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
Two parallel shafts are 240 mm apart. They must be connected by a pair of spur gears with a module of 4 mm and a speed reduction of 3 : 1. How many teeth N1 must the pinion have?
a = 240 mmm = 4 mmi = 3
N1 =
2.Gears & gear trainsHard
In a compound gear train, gear A (20 teeth) on the input shaft turns at 2880 rpm and drives gear B (30 teeth). Gear C (24 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 640 rpm. How many teeth ND must gear D have?
nin = 2880 rpmA = 20B = 30C = 24nout = 640 rpm
ND =
3.Torsion of shaftsMedium
In a torsion test, a solid specimen of diameter 20 mm is twisted by a torque of 20 N·m. The two ends of a gauge length of 400 mm rotate 1.12° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 20 N·md = 20 mmL = 400 mmφ = 1.12°
G = GPa
4.Torsion of shaftsHard
A solid aluminium shaft of diameter 60 mm is fixed at A. A torque of 200 N·m acts at B, 0.5 m from A, and a torque of 400 N·m in the same direction acts at the free end C, a further 0.8 m along. Find the maximum shear stress τmax in the shaft.
d = 60 mmTB = 200 N·mTC = 400 N·mAB = 0.5 mBC = 0.8 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 10 MPa, σᵧ = −50 MPa and τₓᵧ = −10 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 10 MPaσy = −50 MPaτxy = −10 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 0 MPa, σᵧ = −20 MPa and τₓᵧ = 40 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 50° counter-clockwise from the x-axis.
σx = 0 MPaσy = −20 MPaτxy = 40 MPaθ = 50°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 20 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 30 mm, length 500 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.
segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 500 mm, E = 70 GPaδ = 0.2 mm
P = kN
8.Axial stress & strainHard
A steel segment (diameter 30 mm, length 400 mm, E = 200 GPa) and a segment made of aluminium (diameter 60 mm, length 500 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 30 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø30 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø60 mm, 500 mm, E = 70 GPaP = 30 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 1.5 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 90 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 90 kNn = 2.5L = 1.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 100 kg mass on a spring with a stiffness of 40 kN/m has a viscous damper giving a damping ratio of 0.7. Find the damped natural frequency fd of its free vibration in Hz.
m = 100 kgk = 40 kN/mζ = 0.7
fd = Hz
12.Mechanical vibrationsHard
A 40 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 50 kN/m and the lower one k₂ = 10 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 40 kgk1 = 50 kN/mk2 = 10 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: N1 = 30
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.
Total number of teeth
N1 + N2 = 2a/m = 2·240/4 = 120
Split by the ratio
N1 = 120/(1 + 3) = 30
Answer
N1 = 30
2.Gears & gear trains
Answer: ND = 72
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 = 2880·20/30 = 1920 rpm
Teeth on gear D
ND = C·nC/nout = 24·1920/640 = 72
Answer
ND = 72
3.Torsion of shafts
Answer: G = 26.05 GPa
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 moment and angle
J = π·20⁴/32 = 15708 mm⁴
φ = 1.12·π/180 = 0.01955 rad
Shear modulus
T = 20 N·m = 20000 N·mm
G = TL/(Jφ) = 20000·400/(15708·0.01955) = 26054 MPa
G = 26.05 GPa
Answer
G = 26.05 GPa
Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.
4.Torsion of shafts
Answer: τmax = 14.15 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 = 400 N·m
TAB = TC + TB = 400 + 200 = 600 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 600 N·m
τmax = 16·600000/(π·60³) = 14.15 MPa
Answer
τmax = 14.15 MPa
5.Mohr’s circle & stress transformation
Answer: τmax = 31.62 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 = √(((10 − −50)/2)² + −10²) = 31.62 MPa
Answer
τmax = 31.62 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 27.66 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
σθ = −10 + 10·−0.1736 + 40·0.9848 = 27.66 MPa
Answer
σθ = 27.66 MPa
7.Axial stress & strain
Answer: P = 12.14 kN
Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.
Areas
A₁ = 314.2 mm², A₂ = 706.9 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·314.2) = 6.366 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(70000·706.9) = 1.011 × 10⁻⁵ mm/N
Load
P = 0.2/1.647 × 10⁻⁵ = 12142 N = 12.14 kN
Answer
P = 12.14 kN
8.Axial stress & strain
Answer: σ2 = 5.605 MPa
Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.
Stiffnesses
k₁ = 200000·706.9/400 = 353429 N/mm
k₂ = 70000·2827/500 = 395841 N/mm
Share of segment 2
F2 = 30000·395841/749270 = 15849 N
Stress
σ2 = 15849/2827 = 5.605 MPa (compression)
Answer
σ2 = 5.605 MPa
9.Column buckling
Answer: Pcr = 71.19 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 round section has the same I about every axis.
I = πd⁴/64 = π·30⁴/64 = 39761 mm⁴
Effective length
KL = 0.7·1500 = 1050 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·39761/1050² = 71188 N = 71.19 kN
Answer
Pcr = 71.19 kN
10.Column buckling
Answer: d = 67.61 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 = 2.5·90000 = 225000 N
KL = 2·1500 = 3000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·225000·3000²/(π³·200000))^(1/4) = 67.61 mm
Answer
d = 67.61 mm
Check: the slenderness KL/r is 177.5, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 2.273 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 = √(40000/100) = 20 rad/s
fn = 20/(2π) = 3.183 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 3.183·√(1 − 0.7²) = 2.273 Hz
Answer
fd = 2.273 Hz
12.Mechanical vibrations
Answer: fn = 2.297 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
Springs in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
1/keq = 1/50 + 1/10 → keq = 50·10/(50 + 10) = 8.333 kN/m
Natural frequency
ωn = √(8333/40) = 14.43 rad/s
fn = 14.43/(2π) = 2.297 Hz
Answer
fn = 2.297 Hz
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