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 1450 rpm drives a pinion with 30 teeth and a module of 4 mm. Find the tangential force Ft on the pinion teeth.
P = 1.1 kWn = 1450 rpmN1 = 30m = 4 mm
Ft = N
2.Gears & gear trainsHard
A 1.1 kW motor at 960 rpm drives gear A (20 teeth), which meshes with gear B (25 teeth). Gear C (60 teeth) on the same shaft as B drives gear D (72 teeth) on the output shaft. Each gear mesh has an efficiency of 95%. Find the output torque Tout.
P = 1.1 kWnin = 960 rpmA = 20B = 25C = 60D = 72ηmesh = 95%
Tout = N·m
3.Torsion of shaftsMedium
A hollow steel shaft (outer diameter 50 mm, inner diameter 25 mm, length 0.8 m) is fixed at one end and twisted by a torque of 2000 N·m at the other. Find the angle of twist φ in degrees.
T = 2000 N·mdo = 50 mmdi = 25 mmL = 0.8 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 75 kW at 1450 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.
P = 75 kWN = 1450 rpmτallow = 60 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −30 MPa, σᵧ = 100 MPa and τₓᵧ = −10 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = −30 MPaσy = 100 MPaτxy = −10 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 50 MPa, σᵧ = 20 MPa and τₓᵧ = −50 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 45° counter-clockwise from the x-axis.
σx = 50 MPaσy = 20 MPaτxy = −50 MPaθ = 45°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 30 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 40 mm, length 300 mm, E = 100 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, Ø30 mm, 600 mm, E = 200 GPasegment 2: brass, Ø40 mm, 300 mm, E = 100 GPaδ = 0.2 mm
P = kN
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.
P = 90 kNsegment 1: Ø25 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø16 mm, 1 m, copper (E = 117 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 2.5 mE = 70 GPa (aluminium)b × h = 50 × 80 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 150 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 150 kNn = 3L = 2 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 250 kg mass is supported by a spring with a stiffness of 150 kN/m and a viscous damper with a damping coefficient of 6120 N·s/m. Find the damping ratio ζ.
m = 250 kgk = 150 kN/mc = 6120 N·s/m
ζ =
12.Mechanical vibrationsHard
A 100 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 = 100 kN/m and k₂ = 50 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 = 100 kgk1 = 100 kN/mk2 = 50 kN/mk3 = 12 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: Ft = 120.7 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·1450/60 = 151.8 rad/s
T = P/ω = 1100/151.8 = 7.244 N·m
Pitch diameter
d = m·N = 4·30 = 120 mm
Tangential force
Ft = 2T/d = 2·7.244/0.12 = 120.7 N
Answer
Ft = 120.7 N
2.Gears & gear trains
Answer: Tout = 14.81 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·(20/25)·(60/72) = 640 rpm
ωout = 2π·640/60 = 67.02 rad/s
Output power and torque
Pout = 1100·0.95² = 992.8 W
Tout = Pout/ωout = 992.8/67.02 = 14.81 N·m
Answer
Tout = 14.81 N·m
3.Torsion of shafts
Answer: φ = 1.992 °
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(50⁴ − 25⁴)/32 = 575243 mm⁴
Angle of twist
T = 2000 N·m = 2 × 10⁶ N·mm
L = 0.8 m = 800 mm
G = 80 GPa = 80000 MPa
φ = TL/(GJ) = 2 × 10⁶·800/(80000·575243) = 0.03477 rad
φ = 0.03477·180/π = 1.992°
Answer
φ = 1.992°
4.Torsion of shafts
Answer: d = 34.74 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·1450/60 = 151.8 rad/s
T = P/ω = 75000/151.8 = 493.9 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·493929/(π·60)) = 34.74 mm
Answer
d = 34.74 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τmax = 65.76 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 = √(((−30 − 100)/2)² + −10²) = 65.76 MPa
Answer
τmax = 65.76 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = −15 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 = 35 + 15·0 + −50·1 = −15 MPa
Answer
σθ = −15 MPa
7.Axial stress & strain
Answer: P = 30.16 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 = 706.9 mm², A₂ = 1257 mm²
Flexibilities
L₁/(E₁A₁) = 600/(200000·706.9) = 4.244 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(100000·1257) = 2.387 × 10⁻⁶ mm/N
Load
P = 0.2/6.631 × 10⁻⁶ = 30159 N = 30.16 kN
Answer
P = 30.16 kN
8.Axial stress & strain
Answer: σmax = 447.6 MPa
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·25²/4 = 490.9 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 183.3 MPa
σ2 = 447.6 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 447.6 MPa
Answer
σmax = 447.6 MPa
9.Column buckling
Answer: Pcr = 188 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 = 80·50³/12 = 833333 mm⁴
Effective length
KL = 0.7·2500 = 1750 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·833333/1750² = 187992 N = 188 kN
Answer
Pcr = 188 kN
10.Column buckling
Answer: d = 92.84 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 = 3·150000 = 450000 N
KL = 2·2000 = 4000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·450000·4000²/(π³·200000))^(1/4) = 92.84 mm
Answer
d = 92.84 mm
Check: the slenderness KL/r is 172.3, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.4997
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.
Critical damping
cc = 2·√(k·m) = 2·√(150000·250) = 12247 N·s/m
Damping ratio
ζ = c/cc = 6120/12247 = 0.4997
Answer
ζ = 0.4997
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 1.678 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 = 100 + 50 = 150 kN/m
keq = 150·12/(150 + 12) = 11.11 kN/m
Natural frequency
ωn = √(11111/100) = 10.54 rad/s
fn = 10.54/(2π) = 1.678 Hz
Answer
fn = 1.678 Hz
ExamProblems · examproblems.com/courses/machine-design