Lesson 2.4: Torsion of Shafts & Springs

Torsion of shafts and springs is a critical topic in mechanical design. GATE and PSU exams often test torsional stress, angle of twist, shaft design, and spring characteristics.

🔹 1. Torsion of Shafts

• Definition: Torsion is the twisting of a shaft due to applied torque $T$.

• Torsion Equation:

τ=T⋅rJ\tau = \frac{T \cdot r}{J}τ=JT⋅r​

Where:

• $\tau$ = shear stress at radius r

• $T$ = applied torque

• $J$ = polar moment of inertia

• $r$ = distance from shaft axis

• Angle of Twist:

θ=TLJG\theta = \frac{T L}{J G}θ=JGTL​

Where:

• $L$ = shaft length

• $G$ = modulus of rigidity

• Polar Moment of Inertia (for solid shaft):

J=πd432J = \frac{\pi d^4}{32}J=32πd4​

• Polar Moment of Inertia (for hollow shaft):

J=π(do4−di4)32J = \frac{\pi (d_o^4 – d_i^4)}{32}J=32π(do4​−di4​)​

Applications:

• Transmission shafts, axles, drive shafts, turbine shafts

Example:
A solid steel shaft, diameter 40 mm, length 2 m, torque 500 N·m, $G=80\text{ GPa}$. Find max shear stress & angle of twist.
👉 Solution:
τmax=TrJ=500∗0.02/(π∗0.044/32)≈19.89 MPa\tau_{\text{max}} = \frac{T r}{J} = 500 * 0.02 / (\pi * 0.04^4 /32) ≈ 19.89 \text{ MPa}τmax​=JTr​=500∗0.02/(π∗0.044/32)≈19.89 MPa
θ=TL/JG=500∗2/(0.0000804∗80∗109)≈0.156 rad\theta = T L / J G = 500*2 / (0.0000804*80*10^9) ≈ 0.156 \text{ rad}θ=TL/JG=500∗2/(0.0000804∗80∗109)≈0.156 rad

🔹 2. Springs

• Types of Springs:

  • Helical Torsion Springs: Coiled rods under torque

  • Compression/Extension Springs: Store mechanical energy

  • Leaf Springs: Rectangular sections, bending

• Helical Spring Equations (Torsion Coil):

τmax=16WRπd3(d = wire diameter, R = mean radius, W = load)\tau_{\text{max}} = \frac{16 W R}{\pi d^3} \quad (\text{d = wire diameter, R = mean radius, W = load})τmax​=πd316WR​(d = wire diameter, R = mean radius, W = load)

• Deflection of Helical Spring:

δ=8WR3nGd4(n = number of active coils)\delta = \frac{8 W R^3 n}{G d^4} \quad (\text{n = number of active coils})δ=Gd48WR3n​(n = number of active coils)

• Spring Stiffness (k):

k=Wδ=Gd48R3nk = \frac{W}{\delta} = \frac{G d^4}{8 R^3 n}k=δW​=8R3nGd4​

Applications:

• Suspension systems

• Clutches & brakes

• Vibration isolation

🔹 3. Solved Examples (PYQ Style)

Example 1 (GATE ME 2017):
Hollow shaft, do=50 mm,di=30 mm,L=1.5 md_o = 50\text{ mm}, d_i = 30\text{ mm}, L = 1.5 \text{ m}do​=50 mm,di​=30 mm,L=1.5 m, Torque = 800 N·m, $G=80\text{ GPa}$. Find max shear stress.
👉 Solution:
J=π(0.054−0.034)/32≈4.21∗10−6m4J = \pi (0.05^4 – 0.03^4)/32 ≈ 4.21*10^{-6} m^4J=π(0.054−0.034)/32≈4.21∗10−6m4
τmax=Tro/J=800∗0.025/4.21∗10−6≈4.75∗106Pa=4.75MPa\tau_{\text{max}} = T r_o / J = 800*0.025/4.21*10^{-6} ≈ 4.75*10^6 Pa = 4.75 MPaτmax​=Tro​/J=800∗0.025/4.21∗10−6≈4.75∗106Pa=4.75MPa

Example 2 (PSU Exam):
Helical spring: n=10, d=10 mm, R=50 mm, load W=500 N. Find deflection.
δ=8∗500∗0.053∗10/(80∗109∗0.014)≈0.156m\delta = 8*500*0.05^3*10 / (80*10^9*0.01^4) ≈ 0.156 mδ=8∗500∗0.053∗10/(80∗109∗0.014)≈0.156m

🔹 4. Practice Exercises

1. Solid shaft, diameter 30 mm, length 1 m, torque 400 N·m, G = 80 GPa. Find max shear stress & angle of twist.

2. Hollow shaft, do=60 mm,di=40 mmd_o = 60\text{ mm}, d_i = 40\text{ mm}do​=60 mm,di​=40 mm, torque 600 N·m. Find max shear stress.

3. Helical spring, d=8 mm, R=40 mm, n=12, W=300 N. Find spring deflection.

4. Design a shaft to transmit 2 kW at 100 rpm using steel, τ_allow = 40 MPa.

🔹 5. Summary

• Torsion of Shafts: Max shear stress, angle of twist, polar moment of inertia

• Springs: Helical, leaf, compression/extension, stiffness, deflection

• Applications: Power transmission, suspension, vibration isolation

• Key Formulas:

  • τmax=Tr/J\tau_{\text{max}} = T r / Jτmax​=Tr/J

  • $\theta =TL/JG$

  • Spring deflection: δ=8WR3n/Gd4\delta = 8 W R^3 n / G d^4δ=8WR3n/Gd4