4.3 Thermal Stresses & Interference Stresses

4.3 Thermal Stresses & Interference Stresses

Thermal expansion and press/shrink fit interferences represent two primary sources of internal pre-stress in mechanical engineering. When thermal expansion is constrained by adjacent structural boundaries, massive compressive or tensile reaction stresses emerge. Similarly, press and shrink fits intentionally exploit radial dimensional interference to generate high contact pressures, enabling torque transmission across hubs, gears, bearings, and shafts without keyways. Accurately modeling these stresses via Lamé's equations and thermoelastic compatibility is vital to avoid premature yielding, fatigue failure, or joint slippage.

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Thermal Expansion and Indeterminate Restraint

When a homogeneous, isotropic body experiences a uniform temperature change $\Delta T = T_{\text{final}} - T_{\text{initial}}$, it undergoes an unconstrained thermal strain $\epsilon_T$:

$$\epsilon_T = \alpha \Delta T$$

where $\alpha$ is the linear coefficient of thermal expansion (units of $\mu\text{m}/(\text{m}\cdot^\circ\text{C})$ or $10^{-6}\;/^\circ\text{F}$). The unrestrained change in length of a member of length $L$ is:

$$\delta_T = \alpha L \Delta T$$

Statically Indeterminate Fixed Bar

If a uniform bar of cross-sectional area $A$ and modulus $E$ is rigidly confined between immovable walls, the total axial deformation must equal zero:

$$\delta_{\text{total}} = \delta_T + \delta_P = 0$$

Substituting $\delta_T = \alpha L \Delta T$ and mechanical strain displacement $\delta_P = \frac{P L}{A E}$:

$$\alpha L \Delta T + \frac{P L}{A E} = 0 \implies \sigma = \frac{P}{A} = -E \alpha \Delta T$$

A temperature increase ($\Delta T > 0$) generates compressive stress, whereas a temperature drop ($\Delta T < 0$) causes tensile stress.

Restrained Bar with Initial Clearance Gap

When an initial clearance gap $\delta_{\text{gap}}$ exists between the bar and a rigid support:

1. If $\delta_T \le \delta_{\text{gap}}$, the bar expands freely without contact: $\sigma = 0$.
2. If $\delta_T > \delta_{\text{gap}}$, the bar closes the gap, inducing a compressive reaction force:

$$\delta_T - \frac{P L}{A E} = \delta_{\text{gap}} \implies \sigma = -E \left(\alpha \Delta T - \frac{\delta_{\text{gap}}}{L}\right)$$

Bi-Metallic Strips and Thermal Curvature

A bi-metallic strip consists of two dissimilar thin metal strips of equal thickness $t_1 = t_2 = t/2$ and equal width $b$ bonded together along their faces. If material 2 has a higher thermal expansion coefficient than material 1 ($\alpha_2 > \alpha_1$), heating induces differential expansion, forcing the composite beam to bend into an arc of curvature $\kappa = 1/\rho$.

Assuming equal elastic moduli ($E_1 = E_2 = E$), Timoshenko's classical bi-metallic formulation yields the beam curvature:

$$\kappa = \frac{1}{\rho} = \frac{3 (\alpha_2 - \alpha_1) \Delta T}{2 t}$$

where $t$ is the total combined thickness ($t = t_1 + t_2$). For a cantilever strip of length $L$, the tip deflection $v_{\text{tip}}$ is:

$$v_{\text{tip}} \approx \frac{L^2}{2 \rho} = \frac{3 (\alpha_2 - \alpha_1) \Delta T L^2}{4 t}$$

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Thick-Walled Cylinders: Lamé's Equations

When a cylindrical pressure vessel, pipe, or machine bushing has a wall thickness exceeding approximately one-tenth of its inner radius ($t > 0.1 r_i$), the thin-walled membrane assumption breaks down. The hoop (tangential) stress varies significantly across the radial thickness.

Consider a cylinder of inner radius $r_i$ and outer radius $r_o$ subjected to internal pressure $p_i$ and external pressure $p_o$. The general radial equilibrium equation for an axisymmetric differential element is:

$$\frac{d\sigma_r}{dr} + \frac{\sigma_r - \sigma_t}{r} = 0$$

Assuming plane strain ($\epsilon_z = \text{constant}$) or plane stress ($\sigma_z = \text{constant}$), solving the differential equation yields the Lamé equations:

$$\sigma_t(r) = \frac{p_i r_i^2 - p_o r_o^2}{r_o^2 - r_i^2} + \frac{r_i^2 r_o^2 (p_i - p_o)}{r^2 (r_o^2 - r_i^2)}$$ $$\sigma_r(r) = \frac{p_i r_i^2 - p_o r_o^2}{r_o^2 - r_i^2} - \frac{r_i^2 r_o^2 (p_i - p_o)}{r^2 (r_o^2 - r_i^2)}$$

where:

• $\sigma_t(r)$ is the tangential (hoop) normal stress at radius $r$,
• $\sigma_r(r)$ is the radial normal stress at radius $r$.

For closed-end cylinders, axial equilibrium yields uniform longitudinal stress:

$$\sigma_z = \frac{p_i r_i^2 - p_o r_o^2}{r_o^2 - r_i^2}$$

Case 1: Internal Pressure Only ($p_o = 0$)

When external pressure is zero, Lamé's equations simplify to:

$$\sigma_t(r) = \frac{p_i r_i^2}{r_o^2 - r_i^2}\left(1 + \frac{r_o^2}{r^2}\right)$$ $$\sigma_r(r) = \frac{p_i r_i^2}{r_o^2 - r_i^2}\left(1 - \frac{r_o^2}{r^2}\right)$$

• At the inner surface ($r = r_i$):

$$\sigma_t(r_i) = p_i \left(\frac{r_o^2 + r_i^2}{r_o^2 - r_i^2}\right) \quad (\text{maximum tensile hoop stress})$$ $$\sigma_r(r_i) = -p_i \quad (\text{compressive})$$

• At the outer surface ($r = r_o$):

$$\sigma_t(r_o) = \frac{2 p_i r_i^2}{r_o^2 - r_i^2}$$ $$\sigma_r(r_o) = 0$$

Case 2: External Pressure Only ($p_i = 0$)

When internal pressure is zero (e.g., submarine hull or shrink-fit shaft):

$$\sigma_t(r) = -\frac{p_o r_o^2}{r_o^2 - r_i^2}\left(1 + \frac{r_i^2}{r^2}\right)$$ $$\sigma_r(r) = -\frac{p_o r_o^2}{r_o^2 - r_i^2}\left(1 - \frac{r_i^2}{r^2}\right)$$

Maximum compressive hoop stress occurs at the innermost bore ($r = r_i$):

$$\sigma_t(r_i) = -\frac{2 p_o r_o^2}{r_o^2 - r_i^2}$$

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Shrink and Press Fits: Interface Mechanics

In an interference fit, the outer component (hub, sleeve, or gear) is fabricated with an inner diameter $d_{\text{hub}}$ smaller than the outer diameter $d_{\text{shaft}}$ of the inner component (shaft or bushing). The total diametral interference is:

$$\delta = d_{\text{shaft}} - d_{\text{hub}}$$

Correspondingly, radial interference is $\delta_r = \delta / 2$.

Interface Contact Pressure ($p$)

When assembled, the interface must satisfy displacement compatibility: the radial expansion of the hub bore plus the radial compression of the shaft outer surface equals the initial radial interference:

$$\delta_r = u_{r,\text{hub}}(d/2) + |u_{r,\text{shaft}}(d/2)|$$

Applying Hooke's law in polar coordinates with Lamé stress distributions yields the master contact pressure equation:

$$p = \frac{\delta}{d \left[ \frac{1}{E_o}\left(\frac{d_o^2 + d^2}{d_o^2 - d^2} + \nu_o\right) + \frac{1}{E_i}\left(\frac{d^2 + d_i^2}{d^2 - d_i^2} - \nu_i\right) \right]}$$

where:

• $d$ is the nominal mating interface diameter,
• $d_o$ is the outer diameter of the hub,
• $d_i$ is the inner diameter of the shaft ($d_i = 0$ for a solid shaft),
• $E_o, \nu_o$ and $E_i, \nu_i$ are Young's modulus and Poisson's ratio for the outer hub and inner shaft.

Specialized Case: Solid Shaft of Same Material ($E_o = E_i = E, \nu_o = \nu_i = \nu, d_i = 0$)

For a solid shaft mated with a hub of identical material, the contact pressure equation reduces to:

$$p = \frac{E \delta}{2 d} \left(1 - \frac{d^2}{d_o^2}\right)$$

Maximum Stresses in Assembly

1. Outer Hub (Worst-Case Location): The peak stress occurs at the inner bore ($r = d/2$):

$$\sigma_{t,o} = p \left(\frac{d_o^2 + d^2}{d_o^2 - d^2}\right) \quad (\text{tension})$$ $$\sigma_{r,o} = -p \quad (\text{compression})$$

Under the von Mises yield criterion for plane stress ($\sigma_z = 0$):

$$\sigma_{vM,o} = \sqrt{\sigma_{t,o}^2 - \sigma_{t,o}\sigma_{r,o} + \sigma_{r,o}^2} = \sqrt{\sigma_{t,o}^2 + \sigma_{t,o} p + p^2}$$

1. Inner Solid Shaft: Every point in a solid inner shaft experiences uniform hydrostatic in-plane compression:

$$\sigma_{t,i} = \sigma_{r,i} = -p$$

Because $\sigma_1 = \sigma_2 = -p$, the in-plane principal stress difference is zero:

$$\sigma_{vM,i} = \sqrt{(-p)^2 - (-p)(-p) + (-p)^2} = p$$

Consequently, the outer hub almost always yields before the solid inner shaft.

Torque and Axial Load Capacity

The interface contact pressure $p$ creates a normal clamping force over the contact area $A_{\text{contact}} = \pi d L$, where $L$ is the axial engagement length.

• Torque Capacity ($T$):

$$T = F_{\text{friction}} \cdot \left(\frac{d}{2}\right) = (\mu p \pi d L) \left(\frac{d}{2}\right) = \frac{1}{2} \pi \mu p d^2 L$$

• Axial Force Capacity ($F_{ax}$):

$$F_{ax} = \mu p \pi d L$$

where $\mu$ is the static coefficient of friction at the mating interface.

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Thermal Assembly Requirements

To assemble an interference fit without mechanical galling or heavy hydraulic pressing, either the hub is heated to expand its bore, or the shaft is cryogenically cooled.

The required diametral thermal expansion must overcome the interference $\delta$ plus an assembly clearance $\delta_{\text{clearance}}$ (typically $0.001 d$ to allow rapid, unjammed slip-on):

$$\Delta d_{\text{total}} = \delta + \delta_{\text{clearance}} = \alpha d \Delta T$$

Solving for the minimum temperature differential $\Delta T$:

$$\Delta T = \frac{\delta + \delta_{\text{clearance}}}{\alpha d}$$

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Complete Worked Shrink Fit Design Problem

Problem Statement

A solid steel drive shaft ($E_i = 207\,\text{GPa}$, $\nu_i = 0.30$, $\alpha = 11.7 \times 10^{-6}\;/^\circ\text{C}$) with nominal diameter $d = 80.0\,\text{mm}$ is shrink-fitted into an alloy steel gear hub ($E_o = 207\,\text{GPa}$, $\nu_o = 0.30$, yield strength $S_y = 420\,\text{MPa}$) having outer diameter $d_o = 160.0\,\text{mm}$ and engagement length $L = 100.0\,\text{mm}$. The measured diametral interference is $\delta = 0.060\,\text{mm} = 6.0 \times 10^{-5}\,\text{m}$. The interface coefficient of static friction is $\mu = 0.15$.

Calculate:

1. The contact pressure $p$ at the interface.
2. The peak tangential hoop stress and von Mises stress in the gear hub.
3. The maximum torque capacity $T$ transmitted before slipping.
4. The required heating temperature of the hub if ambient temperature is $20^\circ\text{C}$ and an assembly clearance $\delta_{\text{clearance}} = 0.040\,\text{mm}$ is required.

Step 1: Interface Contact Pressure Calculation

Because both members are steel ($E_o = E_i = 207\,\text{GPa}$) and the shaft is solid ($d_i = 0$):

$$p = \frac{E \delta}{2 d} \left(1 - \frac{d^2}{d_o^2}\right)$$

Substitute the values:

$$\frac{E \delta}{2 d} = \frac{(207 \times 10^9\,\text{Pa})(6.0 \times 10^{-5}\,\text{m})}{2 (0.080\,\text{m})} = \frac{1.242 \times 10^7}{0.160} = 77.625 \times 10^6\,\text{Pa} = 77.625\,\text{MPa}$$ $$1 - \frac{d^2}{d_o^2} = 1 - \frac{80^2}{160^2} = 1 - \frac{6400}{25600} = 1 - 0.25 = 0.75$$ $$p = 77.625\,\text{MPa} \times 0.75 = 58.22\,\text{MPa}$$

Step 2: Peak Stresses in Gear Hub

The peak tangential hoop stress develops at the inner bore of the hub ($r = d/2 = 40\,\text{mm}$):

$$\sigma_{t,o} = p \left(\frac{d_o^2 + d^2}{d_o^2 - d^2}\right) = 58.22\,\text{MPa} \times \left(\frac{160^2 + 80^2}{160^2 - 80^2}\right) = 58.22 \times \frac{32{,}000}{19{,}200} = 58.22 \times 1.6667 = 97.03\,\text{MPa}$$

The radial stress at the interface is:

$$\sigma_{r,o} = -p = -58.22\,\text{MPa}$$

Compute the von Mises equivalent stress at the inner bore:

$$\sigma_{vM} = \sqrt{\sigma_{t,o}^2 - \sigma_{t,o}\sigma_{r,o} + \sigma_{r,o}^2} = \sqrt{(97.03)^2 - (97.03)(-58.22) + (-58.22)^2}$$ $$\sigma_{vM} = \sqrt{9414.8 + 5649.1 + 3389.6} = \sqrt{18453.5} \approx 135.84\,\text{MPa}$$

Safety factor against yielding:

$$FS = \frac{S_y}{\sigma_{vM}} = \frac{420\,\text{MPa}}{135.84\,\text{MPa}} \approx 3.09$$

The hub operates safely within the elastic regime.

Step 3: Torque Transmission Capacity

$$T = \frac{1}{2} \pi \mu p d^2 L = \frac{1}{2} \pi (0.15) (58.22 \times 10^6\,\text{N/m}^2) (0.080\,\text{m})^2 (0.100\,\text{m})$$ $$T = 0.5 \times \pi \times 0.15 \times 58.22 \times 10^6 \times 0.0064 \times 0.100 \approx 8780\,\text{N}\cdot\text{m} = 8.78\,\text{kN}\cdot\text{m}$$

Step 4: Thermal Assembly Temperature

Total diametral expansion required:

$$\Delta d_{\text{total}} = \delta + \delta_{\text{clearance}} = 0.060\,\text{mm} + 0.040\,\text{mm} = 0.100\,\text{mm} = 1.0 \times 10^{-4}\,\text{m}$$

Required temperature differential $\Delta T$:

$$\Delta T = \frac{\Delta d_{\text{total}}}{\alpha d} = \frac{1.0 \times 10^{-4}\,\text{m}}{(11.7 \times 10^{-6}\;/^\circ\text{C})(0.080\,\text{m})} = \frac{1.0 \times 10^{-4}}{9.36 \times 10^{-7}} \approx 106.8^\circ\text{C}$$

Required hub heating temperature:

$$T_{\text{hub}} = T_{\text{ambient}} + \Delta T = 20^\circ\text{C} + 106.8^\circ\text{C} \approx 127^\circ\text{C}$$

Heating the hub to $127^\circ\text{C}$ ($261^\circ\text{F}$) allows effortless slip-fit over the cold shaft.