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 2.2 kW motor running at 2900 rpm drives a pinion with 20 teeth and a module of 3 mm. Find the tangential force Ft on the pinion teeth.
P = 2.2 kWn = 2900 rpmN1 = 20m = 3 mm
Ft = N
2.Gears & gear trainsHard
A 5.5 kW motor at 1440 rpm drives gear A (30 teeth), which meshes with gear B (45 teeth). Gear C (24 teeth) on the same shaft as B drives gear D (45 teeth) on the output shaft. Each gear mesh has an efficiency of 95%. Find the output torque Tout.
P = 5.5 kWnin = 1440 rpmA = 30B = 45C = 24D = 45ηmesh = 95%
Tout = N·m
3.Torsion of shaftsMedium
A solid steel 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 = 500 N·md = 25 mmL = 1.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 10 kW at 300 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.
P = 10 kWN = 300 rpmτallow = 40 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a thin plate is in plane stress: σₓ = 120 MPa, σᵧ = 90 MPa and τₓᵧ = −10 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).
σx = 120 MPaσy = 90 MPaτxy = −10 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −50 MPa, σᵧ = −60 MPa and τₓᵧ = −20 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 30° clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = −50 MPaσy = −60 MPaτxy = −20 MPaθ = -30°
τθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the aluminium bar, in microstrain (µε = 10⁻⁶).
P = 70 kNd = 16 mmL = 2 mE = 70 GPa
ε = µε
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, aluminium (E = 70 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 70 GPa (aluminium)b × h = 25 × 100 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at both ends. It must carry 105 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 105 kNn = 2L = 3 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed 4 cycles later by a peak of 23.5 mm. Find the damping ratio ζ.
x0 = 50 mmx4 = 23.5 mmn = 4
ζ =
12.Mechanical vibrationsHard
A 40 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 = 30 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 = 40 kgk1 = 30 kN/mk2 = 50 kN/mk3 = 25 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: Ft = 241.5 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·2900/60 = 303.7 rad/s
T = P/ω = 2200/303.7 = 7.244 N·m
Pitch diameter
d = m·N = 3·20 = 60 mm
Tangential force
Ft = 2T/d = 2·7.244/0.06 = 241.5 N
Answer
Ft = 241.5 N
2.Gears & gear trains
Answer: Tout = 92.58 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 = 1440·(30/45)·(24/45) = 512 rpm
ωout = 2π·512/60 = 53.62 rad/s
Output power and torque
Pout = 5500·0.95² = 4964 W
Tout = Pout/ωout = 4964/53.62 = 92.58 N·m
Answer
Tout = 92.58 N·m
3.Torsion of shafts
Answer: φ = 14.01 °
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·25⁴/32 = 38350 mm⁴
Angle of twist
φ = TL/(GJ) = 500·1000·1500/(80000·38350) = 0.2445 rad
φ = 0.2445·180/π = 14.01°
Answer
φ = 14.01°
4.Torsion of shafts
Answer: d = 34.35 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·300/60 = 31.42 rad/s
T = P/ω = 10000/31.42 = 318.3 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·318310/(π·40)) = 34.35 mm
Answer
d = 34.35 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τabs = 61.51 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 = 105 MPa, R = √(15² + (−10)²) = 18.03 MPa
σ1 = 123 MPa, σ2 = 86.97 MPa, σz = 0
Largest of the three circles
τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(36.06, 123, 86.97)/2 = 61.51 MPa
Answer
τabs = 61.51 MPa
Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.
6.Mohr’s circle & stress transformation
Answer: τθ = −5.67 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
τθ = −5·−0.866 + −20·0.5 = −5.67 MPa
Answer
τθ = −5.67 MPa
7.Axial stress & strain
Answer: ε = 4974 µε
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 = 70·1000/201.1 = 348.2 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 348.2/70000 = 0.004974 = 4974 µε
Answer
ε = 4974 µε
8.Axial stress & strain
Answer: δ = 4.761 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 = 143.2·2000/70000 = 4.093 mm
δ = δ1 + δ2 = 4.761 mm
Answer
δ = 4.761 mm
9.Column buckling
Answer: Pcr = 89.96 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 = 100·25³/12 = 130208 mm⁴
Effective length
KL = 0.5·2000 = 1000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·130208/1000² = 89957 N = 89.96 kN
Answer
Pcr = 89.96 kN
10.Column buckling
Answer: d = 61.1 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 = 2·105000 = 210000 N
KL = 0.5·3000 = 1500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·210000·1500²/(π³·70000))^(1/4) = 61.1 mm
Answer
d = 61.1 mm
Check: the slenderness KL/r is 98.21, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.03003
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 = (1/4)·ln(50/23.5) = 0.1888
Damping ratio
ζ = δ/√(4π² + δ²) = 0.1888/√(39.48 + 0.03563) = 0.03003
Answer
ζ = 0.03003
12.Mechanical vibrations
Answer: fn = 3.473 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 = 30 + 50 = 80 kN/m
keq = 80·25/(80 + 25) = 19.05 kN/m
Natural frequency
ωn = √(19048/40) = 21.82 rad/s
fn = 21.82/(2π) = 3.473 Hz
Answer
fn = 3.473 Hz
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