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Diffstat (limited to 'source/know/concept/cauchy-stress-tensor')
| -rw-r--r-- | source/know/concept/cauchy-stress-tensor/index.md | 80 |
1 files changed, 40 insertions, 40 deletions
diff --git a/source/know/concept/cauchy-stress-tensor/index.md b/source/know/concept/cauchy-stress-tensor/index.md index 1d9ad0d..d83f430 100644 --- a/source/know/concept/cauchy-stress-tensor/index.md +++ b/source/know/concept/cauchy-stress-tensor/index.md @@ -22,10 +22,10 @@ but it is arguably most intuitive for solids. ## Definition In the solid, imagine an infinitesimal cube -whose sides, $\dd{S}_x$, $\dd{S}_y$ and $\dd{S}_z$, -are orthogonal to the $x$, $y$ and $z$ axes, respectively. -There is a force $\dd{\va{F}}_1$ acting on $\dd{S}_x$, -$\dd{\va{F}}_2$ on $\dd{S}_y$, and $\dd{\va{F}}_3$ on $\dd{S}_z$. +whose sides, $$\dd{S}_x$$, $$\dd{S}_y$$ and $$\dd{S}_z$$, +are orthogonal to the $$x$$, $$y$$ and $$z$$ axes, respectively. +There is a force $$\dd{\va{F}}_1$$ acting on $$\dd{S}_x$$, +$$\dd{\va{F}}_2$$ on $$\dd{S}_y$$, and $$\dd{\va{F}}_3$$ on $$\dd{S}_z$$. Then we can decompose each of these forces, for example: $$\begin{aligned} @@ -33,24 +33,24 @@ $$\begin{aligned} = \va{e}_x F_{x1} + \va{e}_y F_{y1} + \va{e}_z F_{z1} \end{aligned}$$ -Where $\va{e}_x$, $\va{e}_y$ and $\va{e}_z$ are the basis unit vectors. -If we divide each of the force components by the area $\dd{S}_x$ +Where $$\va{e}_x$$, $$\va{e}_y$$ and $$\va{e}_z$$ are the basis unit vectors. +If we divide each of the force components by the area $$\dd{S}_x$$ (like in a fluid, in order to get the pressure), -we find the stresses $\sigma_{xx}$, $\sigma_{yx}$ and $\sigma_{zx}$ -that are being "felt" by the $x$ surface element $\dd{S}_x$: +we find the stresses $$\sigma_{xx}$$, $$\sigma_{yx}$$ and $$\sigma_{zx}$$ +that are being "felt" by the $$x$$ surface element $$\dd{S}_x$$: $$\begin{aligned} \dd{\va{F}}_1 = \big( \va{e}_x \sigma_{xx} + \va{e}_y \sigma_{yx} + \va{e}_z \sigma_{zx} \big) \dd{S}_x \end{aligned}$$ -The perpendicular component $\sigma_{xx}$ is called a **tensile stress**, +The perpendicular component $$\sigma_{xx}$$ is called a **tensile stress**, and its sign is always chosen so that a positive value corresponds to a tension, -i.e. the $x$-side is pulled away from the rest of the cube. -The tangential components $\sigma_{yx}$ and $\sigma_{zx}$ +i.e. the $$x$$-side is pulled away from the rest of the cube. +The tangential components $$\sigma_{yx}$$ and $$\sigma_{zx}$$ are called **shear stresses**. -Evidently, the other two forces $\dd{\va{F}}_2$ and $\dd{\va{F}}_3$ +Evidently, the other two forces $$\dd{\va{F}}_2$$ and $$\dd{\va{F}}_3$$ can be decomposed in the exact same way, yielding nine stress components in total: @@ -64,7 +64,7 @@ $$\begin{aligned} = \big( \va{e}_x \sigma_{xz} + \va{e}_y \sigma_{yz} + \va{e}_z \sigma_{zz} \big) \dd{S}_z \end{aligned}$$ -The total force $\dd{\va{F}}$ on the entire infinitesimal cube +The total force $$\dd{\va{F}}$$ on the entire infinitesimal cube is simply the sum of the previous three: $$\begin{aligned} @@ -72,8 +72,8 @@ $$\begin{aligned} = \dd{\va{F}}_1 + \dd{\va{F}}_2 + \dd{\va{F}}_3 \end{aligned}$$ -We can then decompose $\dd{\va{F}}$ into its net components -along the $x$, $y$ and $z$ axes: +We can then decompose $$\dd{\va{F}}$$ into its net components +along the $$x$$, $$y$$ and $$z$$ axes: $$\begin{aligned} \dd{\va{F}} @@ -94,7 +94,7 @@ $$\begin{aligned} \end{aligned}$$ We can write this much more compactly using index notation, -where $i, j \in \{x, y, z\}$: +where $$i, j \in \{x, y, z\}$$: $$\begin{aligned} \boxed{ @@ -103,9 +103,9 @@ $$\begin{aligned} } \end{aligned}$$ -The stress components $\sigma_{ij}$ can be written as a second-rank tensor +The stress components $$\sigma_{ij}$$ can be written as a second-rank tensor (i.e. a matrix that transforms in a certain way), -called the **Cauchy stress tensor** $\hat{\sigma}$: +called the **Cauchy stress tensor** $$\hat{\sigma}$$: $$\begin{aligned} \boxed{ @@ -119,8 +119,8 @@ $$\begin{aligned} } \end{aligned}$$ -Then $\dd{\va{F}}$ is written even more compactly -using the dot product, with $\dd{\va{S}} = (\dd{S}_x, \dd{S}_y, \dd{S}_z)$: +Then $$\dd{\va{F}}$$ is written even more compactly +using the dot product, with $$\dd{\va{S}} = (\dd{S}_x, \dd{S}_y, \dd{S}_z)$$: $$\begin{aligned} \boxed{ @@ -132,13 +132,13 @@ $$\begin{aligned} All forces on the cube's sides can be written in this form. **Cauchy's stress theorem** states that the force on *any* surface element inside the solid can be written like this, -simply by projecting it onto the $x$, $y$ and $z$ zero-planes -to get the areas $\dd{S}_x$, $\dd{S}_y$ and $\dd{S}_z$. +simply by projecting it onto the $$x$$, $$y$$ and $$z$$ zero-planes +to get the areas $$\dd{S}_x$$, $$\dd{S}_y$$ and $$\dd{S}_z$$. -Note that for fluids, the pressure $p$ was defined -such that $\dd{\va{F}} = - p \dd{\va{S}}$. -If we wanted to define $p$ for solids in the same way, -we would need $\hat{\sigma}$ to be diagonal *and* +Note that for fluids, the pressure $$p$$ was defined +such that $$\dd{\va{F}} = - p \dd{\va{S}}$$. +If we wanted to define $$p$$ for solids in the same way, +we would need $$\hat{\sigma}$$ to be diagonal *and* all of its diagonal elements to be identical. Since this is almost never the case, the scalar pressure is ill-defined in solids. @@ -146,9 +146,9 @@ the scalar pressure is ill-defined in solids. ## Equilibrium -The total force $\va{F}$ acting on a (non-infinitesimal) volume $V$ of the solid -is given by the sum of the total body force $\va{F}_b$ and total surface force $\va{F}_s$, -where $\vec{f}$ is the body force density: +The total force $$\va{F}$$ acting on a (non-infinitesimal) volume $$V$$ of the solid +is given by the sum of the total body force $$\va{F}_b$$ and total surface force $$\va{F}_s$$, +where $$\vec{f}$$ is the body force density: $$\begin{aligned} \va{F} @@ -157,7 +157,7 @@ $$\begin{aligned} \end{aligned}$$ We can rewrite the surface term using the divergence theorem, -where $\top$ is the transpose: +where $$\top$$ is the transpose: $$\begin{aligned} \va{F}_s @@ -166,7 +166,7 @@ $$\begin{aligned} \end{aligned}$$ For some people, this equation may be more enlightening in index notation, -where $\nabla_j \equiv \ipdv{}{x_j}$ is the partial derivative with respect to the $j$th coordinate: +where $$\nabla_j \equiv \ipdv{}{x_j}$$ is the partial derivative with respect to the $$j$$th coordinate: $$\begin{aligned} F_{s, i} @@ -174,8 +174,8 @@ $$\begin{aligned} = \int_V \sum_{j} \nabla_{\!j} \sigma_{ij} \dd{V} \end{aligned}$$ -In any case, the total force $\va{F}$ can then be expressed -as a single volume integral over $V$: +In any case, the total force $$\va{F}$$ can then be expressed +as a single volume integral over $$V$$: $$\begin{aligned} \va{F} @@ -183,7 +183,7 @@ $$\begin{aligned} = \int_V \va{f^*} \dd{V} \end{aligned}$$ -Where we have defined the **effective force density** $\va{f^*}$ as follows: +Where we have defined the **effective force density** $$\va{f^*}$$ as follows: $$\begin{aligned} \boxed{ @@ -192,14 +192,14 @@ $$\begin{aligned} } \end{aligned}$$ -The volume $V$ is in **mechanical equilibrium** if the net force acting on it amounts to zero: +The volume $$V$$ is in **mechanical equilibrium** if the net force acting on it amounts to zero: $$\begin{aligned} \va{F} = 0 \end{aligned}$$ -However, because $V$ is abritrary, the equilibrium condition for the whole solid is in fact: +However, because $$V$$ is abritrary, the equilibrium condition for the whole solid is in fact: $$\begin{aligned} \boxed{ @@ -215,7 +215,7 @@ which needs boundary conditions at the object's surface. Newton's third law states that the two sides of the boundary exert opposite forces on each other, so the boundary condition is continuity of the **stress vector** -$\hat{\sigma} \cdot \va{n}$: +$$\hat{\sigma} \cdot \va{n}$$: $$\begin{aligned} \boxed{ @@ -224,10 +224,10 @@ $$\begin{aligned} } \end{aligned}$$ -Where the normal of the outer surface is $\va{n}$, -and the normal of the inner surface is $-\va{n}$. +Where the normal of the outer surface is $$\va{n}$$, +and the normal of the inner surface is $$-\va{n}$$. Note that the above equation does *not* mean -that $-\hat{\sigma}_{\mathrm{inner}}$ equals $\hat{\sigma}_{\mathrm{outer}}$: +that $$-\hat{\sigma}_{\mathrm{inner}}$$ equals $$\hat{\sigma}_{\mathrm{outer}}$$: the tensors are allowed to be very different, as long as the stress vector's three components are equal. |