Torsion
Shear stress and twist in circular shafts, power transmission, indeterminate shafts, and thin-walled tubes.
Chapter Objectives
- Describe how a circular shaft deforms under torque, and derive the torsion formula τ = Tc/J.
- Size shafts that transmit power at a given rotational speed.
- Compute the angle of twist and solve statically indeterminate torque-loaded shafts.
- Analyze noncircular shafts, thin-walled closed tubes (shear flow), and torsional stress concentrations.
5.1Torsional Deformation of a Circular Shaft
Torque and Shaft Kinematics
Torque is a moment that twists a member about its longitudinal axis — the primary loading of drive shafts. For a circular shaft, the deformation has a special simplicity: longitudinal grid lines twist into helices, circles remain circles, cross sections remain flat (no warping), and radial lines remain straight while rotating. For small twist, the shaft's length and radius are essentially unchanged.
If the shaft is fixed at one end, a cross section at distance \(x\) rotates through the angle of twist \(\phi(x)\). A thin disk at \(x\) has its back face rotated by \(\phi\) and its front face by \(\phi + d\phi\); the relative rotation \(d\phi\) shears the material. Matching the arc length: \(\rho\,d\phi = \gamma\,dx\), so
- Since \(d\phi/dx\) is the same for every point of a cross section, the shear strain varies linearly with radial distance \(\rho\): zero at the axis, maximum \(\gamma_{max}\) at the outer surface \(\rho = c\).
- This result is purely geometric — it holds regardless of material behavior.
5.2The Torsion Formula
For linear elastic material, Hooke's law \(\tau = G\gamma\) converts the linear strain variation into a linear stress variation, \(\tau = (\rho/c)\,\tau_{max}\). Requiring the stress distribution to produce the internal torque, \(T = \int_A \rho\,(\tau\,dA) = (\tau_{max}/c)\int_A \rho^2\,dA\), and naming the integral \(J\):
- \(\tau_{max}\) = maximum shear stress, at the outer surface; \(\tau\) = shear stress at interior radius \(\rho\)
- \(T\) = internal torque (method of sections, moment equilibrium about the shaft axis)
- \(J\) = polar moment of inertia of the cross-sectional area; \(c\) = outer radius
- Valid for linear elastic, homogeneous circular shafts and tubes.
- \(J\) is always positive; units mm⁴ or in⁴. \(c_o\), \(c_i\) = outer and inner radii of the tube.
Axial Shear and Wooden Shafts
By the complementary property of shear, the shear stress on the cross section is accompanied by an equal shear stress on axial planes. Wood — far weaker in shear parallel to its grain — therefore tends to split along axial planes when over-torqued.
5.3Power Transmission
A rotating shaft transmits work \(T\,d\theta\) in time \(dt\); power is work per unit time, so:
- \(P\) = power (W = N·m/s in SI; ft·lb/s in FPS, with 1 hp = 550 ft·lb/s)
- \(\omega\) = angular velocity (rad/s); \(f\) = frequency of rotation (Hz), \(\omega = 2\pi f\)
- With \(T = P/\omega\) known, this fixes the geometry. Solid shaft: \(J/c = \pi c^3/2\) gives a unique radius. Tube: choose one of \(c_o, c_i\), solve for the other.
5.4Angle of Twist
Combining the kinematic relation \(d\phi = \gamma\,dx/\rho\), Hooke's law \(\gamma = \tau/G\), and the torsion formula \(\tau = T(x)\rho/J(x)\) gives \(d\phi = T(x)\,dx/[J(x)G(x)]\), the torsional analogue of the axial-bar development (with Saint-Venant's principle again excusing the localized end effects):
- \(\phi\) = angle of twist of one end relative to the other, in radians
- Use \(TL/JG\) per segment when torque, area, or material change abruptly, then sum algebraically.
- Sign convention (right-hand rule): internal torque is positive when the thumb points away from the sectioned face; twist follows the same rule.
5.5Statically Indeterminate Torque-Loaded Members
Indeterminate Shafts
A shaft is statically indeterminate when moment equilibrium about its axis cannot determine all the support torques — e.g. a shaft fixed at both ends. The solution mirrors Section 4.4: equilibrium (\(\Sigma M = 0\) about the axis), a compatibility condition (for two fixed ends, \(\phi_{A/B} = 0\)), and the load–displacement relation \(\phi = TL/JG\) to express compatibility in terms of torques.
5.6Solid Noncircular Shafts
Warping of Noncircular Sections
Only circular (axisymmetric) shafts keep plane cross sections when twisted. Noncircular sections bulge and warp, making the analysis far more complex (theory of elasticity). Two key facts for a square shaft: the shear stress distribution along radial lines is nonlinear, and the corner points carry zero shear stress — the complementary stresses on the free outer surfaces at a corner must vanish, so the stress on the cross section there vanishes too. The maximum shear stress occurs at the edge point closest to the shaft's axis (the midpoint of each flat side for a square).
Elasticity-theory results (tabulated for square, triangular, and elliptical sections) give \(\tau_{max}\) and \(\phi\) for each shape — e.g. for a square of side \(a\): \(\tau_{max} = 4.81\,T/a^3\) and \(\phi = 7.10\,TL/(a^4 G)\). Comparing shapes shows the circular cross section is the most efficient carrier of torque: it develops the smallest maximum stress and the smallest angle of twist for a given torque and area.
5.7Thin-Walled Tubes with Closed Cross Sections
Light frameworks (aircraft structures) use thin-walled noncircular tubes. For a closed section (no slits) with thin walls, the shear stress is taken as uniform across the wall thickness. Longitudinal equilibrium of a wall element shows that the product τ·t is the same at every point of the cross section:
- \(q\) = shear flow — force per unit length along the wall centerline. Because \(q\) is constant, the largest stress occurs where the wall is thinnest.
- \(A_m\) = mean area enclosed by the centerline of the tube's wall.
- \(t\) = wall thickness where the stress is evaluated; the line integral runs around the wall centerline.
5.8Stress Concentration
Torsional Stress Concentrations
The torsion formula fails at sudden changes of cross section. The three common discontinuities are couplings (bolted flange connections), keyways (slots for gears and pulleys), and shoulder fillets on stepped shafts. As with axial loads, the peak stress is captured by a torsional stress concentration factor read from experimental graphs as a function of geometry only (for a stepped shaft: the ratio D/d selects the curve; r/d locates K on it).
- \(Tc/J\) is computed at the smaller cross section. K matters most for brittle materials and fatigue loading; ductile materials under static torque locally yield and redistribute the stress.
5.9Inelastic Torsion
Plastic Analysis of Shafts
Excessive torque yields the material, and a plastic analysis is required. Two facts always hold: the shear strain still varies linearly with radius (pure geometry, Section 5.1), and the stress distribution must still be statically equivalent to the torque: \(T = \int_A \rho\,\tau\,dA = 2\pi\int_0^c \tau\,\rho^2\,d\rho\).
For an elastic perfectly plastic material, yielding begins at the outer surface when \(T_Y = (\pi/2)\,\tau_Y c^3\,/c \cdot c = \tau_Y\,J/c\). Increasing torque spreads an annulus of yielded material inward at constant stress \(\tau_Y\) while an elastic core remains. When the whole section has yielded, the shaft carries its fully plastic torque:
\[ T_p = 2\pi\int_0^c \tau_Y\,\rho^2\,d\rho = \frac{2\pi}{3}\,\tau_Y\,c^3 = \frac{4}{3}\,T_Y \]— one-third more than the torque at first yield, after which the shaft twists freely (no strain hardening assumed).
5.10Residual Stress
Residual Shear Stress After Unloading
When a shaft that has been twisted into the plastic range is released, the unloading is elastic — the material recovers along a line parallel to the initial elastic slope G. The residual stress distribution is found by superposition: subtract from the inelastic (plastic) stress distribution the fully elastic, linear distribution computed for the same torque applied in reverse (using \(\tau = T\rho/J\), which may fictitiously exceed \(\tau_Y\) since it merely represents elastic recovery). The result is a self-equilibrating pattern: material near the surface that yielded ends up with reversed residual shear stress, while the interior retains stress in the original sense. A permanent angle of twist also remains.
Problem-solving strategy
- Draw the whole-member and cut free bodies before choosing a stress relation.
- Declare axes and signs, then determine the internal resultant at the critical location.
- Select a relation whose geometry and material assumptions match the member.
- Check units, directions, strength, instability, and serviceability limits.
Physical meaning
The internal resultants describe how neighboring material resists the applied loading; stress is their local intensity, not the applied force.
Recognition strategy
A transmitted twisting moment and a circular, prismatic shaft indicate Saint-Venant elastic torsion; the requested quantity is stress, not the applied force.
Given information
The shaft is solid and circular, with \(d=32,\text{mm}\), \(T=0.42,\text{kN}\!\cdot\!\text{m}\), and an internal cut taken between the motor and impeller.
Model and interpretation
Take positive torque by the right-hand rule. A cut free body supplies \(T\); assume homogeneous linear elasticity and use \(J=\pi d^4/32\), \(c=d/2\).
Verification
Units: \(\text{N}\!\cdot\!\text{m}^2/\text{m}^4=\text{Pa}\), so the computed stress has correct dimensions.
Signs and direction: The stress reverses if torque reverses; its reported magnitude acts tangentially on the surface.
Plausibility: Halving the radius would reduce \(J\) strongly and increase stress eightfold, consistent with the cubic diameter sensitivity.
Engineering interpretation
This is a local elastic stress prediction away from the coupling and impeller. Compare it with an allowable shear criterion and inspect keyways or shoulders separately, because nominal shaft stress does not capture their fatigue peaks.
Recognition strategy
The same torque passes serially through both circular segments, while changing diameter requires summing \(TL/(JG)\) rather than using one average section.
Given information
Both segments carry \(T=300,\text{N}\!\cdot\!\text{m}\); \(L_{AB}=0.60,\text{m}\), \(L_{BC}=0.40,\text{m}\), and linear elastic steel is supplied.
Model and interpretation
Use a right-hand positive rotation from the fixed end. Compatibility is continuity of rotation at B; neglect local end warping and treat each segment as circular and prismatic.
Verification
Units: Each \(TL/(JG)\) term is dimensionless radians because \(GJ\) has units of \(\text{N}\!\cdot\!\text{m}^2\).
Signs and direction: Both contributions have the applied-torque sense, so they add; an oppositely directed torque would subtract.
Plausibility: The smaller segment supplies most twist despite being shorter, matching torsional stiffness \(GJ/L\).
Engineering interpretation
Rotation, not stress alone, can govern actuator positioning. A serviceability limit should be checked alongside stress; spline backlash, attachments, and nonlinear material behavior are outside this idealized shaft calculation.
Recognition strategy
A thin, closed, single-cell tube is recognized by its enclosed median area; Bredt shear flow, not the circular \(T\rho/J\) formula, distributes torque.
Given information
Use \(A_m=(0.080)(0.050)\,\text{m}^2\), \(T=900,\text{N}\!\cdot\!\text{m}\), and the stated wall thicknesses; the requested result is the controlling wall stress.
Model and interpretation
Assume walls are thin relative to dimensions, the cell is closed, and shear flow is constant. Follow the positive flow direction that produces the positive applied torque.
Verification
Units: \(q\) has \(\text{N/m}\), and division by wall thickness gives \(\text{N/m}^2\), or pascals.
Signs and direction: Flow circulates continuously around the closed cell; only its direction changes with the torque sense, not the reported peak magnitude.
Plausibility: With common flow, the 1.5-mm walls correctly have \(2.5/1.5\) times the stress of the thicker walls.
Engineering interpretation
The thin walls govern static shear and are likely fatigue-critical near joints or cutouts. Aircraft tube sizing additionally requires buckling, fastener load transfer, manufacturing tolerance, and approved allowable-data review.
Chapter recap
- Use an equilibrium free body to obtain the internal torque before calculating stress.
- State coordinates, signs, section properties, material assumptions, and the governing compatibility relation.
- Before moving on: select the applicable model, calculate with consistent units, and audit strength and serviceability.