Torsion

Statement: for the same material, same length and same weight (equal cross-sectional area), a hollow shaft can transmit more torque than a solid shaft, so it is stronger in torsion.

Let the allowable shear stress be $\tau$. Let the solid shaft have diameter $D_s$ and the hollow shaft external diameter $D$ and internal diameter $d$, with $k = d/D$.

Equal weight, material and length means equal area

$$\frac{\pi}{4}D_s^2 = \frac{\pi}{4}(D^2-d^2) \Rightarrow D_s^2 = D^2 - d^2 = D^2(1-k^2)$$

Torque capacity from the torsion formula $T = \dfrac{\tau J}{R}$

Solid shaft:

$$T_s = \frac{\pi}{16}\tau D_s^3$$

Hollow shaft:

$$T_h = \frac{\pi}{16}\tau\,\frac{D^4-d^4}{D} = \frac{\pi}{16}\tau\,\frac{(D^2-d^2)(D^2+d^2)}{D}$$

Ratio

Using $D_s^2 = D^2 - d^2$:

$$\frac{T_h}{T_s} = \frac{(D^2-d^2)(D^2+d^2)}{D\,D_s^3} = \frac{D^2+d^2}{D\,D_s} = \frac{D^2(1+k^2)}{D\cdot D\sqrt{1-k^2}} = \frac{1+k^2}{\sqrt{1-k^2}}$$

For any $0 < k < 1$:

$$(1+k^2)^2 - (1-k^2) = 3k^2 + k^4 > 0 \Rightarrow \frac{1+k^2}{\sqrt{1-k^2}} > 1$$

Therefore $T_h > T_s$: the hollow shaft carries a larger torque for the same weight.

Numerical illustration

(for $k = 0.6$: $\frac{1+0.36}{\sqrt{1-0.36}} = \frac{1.36}{0.8} = 1.70$).

Reason: in a solid shaft the material near the axis is lightly stressed because shear stress varies linearly from zero at the centre to $\tau$ at the surface. In the hollow shaft the same material is moved to the outside, where the lever arm and the stress are greatest, so it is used more efficiently.

Torsional equation: $\dfrac{T}{J} = \dfrac{\tau_s}{R} = \dfrac{G\varphi}{L}$, where $T$ is the torque, $J$ the polar moment of inertia, $\tau_s$ the shear stress at the surface (radius $R$), $G$ the modulus of rigidity, $\varphi$ the angle of twist and $L$ the length.

Assumptions

1. The shaft is straight, of uniform circular section (solid or hollow) and of homogeneous, isotropic material.
2. Plane sections perpendicular to the axis remain plane and circular after twisting; radii remain straight.
3. The twist is uniform along the length and the stress is within the elastic limit (Hooke's law).
4. Shear stress is proportional to shear strain; the shaft is loaded only by torque.

Derivation

  fixed end                 free end
  |---------------------------|  A -> A' (twist)
  |   O ======================|=> torque T
  |      L          B         B'
  |
  section: OB = R, BB' = R*phi, AB' tilts by gamma

Consider a shaft of length $L$ fixed at one end and twisted by torque $T$ at the other. A line AB on the surface moves to AB'. Let $\varphi$ be the angle of twist and $\gamma$ the shear strain at radius $R$. From the geometry:

$$BB' = R\varphi = L\gamma \Rightarrow \gamma = \frac{R\varphi}{L}$$

For a point at radius $r$, the shear strain is $\gamma_r = \dfrac{r\varphi}{L}$.

From Hooke's law in shear, $\tau = G\gamma$:

$$\tau_r = G\frac{r\varphi}{L} \quad\Rightarrow\quad \frac{\tau_r}{r} = \frac{G\varphi}{L} \quad(1)$$

so the shear stress varies linearly with radius, and at the surface $\dfrac{\tau_s}{R} = \dfrac{G\varphi}{L}$.

Consider an elemental ring of radius $r$ and thickness $dr$, area $dA = 2\pi r\,dr$. The force on it is $\tau_r\,dA$ and its moment about the axis is $\tau_r\,dA\cdot r$. The torque resisted by the entire section is

$$T = \int \tau_r\,r\,dA = \int \frac{\tau_s}{R}\,r^2\,dA = \frac{\tau_s}{R}\int r^2\,dA$$

Since $J = \int r^2\,dA$ (polar moment of inertia):

$$T = \frac{\tau_s}{R}J \quad\Rightarrow\quad \frac{T}{J} = \frac{\tau_s}{R} \quad(2)$$

Combining (1) and (2):

$$\boxed{\frac{T}{J} = \frac{\tau_s}{R} = \frac{G\varphi}{L}}$$

Polar moment of inertia: solid shaft $J = \dfrac{\pi D^4}{32}$; hollow shaft $J = \dfrac{\pi}{32}(D^4-d^4)$. Then $\tau_{max} = \dfrac{16T}{\pi D^3}$ for a solid shaft and $\varphi = \dfrac{TL}{GJ}$, where $GJ$ is the torsional rigidity.

Data: $T = 120\ \text{N·m} = 1.2\times10^5\ \text{N·mm}$, $\tau_{allow} = 100\ \text{N/mm}^2$, $G = 10^5\ \text{N/mm}^2$, allowable twist $\theta = 30^\circ = 0.5236$ rad over a length $L = 10d$ (taken as stated in the question). Find the diameter satisfying both strength and stiffness; the larger value governs.

(a) Strength condition

$$T = \frac{\pi}{16}\tau d^3 \Rightarrow d^3 = \frac{16T}{\pi\tau} = \frac{16(1.2\times10^5)}{\pi(100)} = 6111.5\ \text{mm}^3$$ $$d_1 = 18.28\ \text{mm}$$

(b) Stiffness condition

$$\theta = \frac{TL}{GJ} = \frac{T(10d)}{G\frac{\pi}{32}d^4} = \frac{320\,T}{\pi G d^3} \Rightarrow d^3 = \frac{320(1.2\times10^5)}{\pi(10^5)(0.5236)} = 233.44\ \text{mm}^3$$ $$d_2 = 6.16\ \text{mm}$$

Design diameter

The larger of the two governs, so the shaft is designed for strength: $d = 18.28$ mm.

Answer: provide a solid shaft of diameter $d = 18.28\ \text{mm}$, adopt 20 mm.

Data: $D = 100$ mm, $d = 70$ mm, $N = 300$ rpm, $\tau_{allow} = 80\ \text{N/mm}^2$, $T_{max} = 1.3\,T_{mean}$.

Maximum torque the shaft can carry

$$J = \frac{\pi}{32}(D^4 - d^4) = \frac{\pi}{32}(100^4 - 70^4) = 7.4603e+06\ \text{mm}^4$$ $$T_{max} = \frac{\tau J}{R} = \frac{80\times7.4603e+06}{50} = 1.1936e+07\ \text{N·mm} = 11.936\ \text{kN·m}$$

Mean torque

$$T_{mean} = \frac{T_{max}}{1.3} = \frac{11.936}{1.3} = 9.182\ \text{kN·m}$$

Power transmitted

$$P = \frac{2\pi N T_{mean}}{60} = \frac{2\pi(300)(9181.9\ \text{N·m})}{60} = 288.5\ \text{kW}$$

Answer: power transmitted $\approx 288.5\ \text{kW}$.

Torsion and bending

Replacing the hollow shaft by a solid shaft

Equal polar modulus means $Z_p$ (hollow) $= Z_p$ (solid). Hollow shaft: $D = 30$ cm, $d = 20$ cm. Solid alloy shaft: diameter $D_s$.

$$Z_p = \frac{\pi}{16}\,\frac{D^4-d^4}{D} = \frac{\pi}{16}\,\frac{30^4-20^4}{30} = 4254.2\ \text{cm}^3, \qquad Z_p = \frac{\pi}{16}D_s^3$$ $$D_s^3 = \frac{30^4-20^4}{30} = 21666.67 \Rightarrow D_s = 27.88\ \text{cm}$$

Ratio of torsional rigidities

Torsional rigidity is $GJ$. Hollow (steel) $J_h = \dfrac{\pi}{32}(30^4-20^4) = 63813.6\ \text{cm}^4$; solid (alloy) $J_s = \dfrac{\pi}{32}D_s^4 = 59300.2\ \text{cm}^4$, and $G_{steel} = 2.5\,G_{alloy}$:

$$\frac{(GJ)_{hollow}}{(GJ)_{solid}} = \frac{2.5\,G_{alloy}\,J_h}{G_{alloy}\,J_s} = 2.5\times\frac{63813.6}{59300.2} = 2.690$$

Answer: solid shaft diameter $= 27.88$ cm; ratio of torsional rigidities (hollow : solid) $= 2.69$.

Reading of the figure: shaft fixed at A and D; AB and CD are copper ($\phi 80$ mm), BC is steel ($\phi 60$ mm), each 1 m long. A torque $T_B = 20$ kN·m acts at B and $T_C = 4$ kN·m acts at C in the opposite sense. The problem is statically indeterminate (one equation of equilibrium, one of compatibility).

Polar moments

$$J_{80} = \frac{\pi}{32}(80)^4 = 4.0212e+06\ \text{mm}^4, \qquad J_{60} = \frac{\pi}{32}(60)^4 = 1.2723e+06\ \text{mm}^4$$

Compatibility

Let $T_A$ be the reaction at A (positive in the same sense as $T_B$). The torques in the segments are

$$T_{AB} = T_A, \qquad T_{BC} = T_A + 20, \qquad T_{CD} = T_A + 16 \quad(\text{kN·m})$$

The fixed end D does not rotate relative to A, so the total twist is zero ($\varphi = \sum \frac{TL}{GJ} = 0$):

$$\frac{T_A L}{G_cJ_{80}} + \frac{(T_A+20)L}{G_sJ_{60}} + \frac{(T_A+16)L}{G_cJ_{80}} = 0$$

With $\dfrac{L}{G_cJ_{80}} = 5.6518e-09$ and $\dfrac{L}{G_sJ_{60}} = 1.0076e-08$ (per N·mm):

$$T_A = -\frac{20\,b + 16\,a}{2a + b} = -13.656\ \text{kN·m}$$

Torsional reactions

• At A: $|T_A| = 13.656$ kN·m, acting opposite to $T_B$.
• At D: $|T_D| = 2.344$ kN·m, acting opposite to $T_B$.

Check: $13.656 + 2.344 = 16.000 = 20 - 4$ kN·m.

Rotations of the joints (D is fixed, so its rotation is zero)

$$\varphi_B = \frac{|T_{AB}|L}{G_cJ_{80}} = 0.07718\ \text{rad} = 4.422^\circ$$ $$\varphi_C = \varphi_B - \frac{T_{BC}L}{G_sJ_{60}} = 0.01325\ \text{rad} = 0.759^\circ$$

Both are in the direction of the 20 kN·m torque, and $\varphi_D = 0$ (check: $\varphi_C - \frac{T_{CD}L}{G_cJ_{80}} = 0$).

Maximum shear stress ($\tau = \dfrac{16T}{\pi d^3}$)

Data: $P = 300$ kW, $N = 120$ rpm, $\tau_{max} = 100\ \text{MPa}$.

Torque

$$T = \frac{60P}{2\pi N} = \frac{60(300\times10^3)}{2\pi(120)} = 23873.2\ \text{N·m} = 2.3873e+07\ \text{N·mm}$$

Solid shaft

$$T = \frac{\pi}{16}\tau d^3 \Rightarrow d^3 = \frac{16T}{\pi\tau} = \frac{16(2.3873e+07)}{\pi(100)} = 1.2159e+06\ \text{mm}^3$$ $$d = 106.73\ \text{mm}$$

Hollow shaft ($d_i = 0.6D_o$, same $T$ and $\tau$)

$$T = \frac{\pi}{16}\tau D_o^3(1-0.6^4) \Rightarrow D_o^3 = \frac{16T}{\pi\tau(1-0.1296)} = 1.3969e+06\ \text{mm}^3$$ $$D_o = 111.79\ \text{mm}, \qquad d_i = 0.6D_o = 67.07\ \text{mm}$$

Saving in weight

Weight $\propto$ area (same length and material).

$$A_s = \frac{\pi}{4}(106.73)^2 = 8947.0\ \text{mm}^2, \qquad A_h = \frac{\pi}{4}(111.79^2 - 67.07^2) = 6281.2\ \text{mm}^2$$ $$\text{Saving} = \frac{A_s - A_h}{A_s}\times100 = 29.8\%$$

Answer: solid shaft $d = 106.7$ mm; the hollow shaft ($D_o = 111.8$ mm, $d_i = 67.1$ mm) saves about 29.8% of the weight.

Torsional rigidity is the torque required to produce a twist of one radian per unit length of the shaft; it equals the product $GJ$ of the modulus of rigidity and the polar moment of inertia (from $\dfrac{T}{\varphi} = \dfrac{GJ}{L}$). Unit: N·mm² (or N·m²).

Torque of the hollow shaft

Data: $D = 100$ mm, $d = 70$ mm, $\tau = 80\ \text{MN/m}^2 = 80\ \text{N/mm}^2$, $N = 300$ rpm (the speed is not needed for the torque).

$$J = \frac{\pi}{32}(100^4 - 70^4) = 7.4603e+06\ \text{mm}^4$$ $$T_h = \frac{\tau J}{R} = \frac{80\times7.4603e+06}{50} = 1.1936e+07\ \text{N·mm} = 11.936\ \text{kN·m}$$

(Power at 300 rpm would be $P = \frac{2\pi N T}{60} = 375.0$ kW.)

Comparison with a solid shaft of the same weight

Same material, weight and length means equal area:

$$\frac{\pi}{4}D_s^2 = \frac{\pi}{4}(100^2 - 70^2) \Rightarrow D_s = \sqrt{5100} = 71.41\ \text{mm}$$ $$T_s = \frac{\pi}{16}\tau D_s^3 = \frac{\pi}{16}(80)(71.41)^3 = 5.7210e+06\ \text{N·mm} = 5.721\ \text{kN·m}$$ $$\frac{T_h}{T_s} = \frac{11.936}{5.721} = 2.086$$

Answer: torque of the hollow shaft $= 11.94$ kN·m; it is 2.09 times as strong as the solid shaft of equal weight (solid shaft torque $= 5.72$ kN·m).

Data: $P = 300$ kW, $N = 120$ rpm, $\tau_{allow} = 100\ \text{N/mm}^2$, $G = 10^5\ \text{N/mm}^2$, allowable twist $30^\circ$ per diameter length (taken as stated, $L = d$), $T_{max} = 1.3\,T_{mean}$.

Torques

$$T_{mean} = \frac{60P}{2\pi N} = \frac{60(300\times10^3)}{2\pi(120)} = 23873.2\ \text{N·m}, \qquad T_{max} = 1.3T_{mean} = 31035.2\ \text{N·m}$$

Strength

$$d^3 = \frac{16T_{max}}{\pi\tau} = \frac{16(3.1035e+07)}{\pi(100)} = 1.5806e+06\ \text{mm}^3 \Rightarrow d_1 = 116.49\ \text{mm}$$

Stiffness

$$\theta = \frac{T_{max}L}{GJ} = \frac{32\,T_{max}\,d}{\pi G d^4} = \frac{32T_{max}}{\pi G d^3}, \quad \theta = 30^\circ = 0.5236\ \text{rad}$$ $$d^3 = \frac{32(3.1035e+07)}{\pi(10^5)(0.5236)} = 6.0375e+03\ \text{mm}^3 \Rightarrow d_2 = 18.21\ \text{mm}$$

Design

The larger value governs, so strength controls the design: $d = 116.49$ mm.

Answer: diameter of the shaft $\approx 116.5\ \text{mm}$ (adopt 120 mm).

Reading of the figure: shaft of uniform diameter fixed at A and D; $T_B = 15$ kN·m at B and $T_C = 25$ kN·m at C (opposite sense to $T_B$); $AB = 1.5$ m, $BC = 1.2$ m, $CD = 1.5$ m; $\tau_{max} = 50\ \text{MN/m}^2$.

End fixing couples

Let $T_A$ be the reaction at A (positive in the sense of $T_B$). The torques in the segments are

$$T_{AB} = T_A, \qquad T_{BC} = T_A + 15, \qquad T_{CD} = T_A + 15 - 25 = T_A - 10 \quad(\text{kN·m})$$

The ends are fixed, so the total twist is zero. $G$ and $J$ are constant:

$$T_A(1.5) + (T_A+15)(1.2) + (T_A-10)(1.5) = 0 \Rightarrow 4.2\,T_A + 3 = 0 \Rightarrow T_A = -0.714\ \text{kN·m}$$

The negative sign means the reaction at A acts opposite to $T_B$. Also $T_D = 10.714$ kN·m from equilibrium ($T_A + T_D + 15 - 25 = 0$ with the signs above).

Answer (end couples): $|T_A| = 0.714$ kN·m, acting opposite to $T_B$; $|T_D| = 10.714$ kN·m, acting in the same sense as $T_B$.

Diameter

The maximum torque is $T_{BC} = 14.286$ kN·m.

$$d^3 = \frac{16T_{max}}{\pi\tau} = \frac{16(1.4286e+07)}{\pi(50)} = 1.4551e+06\ \text{mm}^3 \Rightarrow d = 113.32\ \text{mm}$$

Section with no angular twist

The rotation of a section relative to A is $\varphi = \sum \dfrac{TL}{GJ}$. At B (taking $GJ = 1$ units, kN·m²): $\varphi_B = 0.714\times1.5 = 1.071$. In BC the twist changes at the rate $T_{BC} = 14.286$ in the opposite direction, so it reaches zero at a distance

$$x = \frac{1.071}{14.286} = 0.075\ \text{m from B}$$

Answer: the section $\approx 0.075$ m from B (i.e. 1.575 m from A, inside BC) suffers no angular twist; $d \approx 113.3$ mm.

Data: $P = 200$ hp $= 200\times746 = 149200$ W, $N = 150$ rpm, $d = 110$ mm, $L = 5$ m, $G = 90\ \text{GN/m}^2 = 90\times10^3\ \text{N/mm}^2$.

Torque

$$T = \frac{60P}{2\pi N} = \frac{60(149200)}{2\pi(150)} = 9498.4\ \text{N·m} = 9.4984e+06\ \text{N·mm}$$

Maximum shear stress

$$\tau_{max} = \frac{16T}{\pi d^3} = \frac{16(9.4984e+06)}{\pi(110)^3} = 36.34\ \text{N/mm}^2$$

Angle of twist in 5 m

$$J = \frac{\pi}{32}(110)^4 = 1.4374e+07\ \text{mm}^4$$ $$\varphi = \frac{TL}{GJ} = \frac{(9.4984e+06)(5000)}{(90\times10^3)(1.4374e+07)} = 0.03671\ \text{rad} = 2.103^\circ$$

Answer: $T = 9.498$ kN·m; $\tau_{max} = 36.34$ MPa; $\varphi = 0.0367$ rad $= 2.10^\circ$.

Data: $L = 1$ m, $P = 1670$ kW, $N = 4500$ rpm, $D = 1.75d$, $\tau_{max} = 210\ \text{MPa}$, $\theta \le 0.5^\circ$, $T_{max} = 1.3\,T_{mean}$, $G = 25.5\ \text{GPa}$. So $k = d/D = 1/1.75 = 0.5714$ and $1-k^4 = 0.8934$.

Torque

$$T_{mean} = \frac{60P}{2\pi N} = \frac{60(1670\times10^3)}{2\pi(4500)} = 3543.9\ \text{N·m}, \qquad T_{max} = 1.3T_{mean} = 4607.0\ \text{N·m}$$

Strength condition

$$T_{max} = \frac{\pi}{16}\tau D^3(1-k^4) \Rightarrow D^3 = \frac{16(4.6070e+06)}{\pi(210)(0.8934)} = 1.2506e+05\ \text{mm}^3$$ $$D_1 = 50.01\ \text{mm}$$

Stiffness condition

$$\theta = \frac{T_{max}L}{GJ}, \quad J = \frac{\pi}{32}D^4(1-k^4), \quad \theta = 0.5^\circ = 0.00873\ \text{rad}$$ $$D^4 = \frac{32\,T_{max}L}{\pi G\theta(1-k^4)} = \frac{32(4.6070e+06)(1000)}{\pi(25\,500)(0.00873)(0.8934)} = 2.3605e+08\ \text{mm}^4$$ $$D_2 = 123.95\ \text{mm}$$

Size of the shaft

The larger value governs (stiffness controls):

$$D = 123.95\ \text{mm}, \qquad d = \frac{D}{1.75} = 70.83\ \text{mm}$$

Answer: external diameter $\approx 124.0$ mm, internal diameter $\approx 70.8$ mm.

Data: $\tau_{allow} = 50\ \text{MPa}$; segment AC: $d_1 = 20$ mm, $L_1 = 0.7$ m; segment CB: $d_2 = 30$ mm, $L_2 = 0.5$ m; same material; torque $T$ applied at C; both ends fixed.

Distribution of torque

Let $T_A$ and $T_B$ be the end reactions: $T_A + T_B = T$ (equilibrium). The rotation of C calculated from either end must be equal:

$$\frac{T_AL_1}{GJ_1} = \frac{T_BL_2}{GJ_2} \Rightarrow \frac{T_A}{T_B} = \frac{L_2/J_2}{L_1/J_1} = \frac{0.5/30^4}{0.7/20^4} = 0.1411$$

So $T_A = 0.1236\,T$ and $T_B = 0.8764\,T$.

Limiting torque

Stress in each segment is $\tau = \dfrac{16T_i}{\pi d^3}$.

• AC reaches 50 MPa when $T_A = \dfrac{\pi(50)(20)^3}{16} = 78.54$ N·m. Then $T = T_A/0.1236 = 635.2$ N·m.
• CB reaches 50 MPa when $T_B = \dfrac{\pi(50)(30)^3}{16} = 265.07$ N·m. Then $T = T_B/0.8764 = 302.5$ N·m.

The smaller value governs (segment CB is stressed to the limit first):

$$T_{max} = 302.5\ \text{N·m}$$

Shear stress at A

With this torque, $T_A = 37399.9$ N·mm and $T_B = 265071.9$ N·mm, so

$$\tau_A = \frac{16T_A}{\pi(20)^3} = 23.81\ \text{MPa}, \qquad \tau_B = \frac{16T_B}{\pi(30)^3} = 50.00\ \text{MPa}$$

Answer: maximum torque at C $= 302.5$ N·m ($\approx 0.302$ kN·m); shear stress at A (segment AC) $= 23.8$ MPa.

Data: length $= 11.25$ m, fixed at both ends. Viewed from A: $T_C = 30$ kN·m clockwise at 4.5 m from A, $T_D = 37.5$ kN·m anticlockwise at 7.5 m from A. Segments: $L_1 = 4.5$ m (A to C), $L_2 = 3.0$ m (C to D), $L_3 = 3.75$ m (D to B). $\tau_{max} = 60\ \text{N/mm}^2$.

End fixing couples

Take clockwise (viewed from A) as positive. Let $T_A$ be the end couple at A. The torques in the segments are

$$T_1 = T_A, \qquad T_2 = T_A + 30, \qquad T_3 = T_A + 30 - 37.5 = T_A - 7.5 \quad(\text{kN·m})$$

For a shaft fixed at both ends the total twist is zero ($G$, $J$ constant):

$$T_A(4.5) + (T_A+30)(3.0) + (T_A-7.5)(3.75) = 0$$ $$11.25\,T_A + 90 - 28.125 = 0 \Rightarrow T_A = -5.50\ \text{kN·m}$$

The negative sign means the couple at A is anticlockwise. From overall equilibrium, $T_A + 30 - 37.5 + T_B = 0$:

$$T_B = 13.00\ \text{kN·m}\ \text{(clockwise)}$$

End couples: at A, $5.50$ kN·m anticlockwise; at B, $13.00$ kN·m clockwise (viewed from A). Check: $13.00 - 5.50 = 7.50 = 37.5 - 30$.

Diameter

The maximum torque is in segment CD: $T_{max} = 24.50$ kN·m.

$$d^3 = \frac{16T_{max}}{\pi\tau} = \frac{16(2.4500e+07)}{\pi(60)} = 2.0796e+06\ \text{mm}^3$$ $$d = 127.64\ \text{mm}$$

Answer: end couples $5.5$ kN·m (A, anticlockwise) and $13$ kN·m (B, clockwise); diameter $\approx 127.6$ mm.

Data: $P = 40$ kW, $N = 240$ rpm, $\tau_{max} = 100\ \text{MPa}$, $D = 2d$ (so $k = d/D = 0.5$ and $1-k^4 = 0.9375$).

Torque

$$T = \frac{60P}{2\pi N} = \frac{60(40\times10^3)}{2\pi(240)} = 1591.5\ \text{N·m} = 1.5915e+06\ \text{N·mm}$$

Diameters

$$T = \frac{\pi}{16}\tau\,D^3(1-k^4) \Rightarrow D^3 = \frac{16T}{\pi\tau(1-k^4)} = \frac{16(1.5915e+06)}{\pi(100)(0.9375)} = 8.6461e+04\ \text{mm}^3$$ $$D = 44.22\ \text{mm}, \qquad d = \frac{D}{2} = 22.11\ \text{mm}$$

Answer: external diameter $\approx 44.2$ mm, internal diameter $\approx 22.1$ mm. (In practice adopt 45 mm and 22.5 mm.)

Reading of the figure: fixed at the left end; three segments each 500 mm long with diameters 80, 60 and 40 mm. Torques of 10 kN·m (at the end of segment 1), 4 kN·m (end of segment 2) and 2 kN·m (free end) act in the same sense. $G = 80$ GPa $= 8\times10^4\ \text{N/mm}^2$.

Torque in each segment (section method from the free end)

Twist of each segment

$$\varphi_i = \frac{T_iL_i}{GJ_i}, \qquad J_i = \frac{\pi}{32}d_i^4$$ $$\varphi_{free} = 0.02487 + 0.02947 + 0.04974 = 0.10408\ \text{rad} = 5.963^\circ$$

Maximum shear stress

$$\tau = \frac{16T}{\pi d^3}: \qquad \tau_1 = 159.2,\quad \tau_2 = 141.5,\quad \tau_3 = 159.2\ \text{N/mm}^2$$

The maximum occurs in segment 2 (the 60 mm part).

Answer: angle of twist at the free end $= 5.96^\circ$ (0.1041 rad); $\tau_{max} = 159.2$ MPa.

Condition: the hollow member must carry the same axial load as the solid one with the same material and the same stress, so the cross-sectional area must be equal. Let the hollow section have outer diameter $D$ and inner diameter $d_i = 0.8D$.

$$A_{solid} = \frac{\pi}{4}(50)^2 = 1963.5\ \text{mm}^2$$ $$A_{hollow} = \frac{\pi}{4}(D^2 - d_i^2) = \frac{\pi}{4}D^2(1-0.8^2) = \frac{\pi}{4}D^2(0.36)$$

Equating areas:

$$0.36D^2 = 50^2 \Rightarrow D = \frac{50}{0.6} = 83.33\ \text{mm}, \qquad d_i = 0.8D = 66.67\ \text{mm}$$

Wall thickness $t = \dfrac{D - d_i}{2} = 8.33$ mm.

Comparison (why the hollow section is better in compression)

For the same weight, the hollow section has a much larger moment of inertia and radius of gyration, so its slenderness ratio is smaller and it resists buckling better.

Answer: outer diameter $\approx 83.3$ mm and inner diameter $\approx 66.7$ mm.

Reading of the figure: fixed at the left; segment 1: $d = 110$ mm, $L = 3$ m, with a 10 kN·m torque at its right end; segment 2: $d = 60$ mm, $L = 2$ m, with a 5 kN·m torque at the free end, both in the same sense. $G = 0.9\times10^5\ \text{N/mm}^2$.

Internal torques

• Segment 2: $T_2 = 5$ kN·m
• Segment 1: $T_1 = 10 + 5 = 15$ kN·m

Polar moments

$$J_1 = \frac{\pi}{32}(110)^4 = 1.4374e+07\ \text{mm}^4, \qquad J_2 = \frac{\pi}{32}(60)^4 = 1.2723e+06\ \text{mm}^4$$

Angle of twist

$$\varphi_1 = \frac{T_1L_1}{GJ_1} = \frac{(15\times10^6)(3000)}{(0.9\times10^5)(1.4374e+07)} = 0.03479\ \text{rad}$$ $$\varphi_2 = \frac{T_2L_2}{GJ_2} = \frac{(5\times10^6)(2000)}{(0.9\times10^5)(1.2723e+06)} = 0.08733\ \text{rad}$$ $$\varphi = \varphi_1 + \varphi_2 = 0.12211\ \text{rad} = 6.997^\circ$$

Answer: angle of twist at the free end $= 0.1221$ rad $= 7.00^\circ$.

Data: brass AB: $d = 80$ mm, $L = 1000$ mm, $\tau_{allow} = 88\ \text{N/mm}^2$, $G = 400\ \text{kN/mm}^2$; steel BC: $d = 60$ mm, $L = 1200$ mm, $\tau_{allow} = 110\ \text{N/mm}^2$, $G = 80\ \text{kN/mm}^2$. The same torque $T$ acts in both segments (applied at the free end C).

Maximum torque

$$T_{brass} = \frac{\pi}{16}\tau d^3 = \frac{\pi}{16}(88)(80)^3 = 8.8467e+06\ \text{N·mm} = 8.847\ \text{kN·m}$$ $$T_{steel} = \frac{\pi}{16}(110)(60)^3 = 4.6653e+06\ \text{N·mm} = 4.665\ \text{kN·m}$$

The smaller governs, so $T_{max} = 4.665$ kN·m (the steel segment reaches its limit first).

Rotation of the free end relative to the fixed end

$$\varphi_{AB} = \frac{TL}{GJ} = \frac{(4.6653e+06)(1000)}{(400\times10^3)\frac{\pi}{32}(80)^4} = 0.00290\ \text{rad}$$ $$\varphi_{BC} = \frac{(4.6653e+06)(1200)}{(80\times10^3)\frac{\pi}{32}(60)^4} = 0.05500\ \text{rad}$$ $$\varphi_C = \varphi_{AB} + \varphi_{BC} = 0.05790\ \text{rad} = 3.317^\circ$$

Answer: $T_{max} = 4.67$ kN·m; total rotation of C $= 0.0579$ rad $= 3.32^\circ$.

Elasto-plastic material: stress rises linearly with strain up to the yield stress in shear $\tau_y$ and then stays constant at $\tau_y$ (perfectly plastic).

Maximum elastic torque

The shaft is on the point of yielding when the surface stress reaches $\tau_y$. For a solid shaft of radius $R$ with $J = \dfrac{\pi R^4}{2}$:

$$T_e = \frac{\tau_y J}{R} = \frac{\tau_y}{R}\cdot\frac{\pi R^4}{2} = \frac{\pi\tau_y R^3}{2}$$

Fully plastic torque

When the whole section yields, the stress is $\tau_y$ at every radius. Take a ring of radius $r$, thickness $dr$, area $2\pi r\,dr$:

$$dT = \tau_y\,(2\pi r\,dr)\,r = 2\pi\tau_y r^2\,dr$$ $$T_p = \int_0^R 2\pi\tau_y r^2\,dr = 2\pi\tau_y\frac{R^3}{3} = \frac{2\pi\tau_y R^3}{3}$$

Ratio

$$\frac{T_p}{T_e} = \frac{2\pi\tau_y R^3/3}{\pi\tau_y R^3/2} = \frac{2}{3}\times 2 = \frac{4}{3}$$

Hence the plastic torque is $\dfrac{4}{3}$ times the maximum elastic torque.