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 at 960 rpm drives a worm with 1 start(s). The worm wheel has 45 teeth and the drive has an efficiency of 55%. Find the output torque Tw on the wheel shaft.
z = 1Nw = 45n = 960 rpmP = 1.1 kWη = 55%
Tw = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (25 teeth) on the input shaft turns at 960 rpm and drives gear B (72 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 120 rpm. How many teeth ND must gear D have?
nin = 960 rpmA = 25B = 72C = 18nout = 120 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 = 20 mmL = 0.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 20 kW at 500 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.
P = 20 kWN = 500 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 0 MPa, σᵧ = 110 MPa and τₓᵧ = −10 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 0 MPaσy = 110 MPaτxy = −10 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = −70 MPa, σᵧ = 10 MPa and τₓᵧ = 20 MPa (σz = 0). The material has a yield strength of 355 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.
σx = −70 MPaσy = 10 MPaτxy = 20 MPaσY = 355 MPa
n =
7.Axial stress & strainMedium
Find the elongation δ of the aluminium bar under the axial load.
P = 75 kNd = 16 mmL = 3 mE = 70 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 = 100 kNsegment 1: Ø40 mm, 0.5 m, steel (E = 200 GPa)segment 2: Ø12 mm, 1 m, aluminium (E = 70 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 4 mE = 70 GPa (aluminium)b × h = 25 × 40 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 5 mE = 70 GPa (aluminium)d = 80 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 15 mm is followed one cycle later by a peak of 10.3 mm. Find the damping ratio ζ.
x0 = 15 mmx1 = 10.3 mmn = 1
ζ =
12.Mechanical vibrationsHard
A 100 kg mass hangs from spring k₃ = 20 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 = 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 = 50 kN/mk2 = 50 kN/mk3 = 20 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: Tw = 270.8 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.
Wheel speed
nw = n·z/Nw = 960·1/45 = 21.33 rpm
Output torque
ωw = 2π·21.33/60 = 2.234 rad/s
Tw = η·P/ωw = 0.55·1100/2.234 = 270.8 N·m
Answer
Tw = 270.8 N·m
2.Gears & gear trains
Answer: ND = 50
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 = 960·25/72 = 333.3 rpm
Teeth on gear D
ND = C·nC/nout = 18·333.3/120 = 50
Answer
ND = 50
3.Torsion of shafts
Answer: φ = 11.69 °
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·20⁴/32 = 15708 mm⁴
Angle of twist
φ = TL/(GJ) = 250·1000·500/(39000·15708) = 0.204 rad
φ = 0.204·180/π = 11.69°
Answer
φ = 11.69°
4.Torsion of shafts
Answer: d = 29.6 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·500/60 = 52.36 rad/s
T = P/ω = 20000/52.36 = 382 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·381972/(π·75)) = 29.6 mm
Answer
d = 29.6 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τmax = 55.9 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 = √(((0 − 110)/2)² + −10²) = 55.9 MPa
Answer
τmax = 55.9 MPa
6.Mohr’s circle & stress transformation
Answer: n = 4.274
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 = −30 MPa, R = 44.72 MPa
σ1 = 14.72 MPa, σ2 = −74.72 MPa, σ3 = 0
Von Mises
σvM = √(14.72² − 14.72·(−74.72) + (−74.72)²) = 83.07 MPa
n = 355/83.07 = 4.274
Answer
n = 4.274
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: δ = 15.99 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·16²/4 = 201.1 mm²
σ = P/A = 75·1000/201.1 = 373 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 373/70000 = 0.005329 = 5329 µε
Elongation
δ = ε·L = 0.005329·3000 mm = 15.99 mm
Answer
δ = 15.99 mm
8.Axial stress & strain
Answer: δ = 12.83 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·12²/4 = 113.1 mm²
σ1 = 79.58 MPa
σ2 = 884.2 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 79.58·500/200000 = 0.1989 mm
δ2 = σ2L2/E2 = 884.2·1000/70000 = 12.63 mm
δ = δ1 + δ2 = 12.83 mm
Answer
δ = 12.83 mm
9.Column buckling
Answer: Pcr = 2.249 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 = 40·25³/12 = 52083 mm⁴
Effective length
KL = 1·4000 = 4000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·52083/4000² = 2249 N = 2.249 kN
Answer
Pcr = 2.249 kN
10.Column buckling
Answer: σcr = 22.56 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·80⁴/64 = 2.011 × 10⁶ mm⁴
Radius of gyration and slenderness
A = 5027 mm²
r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm
KL = 0.7·5000 = 3500 mm
KL/r = 3500/20 = 175
Critical stress
σcr = π²E/(KL/r)² = π²·70000/175² = 22.56 MPa
Answer
σcr = 22.56 MPa
11.Mechanical vibrations
Answer: ζ = 0.05972
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 = (1/1)·ln(15/10.3) = 0.3759
Damping ratio
ζ = δ/√(4π² + δ²) = 0.3759/√(39.48 + 0.1413) = 0.05972
Answer
ζ = 0.05972
12.Mechanical vibrations
Answer: fn = 2.055 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 + 50 = 100 kN/m
keq = 100·20/(100 + 20) = 16.67 kN/m
Natural frequency
ωn = √(16667/100) = 12.91 rad/s
fn = 12.91/(2π) = 2.055 Hz
Answer
fn = 2.055 Hz
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