12.2 Torsion, Flanged Bolt Couplings, and Thin-Walled Pressure Vessels

12.2 Torsion, Flanged Bolt Couplings, and Thin-Walled Pressure Vessels

Torsional shear and biaxial membrane tension represent two critical stress categories encountered in civil and mechanical infrastructure—from circular structural columns subjected to eccentric wind-induced torsional moments to municipal water storage tanks, aqueducts, and pressurized penstocks.

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1. Pure Torsion in Circular Shafts and Polar Moment of Inertia

When a circular cylindrical shaft is subjected to an applied torsional moment (torque $T$), internal shearing stresses develop across every transverse plane.

Fundamental Assumptions of Circular Torsion

1. Axisymmetry: Plane cross-sections perpendicular to the longitudinal axis remain planar and circular after twisting (no warping occurs in circular cross-sections).
2. Radial Line Invariance: Radial lines remain straight and rotate through an angle $\theta$ proportional to distance along the shaft.
3. Homogeneous, Linear Elastic Material: Shearing stress relates to shearing strain via Hooke's Law in shear: $\tau = G \gamma$.

The Torsion Formula

From strain geometry, shearing strain at distance $\rho$ from the shaft center is $\gamma = \rho \frac{d\theta}{dx}$. Applying $\tau = G \gamma$ and equating internal resisting torque to applied torque yields: $\tau = \frac{T \rho}{J}$ $\boxed{\tau_{\max} = \frac{T r}{J} = \frac{T (D/2)}{J}}$ where:

• $T$ is internal torsional moment ($\text{N}\cdot\text{mm}$ or $\text{kN}\cdot\text{m}$).
• $\rho$ is radial distance from the shaft center ($0 \le \rho \le r$).
• $J$ is the polar moment of inertia of the cross-section ($\text{mm}^4$).
• $\tau_{\max}$ occurs at the outermost boundary ($\,\rho = r = D/2$).

Polar Moment of Inertia ($J$)

For circular geometry, polar moment of inertia about the centroidal axis equals the sum of rectangular moments of inertia: $J = I_x + I_y = 2 I$:

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2. Angle of Twist and Shaft Rigidity

For a homogeneous prismatic circular shaft of length $L$, the total relative angle of rotation (twist) $\theta$ between its ends is obtained by integrating along the axis: $\theta = \int_0^L \frac{T(x)}{J(x) G} \, dx \implies \boxed{\theta = \frac{T L}{J G}}$

• Units: $\theta$ is calculated strictly in radians. To convert radians to degrees: $\theta^\circ = \theta_{\text{rad}} \times \left(\frac{180^\circ}{\pi}\right)$.
• Shear Modulus (Modulus of Rigidity, $G$): $G = \frac{E}{2(1 + \nu)}$ For structural steel ($E = 200\text{ GPa}, \nu = 0.30$): $G = \frac{200}{2(1.30)} \approx 76.92\text{ GPa} \approx 77\text{ GPa} = 77,000\text{ MPa}$.
• Torsional Stiffness: $k_t = \frac{T}{\theta} = \frac{J G}{L}$ (torque per unit radian twist).
• Torsional Flexibility: $f_t = \frac{1}{k_t} = \frac{L}{J G}$.

Shafts in Series and Parallel

• Series Shafts: Internal torque $T$ is common throughout all segments; total twist is additive: $\theta_{\text{total}} = \sum \frac{T_i L_i}{J_i G_i}$.
• Parallel / Composite Shafts: Two concentric materials bonded together twist through the identical angle ($\theta_1 = \theta_2$), sharing the total torque: $T = T_1 + T_2 = \theta \left(\frac{J_1 G_1}{L} + \frac{J_2 G_2}{L}\right)$.

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3. Power Transmission in Rotating Shafts

Rotating machinery transmits mechanical power $P$ through drive shafts via rotational torque $T$: $P = T \omega = 2\pi f T$ where:

• $P$ is power in Watts ($\text{W} = \text{N}\cdot\text{m/s}$) or kilowatts ($\text{kW} = 10^3\text{ W}$). (In Imperial units, $1\text{ Horsepower (HP)} = 550\text{ ft}\cdot\text{lb/s} = 745.7\text{ W}$).
• $\omega$ is angular velocity in radians per second: $\omega = \frac{2\pi N}{60}$, with rotational speed $N$ in revolutions per minute ($\text{rpm}$).
• $f$ is rotational frequency in Hertz ($\text{Hz} = \text{rev/s} = N/60$).

Solving for Required Shaft Diameter

$T = \frac{P}{\omega} = \frac{60 P}{2\pi N}$ Substituting $T$ into the solid circular shaft capacity formula ($\tau_{\text{allow}} = \frac{16 T}{\pi d^3}$): $d = \sqrt[3]{\frac{16 T}{\pi \tau_{\text{allow}}}} = \sqrt[3]{\frac{16 (60 P)}{2\pi^2 N \tau_{\text{allow}}}}$

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4. Statically Indeterminate Torsional Systems

When a shaft is fixed rigidly at both ends (supports $A$ and $B$) and subjected to an intermediate applied torque $T_0$ at point $C$:

1. Equilibrium Equation: $T_A + T_B = T_0$
2. Compatibility Equation: The total relative twist between fixed boundaries must vanish: $\theta_{B/A} = \theta_{AC} + \theta_{CB} = 0 \implies \frac{T_A L_{AC}}{J_{AC} G} - \frac{T_B L_{CB}}{J_{CB} G} = 0$ For a uniform prismatic shaft ($J_{AC} = J_{CB}$): $T_A L_{AC} = T_B L_{CB} \implies \boxed{T_A = T_0 \left(\frac{L_{CB}}{L}\right), \quad T_B = T_0 \left(\frac{L_{AC}}{L}\right)}$

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5. Flanged Bolt Couplings

A flanged bolt coupling connects two collinear rotating shafts. Power or torque is transmitted across the joint through shearing forces developed in circumferential bolts positioned on concentric bolt pitch circles.

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Single Pitch Circle

For $n$ identical bolts of cross-sectional area $A_b = \frac{\pi d_b^2}{4}$ positioned on a pitch circle of radius $R$: $T = n P_b R = n (\tau_b A_b) R$

Multiple Concentric Bolt Rings

Consider a coupling with an inner circle of $n_1$ bolts at radius $R_1$ and an outer circle of $n_2$ bolts at radius $R_2$ ($R_2 > R_1$).

• Kinematic Compatibility: Because the flange plates are assumed rigid, shearing strain in each fastener is directly proportional to its radial distance from the center of rotation: $\gamma = R \frac{d\theta}{dx} \implies \frac{\gamma_1}{R_1} = \frac{\gamma_2}{R_2}$
• Constitutive Relation: Assuming all bolts are of identical material ($G_1 = G_2$): $\frac{\tau_1}{R_1} = \frac{\tau_2}{R_2} \implies \tau_1 = \tau_2 \left(\frac{R_1}{R_2}\right)$
• Controlling Stress: Because $R_2 > R_1$, the outer bolts experience higher strain and reach the allowable shearing stress first ($\tau_2 = \tau_{\text{allow}}$). The inner bolts operate at a reduced stress $\tau_1 = \tau_{\text{allow}} (R_1 / R_2)$.
• Bolt Forces: If all bolts share the same diameter ($A_1 = A_2 = A_b$): $P_2 = \tau_{\text{allow}} A_b, \quad P_1 = P_2 \left(\frac{R_1}{R_2}\right)$
• Total Torque Capacity: $\boxed{T = n_1 P_1 R_1 + n_2 P_2 R_2 = n_1 \left(P_2 \frac{R_1}{R_2}\right) R_1 + n_2 P_2 R_2 = P_2 \left(n_1 \frac{R_1^2}{R_2} + n_2 R_2\right)}$

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6. Thin-Walled Pressure Vessels

A pressure vessel is classified as thin-walled when the ratio of wall thickness $t$ to internal radius $r$ is small ($t/r \le 0.10$, or inner diameter to thickness ratio $D/t \ge 20$). Under this condition, radial normal stress across the thickness is negligible compared to membrane tensile stresses, and membrane stress is assumed uniform across the wall thickness.

Cylindrical Pressure Vessels

Consider a thin-walled cylinder of internal diameter $D$, radius $r = D/2$, wall thickness $t$, containing internal fluid at gauge pressure $p$.

1. Tangential (Hoop / Circumferential) Stress ($\sigma_t$): Cutting a half-cylinder of length $L$ through a longitudinal plane exposes the internal fluid pressure acting upward against the projected area $D L$, balanced by tensile forces in two wall cross-sections ($2 t L$): $\sum F_y = 0 \implies p (D L) - 2 (\sigma_t t L) = 0$ $\boxed{\sigma_t = \frac{p D}{2 t} = \frac{p r}{t}}$

2. Longitudinal (Axial) Stress ($\sigma_l$): Cutting through a transverse plane exposes fluid pressure acting against circular end cap area $\frac{\pi D^2}{4}$, balanced by tensile forces across annular wall area $\pi D t$: $\sum F_x = 0 \implies p \left(\frac{\pi D^2}{4}\right) - \sigma_l (\pi D t) = 0$ $\boxed{\sigma_l = \frac{p D}{4 t} = \frac{p r}{2 t} = \frac{1}{2} \sigma_t}$

Spherical Pressure Vessels

Due to complete spherical symmetry, any section cut through the center produces the identical equilibrium balance as the longitudinal cut in a cylinder: $\boxed{\sigma_{\text{sphere}} = \frac{p D}{4 t} = \frac{p r}{2 t}}$ Spherical pressure vessels require only half the wall thickness of a cylindrical vessel of identical diameter to sustain the same internal operating pressure.

In-Plane and Absolute Maximum Shear Stress in Pressure Vessels

On the exterior surface of a cylindrical vessel, principal stresses are $\sigma_1 = \sigma_t = \frac{p D}{2 t}$, $\sigma_2 = \sigma_l = \frac{p D}{4 t}$, and $\sigma_3 = 0$ (atmospheric outer surface):

• Maximum In-Plane Shear Stress: $\tau_{\max, \text{in-plane}} = \frac{\sigma_1 - \sigma_2}{2} = \frac{\sigma_t - \sigma_l}{2} = \frac{p D / 2t - p D / 4t}{2} = \frac{p D}{8 t}$
• Absolute Maximum Shear Stress (3D Mohr's Circle): $\boxed{\tau_{\max, \text{abs}} = \frac{\sigma_1 - \sigma_3}{2} = \frac{\sigma_t - 0}{2} = \frac{p D}{4 t}}$

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7. Joint Efficiency Considerations ($\eta$)

In fabricated steel tanks, boilers, and penstocks, plates are joined by welding or riveting. Fabricated joints possess a joint efficiency $\eta \le 1.0$ (ratio of joint strength to solid plate strength):

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8. Worked Example: Concentric Flanged Bolt Coupling and Drive Shaft Sizing

Problem Statement: A solid circular transmission shaft drives a heavy centrifugal pump delivering $P = 180\text{ kW}$ at $N = 300\text{ rpm}$.

1. Calculate the required shaft diameter if allowable shear stress is $\tau_{\text{allow}, s} = 50\text{ MPa}$ and maximum allowable twist is $0.80^\circ$ per meter length ($G = 80\text{ GPa}$).
2. The shaft connects to the pump via a flanged bolt coupling consisting of two concentric rings of $16\text{ mm}$ diameter bolts: an inner ring of 6 bolts on a pitch diameter of $200\text{ mm}$ ($R_1 = 100\text{ mm}$) and an outer ring of 8 bolts on a pitch diameter of $320\text{ mm}$ ($R_2 = 160\text{ mm}$). If allowable bolt shear stress is $\tau_{\text{allow}, b} = 65\text{ MPa}$, determine the total torque capacity of the coupling and check whether it safely transmits the operating torque.

Step-by-Step Solution:

1. Calculate Transmitted Torque ($T$): $\omega = \frac{2\pi N}{60} = \frac{2\pi (300)}{60} = 10\pi = 31.416\text{ rad/s}$ $T = \frac{P}{\omega} = \frac{180 \times 10^3\text{ W}}{31.416\text{ rad/s}} = 5,729.58\text{ N}\cdot\text{m} = 5.730\text{ kN}\cdot\text{m}$

2. Size Shaft for Shear Strength: $\tau_{\max} = \frac{16 T}{\pi d^3} \le 50\text{ MPa}$ $d_{\text{strength}} = \sqrt[3]{\frac{16 (5,729.58 \times 10^3\text{ N}\cdot\text{mm})}{\pi (50\text{ N/mm}^2)}} = \sqrt[3]{\frac{91,673,280}{157.08}} = \sqrt[3]{583,610} = 83.57\text{ mm}$

3. Size Shaft for Torsional Rigidity: $\theta_{\text{allow}} = 0.80^\circ = 0.80 \times \frac{\pi}{180} = 0.013963\text{ rad per } L = 1000\text{ mm}$ $\theta = \frac{T L}{J G} = \frac{T L}{(\pi d^4 / 32) G} \le \theta_{\text{allow}}$ $d_{\text{rigidity}} = \sqrt[4]{\frac{32 T L}{\pi G \theta_{\text{allow}}}} = \sqrt[4]{\frac{32 (5,729.58 \times 10^3)(1000)}{\pi (80,000)(0.013963)}} = \sqrt[4]{\frac{1.8335 \times 10^{11}}{3,509.3}} = \sqrt[4]{52,246,000} = 85.03\text{ mm}$ Controlling diameter is the larger value: $d = 85.1\text{ mm}$ (governed by torsional stiffness).

4. Coupling Torque Capacity:

   • Bolt cross-sectional area: $A_b = \frac{\pi (16\text{ mm})^2}{4} = 201.06\text{ mm}^2$.
   • Outer bolts control: $\tau_2 = \tau_{\text{allow}, b} = 65\text{ MPa}$. $P_2 = \tau_2 A_b = (65\text{ N/mm}^2)(201.06\text{ mm}^2) = 13,069\text{ N} = 13.069\text{ kN}$
   • Inner bolts carry reduced stress based on radius: $\tau_1 = \tau_2 \left(\frac{R_1}{R_2}\right) = 65 \times \left(\frac{100\text{ mm}}{160\text{ mm}}\right) = 40.625\text{ MPa}$ $P_1 = P_2 \left(\frac{R_1}{R_2}\right) = 13.069 \times \frac{100}{160} = 8.168\text{ kN}$
   • Total coupling torque capacity: $T_{\text{coupling}} = n_1 P_1 R_1 + n_2 P_2 R_2$ $T_{\text{coupling}} = 6 (8.168\text{ kN})(0.100\text{ m}) + 8 (13.069\text{ kN})(0.160\text{ m})$ $T_{\text{coupling}} = 4.901\text{ kN}\cdot\text{m} + 16.728\text{ kN}\cdot\text{m} = \boxed{21.63\text{ kN}\cdot\text{m}}$
   • Capacity Check: $T_{\text{coupling}} = 21.63\text{ kN}\cdot\text{m} > T_{\text{operating}} = 5.73\text{ kN}\cdot\text{m}$. The coupling provides a safety factor of $3.77$, fully safe.

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9. CELE Board Exam Traps & Common Computational Errors