8.11  Torsion

The kinematic assumption

Let \(x\) run along the shaft and let the cross section at \(x\) have rotated by an angle \(\theta(x)\) about that axis. A point at \((y, z)\) in the section is carried around with it, and for small angles a rotation by \(\theta\) moves it by \(\theta\) times its distance from the axis, perpendicular to the radius. The displacement field is therefore

\[ \bm u = \begin{bmatrix} 0 \\ -\theta(x)\, z \\ \theta(x)\, y \end{bmatrix} \tag{8.11.1}\]

with no axial component, which is the statement that the section stays plane. Figure 8.11.2 shows what this looks like: longitudinal lines become helices while every circle stays a circle in its own plane.

Strain and stress

The strain follows by applying 8.7.3 to 8.11.1, which we let SymPy do.

x, y, z = sp.symbols('x y z', real=True)
G = sp.symbols('G', positive=True)

theta = sp.Function('theta')(x)

uu = sp.Matrix([0, -theta*z, theta*y])
XX = sp.Matrix([x, y, z])

grad_u = uu.jacobian(XX)
eps = sp.simplify((grad_u + grad_u.T)/2)

ltx(r"\nabla\bm u =", grad_u, r",\qquad \bm\varepsilon =", eps)

The rigid rotation of the section produces the antisymmetric entries \(-\theta\) and \(+\theta\) in the gradient, and those cancel in the symmetric part exactly as the rigid rotation did in Chapter 8.14. What survives is driven by the rate of twist \(d\theta/dx\),

\[ \varepsilon_{xy} = -\frac{z}{2}\frac{d\theta}{dx}, \qquad \varepsilon_{xz} = \frac{y}{2}\frac{d\theta}{dx} \tag{8.11.2}\]

with every other component zero. A shaft twisted uniformly along its length has a constant rate of twist; a shaft twisted at one end only would have \(d\theta/dx = 0\) and no strain anywhere, which is the correct statement that a rigid rotation of the whole shaft strains nothing.

Since the strain tensor has zero trace, the volumetric term in 8.8.7 drops out and the stress is pure shear,

sigma = sp.simplify(2*G*eps)
ltx(r"\bm\sigma = 2\mu\bm\varepsilon + \lambda\,\mathrm{tr}(\bm\varepsilon)\bm I =", sigma)

\[ \tau_{xy} = -G z \frac{d\theta}{dx}, \qquad \tau_{xz} = G y \frac{d\theta}{dx} \]

The two components combine into a shear stress of magnitude \(G\rho\, d\theta/dx\) at radius \(\rho = \sqrt{y^2+z^2}\), directed tangentially. The stress therefore grows linearly from zero on the axis to a maximum at the surface,

\[ \boxed{\tau = \frac{\rho}{R}\,\tau_{\max}} \tag{8.11.3}\]

which is why a solid shaft wastes its core: the material near the axis has almost no lever arm and carries almost no stress.

tau_mag = sp.simplify(sp.sqrt(sigma[0,1]**2 + sigma[0,2]**2))
ltx(r"\left|\bm\tau\right| = \sqrt{\tau_{xy}^2 + \tau_{xz}^2} =", tau_mag)

The polar second moment of area

The stress distribution must be statically equivalent to the applied torque. Each element of area \(dA\) at radius \(\rho\) carries a tangential force \(\tau\,dA\) acting with lever arm \(\rho\), so

\[ M = \int \rho\,\tau\, dA = \int \rho\,\frac{\rho}{R}\tau_{\max}\, dA = \frac{\tau_{\max}}{R}\int \rho^2\, dA \]

since \(\tau_{\max}/R\) is constant over the section. The remaining integral is a geometric property,

\[ \boxed{J = \int \rho^2\, dA} \tag{8.11.4}\]

the polar second moment of area, the exact counterpart of \(I\) in 8.14.5 with the distance measured from the axis instead of from a neutral plane. Combining the two gives the stress at any radius and its maximum at the surface,

\[ \boxed{\tau = \frac{M\rho}{J}, \qquad \tau_{\max} = \frac{MR}{J} = \frac{M}{W}}, \qquad W := \frac{J}{R} \tag{8.11.5}\]

where \(W\) is the torsional section modulus, playing the same role for torsion that \(I/c\) played for bending. Evaluating 8.11.4 is a one-line integration in polar coordinates.

rho_s, phi, R_o, R_i, t_w = sp.symbols('rho varphi R_o R_i t', positive=True)

def polar_moment(r_inner, r_outer):
    return sp.integrate(sp.integrate(rho_s**3, (rho_s, r_inner, r_outer)), (phi, 0, 2*sp.pi))

J_solid = polar_moment(0, R_o)
J_tube = polar_moment(R_i, R_o)

ltx(r"J_{\text{solid}} &=", J_solid, r"=\frac{\pi d^4}{32}",
    r"\\ W_{\text{solid}} &=", sp.simplify(J_solid/R_o), r"=\frac{\pi d^3}{16}",
    r"\\ J_{\text{tube}} &=", J_tube,
    r"\\ W_{\text{tube}} &=", sp.simplify(J_tube/R_o), aligned=True)

For a thin-walled tube of mean radius \(R\) and wall thickness \(t \ll R\) the integral collapses because every element sits at essentially the same radius, giving \(J = 2\pi R^3 t\) and \(W = 2\pi R^2 t\). That case is worth deriving separately, both as a check on the machinery and because it is the geometry in which torsion is used most efficiently.

A cross-check on the thin-walled tube

Figure 8.11.3 shows the section of a thin-walled shaft, with the shear stress running around the wall. Take a small strip subtending an angle \(d\varphi\).

The strip has area \(dA = R\,d\varphi\, t\) when the wall is thin enough that we may roll it out flat, and it carries a tangential force \(T = \tau\, dA\) with lever arm \(R\), so its contribution to the torque is

\[ dM = TR = \tau\,dA\,R = \tau t R^2\, d\varphi \]

Integrating around the circumference, with \(\tau\) constant because every point of the wall is at the same radius,

\[ M = \int_0^{2\pi} \tau t R^2\, d\varphi = 2\pi R^2 t\,\tau \qquad\Longrightarrow\qquad \tau = \frac{M}{2\pi R^2 t} = \frac{M}{W} \]

which agrees with 8.11.5 and identifies \(W = 2\pi R^2 t\) as before.

The angle of twist

The last arrow of Figure 8.11.1 connects the shear angle to the angle of twist, and it is pure geometry. Figure 8.11.6 shows a slice of length \(dx\) whose far face has rotated by \(d\theta\) relative to the near one. A line on the surface that was parallel to the axis has tilted through the shear angle \(\gamma\), and the same arc length is described either way,

\[ \gamma\, dx = R\, d\theta \tag{8.11.6}\]

Assuming linear elasticity, the material relation for shear is 8.5.7, \(\tau = G\gamma\), where the shear modulus is fixed by \(E\) and \(\nu\) through 8.8.5. Substituting into 8.11.6 and using 8.11.5 for the surface stress,

\[ d\theta = \frac{\gamma}{R}\,dx = \frac{\tau_{\max}}{GR}\,dx = \frac{M}{GJ}\,dx \]

and integrating over the length of a uniform shaft gives

\[ \boxed{\theta = \frac{ML}{GJ}} \tag{8.11.7}\]

The product \(GJ\) is the torsional rigidity, the counterpart of the axial stiffness \(EA\) of 8.9.1 and the bending stiffness \(EI\) of 8.14.6. Where torque, material or section vary along the shaft, the integral must be kept,

\[ \boxed{\theta = \int_0^L \frac{M(x)}{G(x)J(x)}\,dx} \tag{8.11.8}\]

which is the torsional analogue of the tapered rod of Chapter 8.10.

Example: Solid against hollow

A shaft must transmit \(M = 500~\text{Nm}\). Compare a solid shaft of radius \(20~\text{mm}\) with a hollow shaft of the same cross-sectional area, therefore the same mass per unit length, with outer radius \(30~\text{mm}\). Which is stiffer, which is stronger, and by how much?

Equal area requires \(R_o^2 - R_i^2 = R^2\), which fixes the bore.

R_s = 20.0                                  # solid radius [mm]
R_out = 30.0                                # hollow outer radius [mm]
R_in = sp.sqrt(R_out**2 - R_s**2)           # equal area fixes the bore

J_s = sp.pi*R_s**4/2
J_h = sp.pi*(R_out**4 - float(R_in)**4)/2

W_s, W_h = J_s/R_s, J_h/R_out
M_t = 500e3                                  # 500 Nm in Nmm

ltx(r"R_i &=", float(R_in), r"~\text{mm}",
    r"\\ J_{\text{solid}} &=", float(J_s), r"~\text{mm}^4",
    r"\\ J_{\text{hollow}} &=", float(J_h), r"~\text{mm}^4",
    r"\\ J_{\text{hollow}}/J_{\text{solid}} &=", float(J_h/J_s), aligned=True)

G_steel = 81e3                               # shear modulus of steel [MPa]
L_shaft = 1000.0                             # shaft length [mm]

ltx(r"\tau_{\max,\text{solid}} &=", float(M_t/W_s), r"~\text{MPa}",
    r"\\ \tau_{\max,\text{hollow}} &=", float(M_t/W_h), r"~\text{MPa}",
    r"\\ \theta_{\text{solid}} &=", float(M_t*L_shaft/(G_steel*J_s)*180/sp.pi), r"^\circ",
    r"\\ \theta_{\text{hollow}} &=", float(M_t*L_shaft/(G_steel*J_h)*180/sp.pi), r"^\circ", aligned=True)

The hollow shaft is \(3.5\) times stiffer and carries \(2.3\) times less stress for the same mass of steel, because moving material outward increases \(\rho^2\) in 8.11.4 faster than it decreases the amount of material. This is the same argument that puts the flanges of an I-beam far from the neutral axis, and it is why drive shafts, bicycle frames and aircraft spars are tubes.

The gain is not free. A thin wall buckles under torsion long before it yields, so the wall thickness is eventually set by stability rather than by 8.11.5, and the tube must be joined to solid ends somehow. Both limits sit outside the model derived here.

Non-circular sections

Everything above rests on 8.11.1, and that field has no axial component. For a circular section this is exact, because the section has no way to distinguish one radial direction from another and therefore no reason to move out of plane. Any other section does warp, and a rectangular bar in torsion develops axial displacements that are largest at the corners.

The consequences are severe enough that the formulas must not be reused. The shear stress in a rectangular bar is zero at the corners, not maximum, and it peaks at the middle of the long side, which is the opposite of what 8.11.3 would predict. Handbooks tabulate approximate values of \(W\) and an effective \(J\) for common shapes, and the formulary carries a selection. Open thin-walled sections, a slit tube or a channel, are worse still: they are one to two orders of magnitude less stiff in torsion than the closed section of the same dimensions, because the shear flow has no closed path to run around.

Where accuracy matters for a non-circular section, the honest answer is to solve the continuum problem numerically. This is a case where the kinematic assumption that made the chapter possible is the very thing that fails, and where finite element analysis is not a convenience but the only correct route.