Thick cylinders: Lame's equations and press fits

Why it matters

Hydraulic cylinders, fuel-injection lines, gun barrels and high-pressure accumulators have walls too thick for the thin-shell formulas. The same theory also explains press and shrink fits: gears, bearings, flywheels and wheel hubs held on shafts by interference, and cylinder liners pressed into blocks. Lamé's equations give the stress at every radius, show where the wall is most stressed and tell you how much torque a press fit can transmit.

Key ideas

When thin-shell theory fails. If the wall thickness is more than about one-twentieth of the diameter, hoop stress is no longer uniform through the wall and the radial stress (equal to the pressure at a loaded surface) is no longer negligible. Thick-cylinder theory accounts for both.

Assumptions of Lamé's theory.

• Long cylinder, axisymmetric geometry and loading, homogeneous isotropic linear elastic material.
• Plane cross-sections remain plane, so the longitudinal strain is the same at every radius; the longitudinal stress, if present (closed ends), is uniform.
• Stresses depend only on radius r.

Lamé's equations. Equilibrium of a thin ring element plus compatibility gives, with tension positive: σ_r = A − B/r², σ_h = A + B/r². A and B are found from the two boundary conditions: σ_r = −p_i at r = r_i and σ_r = −p_o at r = r_o. Some Indian textbooks write σ_r = b/r² − a with compressive radial stress positive; the physics is the same, but keep one convention throughout.

Internal pressure only (p_o = 0).

• Radial stress is compressive, from −p_i at the bore to zero at the outside.
• Hoop stress is tensile and largest at the bore: σ_h,max = p_i·(r_o² + r_i²)/(r_o² − r_i²), falling to 2p_i·r_i²/(r_o² − r_i²) at the outside.
• The difference σ_h − σ_r = 2B/r² is largest at the bore, so the maximum shear stress, p_i·r_o²/(r_o² − r_i²), is at the bore too. Yielding starts at the inner surface.
• Closed ends: σ_l = p_i·r_i²/(r_o² − r_i²), uniform.

Diminishing returns of thickness. As r_o grows very large, σ_h at the bore tends to p_i, not to zero. So a single-piece cylinder can never hold a pressure above its allowable hoop stress, however thick. That is why compound (shrink-fitted) cylinders, wire winding and autofrettage are used: they put the bore into initial compression so the working hoop stress is more uniform.

Press (shrink) fits. An inner member (shaft or inner cylinder) of outer diameter slightly larger than the bore of an outer member (hub or outer cylinder) is forced or shrunk in. A contact pressure p_f develops at the interface:

• the hub behaves like a thick cylinder under internal pressure p_f: tensile hoop stress, maximum at its bore;
• the shaft is under external pressure p_f: for a solid shaft, σ_r = σ_h = −p_f everywhere (compressive);
• the diametral interference δ equals the increase in the hub's bore plus the decrease in the shaft's diameter, both found from hoop strains at the interface. The fit transmits torque by friction: T = μ·p_f·(π·d·L)·d/2. Heating the hub or cooling the shaft eases assembly. Interferences and fits come from a standard fit system (e.g. ISO H7/p6) chosen in design.

Formulas

σ_r = A − B/r², σ_h = A + B/r² (tension positive) A = (p_i·r_i² − p_o·r_o²)/(r_o² − r_i²), B = (p_i − p_o)·r_i²·r_o²/(r_o² − r_i²)

• σ_r, σ_h: radial and hoop stress at radius r (Pa); p_i, p_o: internal and external pressures (Pa); r_i, r_o: inner and outer radii (m).

Internal pressure only: σ_h,max = p_i·(r_o² + r_i²)/(r_o² − r_i²) (at r_i), σ_h,outer = 2·p_i·r_i²/(r_o² − r_i²) τ_max = p_i·r_o²/(r_o² − r_i²) (at r_i), σ_l = p_i·r_i²/(r_o² − r_i²) (closed ends)

Press fit, same material, hub outside diameter D, interface diameter d, solid shaft: δ = (p_f·d/E)·[2·D²/(D² − d²)]

• δ: diametral interference (m); p_f: contact pressure (Pa); E: Young's modulus (Pa). (For a hollow shaft or different materials, add the separate hub and shaft terms, including ν.)

T = μ·p_f·π·d·L·(d/2)

• T: torque capacity (N·m); μ: coefficient of friction; L: length of fit (m).

Worked examples

Example 1 (standard): thick cylinder under internal pressure. Given: r_i = 100 mm, r_o = 150 mm, p_i = 30 MPa, open ends. Find the hoop stress at the bore and outside, the maximum shear, and compare with thin-shell theory.

1. r_o² − r_i² = 22 500 − 10 000 = 12 500 mm².
2. σ_h,max = p_i·(r_o² + r_i²)/(r_o² − r_i²) = 30 × 32 500/12 500 = 78 MPa at the bore.
3. σ_h,outer = 2·p_i·r_i²/(r_o² − r_i²) = 2 × 30 × 10 000/12 500 = 48 MPa.
4. τ_max at the bore = (78 − (−30))/2 = 54 MPa (check: 30 × 22 500/12 500 = 54 MPa).
5. Thin-shell estimate with the bore diameter: p·d/(2t) = 30 × 200/(2 × 50) = 60 MPa, which underestimates the peak by 23 %. Answer: σ_h = 78 MPa (bore) to 48 MPa (outside); τ_max = 54 MPa; thin theory is unsafe here.

Example 2 (GATE level): hub press-fitted on a shaft. Given: a steel hub, outside diameter D = 100 mm, is pressed onto a solid steel shaft of diameter d = 50 mm over a length L = 60 mm. Diametral interference δ = 0.04 mm, E = 200 GPa, μ = 0.15. Find the contact pressure, the maximum hoop stress in the hub, the stresses in the shaft and the torque the fit can transmit.

1. δ = (p_f·d/E)·2D²/(D² − d²), so p_f = δ·E·(D² − d²)/(2·d·D²) = 0.04 × 200 000 × (10 000 − 2500)/(2 × 50 × 10 000) = 60 MPa.
2. Hub bore hoop stress: p_f·(D² + d²)/(D² − d²) = 60 × 12 500/7500 = 100 MPa (tension).
3. Shaft: σ_r = σ_h = −60 MPa (uniform compression).
4. T = μ·p_f·π·d·L·d/2 = 0.15 × 60 × π × 50 × 60 × 25 = 2.12 × 10⁶ N·mm. Answer: p_f = 60 MPa; hub hoop stress 100 MPa at its bore; shaft −60 MPa; torque capacity ≈ 2.12 kN·m.

Common mistakes

• Mixing the two sign conventions for Lamé's constants within one problem.
• Setting σ_r = +p at the bore; internal pressure gives compressive radial stress.
• Using thin-shell formulas for thick walls, which underestimates the bore stress.
• Taking the interference as radial when it is given as diametral (or vice versa).
• Assuming a thicker wall can always carry more pressure; the bore stress never falls below p_i.
• Forgetting that a solid shaft under external pressure has σ_r = σ_h = −p everywhere.

For GATE ME

Expect maximum hoop stress for given radii and pressure, ratio of hoop stresses at inner and outer surfaces, radial stress at an intermediate radius with both pressures, the contact pressure for a given interference, and press-fit torque capacity. Practise solving for A and B from two boundary conditions quickly.

Quick check

1. Where is the hoop stress maximum in a thick cylinder with internal pressure only?
2. r_i = 0.1 m, r_o = 0.2 m, p_i = 5 MPa, no external pressure. Find σ_h at the bore.
3. What is the radial stress at the outer surface when there is no external pressure?
4. In a press fit, which member is in hoop tension?

Answers: 1. At the inner surface. 2. 5 × (0.04 + 0.01)/0.03 = 8.33 MPa. 3. Zero. 4. The outer member (hub); a solid inner shaft is in compression.