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Diffstat (limited to 'source/know/concept/cauchy-strain-tensor')
| -rw-r--r-- | source/know/concept/cauchy-strain-tensor/index.md | 106 |
1 files changed, 53 insertions, 53 deletions
diff --git a/source/know/concept/cauchy-strain-tensor/index.md b/source/know/concept/cauchy-strain-tensor/index.md index a628820..bae7bb8 100644 --- a/source/know/concept/cauchy-strain-tensor/index.md +++ b/source/know/concept/cauchy-strain-tensor/index.md @@ -11,9 +11,9 @@ layout: "concept" **Strain** quantifies the deformation of a solid object. If the body has been deformed, e.g. by pulling or bending, its constituent particles have moved a bit. -Let $\va{X}$ be the original location of a particle, -and $\va{x}$ its new location after the deformation. -We can thus define the **displacement field** $\va{u}$: +Let $$\va{X}$$ be the original location of a particle, +and $$\va{x}$$ its new location after the deformation. +We can thus define the **displacement field** $$\va{u}$$: $$\begin{aligned} \va{u} @@ -21,7 +21,7 @@ $$\begin{aligned} \end{aligned}$$ We restrict ourselves to **infinitesimal strain**, -where $\va{u}$ is so tiny that the material's properties are unchanged, +where $$\va{u}$$ is so tiny that the material's properties are unchanged, and a **slowly-varying strain**, where the particle's neighbourhood has been distorted, but not completely changed. @@ -31,15 +31,15 @@ is that we need to somehow exclude movements of the *entire* body: for example, you can bend a twig in your hands while walking or dancing, but we are only interested in the twig's shape change, not in your movements. -The above definition of $\vu{u}$ includes both, +The above definition of $$\vu{u}$$ includes both, so we should be careful how we extract the strain from it. ## Definition We use the **Eulerian description** of deformation, -where the new position $\va{x}$ is the reference, -and the old position $\va{X}$ is expressed as a function of $\va{x}$: +where the new position $$\va{x}$$ is the reference, +and the old position $$\va{X}$$ is expressed as a function of $$\va{x}$$: $$\begin{aligned} \va{u}(\va{x}) @@ -47,10 +47,10 @@ $$\begin{aligned} \end{aligned}$$ Let us choose two nearby points in the deformed solid, -and call them $\va{x}$ and $\va{x} + \va{a}$, -where $\va{a}$ is a tiny vector pointing from one to the other. +and call them $$\va{x}$$ and $$\va{x} + \va{a}$$, +where $$\va{a}$$ is a tiny vector pointing from one to the other. Before the displacement, those points respectively had these positions, -where we define $\va{A}$ as the "old" version of $\va{a}$: +where we define $$\va{A}$$ as the "old" version of $$\va{a}$$: $$\begin{aligned} \va{X} = \va{X}(\va{x}) @@ -58,9 +58,9 @@ $$\begin{aligned} \va{X} + \va{A} = \va{X}(\va{x} + \va{a}) \end{aligned}$$ -Because the new positions $\va{x}$ are our reference, -we would like to write $\va{A}$ without $\va{X}$. -To do so, we use the definition of $\va{u}(\va{x})$, yielding: +Because the new positions $$\va{x}$$ are our reference, +we would like to write $$\va{A}$$ without $$\va{X}$$. +To do so, we use the definition of $$\va{u}(\va{x})$$, yielding: $$\begin{aligned} \va{A} @@ -71,8 +71,8 @@ $$\begin{aligned} &= \va{a} - \va{u}(\va{x} + \va{a}) - \va{u}(\va{x}) \end{aligned}$$ -Using the fact that $\va{a}$ is tiny by definition, -we expand the middle term to first order in $\va{a}$: +Using the fact that $$\va{a}$$ is tiny by definition, +we expand the middle term to first order in $$\va{a}$$: $$\begin{aligned} \va{u}(\va{x} + \va{a}) @@ -80,8 +80,8 @@ $$\begin{aligned} = \va{u}(\va{x}) + (\va{a} \cdot \nabla) \va{u}(\va{x}) \end{aligned}$$ -With this, we can now define the "shift" $\delta\va{a}$ -as the difference between $\va{a}$ and $\va{A}$ like so: +With this, we can now define the "shift" $$\delta\va{a}$$ +as the difference between $$\va{a}$$ and $$\va{A}$$ like so: $$\begin{aligned} \delta{\va{a}} @@ -90,24 +90,24 @@ $$\begin{aligned} \end{aligned}$$ In index notation, we write this expression as follows, -with $\nabla_j \equiv \ipdv{}{x_j}$ simply being the partial derivative -with respect to the $j$th coordinate: +with $$\nabla_j \equiv \ipdv{}{x_j}$$ simply being the partial derivative +with respect to the $$j$$th coordinate: $$\begin{aligned} \delta a_i = \sum_{j} a_j \nabla_j u_i \end{aligned}$$ -Where $\nabla_j u_i$ are called the **displacement gradients**, +Where $$\nabla_j u_i$$ are called the **displacement gradients**, and are just one step away from the desired definition of strain. Note that these gradients are dimensionless, -so we can more formally define a *slowly-varying* displacement $\va{u}(\va{x})$ -as one where $|\nabla_j u_i| \ll 1$. +so we can more formally define a *slowly-varying* displacement $$\va{u}(\va{x})$$ +as one where $$|\nabla_j u_i| \ll 1$$. Now, to solve the problem of macroscopic movements, -we take another tiny vector $\va{b}$ starting in the same point $\va{x}$ as $\va{a}$. +we take another tiny vector $$\va{b}$$ starting in the same point $$\va{x}$$ as $$\va{a}$$. Here is the trick: if the whole body is uniformly translated or rotated, -the scalar product $\va{a} \cdot \va{b}$ is unchanged, +the scalar product $$\va{a} \cdot \va{b}$$ is unchanged, but if there is a non-uniform distortion, it changes. We thus define the scalar product's difference like so: @@ -116,7 +116,7 @@ $$\begin{aligned} \equiv \va{a} \cdot \va{b} - \va{A} \cdot \va{B} \end{aligned}$$ -Where $\va{B}$ is the old version of $\va{b}$. +Where $$\va{B}$$ is the old version of $$\va{b}$$. Since these vectors are all tiny, we apply the product rule: $$\begin{aligned} @@ -125,7 +125,7 @@ $$\begin{aligned} \end{aligned}$$ It is more informative to switch to index notation here. -Inserting $\delta\va{a}$ and $\delta\va{b}$ yields: +Inserting $$\delta\va{a}$$ and $$\delta\va{b}$$ yields: $$\begin{aligned} \delta(\va{a} \cdot \va{b}) @@ -136,8 +136,8 @@ $$\begin{aligned} &= \sum_{ij} \big( \nabla_i u_j + \nabla_j u_i \big) \: a_i b_j \end{aligned}$$ -At last, we define the **Cauchy infinitesimal strain tensor** $\hat{u}$ -such that it has $u_{ij}$ as components: +At last, we define the **Cauchy infinitesimal strain tensor** $$\hat{u}$$ +such that it has $$u_{ij}$$ as components: $$\begin{aligned} \boxed{ @@ -154,7 +154,7 @@ $$\begin{aligned} = 2 \va{a} \cdot \hat{u} \cdot \va{b} \end{aligned}$$ -The Cauchy strain tensor $\hat{u}$ is a second-rank tensor, +The Cauchy strain tensor $$\hat{u}$$ is a second-rank tensor, and can alternatively be expressed like so: $$\begin{aligned} @@ -164,17 +164,17 @@ $$\begin{aligned} } \end{aligned}$$ -Where $\top$ is the transpose. Being defined from the scalar product, +Where $$\top$$ is the transpose. Being defined from the scalar product, all macroscopic movements of the body are removed from the tensor, -which turns out to make it symmetric, i.e. $u_{ij} = u_{ji}$. +which turns out to make it symmetric, i.e. $$u_{ij} = u_{ji}$$. ## Geometry So far we have used Cartesian coordinates, -but we can choose any three vectors $\va{a}$, $\va{b}$ and $\va{c}$, -and **project** $\hat{u}$ onto this basis. -For example, the component $u_{ab}$ then becomes: +but we can choose any three vectors $$\va{a}$$, $$\va{b}$$ and $$\va{c}$$, +and **project** $$\hat{u}$$ onto this basis. +For example, the component $$u_{ab}$$ then becomes: $$\begin{aligned} \boxed{ @@ -184,12 +184,12 @@ $$\begin{aligned} \end{aligned}$$ And so forth, for the other eight components. -The basis in which $\hat{u}$ is diagonal is the one formed by its eigenvectors, +The basis in which $$\hat{u}$$ is diagonal is the one formed by its eigenvectors, and their directions are the **principal axes of strain** at that point in the solid. -Because $\hat{u}$ is symmetric, such a basis always exists. +Because $$\hat{u}$$ is symmetric, such a basis always exists. -Given a vector $\va{a}$, its relative length change +Given a vector $$\va{a}$$, its relative length change due to the deformation is simply given by: $$\begin{aligned} @@ -199,8 +199,8 @@ $$\begin{aligned} } \end{aligned}$$ -To find the angle change $\delta\theta$ -between two vectors $\va{a}$ and $\va{b}$, +To find the angle change $$\delta\theta$$ +between two vectors $$\va{a}$$ and $$\va{b}$$, we start with the product rule: $$\begin{aligned} @@ -211,9 +211,9 @@ $$\begin{aligned} - \big|\va{a}\big| \big|\va{b}\big| \sin\theta \: \delta\theta \end{aligned}$$ -We isolate this for $\delta\theta$, using the fact that -$\delta(\va{a} \cdot \va{b}) = 2 \big|\va{a}\big| \big|\va{b}\big| u_{ab}$ -thanks to the projection $u_{ab}$: +We isolate this for $$\delta\theta$$, using the fact that +$$\delta(\va{a} \cdot \va{b}) = 2 \big|\va{a}\big| \big|\va{b}\big| u_{ab}$$ +thanks to the projection $$u_{ab}$$: $$\begin{aligned} \delta\theta @@ -223,7 +223,7 @@ $$\begin{aligned} {\big|\va{a}\big| \big|\va{b}\big| \sin\theta} \end{aligned}$$ -By recognizing the length change $\delta|\va{a}|/|\va{a}| = u_{aa}$, +By recognizing the length change $$\delta|\va{a}|/|\va{a}| = u_{aa}$$, we arrive at the following expression: $$\begin{aligned} @@ -234,8 +234,8 @@ $$\begin{aligned} \end{aligned}$$ Now, everything so far has been about tiny vectors, -so the change of the line element $\dd{\va{l}}$ -is easy to express using the displacement field $\va{u}$: +so the change of the line element $$\dd{\va{l}}$$ +is easy to express using the displacement field $$\va{u}$$: $$\begin{aligned} \boxed{ @@ -244,9 +244,9 @@ $$\begin{aligned} } \end{aligned}$$ -Next, we calculate the change of the differential volume element $\dd{V}$ +Next, we calculate the change of the differential volume element $$\dd{V}$$ by treating it as the volume of a tiny parallelepiped -spanned by $\va{a}$, $\va{b}$ and $\va{c}$: +spanned by $$\va{a}$$, $$\va{b}$$ and $$\va{c}$$: $$\begin{aligned} \delta(\dd{V}) @@ -280,7 +280,7 @@ $$\begin{aligned} &= (\va{a} \cross \va{b} \cdot \va{c}) (\nabla \cdot \va{u}) \end{aligned}$$ -Here, we recognize the definition of $\dd{V}$, +Here, we recognize the definition of $$\dd{V}$$, leading to the following infinitesimal volume change: $$\begin{aligned} @@ -290,8 +290,8 @@ $$\begin{aligned} } \end{aligned}$$ -Finally, for the surface element $\dd{\va{S}} = \va{a} \cross \va{b}$, -we use that the volume element $\dd{V} = \va{c} \cdot \dd{\va{S}}$: +Finally, for the surface element $$\dd{\va{S}} = \va{a} \cross \va{b}$$, +we use that the volume element $$\dd{V} = \va{c} \cdot \dd{\va{S}}$$: $$\begin{aligned} \delta(\dd{V}) @@ -300,7 +300,7 @@ $$\begin{aligned} = (\va{c} \cdot \nabla) \va{u} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}}) \end{aligned}$$ -By comparing this to the previous result for $\delta(\dd{V})$, +By comparing this to the previous result for $$\delta(\dd{V})$$, we arrive at the following equation: $$\begin{aligned} @@ -308,8 +308,8 @@ $$\begin{aligned} = (\va{c} \cdot \nabla) \va{u} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}}) \end{aligned}$$ -Since $\va{c}$ is dot-multiplied at the front of each term, -we remove it, and isolate the rest for $\delta(\dd{\va{S}})$: +Since $$\va{c}$$ is dot-multiplied at the front of each term, +we remove it, and isolate the rest for $$\delta(\dd{\va{S}})$$: $$\begin{aligned} \boxed{ |