8.4 Torsion of Circular Shafts, Springs & Column Buckling

1. Pure Torsion of Circular Shafts

In mining engineering machinery—such as ventilation fan drives, coal crushers, armored face conveyor (AFC) sprockets, and mine hoist winders—rotational power transmission depends heavily on circular shaft design under torsional loading.

Assumptions of Pure Torsion Theory

1. Plane transverse cross-sections remain plane, circular, and unwarped after twisting (strictly valid for circular and annular sections; non-circular sections warp out of plane).
2. Material is linearly elastic, homogeneous, isotropic, and obeys Hooke's Law in shear ($\tau = G\gamma$).
3. Twist angle per unit length is uniform along the shaft.
4. Radial lines remain straight and rotate through angle $\theta$.

  Torsional Shear Stress Distribution across Shaft Radius:

         <---|  τ_max = T R / J (Outer Surface)
             |
             |  τ(r) = (r/R) τ_max (Linear Variation)
  -----------+----------- (Centerline / Neutral Axis, τ = 0)
             |
         |--->  τ_max

The Governing Torsion Equation

$$\mathbf{\frac{T}{J} = \frac{\tau}{r} = \frac{G\theta}{L}}$$

• $T$ = Applied twisting torque ($\text{N}\cdot\text{mm}$)
• $J$ = Polar second moment of area of the cross-section ($\text{mm}^4$)
• $\tau$ = Torsional shear stress at radial distance $r$ from center ($\text{MPa}$)
• $r$ = Radial distance from shaft axis ($0 \le r \le R$)
• $G$ = Modulus of Rigidity / Shear Modulus ($\text{MPa}$)
• $\theta$ = Angle of twist over length $L$ in radians
• $L$ = Length of the shaft ($\text{mm}$)

Polar Section Modulus ($Z_p$)

The maximum torsional shear stress occurs at the outermost boundary ($r = R$):

$$\mathbf{\tau_{\max} = \frac{T}{Z_p}, \qquad \text{where } Z_p = \frac{J}{R}}$$

1. Solid Shaft (Diameter $d$): $$J = \frac{\pi d^4}{32}, \qquad \mathbf{Z_p = \frac{\pi d^3}{16}}$$
2. Hollow Shaft (Outer $D_o$, Inner $D_i$, ratio $k = D_i/D_o$): $$J = \frac{\pi(D_o^4 - D_i^4)}{32} = \frac{\pi D_o^4(1 - k^4)}{32}$$ $$\mathbf{Z_p = \frac{\pi D_o^3(1 - k^4)}{16}}$$

Power Transmission

A shaft rotating at $N$ revolutions per minute (RPM) transmitting power $P$ (in Watts) carries mean torque:

$$\mathbf{P = \frac{2\pi N T}{60} \implies T = \frac{60 P}{2\pi N}}$$

Weight Efficiency: Solid vs. Hollow Shafts

For identical material, length, and torsional strength (equal $T$ and $\tau_{\max}$):

$$\frac{\pi d_s^3}{16} = \frac{\pi D_o^3(1 - k^4)}{16} \implies d_s = D_o(1 - k^4)^{1/3}$$ $$\mathbf{\frac{W_{\text{hollow}}}{W_{\text{solid}}} = \frac{A_{\text{hollow}}}{A_{\text{solid}}} = \frac{D_o^2(1 - k^2)}{d_s^2} = \frac{1 - k^2}{(1 - k^4)^{2/3}}}$$

For a hollow shaft with inner-to-outer diameter ratio $k = 0.6$:

$$\frac{W_{\text{hollow}}}{W_{\text{solid}}} = \frac{1 - (0.6)^2}{(1 - (0.6)^4)^{2/3}} = \frac{0.64}{(0.8704)^{2/3}} = \frac{0.64}{0.9118} \approx \mathbf{0.702}$$

A hollow shaft with $k=0.6$ achieves nearly 30% weight saving for the exact same torque capacity.

2. Shafts in Series and Parallel Connections

  Shafts in Series:                      Shafts in Parallel (Composite):
  +-------------+---------+              +=======================+
  | Segment 1   | Seg 2   | ==(T)        | Outer Sleeve (G1, J1) |
  | (G1, J1, L1)|(G2,J2,L2|              +-----------------------+ ==(T)
  +-------------+---------+              | Inner Core   (G2, J2) |
                                         +=======================+

1. Shafts in Series

Connected end-to-end such that the full torque passes sequentially through each section:

• Torque is identical in all segments: $T_1 = T_2 = T$.
• Total angle of twist is additive: $$\mathbf{\theta_{\text{total}} = \theta_1 + \theta_2 = \frac{T L_1}{G_1 J_1} + \frac{T L_2}{G_2 J_2} = T\left(\frac{L_1}{G_1 J_1} + \frac{L_2}{G_2 J_2}\right)}$$

2. Shafts in Parallel (Composite Shaft)

Connected such that external torque is shared and both segments experience the same twist angle:

• Angle of twist is identical: $\theta_1 = \theta_2 = \theta$.
• Total torque is distributed: $$\mathbf{T = T_1 + T_2 = \theta\left(\frac{G_1 J_1}{L_1} + \frac{G_2 J_2}{L_2}\right)}$$ $$\mathbf{\frac{T_1}{T_2} = \frac{G_1 J_1 / L_1}{G_2 J_2 / L_2}}$$

3. Close-Coiled Helical Springs

A close-coiled helical spring consists of wire of circular diameter $d$ wound into a helix of mean coil diameter $D$ (mean radius $R = D/2$) with $n$ active coils, where the helix angle $\alpha < 10^{\circ}$ (so bending and axial forces are negligible, and wire experiences predominantly torsion under axial load $W$).

        W (Axial Load)
        |
        v
     (@@@@)  Mean Coil Diameter D, Spring Index C = D/d
     (@@@@)  Wire Diameter d, Active Coils n
     (@@@@)
        ^
        |
        W

Governing Equations

1. Nominal Torsional Shear Stress ($\tau_0$): Torque on wire: $T = W \times \frac{D}{2}$. Wire polar modulus $Z_p = \frac{\pi d^3}{16}$. $$\tau_0 = \frac{T}{Z_p} = \frac{W (D/2)}{\pi d^3 / 16} = \mathbf{\frac{8 W D}{\pi d^3}}$$
2. Wahl Stress Factor ($K_w$): Accounts for direct transverse shear stress and stress concentration due to wire curvature as a function of the Spring Index ($C = D/d$): $$\mathbf{K_w = \frac{4C - 1}{4C - 4} + \frac{0.615}{C}}$$ $$\mathbf{\tau_{\max} = K_w \left(\frac{8 W D}{\pi d^3}\right)}$$
3. Axial Deflection ($\delta$): Using strain energy $U = \frac{T^2 L_{\text{wire}}}{2GJ}$ where $L_{\text{wire}} = \pi D n$: $$\mathbf{\delta = \frac{8 W D^3 n}{G d^4}}$$
4. Spring Stiffness / Scale ($k$): $$\mathbf{k = \frac{W}{\delta} = \frac{G d^4}{8 D^3 n}}$$
5. Strain Energy Stored ($U$): $$U = \frac{1}{2} W \delta = \mathbf{\frac{4 W^2 D^3 n}{G d^4}}$$

Cutting and Combining Springs

• Cutting a Spring: Stiffness is inversely proportional to active coils ($k \propto 1/n$). If a spring of stiffness $k$ and $n$ coils is cut into two segments of $n_1$ and $n_2$ coils: $$k_1 n_1 = k_2 n_2 = k n \implies \mathbf{k_1 = k \left(\frac{n}{n_1}\right)}$$ If cut into two equal halves ($n_1 = n/2$), the stiffness of each half is doubled ($k_1 = 2k$).
• Springs in Series: $\frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2}$
• Springs in Parallel: $k_{\text{eq}} = k_1 + k_2$

4. Column Buckling Theory & Euler's Critical Load

Columns and struts are structural members subjected to axial compressive loads. While short compression blocks fail by material crushing/yielding, long slender columns fail by sudden lateral deflection called elastic buckling.

Euler's Buckling Formula

For an ideal, initially straight, linearly elastic column with pinned ends:

$$\mathbf{P_{\text{cr}} = \frac{\pi^2 E I_{\min}}{L_e^2}}$$

• $P_{\text{cr}}$ = Critical Euler buckling load ($\text{N}$)
• $E$ = Modulus of Elasticity ($\text{MPa}$)
• $I_{\min}$ = Minimum second moment of area of cross-section ($\text{mm}^4$)
• $L_e$ = Equivalent / Effective length of the column ($\text{mm}$)

Slenderness Ratio ($\lambda$) & Critical Stress

Defining the minimum radius of gyration $k_{\min} = \sqrt{I_{\min}/A}$, the Slenderness Ratio is $\lambda = \frac{L_e}{k_{\min}}$. The critical buckling stress is:

$$\mathbf{\sigma_{\text{cr}} = \frac{P_{\text{cr}}}{A} = \frac{\pi^2 E I_{\min}}{A L_e^2} = \frac{\pi^2 E}{\left(\frac{L_e}{k_{\min}}\right)^2} = \frac{\pi^2 E}{\lambda^2}}$$

The Four Standard Column End Conditions

  Both Pinned (Le=L)     Both Fixed (Le=0.5L)    Fixed-Pinned (Le=0.707L)   Fixed-Free (Le=2L)
       O                      ///////                  O                         |
      ( )                    |   |                    ( )                        |
      ( )                    (   )                    ( )                       (
      ( )                    |   |                    | |                       (
       O                      ///////                 ///////                   ///////

Validity Limit of Euler's Formula

Euler's formula is valid only when the critical stress $\sigma_{\text{cr}}$ does not exceed the material yield strength $\sigma_y$:

$$\sigma_{\text{cr}} = \frac{\pi^2 E}{\lambda^2} \le \sigma_y \implies \mathbf{\lambda \ge \pi \sqrt{\frac{E}{\sigma_y}}}$$

For standard IS 2062 mild steel ($E = 200\text{ GPa}, \sigma_y = 250\text{ MPa}$):

$$\lambda_{\text{critical}} = \pi \sqrt{\frac{200 \times 10^3}{250}} = \pi \sqrt{800} = 88.85 \approx \mathbf{89}$$

• If $\lambda \ge 89$: Long slender column $\implies$ Euler elastic buckling governs.
• If $\lambda < 89$: Short/intermediate column $\implies$ Material inelastic yielding/crushing governs.

5. Rankine-Gordon Empirical Formula for Intermediate Columns

For short to intermediate columns where buckling and compressive crushing occur simultaneously, Rankine's Empirical Formula provides reliable load capacity:

$$\mathbf{\frac{1}{P_R} = \frac{1}{P_c} + \frac{1}{P_e}}$$

Where $P_c = \sigma_c A$ is the compressive crushing capacity and $P_e = \frac{\pi^2 EI}{L_e^2}$ is the Euler buckling capacity. Rearranging algebraically:

$$\mathbf{P_R = \frac{P_c}{1 + \frac{P_c}{P_e}} = \frac{\sigma_c A}{1 + a \left(\frac{L_e}{k}\right)^2} = \frac{\sigma_c A}{1 + a \lambda^2}}$$

• $\sigma_c$ = Ultimate compressive strength / yield stress
• $a = \frac{\sigma_c}{\pi^2 E}$ = Rankine's Constant
  • For Mild Steel: $a \approx \frac{1}{7500}$ (pinned ends)
  • For Cast Iron: $a \approx \frac{1}{1600}$
  • For Wrought Iron: $a \approx \frac{1}{9000}$
  • For Timber: $a \approx \frac{1}{750}$

Asymptotic Behavioral Checks

1. For very short columns ($\lambda \to 0$): $a \lambda^2 \approx 0 \implies P_R \approx \sigma_c A = P_c$ (Pure crushing failure).
2. For very long columns ($\lambda \to \infty$): $1 \ll a \lambda^2 \implies P_R \approx \frac{\sigma_c A}{a \lambda^2} = \frac{\sigma_c A}{\left(\frac{\sigma_c}{\pi^2 E}\right)\left(\frac{L_e^2}{I/A}\right)} = \frac{\pi^2 EI}{L_e^2} = P_e$ (Pure Euler buckling).

6. Worked Step-by-Step Engineering Calculations

Example 1: Transmission Shaft Sizing (Strength & Stiffness Criteria)

Problem: A solid circular shaft is to transmit $P = 150\text{ kW}$ at $N = 300\text{ RPM}$. The allowable shear stress is $\tau_{\text{allow}} = 50\text{ MPa}$ and the maximum allowable angle of twist is $\theta_{\text{allow}} = 1.0^{\circ}$ per $2.5\text{ m}$ length. Modulus of rigidity $G = 80\text{ GPa}$. Determine the required shaft diameter $d$.

Solution:

• 1. Torque Calculation: $$T = \frac{60 P}{2\pi N} = \frac{60 \times 150 \times 10^3}{2\pi (300)} = \frac{9,000,000}{600\pi} = 4774.65\text{ N}\cdot\text{m} = 4.775 \times 10^6\text{ N}\cdot\text{mm}$$
• 2. Strength Criterion (Shear Stress limit): $$\tau_{\max} = \frac{16 T}{\pi d^3} \le 50 \implies d^3 \ge \frac{16(4.775 \times 10^6)}{\pi (50)} = \frac{7.640 \times 10^7}{157.08} = 4.8637 \times 10^5\text{ mm}^3$$ $$d_{\text{strength}} \ge (4.8637 \times 10^5)^{1/3} = \mathbf{78.64\text{ mm}}$$
• 3. Stiffness Criterion (Angle of Twist limit): $\theta = 1.0^{\circ} = 1.0 \times \frac{\pi}{180} = 0.017453\text{ rad}$ over $L = 2500\text{ mm}$. $$\theta = \frac{T L}{G J} = \frac{T L}{G \left(\frac{\pi d^4}{32}\right)} = \frac{32 T L}{\pi G d^4} \le 0.017453$$ $$d^4 \ge \frac{32(4.775 \times 10^6)(2500)}{\pi (80 \times 10^3)(0.017453)} = \frac{3.820 \times 10^{11}}{4386.4} = 8.7087 \times 10^7\text{ mm}^4$$ $$d_{\text{stiffness}} \ge (8.7087 \times 10^7)^{1/4} = \mathbf{96.65\text{ mm}}$$
• Conclusion: The governing diameter is the larger value: select $d = \mathbf{100\text{ mm}}$ (standard metric shaft size).

---

Example 2: Euler Buckling Load Across Boundary Conditions

Problem: A structural steel pipe column ($E = 200\text{ GPa}$, outside diameter $D_o = 100\text{ mm}$, thickness $t = 10\text{ mm}$, inside diameter $D_i = 80\text{ mm}$) of length $L = 4.0\text{ m}$ is used in a mine gantry. Calculate the Euler critical buckling load $P_{\text{cr}}$ if:

1. Both ends are pinned
2. Both ends are fixed

Solution:

• Moment of inertia: $$I = \frac{\pi}{64}(D_o^4 - D_i^4) = \frac{\pi}{64}(100^4 - 80^4) = \frac{\pi}{64}(100,000,000 - 40,960,000) = \frac{\pi}{64}(59,040,000) = 2.898 \times 10^6\text{ mm}^4 = 2.898 \times 10^{-6}\text{ m}^4$$
• 1. Both Ends Pinned ($L_e = L = 4.0\text{ m}$): $$P_{\text{cr}} = \frac{\pi^2 EI}{L^2} = \frac{\pi^2 (200 \times 10^9)(2.898 \times 10^{-6})}{(4.0)^2} = \frac{5.7205 \times 10^6}{16} = 357,531\text{ N} = \mathbf{357.53\text{ kN}}$$
• 2. Both Ends Fixed ($L_e = 0.5L = 2.0\text{ m}$): $$P_{\text{cr}} = 4 \times P_{\text{pinned}} = 4 \times 357.53\text{ kN} = \mathbf{1430.13\text{ kN}}$$