## Derivation of the equation

In this section, we derive the differential equation for torsion of an elastic bar. The derivation follows closely the one in Differential equation for axial loading of a bar.

### Equilibrium

Consider a bar loaded at its ends by the torque $T_1$ och $T_2$ according to the figure below. Along the bar, there is an additional loading in terms of a distributed torque $\qv(x)$ per unit length [Nm/m]. The bar is allowed to have a varying cross-sectional area implying a variation of $\Kv(x)$. Finally, the bar is modeled as linear elastic with a possible variation in shear modulus $G(x)$.

Study a free body diagram of a small volume element with length $\Delta x$:

Equilibrium around the $x$-axis gives:

$$\twoheadrightarrow: \qquad \Mv(x + \Dx) + \int_x^{x+\Delta x} {\qv(x)}

\cdot {\rm d} x – \Mv(x) = 0$$

which for small $\Delta x$ can be written

$$\Mv(x + \Dx) + \qv(x) \cdot \Delta x – \Mv(x) = 0$$

Dividing by $\Dx$ and taking the limit as $\Dx \rightarrow 0$ gives

$$\lim_{\Dx \rightarrow 0} \frac{\Mv (x + \Dx) – \Mv(x) }{\Dx} + \qv =

\Mv’ + \qv = 0$$

Equilibirum of this bar is summarized as

$$\Mv’ + \qv = 0$$

One should compare this equation with the equilibrium equation for axial loading of a bar $N’ + \Kx A = 0$.

As was shown earlier, the relation between the internal torque $\Mv$ and the shear stress $\tau$, can be written as

$$\Mv=\frac{\Kv \, \tau}{r}$$

which is valid for both thin-walled and thick-walled circular cross-sections.

### Compatability equation

Just as for torsion of uniform bars, we assume that each cross-section only rotates and remain plane during deformation. For a bar of length $L$ we derived:

$$\gamma=r \, \frac{\varphi(L)-\varphi(0)}{L}$$

Ggeneralizing the above expression by studying a small volume element of length $\Delta x$, one obtains

$$\gamma(x)=r \, \lim_{\Dx \rightarrow 0}\frac{\varphi(x+\Delta x)-\varphi(x)}{\Delta x}=r \, \varphi'(x)$$

### Material relation

We assume Hooke’s law in shear is valid:

$$\tau= G \, \gamma$$

### Differential equation for torsion of a bar

The differential equation is obtained by combining equilibrium, compatibility and the material relation.

Start with $\Mv$ and insert Hooke’s law into the compatibility equation.

$$\Mv=\frac{\Kv \, \tau}{r} =

\frac{\Kv \, G \, \gamma}{r} =

G \,\Kv \, \varphi’$$

which finally is inserted into the equilibrium equation to give

$$\Mv’ + \qv = \frac{{\rm d}}{{\rm d}x}(G \,\Kv \, \varphi’) + \qv =0$$

If we introduce the notation $\DGK = G\cdot \Kv$ to symbolize the torsional rigidity the differential equation can be summarized as

$$\frac{{\rm d}}{{\rm d}x} \paren{\DGK \, \od{\varphi}{x}} + \qv=0$$

- $\varphi$ — angle of rotation/twist [-]
- $\DGK = G\cdot \Kv$ — torsional rigidity [N m^2]
- $G$ — shear modulus [Pa]
- $\Kv$ — polar moment of inertia [m^4]
- $\qv$ — distributed torque per unit length [Nm/m]

Two boundary conditions are required to solve the equation. The two most common types are prescribed angle of twist or prescribed internal torque.