Energy Methods, Torsion of Shafts, Euler Buckling and Thin-Walled Pressure Vessels

This chapter belongs to the Engineering Sciences (XE) paper and answers the closing lines of Section XE3.2, Mechanics of Deformable Bodies, of XE3 Solid Mechanics, one of the nine optional sections, any two of which a candidate chooses in the exam. The lines it covers read: “energy methods (Castigliano’s theorems); torsion of circular shaft; Euler Buckling; thin-walled pressure vessels.” It teaches strain energy and Castigliano’s theorems for deflections and for the redundant reaction, torsion of solid and hollow circular shafts including power transmission and shafts in series and parallel, Euler’s critical load with effective lengths and the limit of the formula, and the stresses and volume change of thin cylinders and spheres. The numericals are Castigliano deflections, shaft diameter and twist, series shafts, critical loads and slenderness limits, and hoop stress with volumetric strain.

1. Strain energy and Castigliano’s theorems

For a linear elastic body the work done by the loads is stored as strain energy U. The standard expressions are: axial U = ∫N²dx/(2AE) (for a uniform bar U = P²L/2AE); bending U = ∫M²dx/(2EI); torsion U = ∫T²dx/(2GJ); transverse shear U = ∫αV²dx/(2GA) with a section factor α. Energy is quadratic in the load, so it can never be superposed: U of two loads is not the sum of the two separate energies, because of the cross term.

Castigliano’s first theorem: if U is written as a function of the displacements δᵢ, the load at i is Pᵢ = ∂U/∂δᵢ. Castigliano’s second theorem: if U is written as a function of the loads, the displacement at i, in the direction of Pᵢ, is δᵢ = ∂U/∂Pᵢ. For a point where there is no load, add a dummy load Q, differentiate, and then set Q = 0. The same applies to rotations with a couple in place of Q.

Worked example: cantilever tip deflection. Take x from the free end, M = −Px. U = ∫₀ᴸ P²x²dx/(2EI) = P²L³/(6EI), so δ = ∂U/∂P = PL³/(3EI), matching the standard result. For a tip load P plus a uniform load w, with a dummy load Q kept at the tip, M = −(P + Q)x − wx²/2, ∂M/∂Q = −x, and δ = ∫(M)(∂M/∂Q)dx/EI = [PL³/3 + wL⁴/8]/EI. Numerically, for L = 2 m, EI = 8000 kN·m², P = 5 kN and w = 10 kN/m: δ = (5 × 8/3 + 10 × 16/8)/8000 = 33.33/8000 = 4.17 mm.

⚠️ Dummy load and the direction of the displacement
The theorem gives the displacement of the point of application in the direction of that load. Differentiate before substituting numbers, and set the dummy load to zero only after differentiating. In bending-dominated beams and frames shear and axial energy are normally ignored, so state which terms are kept; for a short, deep beam the shear term is not negligible.

2. Castigliano for the redundant reaction

If a support is a rigid prop, the deflection at the prop is zero, so with the redundant reaction R as the load variable ∂U/∂R = 0 (and, for a spring support of stiffness k, ∂U/∂R = −R/k). The beam is first made statically determinate by regarding R as an unknown external load, M is written in terms of the real loads and R, and U is differentiated.

Worked example: propped cantilever. Fixed at one end, prop at the other, span L, uniform load w. Measure x from the prop: M = Rx − wx²/2. Then ∂U/∂R = (1/EI)∫₀ᴸ M (∂M/∂R) dx = (1/EI)∫₀ᴸ (Rx² − wx³/2) dx = (1/EI)(RL³/3 − wL⁴/8) = 0, so R = 3wL/8, in agreement with the compatibility method of the previous chapter. The same calculation with a central point load P gives R = 5P/16.

🎯 Why the deflection at a rigid support is zero
The prop does no work when the beam deforms, because it does not move. Hence the energy stored is a minimum with respect to the redundant, which is exactly the statement ∂U/∂R = 0 (the theorem of least work). The condition is geometric, so it is the same compatibility requirement as before, only expressed through energy.

3. Torsion of circular shafts

For a circular shaft (solid or hollow) in elastic torsion, cross-sections remain plane and circular, and the shear strain varies linearly with radius. This gives τ = Tr/J and θ = TL/(GJ) (radians), with T/J = τ/r = Gθ/L. The polar second moment is J = πd⁴/32 for a solid shaft and J = π(D⁴ − d⁴)/32 for a hollow one. The maximum stress τmax = TD/2J = 16T/(πd³) acts at the outer surface; in a hollow shaft the stress at the inner radius is τmax × (d/D). The stresses on the planes at 45° are principal, ±τ, which is why brittle shafts crack on helices.

Worked example. A solid steel shaft, d = 50 mm, G = 80 GPa, length 1.5 m, transmits 30 kW at 300 rpm. ω = 2π × 300/60 = 31.42 rad/s, so T = P/ω = 30 000/31.42 = 954.9 N·m. τmax = 16 × 954.9 × 10³/(π × 50³) = 38.9 MPa. J = π × 50⁴/32 = 6.136 × 10⁵ mm⁴ and θ = 954.9 × 10³ × 1500/(80 000 × 6.136 × 10⁵) = 0.02918 rad = 1.67°. Power is P = Tω = 2πNT/60 with N in rpm and T in N·m giving watts; at fixed power, torque falls as speed rises, so a high-speed shaft can be thinner.

⚠️ rpm and rad/s
Convert rpm to rad/s with 2π/60 before computing T from power. A common slip is to use N directly in P = Tω, which overstates ω by 9.55 times. Also, J for a hollow shaft is a difference of fourth powers, not (D − d)⁴.

4. Hollow versus solid shafts, series and parallel shafts

Material near the axis is lightly stressed, so removing it costs little strength. For equal weight (equal area) and the same material, a hollow shaft of larger diameter has a larger J/R (polar section modulus) and carries more torque for the same τmax. Example: the solid shaft of 50 mm diameter (area 1963 mm²) is replaced by a tube of outer diameter 60 mm and inner diameter 33.17 mm, which has the same area. Its polar section modulus is π(D⁴ − d⁴)/(16D) = 38 452 mm³ against πd³/16 = 24 544 mm³ for the solid shaft, i.e. 1.57 times the torque at the same τmax.

Shafts in series (end to end, one fixed end, torque applied at the free end) carry the same torque T, and the angles add: θ = ΣTLᵢ/(GᵢJᵢ). Shafts in parallel (a bar inside a tube, or two shafts fixed at both ends and joined to a common end plate) twist through the same angle, and the torques share in proportion to GJ/L: T₁ + T₂ = T, T₁L₁/G₁J₁ = T₂L₂/G₂J₂. A statically indeterminate shaft fixed at both ends under a torque at an intermediate section is solved the same way.

Worked example. Shaft AB, d = 40 mm, 1.0 m long, is joined to shaft BC, d = 30 mm, 0.8 m long; A is fixed and a torque of 400 N·m acts at C (G = 80 GPa). J₁ = 2.513 × 10⁵ mm⁴ and J₂ = 7.952 × 10⁴ mm⁴. θ = 400 × 10³[1000/(80 000 × 2.513 × 10⁵) + 800/(80 000 × 7.952 × 10⁴)] = 0.01989 + 0.05030 = 0.0702 rad = 4.02°. The thinner part, though shorter, supplies most of the twist because J varies as d⁴.

5. Euler buckling of columns

A slender straight column under axial compression is stable only up to the critical load. Solving EI y″ = −Py for the pinned–pinned column gives Pcr = π²EI/L², and for other ends Pcr = π²EI/Le² with the effective length Le (the distance between inflection points, or the length of the equivalent pinned column). The column buckles about the axis of least I, i.e. with the least radius of gyration r = √(I/A).

Effective length for standard end conditions (theoretical values)
End conditionsLePcr relative to pinned–pinned
Pinned–pinnedL1
Fixed–free (flagpole)2L1/4
Fixed–fixedL/24
Fixed–pinned≈ 0.7L (0.699L)≈ 2.05

Worked example. A steel column (E = 200 GPa) of solid circular section, d = 40 mm, pinned at both ends, L = 1.5 m. I = π × 40⁴/64 = 1.2566 × 10⁵ mm⁴, so Pcr = π² × 200 000 × 1.2566 × 10⁵/1500² = 110.2 kN. Its radius of gyration is r = d/4 = 10 mm, so the slenderness ratio is λ = Le/r = 150. Euler stress is σcr = π²E/λ², and the slenderness ratio is the single parameter that measures the column. Fixed at the base and free at the top, the same column carries only a quarter of this.

⚠️ Euler’s limit: stocky columns do not buckle elastically
Euler’s formula assumes the column stays elastic, so it is valid only while σcr = π²E/λ² ≤ σy, i.e. λ ≥ π√(E/σy). For E = 200 GPa and σy = 250 MPa this is λ ≥ 88.9. For λ = 60 the formula would give 548 MPa, above the yield strength, so the column fails by yielding or by inelastic buckling at a lower load. Pcr does not depend on the yield strength inside the Euler range, and a stronger steel of the same E buckles at the same load.

6. Thin-walled pressure vessels

A vessel is thin-walled when t is small compared with the radius (t/d below about 1/20), so the stress is taken as uniform through the thickness and the radial stress (between 0 and −p) is neglected. For a closed-ended cylinder of diameter d under gauge pressure p, a cut along the axis gives the hoop stress σh = pd/(2t), and a cut across the section gives the longitudinal stress σL = pd/(4t), so σh = 2σL. For a sphere, both in-plane stresses are σ = pd/(4t), equal to the cylinder’s longitudinal stress and half its hoop stress, which is why spherical vessels are the most economical.

Strains and volume change. With Hooke’s law in plane stress, εh = (σh − νσL)/E, εL = (σL − νσh)/E, and for a cylinder the volumetric strain is εv = 2εh + εL. For a sphere, ε = σ(1 − ν)/E in every direction and εv = 3ε. Worked example: a cylinder of diameter 1.2 m, wall 10 mm, p = 2 MPa gives σh = 120 MPa and σL = 60 MPa. With E = 200 GPa and ν = 0.3: εh = (120 − 18)/200 000 = 5.1 × 10⁻⁴, εL = (60 − 36)/200 000 = 1.2 × 10⁻⁴, εv = 2 × 5.1 × 10⁻⁴ + 1.2 × 10⁻⁴ = 1.14 × 10⁻³. The corresponding sphere has σ = 60 MPa, ε = 2.1 × 10⁻⁴ and εv = 6.3 × 10⁻⁴.

⚠️ Which stress is which, and what the maximum shear is
The hoop stress is the larger one, and the seam along the axis (longitudinal joint) is the one that must be stronger. For the cylinder the in-plane maximum shear is (σh − σL)/2 = pd/(8t), while the absolute maximum shear, which includes the radial stress taken as zero, is σh/2 = pd/(4t). Use gauge pressure and the diameter, not the radius, in these formulas.

Key takeaways

  • U = ∫N²/2AE, ∫M²/2EI, ∫T²/2GJ; Castigliano’s second theorem gives δ = ∂U/∂P (dummy load if none), and ∂U/∂R = 0 for a rigid prop.
  • Torsion: τ = Tr/J, θ = TL/GJ, J = πd⁴/32 (solid) or π(D⁴ − d⁴)/32 (hollow); at equal weight a hollow shaft carries more torque.
  • Power P = 2πNT/60: always convert rpm to rad/s; series shafts share the torque and add angles, parallel shafts share the angle and split the torque by GJ/L.
  • Euler: Pcr = π²EI/Le² with Le = L, 2L, L/2, 0.7L; it depends on E and the least I, not on yield strength.
  • The Euler formula holds only for λ ≥ π√(E/σy); for stockier columns σcr from the formula exceeds yield and is invalid.
  • Thin cylinder σh = pd/2t, σL = pd/4t; sphere σ = pd/4t; εv = 2εh + εL for the cylinder and 3ε for the sphere.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. For a linear elastic structure, Castigliano’s second theorem states that the displacement of the point of application of a load Pi, measured in the direction of Pi, is:

    1. the partial derivative of the strain energy, written in terms of the loads, with respect to Pi
    2. the partial derivative of the strain energy, written in terms of the displacements, with respect to Pi
    3. the strain energy divided by Pi
    4. the integral of the strain energy over the length of the member
    Show answer

    Answer: A — the partial derivative of the strain energy, written in terms of the loads, with respect to Pi

    With U expressed through the loads, δi = ∂U/∂Pi. If U is expressed through the displacements the derivative with respect to δi gives the load Pi, which is the first theorem. Dividing U by Pi gives half the displacement, not the displacement, since U = ½Pδ for one load.

  2. The strain energy stored in bending in a beam of length L, flexural rigidity EI and bending moment M(x), neglecting shear and axial effects, is:

    1. ∫₀ᴸ M² dx/(2EI)
    2. ∫₀ᴸ M dx/(2EI)
    3. ∫₀ᴸ EI M² dx/2
    4. ∫₀ᴸ M² dx/(EI)
    Show answer

    Answer: A — ∫₀ᴸ M² dx/(2EI)

    The energy per unit length is ½Mκ with κ = M/EI, giving M²/2EI. The expression without the square is not an energy, the form with EI in the numerator has wrong dimensions, and dropping the factor of one-half doubles the energy.

  3. A cantilever of length 3 m (EI = 12 000 kN·m², uniform) carries a downward point load of 8 kN at its free end and a uniformly distributed downward load of 6 kN/m over its entire length. Using Castigliano’s second theorem (bending energy only), the deflection of the free end, in mm, is ____ (give two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 11.06

    From the free end M = −Px − wx²/2 and ∂M/∂P = −x, so δ = (1/EI)∫₀ᴸ (Px² + wx³/2) dx = (PL³/3 + wL⁴/8)/EI = (8 × 27/3 + 6 × 81/8)/12 000 = (72 + 60.75)/12 000 = 0.011063 m = 11.06 mm. Using only the point load would give 6.00 mm.

  4. A propped cantilever has a rigid prop at its right end. If the prop reaction R is treated as the redundant and U is the bending energy written in terms of the real loads and R, the compatibility condition given by Castigliano’s theorem is:

    1. ∂U/∂R = R, since the prop carries the load
    2. ∂U/∂R = 0, since the prop point does not deflect
    3. U = 0, since no energy is stored with a rigid support
    4. ∂U/∂R equals the fixed-end moment divided by L
    Show answer

    Answer: B — ∂U/∂R = 0, since the prop point does not deflect

    The derivative ∂U/∂R is the deflection at the prop in the direction of R, which is zero for a rigid prop. The beam still stores energy as it bends, so U is not zero, and R is a force, not a displacement, so it cannot equal ∂U/∂R.

  5. A solid circular steel shaft transmits 50 kW at 600 rpm. The allowable shear stress is 40 MPa and the shaft is in pure torsion. Taking τ = 16T/(πd³), the minimum shaft diameter, in mm, is ____ (give one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 46.6

    ω = 2π × 600/60 = 62.83 rad/s, T = 50 000/62.83 = 795.8 N·m. d³ = 16T/(πτ) = 16 × 795.8 × 10³/(π × 40) = 1.0133 × 10⁵ mm³, so d = 46.6 mm. Using 600 as ω directly would give a torque 9.55 times too small and d about 22 mm.

  6. A solid circular shaft of diameter 50 mm and length 2 m (G = 80 GPa) carries a constant torque of 1.0 kN·m along its length. The angle of twist of one end relative to the other, in degrees, is ____ (give two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 2.33

    J = π × 50⁴/32 = 6.136 × 10⁵ mm⁴ and θ = TL/(GJ) = 1.0 × 10⁶ × 2000/(80 000 × 6.136 × 10⁵) = 0.04075 rad. In degrees, 0.04075 × 180/π = 2.33°. Leaving the answer in radians is the usual slip.

  7. Shaft AB (solid, diameter 60 mm, length 1.2 m) is joined end to end to shaft BC (solid, diameter 40 mm, length 0.9 m); both are steel with G = 80 GPa. End A is fixed and a torque of 1.5 kN·m is applied at end C. The total angle of twist of C relative to A, in degrees, is ____ (give two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 4.86

    Both parts carry the same torque T = 1.5 × 10⁶ N·mm. J1 = π × 60⁴/32 = 1.272 × 10⁶ mm⁴, J2 = π × 40⁴/32 = 2.513 × 10⁵ mm⁴. θ1 = 1.5 × 10⁶ × 1200/(80 000 × 1.272 × 10⁶) = 0.01768 rad; θ2 = 1.5 × 10⁶ × 900/(80 000 × 2.513 × 10⁵) = 0.06715 rad. Total = 0.08483 rad = 4.86°. Angles in series add; torques do not.

  8. A hollow circular shaft has outer diameter 100 mm and inner diameter 60 mm and is in elastic torsion with a maximum shear stress of 80 MPa at the outer surface. The shear stress at the inner surface is:

    1. 28.8 MPa
    2. 64.0 MPa
    3. 80.0 MPa
    4. 48.0 MPa
    Show answer

    Answer: D — 48.0 MPa

    τ varies linearly with radius, so τinner = 80 × 60/100 = 48 MPa. The value 28.8 MPa comes from squaring the ratio 0.6, which is wrong, and 80 MPa would apply only if the stress were uniform through the wall.

  9. A solid shaft of fixed length and material must transmit a fixed power P at the allowable shear stress. If the operating speed is doubled, the diameter required, relative to the original, becomes:

    1. 0.500 of the original
    2. 0.630 of the original
    3. 0.794 of the original
    4. 0.707 of the original
    Show answer

    Answer: C — 0.794 of the original

    T = P/ω halves, and d³ ∝ T, so d falls by 2^(1/3) = 1.26, i.e. to 0.794 of the original. A factor of 0.5 would hold if d were proportional to T, 0.707 would result if d² were proportional to T, and 0.630 = 2^(−2/3) would need d^1.5 proportional to T.

  10. A steel column (E = 200 GPa) of solid circular section, diameter 40 mm and length 1.0 m, is fixed at its base and free at its top. Take the theoretical effective length for these end conditions and assume Euler’s formula applies. The critical buckling load, in kN, is ____ (give one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 62.0

    I = π × 40⁴/64 = 1.2566 × 10⁵ mm⁴ and Le = 2L = 2000 mm. Pcr = π²EI/Le² = 9.8696 × 200 000 × 1.2566 × 10⁵/2000² = 62.0 kN. Using Le = L would give 248 kN, four times too much.

  11. Two columns have the same length, section and material. One is fixed at both ends and the other is fixed at the base and free at the top. Taking the theoretical effective lengths, the ratio of the Euler critical load of the first to that of the second is:

    1. 2
    2. 4
    3. 8
    4. 16
    Show answer

    Answer: D — 16

    Le is L/2 for the fixed–fixed column and 2L for the fixed–free one, and Pcr ∝ 1/Le², so the ratio is (2L/(L/2))² = 4² = 16. The value 4 compares the fixed–fixed column with the pinned–pinned one, which is a different pair.

  12. A pinned–pinned steel column (E = 200 GPa, yield strength 250 MPa) has a slenderness ratio Le/r = 60. Which statement is correct?

    1. Euler’s formula gives 548 MPa, above yield, so the formula is invalid and the column fails earlier by yielding or inelastic buckling
    2. Euler’s formula applies, and the column buckles elastically at 548 MPa, since λ = 60 is in the long-column range
    3. Euler’s formula applies and the critical stress equals the yield strength of 250 MPa, since the column is slender
    4. the column cannot fail, because λ below 89 means it is a short column that always stays elastic under axial load
    Show answer

    Answer: A — Euler’s formula gives 548 MPa, above yield, so the formula is invalid and the column fails earlier by yielding or inelastic buckling

    σcr = π²E/λ² = 9.8696 × 200 000/3600 = 548 MPa, more than σy = 250 MPa, and the Euler range requires λ ≥ π√(E/σy) = 88.9. At λ = 60 the column yields or buckles inelastically at a lower load. The statement that a short column cannot fail confuses the lack of elastic buckling with the lack of failure, since it fails by crushing.

  13. Which of the following statements about the Euler critical load of a long, straight, elastic column are correct?

    1. P: For given length and end conditions, it depends on E and I but not on the yield strength.
    2. Q: Doubling the length of a pinned–pinned column doubles the critical load.
    3. R: For the same cross-sectional area and material, a thin-walled tube has a larger critical load than a solid bar.
    4. S: The column buckles about the principal axis of the section with the larger second moment of area.
    Show answer

    Answer: A — P: For given length and end conditions, it depends on E and I but not on the yield strength.; C — R: For the same cross-sectional area and material, a thin-walled tube has a larger critical load than a solid bar.

    Pcr = π²EI/Le² has no yield strength in it (P), and at equal area a tube has the larger I, hence the larger Pcr (R). Doubling L divides Pcr by four, so Q is wrong, and the column buckles about the axis of smaller I, so S is wrong.

  14. A thin spherical vessel and a thin closed-ended cylindrical vessel have the same diameter, wall thickness and internal gauge pressure. The membrane stress in the sphere compared with the hoop stress in the cylinder is:

    1. the same as the hoop stress in the cylinder
    2. twice the hoop stress in the cylinder
    3. half the hoop stress in the cylinder
    4. one-fourth of the hoop stress in the cylinder
    Show answer

    Answer: C — half the hoop stress in the cylinder

    The sphere has σ = pd/4t in every direction and the cylinder has σh = pd/2t, so the sphere’s stress is half of the cylinder’s hoop stress. It equals the cylinder’s longitudinal stress, which is the tempting wrong comparison for the same-as option.

  15. A thin-walled closed-ended cylindrical vessel of diameter 900 mm and wall thickness 6 mm is subjected to an internal gauge pressure of 2 MPa. The material has E = 200 GPa and ν = 0.3, and the radial stress is neglected. The volumetric strain, in microstrain (10⁻⁶), to the nearest integer, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 1425

    σh = pd/2t = 2 × 900/12 = 150 MPa and σL = 75 MPa. εh = (150 − 0.3 × 75)/200 000 = 6.375 × 10⁻⁴ and εL = (75 − 0.3 × 150)/200 000 = 1.5 × 10⁻⁴. εv = 2εh + εL = 1.275 × 10⁻³ + 1.5 × 10⁻⁴ = 1.425 × 10⁻³, i.e. 1425 microstrain. Taking εv = 3εh would give about 1910.

  16. Which of the following statements about a thin-walled closed-ended cylindrical vessel under internal gauge pressure p are correct (diameter d, wall thickness t, with t much smaller than d)?

    1. P: The hoop stress is twice the longitudinal stress.
    2. Q: Doubling the wall thickness at the same p and d halves both the hoop and the longitudinal stresses.
    3. R: The radial stress through the wall is small compared with the hoop stress and is neglected.
    4. S: The maximum in-plane shear stress equals the hoop stress.
    Show answer

    Answer: A — P: The hoop stress is twice the longitudinal stress.; B — Q: Doubling the wall thickness at the same p and d halves both the hoop and the longitudinal stresses.; C — R: The radial stress through the wall is small compared with the hoop stress and is neglected.

    σh = pd/2t, σL = pd/4t (P), both vary as 1/t (Q), and the radial stress lies between −p and 0, small against pd/2t when t ≪ d (R). The in-plane maximum shear is (σh − σL)/2 = pd/8t, a quarter of σh, so S is wrong.

  17. Which of the following statements about circular shafts in elastic torsion are correct?

    1. P: Two shafts joined end to end and carrying a torque at the free end transmit the same torque.
    2. Q: A solid bar inside a tube, rigidly joined at both ends, twists through the same angle as the tube.
    3. R: In the bar-in-tube arrangement the torque is shared equally by the two members.
    4. S: The shear stress is uniform over the cross-section of a solid shaft.
    Show answer

    Answer: A — P: Two shafts joined end to end and carrying a torque at the free end transmit the same torque.; B — Q: A solid bar inside a tube, rigidly joined at both ends, twists through the same angle as the tube.

    In series the torque is common (P), and in parallel the angle is common (Q). The torque is shared in proportion to GJ/L, not equally, so R is wrong, and τ = Tr/J grows linearly with r, so S is wrong.

  18. A solid shaft and a hollow shaft of the same material and the same length have the same mass. The hollow one has a larger outer diameter. When both are loaded to the same maximum shear stress, the hollow shaft:

    1. carries a smaller torque, because the material near the axis, which is removed, carries the most torque
    2. carries the same torque, because the area is the same
    3. carries a larger torque, because material is moved to larger radius where the stress is higher
    4. carries a larger torque, because its maximum shear stress occurs at the inner surface
    Show answer

    Answer: C — carries a larger torque, because material is moved to larger radius where the stress is higher

    The limiting torque is τmax J/R, and for equal area the tube has the larger J/R (the example in the text gives 1.57 times). The material near the axis is lightly stressed, so removing it hardly reduces capacity. The maximum stress is at the outer surface for a hollow shaft as well, so the claim about the inner surface is incorrect.