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 960 rpm drives a pinion with 15 teeth, which meshes with a gear with 16 teeth. The gear pair has an efficiency of 98%. Find the output torque T2 on the gear shaft.
P = 1.1 kWn1 = 960 rpmN1 = 15N2 = 16η = 98%
T2 = N·m
2.Gears & gear trainsHard
A 4 kW motor at 1450 rpm drives gear A (16 teeth), which meshes with gear B (40 teeth). Gear C (30 teeth) on the same shaft as B drives gear D (36 teeth) on the output shaft. Each gear mesh has an efficiency of 95%. Find the output torque Tout.
P = 4 kWnin = 1450 rpmA = 16B = 40C = 30D = 36ηmesh = 95%
Tout = N·m
3.Torsion of shaftsMedium
A hollow steel shaft (outer diameter 60 mm, inner diameter 36 mm, length 1.5 m) is fixed at one end and twisted by a torque of 2000 N·m at the other. Find the angle of twist φ in degrees.
T = 2000 N·mdo = 60 mmdi = 36 mmL = 1.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid steel shaft of diameter 60 mm is fixed at A. A torque of 400 N·m acts at B, 3 m from A, and a torque of 750 N·m in the same direction acts at the free end C, a further 1 m along. Find the maximum shear stress τmax in the shaft.
d = 60 mmTB = 400 N·mTC = 750 N·mAB = 3 mBC = 1 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 10 MPa, σᵧ = −10 MPa and τₓᵧ = 30 MPa (tension positive). The element is rotated 30° clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.
σx = 10 MPaσy = −10 MPaτxy = 30 MPaθ = -30°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 130 MPa, σᵧ = 70 MPa and τₓᵧ = 30 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 45° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 130 MPaσy = 70 MPaτxy = 30 MPaθ = 45°
τθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the steel bar under the axial load.
P = 45 kNd = 30 mmL = 2 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 total elongation δ.
P = 15 kNsegment 1: Ø35 mm, 0.5 m, copper (E = 117 GPa)segment 2: Ø16 mm, 1.5 m, aluminium (E = 70 GPa)
δ = mm
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 = 25 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at both ends. It must carry 15 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 15 kNn = 2.5L = 4 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 30 mm is followed one cycle later by a peak of 21.9 mm. Find the damping ratio ζ.
x0 = 30 mmx1 = 21.9 mmn = 1
ζ =
12.Mechanical vibrationsHard
A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 25 kN/m and k₂ = 60 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 100 kgk1 = 25 kN/mk2 = 60 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 11.44 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·960/60 = 100.5 rad/s
T1 = P/ω₁ = 1100/100.5 = 10.94 N·m
Output torque
T2 = 10.94·16/15·0.98 = 11.44 N·m
Answer
T2 = 11.44 N·m
2.Gears & gear trains
Answer: Tout = 71.32 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 = 1450·(16/40)·(30/36) = 483.3 rpm
ωout = 2π·483.3/60 = 50.61 rad/s
Output power and torque
Pout = 4000·0.95² = 3610 W
Tout = Pout/ωout = 3610/50.61 = 71.32 N·m
Answer
Tout = 71.32 N·m
3.Torsion of shafts
Answer: φ = 1.94 °
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(60⁴ − 36⁴)/32 = 1.107 × 10⁶ mm⁴
Angle of twist
T = 2000 N·m = 2 × 10⁶ N·mm
L = 1.5 m = 1500 mm
G = 80 GPa = 80000 MPa
φ = TL/(GJ) = 2 × 10⁶·1500/(80000·1.107 × 10⁶) = 0.03386 rad
φ = 0.03386·180/π = 1.94°
Answer
φ = 1.94°
4.Torsion of shafts
Answer: τmax = 27.12 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.
Internal torques
Sense of TC taken as positive.
TBC = TC = 750 N·m
TAB = TC + TB = 750 + 400 = 1150 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 1150 N·m
τmax = 16·1.15 × 10⁶/(π·60³) = 27.12 MPa
Answer
τmax = 27.12 MPa
5.Mohr’s circle & stress transformation
Answer: σy′ = 20.98 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′ = 0 − 10·0.5 − 30·(−0.866) = 20.98 MPa
Check
The sum of the normal stresses does not change with rotation.
σx′ = −20.98 MPa, σx′ + σy′ = 0 = σx + σy
Answer
σy′ = 20.98 MPa
6.Mohr’s circle & stress transformation
Answer: τθ = −30 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 = −30·1 + 30·0 = −30 MPa
Answer
τθ = −30 MPa
7.Axial stress & strain
Answer: δ = 0.6366 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·30²/4 = 706.9 mm²
σ = P/A = 45·1000/706.9 = 63.66 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 63.66/200000 = 0.0003183 = 318.3 µε
Elongation
δ = ε·L = 0.0003183·2000 mm = 0.6366 mm
Answer
δ = 0.6366 mm
8.Axial stress & strain
Answer: δ = 1.665 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·16²/4 = 201.1 mm²
σ1 = 15.59 MPa
σ2 = 74.6 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 15.59·500/117000 = 0.06663 mm
δ2 = σ2L2/E2 = 74.6·1500/70000 = 1.599 mm
δ = δ1 + δ2 = 1.665 mm
Answer
δ = 1.665 mm
9.Column buckling
Answer: Pcr = 8.583 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 = 0.7·3000 = 2100 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·19175/2100² = 8583 N = 8.583 kN
Answer
Pcr = 8.583 kN
10.Column buckling
Answer: d = 45.86 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 = 2.5·15000 = 37500 N
KL = 0.5·4000 = 2000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·37500·2000²/(π³·70000))^(1/4) = 45.86 mm
Answer
d = 45.86 mm
Check: the slenderness KL/r is 174.4, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.05003
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(30/21.9) = 0.3147
Damping ratio
ζ = δ/√(4π² + δ²) = 0.3147/√(39.48 + 0.09904) = 0.05003
Answer
ζ = 0.05003
12.Mechanical vibrations
Answer: fn = 4.64 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
Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 25 + 60 = 85 kN/m
Natural frequency
ωn = √(85000/100) = 29.15 rad/s
fn = 29.15/(2π) = 4.64 Hz
Answer
fn = 4.64 Hz
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