# Eccentric hollow circular section

## Values for calculation

$T$ $\mathrm{Nm}$
$D$ $\mathrm{mm}$
$d$ $\mathrm{mm}$
$e$ $\mathrm{mm}$
$L$ $\mathrm{mm}$
$G$ $\mathrm{MPa}$

## Calculation

### Coefficient $λ$

$$λ=\cfrac{e}{D}$$

### Coefficient $n$

$$n=\cfrac{d}{D}$$

### Coefficient $C$

$$C=1+\cfrac{16\cdot n^2}{\left(1-n^2\right)\cdot\left(1-n^4\right)}\cdot λ^2+\cfrac{384\cdot n^4}{\left(1-n^2\right)^2\cdot\left(1-n^4\right)^4}\cdot λ^4$$

### Coefficient $F$

$$F=1+\cfrac{4\cdot n^2}{1-n^2}\cdot λ+\cfrac{32\cdot n^2}{\left(1-n^2\right)\cdot\left(1-n^4\right)}\cdot λ^2+\cfrac{48\cdot n^2\cdot\left(1+2\cdot n^2+3\cdot n^4+2\cdot n^6\right)}{\left(1-n^2\right)\cdot\left(1-n^4\right)\cdot\left(1-n^6\right)}\cdot λ^3+\cfrac{64\cdot n^2\cdot\left(2+12\cdot n^2+19\cdot n^4+28\cdot n^6+18\cdot n^8+14\cdot n^{10}+3\cdot n^{12}\right)}{\left(1-n^2\right)\cdot\left(1-n^4\right)\cdot\left(1-n^6\right)\cdot\left(1-n^8\right)}\cdotλ^4$$

### Polar moment of inertia

$$K=\cfrac{π\cdot\left(D^4-d^4\right)}{32\cdot C}$$

### Angle of twist

$$θ=\cfrac{T\cdot 10^3\cdot L}{K\cdot G}$$

### Torsion stress

$$τ_{max}=\cfrac{16\cdot{10}^3\cdot T\cdot D\cdot F}{π\cdot \left(D^4-d^4\right)}$$