Unit 1
Design for static and dynamic loading
Static loads are applied slowly and stay constant, so the part only needs to survive them once: compare the stress with yield (ductile) or ultimate strength (brittle). Dynamic loads change with time. They either hit suddenly (impact) or repeat millions of times (fatigue, Unit 2), and both make a part fail well below its static strength.
The basic stresses
Direct σ = F / A
Bending σ_b = M·y / I solid round: σ_b = 32 M / (π d³)
Torsion τ = T·r / J solid round: τ = 16 T / (π d³)
Direct shear τ = F / A (average; peak is 4/3 × for a round section)
Bearing/crushing σ_c = F / (d · t)
Factor of safety
n = strength / stress ductile: S_y / σ brittle: S_ut / σ
| Situation | Typical n |
|---|
| Well-known materials, loads and environment | 1.25 – 1.5 |
| Average materials, loads that can be estimated | 2 – 3 |
| Uncertain loads or materials, brittle materials | 3 – 4 |
| Lifting gear, pressure parts, anything that can hurt people | Follow the code (often 4 – 6) |
Stress concentration
Holes, grooves, shoulders and keyways raise the local stress to Kt × nominal (Kt from charts, usually 1.5–3). For ductile parts under static load the peak yields locally and redistributes, so Kt is usually ignored. For brittle parts, and for every part under fatigue, it must be applied (for fatigue as Kf, see Unit 2). Reduce it with generous fillet radii, relief grooves and by moving holes out of high-stress areas.
Impact loading
Falling weight W from height h onto a part whose static deflection under W is δ_st:
impact factor = 1 + √(1 + 2h / δ_st)
σ_impact = σ_static × impact factor
Suddenly applied load (h = 0): σ = 2 × σ_static
The impact factor depends on stiffness: a more flexible part (bigger δst) absorbs the same energy at lower stress. That is why bolts for impact are turned down to the root diameter along the shank, and why rubber mounts and springs protect machines.
Typical material properties
| Material | Sut | Sy | E, GPa | Notes |
|---|
| Plain carbon steel 30C8 / AISI 1030 | 500 | 300 | 207 | General parts, welded frames |
| Medium carbon steel 40C8 / EN8 / AISI 1040 | 580 | 380 | 207 | Shafts, pins, keys |
| Alloy steel 40Cr4Mo2 / EN19 / 4140 (Q&T) | 900–1000 | 700–850 | 207 | Gears, highly loaded shafts |
| Stainless 304 | 515 | 205 | 193 | Food and wet machines |
| Aluminium 6061-T6 | 310 | 275 | 69 | No true endurance limit |
| Grey cast iron FG 200 | 200 | — | 100 | Brittle; compressive strength about 3–4 × Sut |
Typical values only; the mill certificate or the standard for your grade governs.
Theories of failure
A tensile test gives one number for one stress state. Real parts see combined stresses. A failure theory turns the combined state into one equivalent value you can compare with the tensile test.
Principal stresses (plane stress)
σ₁, σ₂ = (σx + σy)/2 ± √( ((σx − σy)/2)² + τxy² ) σ₃ = 0
τ_max = largest of |σ₁ − σ₂|/2, |σ₁|/2, |σ₂|/2
The theories
| Theory | Failure when | Use for |
|---|
| Maximum principal stress (Rankine) | σ₁ = Sut or σ₃ = −Suc | Brittle materials, simple estimate |
| Maximum shear stress (Tresca, Guest) | σ₁ − σ₃ = Sy | Ductile; conservative, easy hand calculation |
| Distortion energy (von Mises, Hencky) | σ′ = Sy | Ductile; the most accurate, used in FEA |
| Maximum principal strain (St. Venant) | ε₁ = Sy / E | Historical; rarely used now |
| Strain energy (Haigh) | Total strain energy = that at yield | Historical; superseded by distortion energy |
| Coulomb–Mohr (brittle) | σ₁/Sut − σ₃/Suc = 1 | Brittle materials with Suc ≠ Sut (cast iron) |
von Mises: σ′ = √(σx² − σx·σy + σy² + 3τxy²) shaft (σ, τ): σ′ = √(σ² + 3τ²)
Tresca: shaft: τ_max = √((σ/2)² + τ²) ≤ S_y / (2n)
Shear yield: S_sy = 0.577 S_y (von Mises) or 0.5 S_y (Tresca)
Rule of thumb: ductile (elongation > 5%) → von Mises, or Tresca for a conservative quick check. Brittle → Coulomb–Mohr or Rankine.
For a brittle material, set Suc to its real compressive strength (about 3–4 × Sut for grey cast iron) and read the Rankine and Coulomb–Mohr results.
Unit 2
Fatigue strength and design against fluctuating loads
Under repeated stress a crack starts at a surface flaw or notch, grows a little each cycle, and the part snaps without warning at a stress well below yield. Most machine failures are fatigue failures. The fracture surface shows smooth beach marks from the crack growth and a rough final break.
Describing the cycle
Alternating σ_a = (σ_max − σ_min) / 2
Mean σ_m = (σ_max + σ_min) / 2
Stress ratio R = σ_min / σ_max fully reversed R = −1, repeated R = 0
The S-N diagram
Plot alternating stress against cycles to failure on log–log axes. For steel the line runs from about 0.9·Sut at 10³ cycles down to the endurance limit Se at about 10⁶ cycles, then flattens: below Se steel lasts indefinitely. Aluminium and copper alloys have no knee; quote their fatigue strength at a given life (often 5×10⁸ cycles).
Between 10³ and 10⁶ cycles: S_f = a · N^b
a = (f · S_ut)² / S_e b = −(1/3) · log₁₀(f · S_ut / S_e) f ≈ 0.9
Cycles to failure: N = (σ_a / a)^(1/b)
Endurance limit: from test bar to real part
Rotating-beam specimen (steel): S_e′ = 0.5 S_ut (S_ut ≤ 1400 MPa), 700 MPa above that
Real part (Marin): S_e = k_a · k_b · k_c · k_d · k_e · S_e′
| Factor | Accounts for | Value |
|---|
| ka surface | Roughness | a · Sutb: ground 1.58, −0.085; machined or cold drawn 4.51, −0.265; hot rolled 57.7, −0.718; as forged 272, −0.995 |
| kb size | Bigger parts have more flaws | Bending/torsion: 1.24 d−0.107 (2.79–51 mm), 1.51 d−0.157 (51–254 mm). Axial: 1 |
| kc load | Type of loading | Bending 1, axial 0.85, torsion 0.59 |
| kd temperature | Heat | 1 up to about 250 °C for steel |
| ke reliability | Scatter | 50% 1, 90% 0.897, 95% 0.868, 99% 0.814, 99.9% 0.753 |
Fatigue stress concentration: K_f = 1 + q (K_t − 1) q = notch sensitivity, 0 to 1 (≈ 0.7–0.95 for steels)
Mean stress
A tensile mean stress opens cracks and shortens life. The criteria join Se on the alternating axis to a strength on the mean axis:
Goodman σ_a / S_e + σ_m / S_ut = 1 / n (standard choice)
Soderberg σ_a / S_e + σ_m / S_y = 1 / n (conservative, guards against yield)
Gerber n σ_a / S_e + (n σ_m / S_ut)² = 1 (fits test data best)
Yield check n_y = S_y / (σ_a + σ_m) (Langer)
Varying loads: Miner's rule
Σ (n_i / N_i) = 1 at failure n_i = cycles at stress level i, N_i = life at that level from the S-N line
Combined fluctuating stresses
When a part sees fluctuating bending, axial and torsional stress together, combine the alternating parts and the mean parts separately with von Mises, then use them in the Goodman line:
σ_a′ = √( (K_f σ_a,bending + K_f σ_a,axial / 0.85)² + 3 (K_fs τ_a)² )
σ_m′ = √( (K_f σ_m,bending + K_f σ_m,axial)² + 3 (K_fs τ_m)² )
then σ_a′ / S_e + σ_m′ / S_ut = 1 / n (use k_c = 1 for S_e: the load type is already in σ′)
Designing against fluctuating loads
- Find the real load cycle first: σmax and σmin at the critical point, not just the peak load.
- Remove stress raisers: large fillet radii, relief grooves beside shoulders, holes away from high-stress zones, no sharp tool marks.
- Improve the surface: grinding or polishing raises ka; shot peening, surface rolling and case hardening add compressive residual stress that slows crack growth.
- Avoid tensile mean stress where possible, and keep bolted joints preloaded so the bolt sees only a small stress swing.
- Welds, threads and fretting surfaces have their own low fatigue strengths: check them against the relevant code curves.
Enter nominal stresses; the calculator multiplies both by Kf. For combined bending and torsion, enter von Mises alternating and mean stresses and pick Bending. The dot on the S-N plot is the equivalent fully reversed stress (Goodman).
Unit 3
Design of joints
Bolted joints
Preload, tightening torque, joint stiffness and slip are covered with a calculator in the Bolts and nuts guide. For a group of bolts loaded off-centre (a bracket on a column):
Primary shear (each bolt) F′ = F / n
Secondary shear (bolt i) F″_i = F · e · r_i / Σ r_j² r = distance from the group centroid
Add F′ and F″ as vectors; the bolt with the largest resultant governs.
Bracket in tension about an edge: bolt load P_i = F · e · l_i / Σ l_j² (l from the tilting edge)
Riveted joints
Lap joints have one plate over the other (rivets in single shear). Butt joints use one or two cover straps (two straps put the rivets in double shear). Check each way the joint can fail over one pitch length p:
Tearing of plate between rivets P_t = (p − d) · t · σ_t
Shearing of rivets P_s = n · (π/4) d² · τ × 2 for double shear (some codes use 1.875)
Crushing of plate or rivet P_c = n · d · t · σ_c
Solid plate strength P = p · t · σ_t
Efficiency η = min(P_t, P_s, P_c) / P
Rivet diameter (Unwin, t > 8 mm) d = 6 √t margin m = 1.5 d
n = rivets per pitch length, d = hole diameter, t = plate thickness
Welded joints
Butt weld (full penetration) P = σ_t · t · l
Fillet weld throat t_throat = 0.707 h h = leg size
Parallel (side) fillet P = 0.707 h · l · τ per weld
Transverse (end) fillet P = 0.707 h · l · σ_t per weld
Circular fillet in torsion τ = 2.83 T / (π h d²)
Eccentric load on a weld group τ″ = M · r / J, J = 0.707 h · J_u (J_u = unit polar moment of the weld lines)
- Add about 15 mm to each weld length for the start and crater, which are weak.
- Weld toes are fatigue hot spots. For cyclic loads use the weld class S-N curves in a code (IIW, BS 7608, AWS D1.1) rather than parent metal strength.
- Put welds away from bends and high-stress corners, and keep the load path through the throat simple.
Unit 4
Design of shafts and keys
A shaft transmits torque and carries the bending from gears, pulleys and sprockets. Draw the bending moment diagrams in two planes, combine them (M = √(MH² + MV²)), and size the shaft at the worst section, usually under a gear or at a shoulder near a bearing.
Torque from power T = 9550 · P / n (T in N·m, P in kW, n in rpm)
Belt pull T = (T₁ − T₂) · r bending load = T₁ + T₂
Gear forces F_t = 2T / d, F_r = F_t · tan φ
Static strength (ASME code)
d³ = 16 / (π τ_allow) · √( (K_b · M)² + (K_t · T)² )
τ_allow = 0.30 S_y or 0.18 S_ut, whichever is lower; × 0.75 if there is a keyway
Equivalent torque T_e = √(M² + T²) Equivalent moment M_e = ½ (M + √(M² + T²))
| Load | Kb (bending) | Kt (torsion) |
|---|
| Gradually applied, rotating shaft | 1.5 | 1.0 |
| Suddenly applied, minor shock | 1.5 – 2.0 | 1.0 – 1.5 |
| Suddenly applied, heavy shock | 2.0 – 3.0 | 1.5 – 3.0 |
Fatigue (DE-Goodman)
A rotating shaft sees fully reversed bending (Ma = M) with steady torque (Tm = T):
d = [ (16 n / π) · ( 2 K_f M_a / S_e + √3 K_fs T_m / S_ut ) ]^(1/3)
Get Se from the fatigue calculator. Kf is about 1.7 at a sharp shoulder, 1.5 at a well-rounded one, 2.2 at an end-milled keyway (first estimates; check with charts once the geometry is known).
Rigidity and speed
Twist θ = T · L / (G · J) J = π d⁴ / 32, G ≈ 79,300 MPa; limit about 0.25°/m for line shafts
Deflection at gears ≤ 0.13 mm; slope at ball bearings ≤ 0.001 rad (deep groove), 0.0005 at gears
Critical speed keep running speed ≤ 0.5–0.7 × first critical speed
Keys
A key locks a hub (gear, pulley, coupling) to a shaft so torque passes between them. It is deliberately the weakest part of the joint: a sheared key is cheap to replace, a cracked shaft is not.
| Type | Description | Use |
|---|
| Sunk parallel key (rectangular or square) | Half in the shaft, half in the hub, sides fit tightly | The standard choice for gears, pulleys and couplings |
| Taper and gib-head key | 1:100 taper on top, driven in; gib head for removal | Heavy, shock-loaded hubs; also holds the hub axially |
| Feather key | Parallel key fixed to the shaft or hub, hub slides along it | Sliding gears in gearboxes, clutches |
| Woodruff key | Half-disc in a circular seat | Small shafts, tapered shaft ends, automotive |
| Saddle and flat keys | No keyway (saddle) or a flat on the shaft | Light duty only, relies on friction |
| Splines | Many integral keys around the shaft | High torque, sliding hubs, even load sharing |
Standard parallel key sizes (DIN 6885 / ISO R773)
| Shaft diameter, mm | Width b × height h | Shaft keyway depth t₁ |
|---|
| over 10 to 12 | 4 × 4 | 2.5 |
| 12 – 17 | 5 × 5 | 3.0 |
| 17 – 22 | 6 × 6 | 3.5 |
| 22 – 30 | 8 × 7 | 4.0 |
| 30 – 38 | 10 × 8 | 5.0 |
| 38 – 44 | 12 × 8 | 5.0 |
| 44 – 50 | 14 × 9 | 5.5 |
| 50 – 58 | 16 × 10 | 6.0 |
| 58 – 65 | 18 × 11 | 7.0 |
| 65 – 75 | 20 × 12 | 7.5 |
| 75 – 85 | 22 × 14 | 9.0 |
Textbook proportions for a first guess: width b = d/4, height h = 2b/3 (rectangular) or h = b (square).
Strength of a sunk key
Force on the key F = 2T / d
Shear τ = F / (b · l) = 2T / (d · b · l)
Crushing σ_c = F / (l · h/2) = 4T / (d · h · l)
Square key, same material as the shaft: key as strong as the shaft in torsion when l ≈ 1.57 d
Keyway weakening (Moore): e = 1 − 0.2 (b/d) − 1.1 (t₁/d) shaft torque capacity × e
- Key material is usually 45C8 / C45 or similar, slightly weaker than the shaft. Allowable crushing stress is about twice the allowable shear.
- The hub should be at least as long as the key; 1.5 d is a common hub length.
- If one key is too long, use two keys at 180° (each takes about 75% of the torque in practice) or move to splines.
Unit 5
Design of springs
Helical compression and extension springs
Spring index C = D / d keep 4 – 12 (D = mean coil diameter, d = wire diameter)
Wahl factor K = (4C − 1)/(4C − 4) + 0.615 / C (curvature + direct shear)
Shear stress τ = K · 8 P D / (π d³)
Deflection δ = 8 P D³ N / (G d⁴) G ≈ 81,000 MPa for spring steel
Stiffness k = G d⁴ / (8 D³ N)
Springs in series 1/k = 1/k₁ + 1/k₂ in parallel k = k₁ + k₂
| End type | Total coils Nt | Solid length |
|---|
| Plain | N | (N + 1) d |
| Plain and ground | N + 1 | (N + 1) d |
| Squared (closed) | N + 2 | (N + 3) d |
| Squared and ground | N + 2 | (N + 2) d |
- Free length = solid length + maximum working deflection + clash allowance (about 15% of the deflection).
- Buckling: if free length / D is more than about 4 (ends free to tilt) or 5 (squared ends on parallel plates), guide the spring on a rod or in a bore.
- Allowable shear stress, static: about 0.45 Sut for music wire and oil-tempered wire. For fatigue use the spring Goodman line with Ssu = 0.67 Sut, and shot-peen the wire.
- Surge: keep the operating frequency well below the spring's natural frequency f = (d / (2π D² N)) · √(G / (2ρ)).
Leaf springs
Semi-elliptic, total load 2W on span 2L: σ = 6 W L / (n b t²)
Full and graduated leaves (n_f full, n_g graduated), pre-stressed (nipped) to equalise:
σ = 6 W L / (n b t²) δ = 12 W L³ / (E b t³ (3 n_f + 2 n_g))
Torsion springs
σ = K_i · 32 M / (π d³) K_i = (4C² − C − 1) / (4C (C − 1)) angle θ = 64 M D N / (E d⁴) (rad)
Unit 6
Design of gears
| Type | Shafts | Notes |
|---|
| Spur | Parallel | Simplest, noisy at high speed, no axial thrust |
| Helical | Parallel (or crossed) | Quieter, stronger, creates axial thrust (use double helical to cancel it) |
| Bevel | Intersecting, usually 90° | Straight or spiral; needs accurate mounting |
| Worm | Crossed, 90° | Ratios 5–100 in one stage, can be self-locking, low efficiency at high ratio |
Geometry (metric, 20° full depth)
Module m = d / z (standard: 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10)
Addendum = m, dedendum = 1.25 m, circular pitch p = π m
Centre distance a = m (z₁ + z₂) / 2 Ratio i = z₂ / z₁
Minimum teeth 18 on the pinion to avoid interference (14 with 25° pressure angle)
Face width b ≈ 9.5 m – 12.5 m (10 m is a common start)
Bending strength (Lewis) and dynamic load
Tangential load P_t = 2T / d = 1000 · P(kW) / v(m/s)
Beam strength S_b = m · b · σ_b · Y Y = 0.484 − 2.87 / z (20° full depth)
Velocity factor C_v = 3 / (3 + v) (ordinary cut, v < 10 m/s) 6 / (6 + v) (generated, v < 20 m/s)
Effective load P_eff = C_s · P_t / C_v C_s service factor 1.0 – 2.0
Requirement S_b ≥ n · P_eff
If pinion and gear share a material, the pinion (fewer teeth, smaller Y) is weaker. If materials differ, check the one with the lower σb·Y.
Wear strength (Buckingham)
S_w = b · Q · d_p · K Q = 2 z_g / (z_g + z_p)
K = 0.16 (BHN / 100)² (steel pinion and gear, BHN = surface hardness, N/mm²)
Pitting from contact stress, not tooth bending, is what ends most gear lives; hardening the teeth raises K with the square of hardness. Lubricate, and use AGMA or ISO 6336 for final rating.
Default σb = Sut/3 for 40C8-class steel. The factor of safety reported is strength / effective load.
Unit 7
Rolling and sliding contact bearings
Rolling contact
Type selection, equivalent load, L10 life, static safety and fits, with a life calculator, are in the Bearing selection guide. The key result:
L₁₀ = (C / P)^p million rev p = 3 ball, 10/3 roller
For a cyclic load: P_mean = ( Σ P_i^p · n_i / Σ n_i )^(1/p)
Sliding contact (journal) bearings
In a hydrodynamic bearing the rotating journal drags oil into a converging wedge, building pressure that lifts the shaft clear of the bush. There is no metal contact once it is running, so life is very long, but it needs oil supply, and it wears at start and stop.
Unit load p = W / (l · d) l/d usually 0.5 – 1.5
Clearance ratio c / r ≈ 0.001 (radial clearance / radius)
Petroff friction f = 2π² · (μ N / p) · (r / c) μ = viscosity Pa·s, N = rev/s
Sommerfeld number S = (r / c)² · μ N / p → read eccentricity, film thickness, f from Raimondi–Boyd charts
Heat generated H = f · W · V (W, V = journal surface speed m/s)
- Bearing modulus Z·N/p: below a minimum value the film breaks down. Run at 3 times the minimum or more so speed or load changes do not cause metal contact.
- Minimum film thickness must exceed the combined surface roughness (often 0.005 × c to a few µm); check temperature rise against the oil's limit.
- Materials: babbitt (white metal) for large hydrodynamic bearings, bronze for bushes, sintered oil-impregnated bronze and PTFE-lined bushes for dry or lightly lubricated slow motion.
Dry and boundary-lubricated bushes: the pV limit
p · V ≤ (pV)_max p = W / (l · d) MPa, V = π d n / 60000 m/s
Rough (pV)max: acetal and nylon about 0.1 MPa·m/s, sintered bronze about 1.8, PTFE composite bushes up to about 3.5. Check the maker's data, which also limits p and V separately.
Unit 8
Brakes and clutches
Both transmit torque through friction. A clutch connects two rotating shafts; a brake connects a rotating part to a fixed one. Design is about torque capacity, contact pressure (wear) and heat.
Plate (disc) clutches
Uniform wear (worn-in clutches, used for design):
T = n · μ · W · (R₁ + R₂) / 2 p_max = W / (2π R₂ (R₁ − R₂)) at the inner radius R₂
Uniform pressure (new clutches):
T = n · μ · W · (2/3)(R₁³ − R₂³) / (R₁² − R₂²) p = W / (π (R₁² − R₂²))
n = number of friction surfaces (pairs of faces in contact), W = axial spring force
Best torque for a given outer radius: R₂ ≈ 0.58 R₁
Cone clutch T = μ · W · R_m / sin α α = semi-cone angle (≥ 12° to avoid jamming)
Centrifugal clutch F = m · ω² · r − spring force, T = n · μ · F · R
Brakes
Short block brake T = μ · N · r
Long shoe (2θ > 60°) μ′ = 4 μ sin θ / (2θ + sin 2θ) use μ′ in place of μ
Band brake T₁ / T₂ = e^(μ θ), T = (T₁ − T₂) · r θ = wrap angle in radians
Internal expanding T = μ p_max b r² (cos θ₁ − cos θ₂) / sin θ_max per shoe
Energy to absorb E = ½ m v² + ½ I ω² (+ m g h when lowering a load)
Temperature rise ΔT = E / (m_brake · c) c ≈ 460 J/kg·K for steel and iron
Check direction: on a block or band brake one direction of rotation can be self-energising (the friction helps apply the brake) and the other is not. Avoid self-locking unless you want it.
| Lining | μ (dry) | pmax, MPa |
|---|
| Woven asbestos-free / organic on steel or cast iron | 0.3 – 0.45 | 0.35 – 0.7 |
| Sintered metal on cast iron (dry) | 0.3 – 0.4 | 1.0 |
| Sintered metal in oil (wet clutch) | 0.05 – 0.1 | 2.0 |
| Cork on metal | 0.3 – 0.5 | 0.06 |
Formulas follow standard machine design texts (Shigley, Bhandari) and the ASME shaft code. They are for learning and first sizing; use the relevant design code and your supplier's data for final numbers, and have anything safety-critical checked by a qualified engineer.