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 4 kW motor running at 1440 rpm drives a pinion with 25 teeth and a module of 3 mm. Find the tangential force Ft on the pinion teeth.
P = 4 kWn = 1440 rpmN1 = 25m = 3 mm
Ft = N
2.Gears & gear trainsHard
In a compound gear train, gear A (30 teeth) on the input shaft turns at 1500 rpm and drives gear B (36 teeth). Gear C (18 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 750 rpm. How many teeth ND must gear D have?
nin = 1500 rpmA = 30B = 36C = 18nout = 750 rpm
ND =
3.Torsion of shaftsMedium
A solid brass shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.
T = 400 N·md = 40 mmL = 2 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 10 kW at 750 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.
P = 10 kWN = 750 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −30 MPa, σᵧ = 10 MPa and τₓᵧ = −10 MPa (tension positive). The element is rotated 50° counter-clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.
σx = −30 MPaσy = 10 MPaτxy = −10 MPaθ = 50°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 10 MPa, σᵧ = 60 MPa and τₓᵧ = −50 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 40° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 10 MPaσy = 60 MPaτxy = −50 MPaθ = 40°
τθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the steel bar under the axial load.
P = 45 kNd = 12 mmL = 3 mE = 200 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 = 105 kNsegment 1: Ø40 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 1.5 m, copper (E = 117 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 4 mE = 70 GPa (aluminium)b × h = 40 × 50 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 1 mE = 70 GPa (aluminium)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 2 kg mass on a spring with a stiffness of 12 kN/m has a viscous damper giving a damping ratio of 0.6. Find the damped natural frequency fd of its free vibration in Hz.
m = 2 kgk = 12 kN/mζ = 0.6
fd = Hz
12.Mechanical vibrationsHard
A 10 kg mass hangs from spring k₃ = 12 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 15 kN/m and k₂ = 40 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.
m = 10 kgk1 = 15 kN/mk2 = 40 kN/mk3 = 12 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: Ft = 707.4 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·1440/60 = 150.8 rad/s
T = P/ω = 4000/150.8 = 26.53 N·m
Pitch diameter
d = m·N = 3·25 = 75 mm
Tangential force
Ft = 2T/d = 2·26.53/0.075 = 707.4 N
Answer
Ft = 707.4 N
2.Gears & gear trains
Answer: ND = 30
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.
Intermediate shaft speed
nB = nC = 1500·30/36 = 1250 rpm
Teeth on gear D
ND = C·nC/nout = 18·1250/750 = 30
Answer
ND = 30
3.Torsion of shafts
Answer: φ = 4.676 °
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·40⁴/32 = 251327 mm⁴
Angle of twist
φ = TL/(GJ) = 400·1000·2000/(39000·251327) = 0.08162 rad
φ = 0.08162·180/π = 4.676°
Answer
φ = 4.676°
4.Torsion of shafts
Answer: d = 20.52 mm
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.
Torque
ω = 2πN/60 = 2π·750/60 = 78.54 rad/s
T = P/ω = 10000/78.54 = 127.3 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·127324/(π·75)) = 20.52 mm
Answer
d = 20.52 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: σy′ = −3.625 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.
Transformation equation for the y′ face
σy′ = −10 − (−20)·(−0.1736) − (−10)·0.9848 = −3.625 MPa
Check
The sum of the normal stresses does not change with rotation.
σx′ = −16.38 MPa, σx′ + σy′ = −20 = σx + σy
Answer
σy′ = −3.625 MPa
6.Mohr’s circle & stress transformation
Answer: τθ = 15.94 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.
Transformation equation
τθ = −−25·0.9848 + −50·0.1736 = 15.94 MPa
Answer
τθ = 15.94 MPa
7.Axial stress & strain
Answer: δ = 5.968 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·12²/4 = 113.1 mm²
σ = P/A = 45·1000/113.1 = 397.9 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 397.9/200000 = 0.001989 = 1989 µε
Elongation
δ = ε·L = 0.001989·3000 mm = 5.968 mm
Answer
δ = 5.968 mm
8.Axial stress & strain
Answer: δ = 4.882 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 = 83.56 MPa
σ2 = 334.2 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 83.56·500/70000 = 0.5968 mm
δ2 = σ2L2/E2 = 334.2·1500/117000 = 4.285 mm
δ = δ1 + δ2 = 4.882 mm
Answer
δ = 4.882 mm
9.Column buckling
Answer: Pcr = 2.879 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 = 50·40³/12 = 266667 mm⁴
Effective length
KL = 2·4000 = 8000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·266667/8000² = 2879 N = 2.879 kN
Answer
Pcr = 2.879 kN
10.Column buckling
Answer: σcr = 17.27 MPa
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 round section has the same I about every axis.
I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴
Radius of gyration and slenderness
A = 1257 mm²
r = √(I/A) = √(125664/1257) = 10 mm
KL = 2·1000 = 2000 mm
KL/r = 2000/10 = 200
Critical stress
σcr = π²E/(KL/r)² = π²·70000/200² = 17.27 MPa
Answer
σcr = 17.27 MPa
11.Mechanical vibrations
Answer: fd = 9.862 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 = √(12000/2) = 77.46 rad/s
fn = 77.46/(2π) = 12.33 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 12.33·√(1 − 0.6²) = 9.862 Hz
Answer
fd = 9.862 Hz
12.Mechanical vibrations
Answer: fn = 4.995 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.
Equivalent stiffness
Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.
k12 = 15 + 40 = 55 kN/m
keq = 55·12/(55 + 12) = 9.851 kN/m
Natural frequency
ωn = √(9851/10) = 31.39 rad/s
fn = 31.39/(2π) = 4.995 Hz
Answer
fn = 4.995 Hz
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