21 时,
e
(
−
1
2
)
=
I
−
V
−
1
\bold{e}^{(-\frac{1}{2})}=\bold{I-V^{-1}}
e(−21)=I−V−1
3.4. Almansi 应变张量
当
n
=
−
1
n=-1
n=−1 时,
e
(
−
1
)
=
1
2
(
I
−
B
−
1
)
=
1
2
(
I
−
c
)
\bold{e}^{(-1)} =\dfrac{1}{2}(\bold{I-\overset{-1}{B}}) =\dfrac{1}{2}(\bold{I-\bold c})
e(−1)=21(I−B−1)=21(I−c)
对 Almansi 应变张量的几何意义 进行说明:
立足于当前构型,经历变形后,任意方向线元
l
⃗
\vec{l}
l 的长度比
λ
l
\lambda_{l}
λl 为:
(
d
s
d
s
0
)
2
=
l
⃗
⋅
c
⋅
l
⃗
\left(\dfrac{ds}{ds_0}\right)^2=\vec{l}\cdot\bold c\cdot\vec{l}
(ds0ds)2=l⋅c⋅l
则,
l
⃗
⋅
e
(
−
1
)
⋅
l
⃗
=
l
⃗
⋅
[
1
2
(
I
−
c
)
]
⋅
l
⃗
=
1
2
[
1
−
(
d
s
d
s
0
)
2
]
\vec{l}\cdot\bold e^{(-1)}\cdot\vec{l} =\vec{l}\cdot\left[\dfrac{1}{2}(\bold{I-c})\right]\cdot\vec{l} =\dfrac{1}{2}\left[1-\left(\dfrac{ds}{ds_0}\right)^2\right]
l⋅e(−1)⋅l=l⋅[21(I−c)]⋅l=21[1−(ds0ds)2]
在 空间坐标系 下,Almansi 应变张量可由位移进行表示:
e
(
−
1
)
=
1
2
(
I
−
F
−
T
⋅
F
−
1
)
=
1
2
[
I
−
(
I
−
∂
u
⃗
∂
x
i
⊗
g
⃗
i
)
T
⋅
(
I
−
∂
u
⃗
∂
x
i
⊗
g
⃗
i
)
]
=
1
2
(
▽
u
⃗
+
u
⃗
▽
−
▽
u
⃗
⋅
u
⃗
▽
)
\begin{aligned} & \bold{e}^{(-1)} =\dfrac{1}{2}(\bold{I}-\overset{-T}{\bold F}\cdot\overset{-1}{\bold F})\\\\ &\quad\ \ \ =\dfrac{1}{2}\left[\bold{I}-\left(\bold I-\dfrac{\partial \vec{u}}{\partial x^i}\otimes\vec{g}^i\right)^T\cdot\left(\bold I-\dfrac{\partial \vec{u}}{\partial x^i}\otimes\vec{g}^i\right)\right] \\\\ &\quad\ \ \ =\dfrac{1}{2}\left(\triangledown\vec{u}+\vec{u}\triangledown-\triangledown\vec{u}\cdot\vec{u}\triangledown\right) \end{aligned}
e(−1)=21(I−F−T⋅F−1) =21[I−(I−∂xi∂u⊗gi)T⋅(I−∂xi∂u⊗gi)] =21(▽u+u▽−▽u⋅u▽)
分量形式为:
e
i
j
(
−
1
)
=
1
2
(
u
i
∣
j
+
u
j
∣
i
−
u
k
∣
i
u
k
∣
j
)
e^{(-1)}_{ij}=\dfrac{1}{2}(u_i|_j+u_j|_i-u^k|_iu_k|j)
eij(−1)=21(ui∣j+uj∣i−uk∣iuk∣j)