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 5.5 kW motor running at 1440 rpm drives a pinion with 18 teeth and a module of 2.5 mm. Find the tangential force Ft on the pinion teeth.
P = 5.5 kWn = 1440 rpmN1 = 18m = 2.5 mm
Ft = N
2.Gears & gear trainsHard
A 1.5 kW motor at 960 rpm drives gear A (24 teeth), which meshes with gear B (60 teeth). Gear C (40 teeth) on the same shaft as B drives gear D (60 teeth) on the output shaft. Each gear mesh has an efficiency of 97%. Find the output torque Tout.
P = 1.5 kWnin = 960 rpmA = 24B = 60C = 40D = 60ηmesh = 97%
Tout = N·m
3.Torsion of shaftsMedium
A solid aluminium 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 = 1000 N·md = 30 mmL = 0.8 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 20 kW at 1800 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.
P = 20 kWN = 1800 rpmτallow = 50 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −50 MPa, σᵧ = 10 MPa and τₓᵧ = −60 MPa (tension positive). The element is rotated 35° counter-clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.
σx = −50 MPaσy = 10 MPaτxy = −60 MPaθ = 35°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 70 MPa, σᵧ = −80 MPa and τₓᵧ = −10 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° clockwise from the x-axis.
σx = 70 MPaσy = −80 MPaτxy = −10 MPaθ = -30°
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 60 mm, length 200 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.
segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø60 mm, 200 mm, E = 70 GPaδ = 0.5 mm
P = kN
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 45 kNsegment 1: Ø35 mm, 1 m, aluminium (E = 70 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 both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 6 mE = 200 GPa (steel)d = 25 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at both ends. It must carry 150 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 150 kNn = 3L = 5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed 5 cycles later by a peak of 3.2 mm. Find the damping ratio ζ.
x0 = 40 mmx5 = 3.2 mmn = 5
ζ =
12.Mechanical vibrationsHard
A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 15 kN/m and k₂ = 80 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 100 kgk1 = 15 kN/mk2 = 80 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: Ft = 1621 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 = 5500/150.8 = 36.47 N·m
Pitch diameter
d = m·N = 2.5·18 = 45 mm
Tangential force
Ft = 2T/d = 2·36.47/0.045 = 1621 N
Answer
Ft = 1621 N
2.Gears & gear trains
Answer: Tout = 52.65 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·(24/60)·(40/60) = 256 rpm
ωout = 2π·256/60 = 26.81 rad/s
Output power and torque
Pout = 1500·0.97² = 1411 W
Tout = Pout/ωout = 1411/26.81 = 52.65 N·m
Answer
Tout = 52.65 N·m
3.Torsion of shafts
Answer: φ = 22.17 °
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·30⁴/32 = 79522 mm⁴
Angle of twist
φ = TL/(GJ) = 1000·1000·800/(26000·79522) = 0.3869 rad
φ = 0.3869·180/π = 22.17°
Answer
φ = 22.17°
4.Torsion of shafts
Answer: d = 22.11 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·1800/60 = 188.5 rad/s
T = P/ω = 20000/188.5 = 106.1 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·106103/(π·50)) = 22.11 mm
Answer
d = 22.11 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: σy′ = 46.64 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′ = −20 − (−30)·0.342 − (−60)·0.9397 = 46.64 MPa
Check
The sum of the normal stresses does not change with rotation.
σx′ = −86.64 MPa, σx′ + σy′ = −40 = σx + σy
Answer
σy′ = 46.64 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 41.16 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 = −5 + 75·0.5 + −10·−0.866 = 41.16 MPa
Answer
σθ = 41.16 MPa
7.Axial stress & strain
Answer: P = 67.78 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 = 2827 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·314.2) = 6.366 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(70000·2827) = 1.011 × 10⁻⁶ mm/N
Load
P = 0.5/7.377 × 10⁻⁶ = 67781 N = 67.78 kN
Answer
P = 67.78 kN
8.Axial stress & strain
Answer: δ = 3.533 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·35²/4 = 962.1 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 46.77 MPa
σ2 = 143.2 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 46.77·1000/70000 = 0.6682 mm
δ2 = σ2L2/E2 = 143.2·2000/100000 = 2.865 mm
δ = δ1 + δ2 = 3.533 mm
Answer
δ = 3.533 mm
9.Column buckling
Answer: Pcr = 4.205 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·25⁴/64 = 19175 mm⁴
Effective length
KL = 0.5·6000 = 3000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·19175/3000² = 4205 N = 4.205 kN
Answer
Pcr = 4.205 kN
10.Column buckling
Answer: d = 73.4 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 = 0.5·5000 = 2500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·450000·2500²/(π³·200000))^(1/4) = 73.4 mm
Answer
d = 73.4 mm
Check: the slenderness KL/r is 136.2, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.08014
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.
Logarithmic decrement
δ = (1/5)·ln(40/3.2) = 0.5051
Damping ratio
ζ = δ/√(4π² + δ²) = 0.5051/√(39.48 + 0.2552) = 0.08014
Answer
ζ = 0.08014
12.Mechanical vibrations
Answer: keq = 95 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 = 15 + 80 = 95 kN/m
Answer
keq = 95 kN/m
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