Gears & gear trains
- Meshing gears
- n1·N1 = n2·N2
- Torque
- T2 = T1·(N2/N1)·η
- Power
- P = T·ω, ω = 2π·n/60
- Compound train
- total ratio = product of the stage ratios
Machine Design: cram pack
Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.
1Formulas and methods
Torsion of shafts
- Shear stress
- τ = T·r/J
- Solid shaft
- J = πd⁴/32
- Hollow shaft
- J = π(D⁴ − d⁴)/32
- Angle of twist
- φ = TL/(GJ)
- Power
- P = T·ω = 2πnT/60
Method
- Find the polar moment of area: J = πd⁴/32 for a solid shaft.
- Shear stress at the surface: τ = T·r/J, with r the outer radius.
- Angle of twist: φ = TL/(GJ) in radians.
- If power is given, find the torque first with P = T·ω, ω = 2πn/60.
Mohr’s circle & stress transformation
- Centre
- C = (σx + σy)/2
- Radius
- R = √(((σx − σy)/2)² + τxy²)
- Principal stresses
- σ1,2 = C ± R
- Principal direction
- tan 2θp = 2τxy/(σx − σy)
- Inclined plane
- σθ = C + (σx − σy)/2·cos 2θ + τxy·sin 2θ
Method
- Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
- Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
- The principal stresses are C ± R and the largest in-plane shear stress is R.
- Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).
Axial stress & strain
- Normal stress
- σ = F/A
- Strain
- ε = ΔL/L
- Hooke’s law
- σ = E·ε
- Elongation
- ΔL = FL/(EA)
- Thermal strain
- εT = α·ΔT
Column buckling
- Euler load
- Pcr = π²EI/(KL)²
- Effective length factor
- K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
- Critical stress
- σcr = Pcr/A
- Slenderness
- λ = KL/r, r = √(I/A)
Mechanical vibrations
- Natural frequency
- ωn = √(k/m), fn = ωn/(2π)
- From static deflection
- ωn = √(g/δst)
- Critical damping
- cc = 2√(km) = 2mωn
- Damping ratio
- ζ = c/cc
- Damped frequency
- ωd = ωn√(1 − ζ²)
- Logarithmic decrement
- δ = (1/n)·ln(x0/xn), ζ = δ/√(4π² + δ²)
- Springs in parallel
- k = k1 + k2
- Springs in series
- 1/k = 1/k1 + 1/k2
2Practice problems
1.Gears & gear trainsMedium
A 1.1 kW motor running at 960 rpm drives a pinion with 20 teeth and a module of 3 mm. Find the tangential force Ft on the pinion teeth.
P = 1.1 kWn = 960 rpmN1 = 20m = 3 mm
Ft = N
2.Gears & gear trainsHard
A 15 kW motor at 960 rpm drives gear A (40 teeth), which meshes with gear B (48 teeth). Gear C (18 teeth) on the same shaft as B drives gear D (45 teeth) on the output shaft. Each gear mesh has an efficiency of 95%. Find the output torque Tout.
P = 15 kWnin = 960 rpmA = 40B = 48C = 18D = 45ηmesh = 95%
Tout = N·m
3.Torsion of shaftsMedium
A hollow steel shaft (outer diameter 30 mm, inner diameter 15 mm, length 1.5 m) is fixed at one end and twisted by a torque of 1500 N·m at the other. Find the angle of twist φ in degrees.
T = 1500 N·mdo = 30 mmdi = 15 mmL = 1.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid steel shaft of diameter 50 mm is fixed at A. A torque of 200 N·m acts at B, 0.8 m from A, and a torque of 200 N·m in the same direction acts at the free end C, a further 1.5 m along. Find the maximum shear stress τmax in the shaft.
d = 50 mmTB = 200 N·mTC = 200 N·mAB = 0.8 mBC = 1.5 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 60 MPa, σᵧ = 120 MPa and τₓᵧ = 20 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 60 MPaσy = 120 MPaτxy = 20 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = 120 MPa, σᵧ = 0 MPa and τₓᵧ = 10 MPa (σz = 0). The material has a yield strength of 275 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.
σx = 120 MPaσy = 0 MPaτxy = 10 MPaσY = 275 MPa
n =
7.Axial stress & strainMedium
Find the elongation δ of the copper bar under the axial load.
P = 25 kNd = 25 mmL = 2.5 mE = 117 GPa
δ = mm
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 20 kNsegment 1: Ø40 mm, 1.5 m, steel (E = 200 GPa)segment 2: Ø20 mm, 2 m, brass (E = 100 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 1.5 mE = 200 GPa (steel)b × h = 40 × 80 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 105 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 105 kNn = 2.5L = 2 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 100 kg mass on a spring with a stiffness of 200 kN/m has a viscous damper giving a damping ratio of 0.35. Find the damped natural frequency fd of its free vibration in Hz.
m = 100 kgk = 200 kN/mζ = 0.35
fd = Hz
12.Mechanical vibrationsHard
A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 50 kN/m and k₂ = 50 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 10 kgk1 = 50 kN/mk2 = 50 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: Ft = 364.7 N
Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.
Torque on the pinion
ω = 2π·960/60 = 100.5 rad/s
T = P/ω = 1100/100.5 = 10.94 N·m
Pitch diameter
d = m·N = 3·20 = 60 mm
Tangential force
Ft = 2T/d = 2·10.94/0.06 = 364.7 N
Answer
Ft = 364.7 N
2.Gears & gear trains
Answer: Tout = 404 N·m
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
Output speed
nout = 960·(40/48)·(18/45) = 320 rpm
ωout = 2π·320/60 = 33.51 rad/s
Output power and torque
Pout = 15000·0.95² = 13538 W
Tout = Pout/ωout = 13538/33.51 = 404 N·m
Answer
Tout = 404 N·m
3.Torsion of shafts
Answer: φ = 21.62 °
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
Polar second moment of area
J = π(30⁴ − 15⁴)/32 = 74551 mm⁴
Angle of twist
T = 1500 N·m = 1.5 × 10⁶ N·mm
L = 1.5 m = 1500 mm
G = 80 GPa = 80000 MPa
φ = TL/(GJ) = 1.5 × 10⁶·1500/(80000·74551) = 0.3773 rad
φ = 0.3773·180/π = 21.62°
Answer
φ = 21.62°
4.Torsion of shafts
Answer: τmax = 16.3 MPa
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
Internal torques
Sense of TC taken as positive.
TBC = TC = 200 N·m
TAB = TC + TB = 200 + 200 = 400 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 400 N·m
τmax = 16·400000/(π·50³) = 16.3 MPa
Answer
τmax = 16.3 MPa
5.Mohr’s circle & stress transformation
Answer: τmax = 36.06 MPa
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.
Radius of Mohr’s circle
τmax = R = √(((60 − 120)/2)² + 20²) = 36.06 MPa
Answer
τmax = 36.06 MPa
6.Mohr’s circle & stress transformation
Answer: n = 2.26
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.
Principal stresses
C = 60 MPa, R = 60.83 MPa
σ1 = 120.8 MPa, σ2 = −0.8276 MPa, σ3 = 0
Tresca
σmax − σmin = 121.7 MPa
n = 275/121.7 = 2.26
Answer
n = 2.26
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: δ = 1.088 mm
Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).
Stress
A = π·25²/4 = 490.9 mm²
σ = P/A = 25·1000/490.9 = 50.93 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 50.93/117000 = 0.0004353 = 435.3 µε
Elongation
δ = ε·L = 0.0004353·2500 mm = 1.088 mm
Answer
δ = 1.088 mm
8.Axial stress & strain
Answer: δ = 1.393 mm
Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.
Same force in both segments
The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.
A1 = π·40²/4 = 1257 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 15.92 MPa
σ2 = 63.66 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 15.92·1500/200000 = 0.1194 mm
δ2 = σ2L2/E2 = 63.66·2000/100000 = 1.273 mm
δ = δ1 + δ2 = 1.393 mm
Answer
δ = 1.393 mm
9.Column buckling
Answer: Pcr = 93.58 kN
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Smallest second moment of area
A rectangle buckles about its weak axis, using the smaller dimension cubed.
Imin = h·b³/12 = 80·40³/12 = 426667 mm⁴
Effective length
KL = 2·1500 = 3000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·426667/3000² = 93578 N = 93.58 kN
Answer
Pcr = 93.58 kN
10.Column buckling
Answer: d = 105.5 mm
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Required critical load
Pcr = n·P = 2.5·105000 = 262500 N
KL = 2·2000 = 4000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·262500·4000²/(π³·70000))^(1/4) = 105.5 mm
Answer
d = 105.5 mm
Check: the slenderness KL/r is 151.7, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 6.667 Hz
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Undamped natural frequency
ωn = √(200000/100) = 44.72 rad/s
fn = 44.72/(2π) = 7.118 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 7.118·√(1 − 0.35²) = 6.667 Hz
Answer
fd = 6.667 Hz
12.Mechanical vibrations
Answer: keq = 100 kN/m
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Equivalent stiffness
Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 50 + 50 = 100 kN/m
Answer
keq = 100 kN/m
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