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[mathematics] [continuum mechanics] symmetry tensor, strain tensor and stress tensor in fluid mechanics

2022-06-10 23:52:00 beidou111

The origin of the problem

stay splishsplash Thesis tutorial among , When it comes to stickiness , There is such a page PPT
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This is the basic knowledge of fluid mechanics , That is, it is constructed by Newton constitutive model NS equation .

Let's first look at the formula on the left

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It is called Cauchy momentum equation . This equation is applicable to both fluid mechanics and solid mechanics . Therefore, it can be regarded as NS The existence of more fundamental equations . It's actually a linear momentum equation . The acceleration on the left , On the right is internal power ( Expressed by stress , Is surface force ) And external forces ( Usually physical strength ).

among T It's the stress tensor .

Let's look at the formula on the right .

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This formula is Newton's constitutive equation . Equations constructed for Newtonian fluids . If it wasn't Newtonian fluid , Not applicable . The so-called constitutive equation , Is to find the relationship between stress and strain . actually , In terms of Solid Mechanics , The second equation is a geometric equation . The first equation is the physical equation , Or constitutive equation . We don't distinguish between , It's called his constitutive equation .

however , Our focus is not on the equation itself . Our aim is to make readers Be familiar with these mathematical signs The meaning of . in other words , We from Tensor mathematics Look at these quantities from the angle of What is the form of a matrix that we are familiar with

Nothing else is difficult . The only thing we are not familiar with is the velocity gradient . namely
∇ v \nabla \mathbf{v} v

We previously wrote a popular introduction to tensors , as follows :
https://blog.csdn.net/weixin_43940314/article/details/123559800

velocity gradient

Velocity is obviously a vector
v = ( u , v , w ) T \mathbf{v} = (u, v, w)^T v=(u,v,w)T

and nabla The operator is just a vector
∇ = ( ∂ ∂ x , ∂ ∂ y , ∂ ∂ z ) T \nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})^T =(x,y,z)T

Vector and vector write together , There is no sign in the middle , It's actually a multiplication of ( Also called dyadic ). Sometimes we take the initiative to add a symbol , It's written in ⨂ \bigotimes .

Actually , gradient Namely nabla Multiplication of operators and physical quantities .

What rule does the multiplication follow ?

Union multiplication and dot multiplication 、 Cross multiplication is different , It does nothing ! Just simply write the vectors side by side .

Besides , remember : And multiplication is ascending , The order of the result is to add up the original order !

Let's write the union multiplication in the familiar matrix form .

We write line by line
∇ ⊗ v = ( ∂ ∂ x ∂ ∂ y ∂ ∂ z ) ⊗ ( u v w ) \nabla \otimes \mathbf{v} = \begin{pmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y}\\ \frac{\partial}{\partial z} \end{pmatrix} \otimes \begin{pmatrix} u \\ v\\ w \end{pmatrix} v=xyzuvw

1 first line

The corresponding items are simply written side by side

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1.1 Component and form

Write in the form of component sum
∂ u ∂ x i i + ∂ v ∂ x i j + ∂ w ∂ x i k \frac{\partial u}{\partial x} \mathbf{ii} + \frac{\partial v}{\partial x }\mathbf{ij} + \frac{\partial w}{\partial x} \mathbf{ik} xuii+xvij+xwik

1.2 Matrix form

Or in the form of a matrix

That is to say

( ∂ u ∂ x ∂ v ∂ x ∂ w ∂ x ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ) \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial v}{\partial x } & \frac{\partial w}{\partial x} \\ \cdots & \cdots & \cdots\\ \cdots & \cdots & \cdots\\ \end{pmatrix} xuxvxw

2 The second line

Empathy

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2.1 Component and form

Write in the form of component sum
∂ u ∂ y i i + ∂ v ∂ y i j + ∂ w ∂ y i k \frac{\partial u}{\partial y} \mathbf{ii} + \frac{\partial v}{\partial y }\mathbf{ij} + \frac{\partial w}{\partial y} \mathbf{ik} yuii+yvij+ywik

2.2 Matrix form

Or in the form of a matrix

That is to say

( ⋯ ⋯ ⋯ ∂ u ∂ y ∂ v ∂ y ∂ w ∂ y ⋯ ⋯ ⋯ ) \begin{pmatrix} \cdots & \cdots & \cdots\\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y} & \frac{\partial w}{\partial y}\\ \cdots & \cdots & \cdots\\ \end{pmatrix} yuyvyw

3 The third line

Empathy
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3.1 Component and form

Write in the form of component sum
∂ u ∂ z i i + ∂ v ∂ z i j + ∂ w ∂ z i k \frac{\partial u}{\partial z} \mathbf{ii} + \frac{\partial v}{\partial z }\mathbf{ij} + \frac{\partial w}{\partial z} \mathbf{ik} zuii+zvij+zwik

3.2 Matrix form

Or in the form of a matrix

That is to say

( ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ∂ u ∂ z ∂ v ∂ z ∂ w ∂ z ) \begin{pmatrix} \cdots & \cdots & \cdots\\ \cdots & \cdots & \cdots\\ \frac{\partial u}{\partial z} & \frac{\partial v}{\partial z} & \frac{\partial w}{\partial z}\\ \end{pmatrix} zuzvzw

4 Close together

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Component and form
∇ v = ∂ u ∂ x i i + ∂ v ∂ x i j + ∂ w ∂ x i k + ∂ u ∂ y i i + ∂ v ∂ y i j + ∂ w ∂ y i k + ∂ u ∂ z i i + ∂ v ∂ z i j + ∂ w ∂ z i k \nabla \mathbf{v} = \\ \frac{\partial u}{\partial x} \mathbf{ii} + \frac{\partial v}{\partial x }\mathbf{ij} + \frac{\partial w}{\partial x} \mathbf{ik} + \\ \frac{\partial u}{\partial y} \mathbf{ii} + \frac{\partial v}{\partial y }\mathbf{ij} + \frac{\partial w}{\partial y} \mathbf{ik} + \\ \frac{\partial u}{\partial z} \mathbf{ii} + \frac{\partial v}{\partial z }\mathbf{ij} + \frac{\partial w}{\partial z} \mathbf{ik} v=xuii+xvij+xwik+yuii+yvij+ywik+zuii+zvij+zwik

Matrix form

( ∂ u ∂ x ∂ v ∂ x ∂ w ∂ x ∂ u ∂ y ∂ v ∂ y ∂ w ∂ y ∂ u ∂ z ∂ v ∂ z ∂ w ∂ z ) \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial v}{\partial x } & \frac{\partial w}{\partial x} \\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y} & \frac{\partial w}{\partial y}\\ \frac{\partial u}{\partial z} & \frac{\partial v}{\partial z} & \frac{\partial w}{\partial z}\\ \end{pmatrix} xuyuzuxvyvzvxwywzw

Symmetric tensor

We write the velocity gradient , Now let's look at the symmetric tensor
∇ v + ( ∇ v ) T \nabla \mathbf{v} + (\nabla \mathbf{v})^T v+(v)T

It's just the velocity gradient plus its own transpose .

Transpose is to write rows into columns , Columns are written in rows .

Let's write transpose first
( ∇ v ) T = ( ∂ u ∂ x ∂ u ∂ y ∂ u ∂ z ∂ v ∂ x ∂ v ∂ y ∂ v ∂ z ∂ w ∂ x ∂ w ∂ y ∂ w ∂ z ) (\nabla \mathbf{v})^T= \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y } & \frac{\partial u}{\partial z} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z}\\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} & \frac{\partial w}{\partial z}\\ \end{pmatrix} (v)T=xuxvxwyuyvywzuzvzw

Then add them up

( ∂ u ∂ x ∂ v ∂ x ∂ w ∂ x ∂ u ∂ y ∂ v ∂ y ∂ w ∂ y ∂ u ∂ z ∂ v ∂ z ∂ w ∂ z ) + ( ∂ u ∂ x ∂ u ∂ y ∂ u ∂ z ∂ v ∂ x ∂ v ∂ y ∂ v ∂ z ∂ w ∂ x ∂ w ∂ y ∂ w ∂ z ) = ( 2 ∂ u ∂ x ∂ v ∂ x + ∂ u ∂ y ∂ w ∂ x + ∂ u ∂ z ∂ u ∂ y + ∂ v ∂ x 2 ∂ v ∂ y ∂ w ∂ y + ∂ v ∂ z ∂ u ∂ z + ∂ w ∂ x ∂ v ∂ z + ∂ w ∂ y 2 ∂ w ∂ z ) \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial v}{\partial x } & \frac{\partial w}{\partial x} \\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y} & \frac{\partial w}{\partial y}\\ \frac{\partial u}{\partial z} & \frac{\partial v}{\partial z} & \frac{\partial w}{\partial z}\\ \end{pmatrix} + \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y } & \frac{\partial u}{\partial z} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z}\\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} & \frac{\partial w}{\partial z}\\ \end{pmatrix} =\\ \begin{pmatrix} 2\frac{\partial u}{\partial x} &\frac{\partial v}{\partial x }+\frac{\partial u}{\partial y } & \frac{\partial w}{\partial x}+\frac{\partial u}{\partial z} \\ \frac{\partial u}{\partial y}+\frac{\partial v}{\partial x} & 2\frac{\partial v}{\partial y} & \frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}\\ \frac{\partial u}{\partial z}+\frac{\partial w}{\partial x} & \frac{\partial v}{\partial z}+\frac{\partial w}{\partial y} & 2 \frac{\partial w}{\partial z}\\ \end{pmatrix} xuyuzuxvyvzvxwywzw+xuxvxwyuyvywzuzvzw=2xuyu+xvzu+xwxv+yu2yvzv+ywxw+zuyw+zv2zw

We look at this matrix
∇ v + ( ∇ v ) T = ( 2 ∂ u ∂ x ∂ v ∂ x + ∂ u ∂ y ∂ w ∂ x + ∂ u ∂ z ∂ u ∂ y + ∂ v ∂ x 2 ∂ v ∂ y ∂ w ∂ y + ∂ v ∂ z ∂ u ∂ z + ∂ w ∂ x ∂ v ∂ z + ∂ w ∂ y 2 ∂ w ∂ z ) \nabla \mathbf{v} + (\nabla \mathbf{v})^T= \begin{pmatrix} 2\frac{\partial u}{\partial x} &\frac{\partial v}{\partial x }+\frac{\partial u}{\partial y } & \frac{\partial w}{\partial x}+\frac{\partial u}{\partial z} \\ \frac{\partial u}{\partial y}+\frac{\partial v}{\partial x} & 2\frac{\partial v}{\partial y} & \frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}\\ \frac{\partial u}{\partial z}+\frac{\partial w}{\partial x} & \frac{\partial v}{\partial z}+\frac{\partial w}{\partial y} & 2 \frac{\partial w}{\partial z}\\ \end{pmatrix} v+(v)T=2xuyu+xvzu+xwxv+yu2yvzv+ywxw+zuyw+zv2zw

Two features can be easily found :

  1. It's symmetrical
  2. The diagonal element has a coefficient 2

actually , Any matrix plus its transpose is symmetric . It's easy to figure that out .

therefore , In hydrodynamics , We call it Symmetric tensor Okay .( Wild but simple names )

By the way , stay OpenFOAM among , Its code representation is

twoSymm(gradU)

Why add a two Well ? Because it is 2 Times the strain tensor . We can actually remember this way : Its diagonal element has a coefficient 2.

Strain tensor

Said just now , The strain tensor is actually 1/2 The symmetric tensor of .

E = 1 2 ( ∇ v + ( ∇ v ) T ) E = \frac{1}{2}(\nabla \mathbf{v} + (\nabla \mathbf{v})^T) E=21(v+(v)T)

Simple and clear .

Stress tensor

Up to the top , We don't use Newton's constitutive law . in other words , The above applies to any fluid .

At this time, Newton's constitutive equation is inserted .

Suppose it is a Newtonian fluid
T = − p 1 + 2 μ E \mathbf{T}=-p \mathbb{1}+2 \mu \mathbf{E} T=p1+2μE
there 1 It's a unit array .

So it's nothing more than multiplying by a few coefficients .

T = − p 1 + 2 μ E = ( 2 μ ∂ u ∂ x − p μ ( ∂ v ∂ x + ∂ u ∂ y ) μ ( ∂ w ∂ x + ∂ u ∂ z ) μ ( ∂ u ∂ y + ∂ v ∂ x ) 2 μ ∂ v ∂ y − p μ ( ∂ w ∂ y + ∂ v ∂ z ) μ ( ∂ u ∂ z + ∂ w ∂ x ) μ ( ∂ v ∂ z + ∂ w ∂ y ) 2 μ ∂ w ∂ z − p ) \mathbf{T}=-p \mathbb{1}+2 \mu \mathbf{E}=\\ \begin{pmatrix} 2\mu \frac{\partial u}{\partial x} -p & \mu ( \frac{\partial v}{\partial x }+\frac{\partial u}{\partial y }) & \mu ( \frac{\partial w}{\partial x}+\frac{\partial u}{\partial z}) \\ \mu ( \frac{\partial u}{\partial y}+ \frac{\partial v}{\partial x}) & 2\mu \frac{\partial v}{\partial y}-p & \mu ( \frac{\partial w}{\partial y}+\frac{\partial v}{\partial z})\\ \mu ( \frac{\partial u}{\partial z}+ \frac{\partial w}{\partial x}) & \mu ( \frac{\partial v}{\partial z}+\frac{\partial w}{\partial y}) & 2\mu \frac{\partial w}{\partial z}-p\\ \end{pmatrix} T=p1+2μE=2μxupμ(yu+xv)μ(zu+xw)μ(xv+yu)2μyvpμ(zv+yw)μ(xw+zu)μ(yw+zv)2μzwp

end
2022-6-10

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