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 24 teeth, which meshes with a gear with 40 teeth. The gear pair has an efficiency of 95%. Find the output torque T2 on the gear shaft.
P = 1.1 kWn1 = 1450 rpmN1 = 24N2 = 40η = 95%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (30 teeth) on the input shaft turns at 2880 rpm and drives gear B (40 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 1296 rpm. How many teeth ND must gear D have?
nin = 2880 rpmA = 30B = 40C = 15nout = 1296 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 = 250 N·md = 50 mmL = 1.2 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft transmits 100 kW at 1450 rpm. Find the maximum shear stress τmax in the shaft.
P = 100 kWN = 1450 rpmdo = 100 mmdi = 70 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a thin plate is in plane stress: σₓ = 70 MPa, σᵧ = −70 MPa and τₓᵧ = −30 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).
σx = 70 MPaσy = −70 MPaτxy = −30 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = 160 MPa, σᵧ = −20 MPa and τₓᵧ = −60 MPa (σz = 0). Find the von Mises equivalent stress σvM.
σx = 160 MPaσy = −20 MPaτxy = −60 MPa
σvM = MPa
7.Axial stress & strainMedium
A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 80 mm, inner diameter 50 mm, E = 120 GPa). Both are 400 mm long and are compressed together between rigid plates by 100 kN. Find the shortening δ.
P = 100 kNrod: Ø40 mm, E = 200 GPatube: Ø80/Ø50 mm, E = 120 GPaL = 400 mm
δ = mm
8.Axial stress & strainHard
A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of aluminium (outer diameter 120 mm, inner diameter 70 mm, E = 70 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø25 mm, E = 200 GPa, σallow = 120 MPatube: Ø120/Ø70 mm, E = 70 GPa, σallow = 50 MPa
Pmax = kN
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 = 3 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is pinned at both ends. It must carry 30 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 30 kNn = 3L = 3 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 150 kg mass on a spring with a stiffness of 20 kN/m has a viscous damper giving a damping ratio of 0.5. Find the damped natural frequency fd of its free vibration in Hz.
m = 150 kgk = 20 kN/mζ = 0.5
fd = Hz
12.Mechanical vibrationsHard
A 10 kg mass hangs from spring k₃ = 25 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 50 kN/m and k₂ = 20 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 = 50 kN/mk2 = 20 kN/mk3 = 25 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 11.47 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.
Motor torque
ω₁ = 2π·1450/60 = 151.8 rad/s
T1 = P/ω₁ = 1100/151.8 = 7.244 N·m
Output torque
T2 = 7.244·40/24·0.95 = 11.47 N·m
Answer
T2 = 11.47 N·m
2.Gears & gear trains
Answer: ND = 25
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 = 2880·30/40 = 2160 rpm
Teeth on gear D
ND = C·nC/nout = 15·2160/1296 = 25
Answer
ND = 25
3.Torsion of shafts
Answer: φ = 0.7183 °
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·50⁴/32 = 613592 mm⁴
Angle of twist
φ = TL/(GJ) = 250·1000·1200/(39000·613592) = 0.01254 rad
φ = 0.01254·180/π = 0.7183°
Answer
φ = 0.7183°
4.Torsion of shafts
Answer: τmax = 4.414 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.
Torque
ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s
T = P/ω = 100000/151.8 = 658.6 N·m
Polar second moment of area
J = π(100⁴ − 70⁴)/32 = 7.46 × 10⁶ mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 658572·50/7.46 × 10⁶ = 4.414 MPa
Answer
τmax = 4.414 MPa
5.Mohr’s circle & stress transformation
Answer: τabs = 76.16 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. In three dimensions the absolute maximum shear stress is (σmax − σmin)/2 over all three principal stresses; in plane stress the third one is zero.
In-plane principal stresses
C = 0 MPa, R = √(70² + (−30)²) = 76.16 MPa
σ1 = 76.16 MPa, σ2 = −76.16 MPa, σz = 0
Largest of the three circles
τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(152.3, 76.16, 76.16)/2 = 76.16 MPa
Answer
τabs = 76.16 MPa
Here σ₁ and σ₂ have opposite signs, so the in-plane maximum R is also the absolute maximum.
6.Mohr’s circle & stress transformation
Answer: σvM = 200 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. 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 = 70 MPa, R = 108.2 MPa
σ1 = 178.2 MPa, σ2 = −38.17 MPa, σ3 = 0
Von Mises
σvM = √(178.2² − 178.2·(−38.17) + (−38.17)²) = 200 MPa
Answer
σvM = 200 MPa
7.Axial stress & strain
Answer: δ = 0.06463 mm
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
Ar = π·40²/4 = 1257 mm²
At = π(80² − 50²)/4 = 3063 mm²
Combined axial stiffness
ΣEA = 200·10³·1257 + 120·10³·3063 = 6.189 × 10⁸ N
Shortening
δ = 100000·400/6.189 × 10⁸ = 0.06463 mm
Answer
δ = 0.06463 mm
8.Axial stress & strain
Answer: Pmax = 372.3 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
Ar = π·25²/4 = 490.9 mm²
At = π(120² − 70²)/4 = 7461 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 70·10³·7461 = 6.205 × 10⁸ N
Governing strain
ε = min(0.0006, 0.0007143) = 0.0006 (steel)
Largest load
Pmax = 0.0006·6.205 × 10⁸ = 372279 N = 372.3 kN
Answer
Pmax = 372.3 kN
9.Column buckling
Answer: Pcr = 17.8 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 = 0.7·3000 = 2100 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·39761/2100² = 17797 N = 17.8 kN
Answer
Pcr = 17.8 kN
10.Column buckling
Answer: d = 53.77 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 = 3·30000 = 90000 N
KL = 1·3000 = 3000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·90000·3000²/(π³·200000))^(1/4) = 53.77 mm
Answer
d = 53.77 mm
Check: the slenderness KL/r is 223.2, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 1.592 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 = √(20000/150) = 11.55 rad/s
fn = 11.55/(2π) = 1.838 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 1.838·√(1 − 0.5²) = 1.592 Hz
Answer
fd = 1.592 Hz
12.Mechanical vibrations
Answer: fn = 6.831 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 = 50 + 20 = 70 kN/m
keq = 70·25/(70 + 25) = 18.42 kN/m
Natural frequency
ωn = √(18421/10) = 42.92 rad/s
fn = 42.92/(2π) = 6.831 Hz
Answer
fn = 6.831 Hz
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