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 1450 rpm drives a worm with 3 start(s). The worm wheel has 72 teeth and the drive has an efficiency of 50%. Find the output torque Tw on the wheel shaft.
z = 3Nw = 72n = 1450 rpmP = 1.1 kWη = 50%
Tw = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (18 teeth) on the input shaft turning at 2900 rpm drives gear B (36 teeth). Gear C (72 teeth) is fixed to the same shaft as B and drives gear D (80 teeth) on the output shaft. Find the output speed nout.
nin = 2900 rpmA = 18B = 36C = 72D = 80
nout = rpm
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 = 400 N·md = 40 mmL = 2.5 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, 2 m from A, and a torque of 500 N·m in the opposite direction acts at the free end C, a further 0.8 m along. Find the maximum shear stress τmax in the shaft.
d = 60 mmTB = 400 N·mTC = 500 N·mAB = 2 mBC = 0.8 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 50 MPa, σᵧ = −60 MPa and τₓᵧ = 10 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 50 MPaσy = −60 MPaτxy = 10 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 70 MPa, σᵧ = −30 MPa and τₓᵧ = −50 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 20° counter-clockwise from the x-axis.
σx = 70 MPaσy = −30 MPaτxy = −50 MPaθ = 20°
σθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the copper bar, in microstrain (µε = 10⁻⁶).
P = 35 kNd = 30 mmL = 2 mE = 117 GPa
ε = µε
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 = 70 kNsegment 1: Ø40 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1.5 m, steel (E = 200 GPa)
σmax = MPa
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 = 6 mE = 70 GPa (aluminium)b × h = 30 × 60 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 1.5 mE = 200 GPa (steel)d = 25 mm
σcr = MPa
11.Mechanical vibrationsMedium
An 80 kg mass is supported by a spring with a stiffness of 200 kN/m and a viscous damper with a damping coefficient of 1600 N·s/m. Find the damping ratio ζ.
m = 80 kgk = 200 kN/mc = 1600 N·s/m
ζ =
12.Mechanical vibrationsHard
A 50 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 25 kN/m and k₂ = 20 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 = 50 kgk1 = 25 kN/mk2 = 20 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: Tw = 86.93 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 = 1450·3/72 = 60.42 rpm
Output torque
ωw = 2π·60.42/60 = 6.327 rad/s
Tw = η·P/ωw = 0.5·1100/6.327 = 86.93 N·m
Answer
Tw = 86.93 N·m
2.Gears & gear trains
Answer: nout = 1305 rpm
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.
First stage
nB = 2900·18/36 = 1450 rpm = nC
Second stage
nout = 1450·72/80 = 1305 rpm
Answer
nout = 1305 rpm
total ratio = 2.222 : 1
3.Torsion of shafts
Answer: φ = 2.85 °
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·40⁴/32 = 251327 mm⁴
Angle of twist
φ = TL/(GJ) = 400·1000·2500/(80000·251327) = 0.04974 rad
φ = 0.04974·180/π = 2.85°
Answer
φ = 2.85°
4.Torsion of shafts
Answer: τmax = 11.79 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 = 500 N·m
TAB = TC − TB = 500 − 400 = 100 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 500 N·m
τmax = 16·500000/(π·60³) = 11.79 MPa
Answer
τmax = 11.79 MPa
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 = √(((50 − −60)/2)² + 10²) = 55.9 MPa
Answer
τmax = 55.9 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 26.16 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.
Transformation equation
σθ = 20 + 50·0.766 + −50·0.6428 = 26.16 MPa
Answer
σθ = 26.16 MPa
7.Axial stress & strain
Answer: ε = 423.2 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·30²/4 = 706.9 mm²
σ = P/A = 35·1000/706.9 = 49.51 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 49.51/117000 = 0.0004232 = 423.2 µε
Answer
ε = 423.2 µε
8.Axial stress & strain
Answer: σmax = 348.2 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·40²/4 = 1257 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 55.7 MPa
σ2 = 348.2 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 348.2 MPa
Answer
σmax = 348.2 MPa
9.Column buckling
Answer: Pcr = 10.36 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 = 60·30³/12 = 135000 mm⁴
Effective length
KL = 0.5·6000 = 3000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·135000/3000² = 10363 N = 10.36 kN
Answer
Pcr = 10.36 kN
10.Column buckling
Answer: σcr = 8.567 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·25⁴/64 = 19175 mm⁴
Radius of gyration and slenderness
A = 490.9 mm²
r = √(I/A) = √(19175/490.9) = 6.25 mm
KL = 2·1500 = 3000 mm
KL/r = 3000/6.25 = 480
Critical stress
σcr = π²E/(KL/r)² = π²·200000/480² = 8.567 MPa
Answer
σcr = 8.567 MPa
11.Mechanical vibrations
Answer: ζ = 0.2
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.
Critical damping
cc = 2·√(k·m) = 2·√(200000·80) = 8000 N·s/m
Damping ratio
ζ = c/cc = 1600/8000 = 0.2
Answer
ζ = 0.2
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 4.775 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 + 20 = 45 kN/m
Natural frequency
ωn = √(45000/50) = 30 rad/s
fn = 30/(2π) = 4.775 Hz
Answer
fn = 4.775 Hz
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