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 7.5 kW motor running at 1500 rpm drives a pinion with 18 teeth, which meshes with a gear with 24 teeth. Ignore losses. Find the output torque T2 on the gear shaft.
P = 7.5 kWn1 = 1500 rpmN1 = 18N2 = 24η = 100%
T2 = N·m
2.Gears & gear trainsHard
A 3 kW motor at 1440 rpm drives gear A (40 teeth), which meshes with gear B (48 teeth). Gear C (16 teeth) on the same shaft as B drives gear D (80 teeth) on the output shaft. Each gear mesh has an efficiency of 97%. Find the output torque Tout.
P = 3 kWnin = 1440 rpmA = 40B = 48C = 16D = 80ηmesh = 97%
Tout = N·m
3.Torsion of shaftsMedium
A hollow steel shaft (outer diameter 30 mm, inner diameter 21 mm, length 1.5 m) is fixed at one end and twisted by a torque of 200 N·m at the other. Find the angle of twist φ in degrees.
T = 200 N·mdo = 30 mmdi = 21 mmL = 1.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid aluminium shaft transmits 10 kW at 1450 rpm. The shear stress must not exceed 40 MPa and the twist must not exceed 0.75° per m of length. Find the smallest diameter d that satisfies both limits.
P = 10 kWN = 1450 rpmτallow = 40 MPaθallow = 0.75°/mG = 26 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −30 MPa and σᵧ = −40 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches −20 MPa?
σx = −30 MPaσy = −40 MPaσ1 = −20 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 70 MPa, σᵧ = −40 MPa and τₓᵧ = −20 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 40° counter-clockwise from the x-axis.
σx = 70 MPaσy = −40 MPaτxy = −20 MPaθ = 40°
σθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the steel bar under the axial load.
P = 75 kNd = 10 mmL = 0.5 mE = 200 GPa
δ = mm
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 = 110 kNsegment 1: Ø35 mm, 1.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 2 m, steel (E = 200 GPa)
σmax = MPa
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 = 2.5 mE = 70 GPa (aluminium)b × h = 20 × 40 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 50 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 50 kNn = 2L = 5 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed one cycle later by a peak of 31.1 mm. Find the damping ratio ζ.
x0 = 40 mmx1 = 31.1 mmn = 1
ζ =
12.Mechanical vibrationsHard
A 100 kg mass hangs from spring k₃ = 30 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 80 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 = 80 kN/mk2 = 50 kN/mk3 = 30 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 63.66 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·1500/60 = 157.1 rad/s
T1 = P/ω₁ = 7500/157.1 = 47.75 N·m
Output torque
T2 = 47.75·24/18·1 = 63.66 N·m
Answer
T2 = 63.66 N·m
2.Gears & gear trains
Answer: Tout = 112.3 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·(40/48)·(16/80) = 240 rpm
ωout = 2π·240/60 = 25.13 rad/s
Output power and torque
Pout = 3000·0.97² = 2823 W
Tout = Pout/ωout = 2823/25.13 = 112.3 N·m
Answer
Tout = 112.3 N·m
3.Torsion of shafts
Answer: φ = 3.556 °
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 − 21⁴)/32 = 60428 mm⁴
Angle of twist
T = 200 N·m = 200000 N·mm
L = 1.5 m = 1500 mm
G = 80 GPa = 80000 MPa
φ = TL/(GJ) = 200000·1500/(80000·60428) = 0.06206 rad
φ = 0.06206·180/π = 3.556°
Answer
φ = 3.556°
4.Torsion of shafts
Answer: d = 37.47 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 = 10000/151.8 = 65.86 N·m
Diameter for strength
T = 65.86 N·m = 65857 N·mm
dτ = ∛(16·65857/(π·40)) = 20.32 mm
Diameter for stiffness
θ = 0.75·π/180/1000 = 1.309 × 10⁻⁵ rad/mm
G = 26 GPa = 26000 MPa
dφ = (32·65857/(π·26000·1.309 × 10⁻⁵))1/4 = 37.47 mm
Governing diameter
Both limits must hold, so the larger diameter governs.
d = max(20.32, 37.47) = 37.47 mm
Answer
d = 37.47 mm
Here stiffness governs.
5.Mohr’s circle & stress transformation
Answer: τxy = 14.14 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.
Required radius
C = (−30 + (−40))/2 = −35 MPa
R = σ1 − C = −20 − (−35) = 15 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(15² − 5²) = 14.14 MPa
Answer
τxy = 14.14 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 4.854 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 = 15 + 55·0.1736 + −20·0.9848 = 4.854 MPa
Answer
σθ = 4.854 MPa
7.Axial stress & strain
Answer: δ = 2.387 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·10²/4 = 78.54 mm²
σ = P/A = 75·1000/78.54 = 954.9 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 954.9/200000 = 0.004775 = 4775 µε
Elongation
δ = ε·L = 0.004775·500 mm = 2.387 mm
Answer
δ = 2.387 mm
8.Axial stress & strain
Answer: σmax = 972.6 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·35²/4 = 962.1 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 114.3 MPa
σ2 = 972.6 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 972.6 MPa
Answer
σmax = 972.6 MPa
9.Column buckling
Answer: Pcr = 2.948 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 = 40·20³/12 = 26667 mm⁴
Effective length
KL = 1·2500 = 2500 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·26667/2500² = 2948 N = 2.948 kN
Answer
Pcr = 2.948 kN
10.Column buckling
Answer: d = 131 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 = 2·50000 = 100000 N
KL = 2·5000 = 10000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·100000·10000²/(π³·70000))^(1/4) = 131 mm
Answer
d = 131 mm
Check: the slenderness KL/r is 305.2, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.04002
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(40/31.1) = 0.2517
Damping ratio
ζ = δ/√(4π² + δ²) = 0.2517/√(39.48 + 0.06334) = 0.04002
Answer
ζ = 0.04002
12.Mechanical vibrations
Answer: fn = 2.485 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 = 80 + 50 = 130 kN/m
keq = 130·30/(130 + 30) = 24.38 kN/m
Natural frequency
ωn = √(24375/100) = 15.61 rad/s
fn = 15.61/(2π) = 2.485 Hz
Answer
fn = 2.485 Hz
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