Machine Design
GATE ME syllabus for machine design, with theories of failure, fatigue and the Goodman and Soderberg lines, design of shafts, joints, springs, bearings, gears, brakes and clutches, with solved examples.
Syllabus
Design for static and dynamic loading; failure theories; fatigue strength and the S-N diagram; principles of design of bolted, riveted and welded joints, shafts, gears, rolling and sliding contact bearings, brakes and clutches, and springs.
Key concepts and formulas
Theories of failure (σy = yield strength, N = factor of safety):
| Theory | Failure when | Suited to |
|---|---|---|
| Maximum principal stress (Rankine) | σ₁ = σy/N | Brittle materials |
| Maximum shear stress (Tresca) | (σ₁ − σ₂)/2 = σy/(2N) (in-plane, with σ₃ = 0 and σ₁, σ₂ of opposite sign) | Ductile, conservative |
| Distortion energy (von Mises) | √(σ₁² − σ₁σ₂ + σ₂²) = σy/N | Ductile, most accurate |
In pure shear, Tresca predicts yield at τ = 0.5σy and von Mises at τ = σy/√3 ≈ 0.577σy.
Shafts under combined bending and torsion. Equivalent torque Te = √(M² + T²) (max shear stress theory), equivalent moment Me = ½(M + √(M² + T²)) (max principal stress theory). Then τmax = 16Te/(πd³) and σmax = 32Me/(πd³).
Fatigue. For steels, the endurance limit Se′ ≈ 0.5Sut, reduced by surface, size, reliability and other factors. With alternating stress σa and mean stress σm:
- Goodman: σa/Se + σm/Sut = 1/N
- Soderberg: σa/Se + σm/Sy = 1/N (more conservative)
- Gerber: σa/Se + (σm/Sut)² = 1 (with N = 1)
Joints. Fillet weld: throat = 0.707 × leg size; shear stress = load/(0.707 s l). Bolts: tensile stress area, preload.
Helical springs. Spring index C = D/d. Shear stress τ = 8PDK/(πd³) (K = Wahl factor); deflection δ = 8PD³n/(Gd⁴); stiffness k = Gd⁴/(8D³n).
Rolling bearings. Rating life L₁₀ = (C/P)ᵖ million revolutions, with p = 3 for ball and 10/3 for roller bearings.
Gears. Lewis bending strength F = σbbmY (Y = Lewis form factor).
Brakes and clutches. Single-plate clutch with uniform wear: T = μW(r₁ + r₂)/2 per friction surface; with uniform pressure: T = (2/3)μW(r₁³ − r₂³)/(r₁² − r₂²). Band brake: T₁/T₂ = eμθ.
Worked example
A ductile shaft is under σ₁ = 120 MPa and σ₂ = −60 MPa, with σy = 300 MPa. Find the factor of safety by the Tresca and von Mises theories. Then find the Goodman factor of safety for σa = 100 MPa, σm = 150 MPa, Se = 250 MPa and Sut = 600 MPa, and the life of a ball bearing with C = 30 kN and P = 6 kN.
import math
s1, s2, sy = 120, -60, 300
tresca = sy / (s1 - s2)
vm = sy / math.sqrt(s1 ** 2 - s1 * s2 + s2 ** 2)
print(f"Tresca N = {tresca:.3f}, von Mises N = {vm:.3f}")
sa, sm, se, sut = 100, 150, 250, 600
print(f"Goodman N = {1 / (sa / se + sm / sut):.4f}")
C, P = 30, 6
print(f"ball bearing life = {(C / P) ** 3:.0f} million revolutions")
G, d, D, n, F = 80e3, 8, 48, 10, 500
print(f"spring stiffness = {G * d ** 4 / (8 * D ** 3 * n):.3f} N/mm")
Tresca N = 1.667, von Mises N = 1.890 Goodman N = 1.5385 ball bearing life = 125 million revolutions spring stiffness = 37.037 N/mm
Practice questions
A ball bearing has a dynamic load rating of 20 kN and carries 4 kN. What is its rated life in million revolutions?
Show answer and explanation
Answer: 125L₁₀ = (C/P)³ = 5³ = 125 million revolutions.
In pure shear, what shear stress does the maximum shear stress (Tresca) theory allow, as a fraction of the yield strength?
Show answer and explanation
Answer: 0.5Tresca predicts yielding when τmax = σy/2, so the allowed shear stress is 0.5σy (von Mises gives 0.577σy).
A shaft carries a bending moment of 3 kN m and a torque of 4 kN m. What is the equivalent torque in kN m (maximum shear stress theory)?
Show answer and explanation
Answer: 5Te = √(M² + T²) = √(9 + 16) = 5 kN m.
A component has Se = 200 MPa and Sut = 500 MPa, with σa = 80 MPa and σm = 100 MPa. What is the factor of safety by the Goodman criterion? Give the answer to three decimal places.
Show answer and explanation
Answer: 1.6671/N = 80/200 + 100/500 = 0.4 + 0.2 = 0.6, so N ≈ 1.667.
Which failure theory is generally recommended for ductile materials as the most accurate?
Show answer and explanation
Answer: CThe distortion energy theory agrees best with experiments on ductile materials. The maximum principal stress theory suits brittle materials.