Irodov 1.305: One Relation, Two Polar Moments – Amit & Deepali Sinha Foundation

Problem Irodov 1.305

A tube of length $l$ , radius $r$ and wall thickness $\Delta r$ , with $\Delta r \ll r$ , and a solid rod of the same length $l$ and radius $r$ , both made of a material of shear modulus $G$ , are each clamped at one end and twisted about their own axis by a torque $N$ applied at the free end, producing a torsion (twist) angle $\varphi$ between the two ends. Find the relationship between $N$ and $\varphi$ for (a) the tube and (b) the solid rod.

A circular shaft of length l, clamped at one end, twisted by a torque N at the free end through a torsion angle φ: the shared geometry of both cases below. — A circular shaft of length $l$ , clamped at one end, twisted by a torque $N$ at the free end through a torsion angle $\varphi$ : the shared geometry of both cases below.

Given

$l$
$r$
$G$
$\Delta r \ll r$

Find

(a) the relation between

N

and

\varphi

for the tube; (b) the same for the solid rod

Solution

Step 1Shear strain and stress across a section

Take the standard picture of torsion: cross-sections stay plane and simply rotate rigidly about the shaft’s axis, so a generator that ran straight along the unloaded shaft is sheared through an angle growing linearly with radius. Over the length $l$ , a total twist $\varphi$ between the two ends gives, at radius $\rho$ , a shear strain and, by Hooke’s law for shear, a shear stress.

\begin{align*}\gamma(\rho) &= \frac{\rho\varphi}{l} \\ \tau(\rho) &= G\gamma(\rho) = \frac{G\rho\varphi}{l}\end{align*}

Step 2Torque carried by a ring element

A thin ring of radius $\rho$ and width $d\rho$ (the highlighted element in the cross-section figures below) carries the shear stress $\tau(\rho)$ tangentially over its own area $dA = 2\pi\rho\,d\rho$ ; the torque it contributes about the axis is that tangential force times its own radius.

dN = \rho\,\tau(\rho)\,dA = \rho\cdot\frac{G\rho\varphi}{l}\cdot 2\pi\rho\,d\rho

dN = \dfrac{2\pi G\varphi}{l}\,\rho^3\,d\rho

(1)

Cross-section of the tube: outer radius r, wall thickness Δ r r. Since the whole wall lies at essentially one radius, it is itself a single thin ring element. — Cross-section of the tube: outer radius $r$ , wall thickness $\Delta r \ll r$ . Since the whole wall lies at essentially one radius, it is itself a single thin ring element.

Cross-section of the solid rod: a ring element at radius ρ, width dρ, swept out to the rod's own radius r. — Cross-section of the solid rod: a ring element at radius $\rho$ , width $d\rho$ , swept out to the rod’s own radius $r$ .

Step 3Case (a): the thin-walled tube

The tube’s material occupies only the thin band from $\rho = r-\Delta r$ to $\rho = r$ , so integrating the ring torque (1) over that band gives the total torque.

N = \frac{2\pi G\varphi}{l}\int_{r-\Delta r}^{r}\rho^3\,d\rho = \frac{2\pi G\varphi}{l}\cdot\frac{r^4-(r-\Delta r)^4}{4}

N \approx \frac{2\pi G\varphi}{l}\cdot r^3\Delta r

Solving for $\varphi$ :

\varphi = \dfrac{Nl}{2\pi G r^3\Delta r}

(2)

Step 4Case (b): the solid rod

Now the material fills the whole disc from the axis out to $\rho = r$ , so integrating the same (1) from $0$ to $r$ gives the total torque.

N = \frac{2\pi G\varphi}{l}\int_0^r \rho^3\,d\rho = \frac{2\pi G\varphi}{l}\cdot\frac{r^4}{4} = \frac{\pi G\varphi r^4}{2l}

Solving for $\varphi$ :

\varphi = \dfrac{2Nl}{\pi G r^4}

(3)

Check

Expanding exactly, $r^4-(r-\Delta r)^4 = 4r^3\Delta r - 6r^2\Delta r^2 + 4r\Delta r^3 - \Delta r^4$ , so the correction to the leading term used in Case (a) is of relative order $\Delta r/r$ , exactly the smallness the problem assumes. Both boxed results are also dimensionless in $\varphi$ : $[Nl]/[Gr^3\Delta r] = (\text{N·m})(\text{m})/(\text{Pa}\cdot\text{m}^4) = 1$ , as an angle must be, and likewise for $[Nl]/[Gr^4]$ .

Remark

Both cases obey the same form $\varphi = Nl/(GJ)$ with a polar second moment $J = 2\pi\int\rho^3\,d\rho$ : $J_{\text{tube}} \approx 2\pi r^3\Delta r$ against $J_{\text{rod}} = \pi r^4/2$ . Comparing them at equal cross-sectional area (equal mass for the same length $l$ ), a solid rod of area $2\pi r\Delta r$ has radius $r_s = \sqrt{2r\Delta r}$ and $J_s = \pi r_s^4/2 = 2\pi r^2\Delta r^2$ , so $J_{\text{tube}}/J_s = r/\Delta r \gg 1$ : moving material away from the axis into a thin tube buys far more torsional stiffness than a solid rod of the same mass, the reason drive shafts are so often built hollow.