{{ :wikifemfuchde2019:profile_in_torsion_a2019_v003.mfd |}}


{{ :wikifemfuchde2019:chapter_9_torsion_of_thin-walled_tubes.pdf |}}

Curiosità: {{ :restricted:griffith1917.pdf | Griffith, A.A, Taylor, G.I., Use of Soap Films in Solving Torsion Problems, 1917}}

{{ :wikifemfuchde2019:dispensa_2018_v2019-03-26.pdf |}}

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**Prandtl stress function** $\phi$

shear stresses: $\tau_{zx}=\frac{\partial \phi}{\partial y}$ , $\tau_{yz}=-\frac{\partial \phi}{\partial x}$

shear stresses are hence defined as the 90° clockwise rotated $\phi$  gradient. 


equilibrium equation $\frac{\partial \tau_{zx}}{\partial x} + \frac{\partial \tau_{yz}}{\partial y}=0$ satisfied due to Schwarz.

Compatibility relationship, for an homogeneous, isotropic hookean material: $\nabla^2\phi=-2G\beta$ where $\beta$ is the rate of twist.

Here, $\nabla^2\phi=\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2}$.

Origin of the compatibility equations: the twist rate defines the derivatives
$\frac{\partial u}{\partial z}=-\beta y$ and
$\frac{\partial v}{\partial z}=+\beta x$; the two terms 
$\frac{\partial w}{\partial x}$ and $\frac{\partial w}{\partial y}$ complete the $\gamma_{zx}$ and the $\gamma_{yz}$ shear strain component definition by introducing the $w$ warping displacement. 
Those shear strains are related to the shear stresses by the constitutive equations $G\gamma_{zx}=\tau_{zx}$ and $G\gamma_{yz}=\tau_{yz}$.

Let's consider a closed path along the section consisting of four segments defined as
{{tablelayout?rowsHeaderSource=Auto}}
^ from:         ^ to:           ^ $\frac{d x}{d s}=\frac{d s}{d x}$  ^ $\frac{d y}{d s}=\frac{d s}{d y}$  ^
|  $x   ,y   $  |  $x+dx,y   $  |  +1                                |  0                                 |
|  $x+dx,y   $  |  $x+dx,y+dy$  |  0                                 |  +1                                |
|  $x+dx,y+dy$  |  $x   ,y+dy$  |  -1                                |  0                                 |
|  $x   ,y+dy$  |  $x   ,y   $  |  0                                 |  -1                                |

Then, the circuital integral along such a loop
$$
\oint \frac{\partial w}{\partial s} ds = 0
$$
must be zero to enforce a single valued $w$ displacement at the $x,y$ point, thus avoiding a [[https://en.wikipedia.org/wiki/Penrose_stairs|Penrose stairs]]-like impossible behaviour.

{{:wikifemfuchde2019:372px-impossible_staircase.svg.png?100|}}

Then we have
$$
0=
\oint \left(
\frac{\partial w}{\partial x}\frac{d x}{d s}+
\frac{\partial w}{\partial y}\frac{d y}{d s}
\right) ds, 
$$
whose warping slope components may be expressed based on the twist rate and the shear stresses, thus obtaining
$$
0= 
\oint \left(
\left(\frac{\tau_{zx}}{G}+\beta y\right)\frac{d x}{d s}+
\left(\frac{\tau_{yz}}{G}-\beta x\right)\frac{d y}{d s}
\right) ds.
$$
By accumulating the contributions on the four segments that constitute the loop we obtain
$$
\left(-\frac{1}{G}\frac{\partial\tau_{zx}}{\partial y}dy-\beta dy\right)dx
+
\left(+\frac{1}{G}\frac{\partial\tau_{yz}}{\partial x}dx-\beta dx\right)dy
=0
$$
and
$$
+\frac{\partial\tau_{zx}}{\partial y}
-\frac{\partial\tau_{yz}}{\partial x}
=-2 G \beta.
$$
By substituting the shear stress component definition based on $\phi$, the compatibility equation is finally obtained.

boundary conditions: $\phi=c_i$, where $c_i$ is a constant on the $i$-th boundary; such condition, along with the way the $\tau_{zx},\tau_{yz}$ components based are derived from the $\phi$ Prandtl stress function, enforces the normal shear component with respect to the boundary to be zero.

Torque: $T=2\iint_A \phi dA$