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 2900 rpm drives a pinion with 25 teeth and a module of 2 mm. Find the tangential force Ft on the pinion teeth.
P = 1.1 kWn = 2900 rpmN1 = 25m = 2 mm
Ft = N
2.Gears & gear trainsHard
In a compound gear train, gear A (18 teeth) on the input shaft turns at 1500 rpm and drives gear B (60 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 225 rpm. How many teeth ND must gear D have?
nin = 1500 rpmA = 18B = 60C = 12nout = 225 rpm
ND =
3.Torsion of shaftsMedium
A solid aluminium shaft 2.5 m long carries a torque of 500 N·m. The angle of twist between its ends must not exceed 3°. Find the smallest diameter d that meets this limit.
T = 500 N·mL = 2.5 mG = 26 GPaφallow = 3°
d = mm
4.Torsion of shaftsHard
A hollow shaft transmits 20 kW at 500 rpm. Find the maximum shear stress τmax in the shaft.
P = 20 kWN = 500 rpmdo = 80 mmdi = 40 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 50 MPa, σᵧ = 10 MPa and τₓᵧ = 50 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).
σx = 50 MPaσy = 10 MPaτxy = 50 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 60 mm carries an axial compressive force of 190 kN together with a torque of 1500 N·m. At the surface of the shaft, find the largest principal stress σ1.
d = 60 mmF = −190 kNT = 1500 N·m
σ1 = MPa
7.Axial stress & strainMedium
Find the elongation δ of the copper bar under the axial load.
P = 85 kNd = 30 mmL = 3 mE = 117 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 = 15 kNsegment 1: Ø35 mm, 2 m, steel (E = 200 GPa)segment 2: Ø12 mm, 0.5 m, aluminium (E = 70 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 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 = 2 mE = 200 GPa (steel)d = 25 mm
σcr = MPa
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 27.4 mm. Find the damping ratio ζ.
x0 = 40 mmx1 = 27.4 mmn = 1
ζ =
12.Mechanical vibrationsHard
A 50 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 = 10 kN/m and k₂ = 40 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 = 50 kgk1 = 10 kN/mk2 = 40 kN/mk3 = 20 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: Ft = 144.9 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 = 1100/303.7 = 3.622 N·m
Pitch diameter
d = m·N = 2·25 = 50 mm
Tangential force
Ft = 2T/d = 2·3.622/0.05 = 144.9 N
Answer
Ft = 144.9 N
2.Gears & gear trains
Answer: ND = 24
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 = 1500·18/60 = 450 rpm
Teeth on gear D
ND = C·nC/nout = 12·450/225 = 24
Answer
ND = 24
3.Torsion of shafts
Answer: d = 55.3 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.
Required polar moment
T = 500 N·m = 500000 N·mm
L = 2.5 m = 2500 mm
G = 26 GPa = 26000 MPa
φ = 3·π/180 = 0.05236 rad
J = TL/(Gφ) = 500000·2500/(26000·0.05236) = 918202 mm⁴
Diameter
d = (32J/π)1/4 = 55.3 mm
Answer
d = 55.3 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: τmax = 4.053 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.
Torque
ω = 2πN/60 = 2π·500/60 = 52.36 rad/s
T = P/ω = 20000/52.36 = 382 N·m
Polar second moment of area
J = π(80⁴ − 40⁴)/32 = 3.77 × 10⁶ mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 381972·40/3.77 × 10⁶ = 4.053 MPa
Answer
τmax = 4.053 MPa
5.Mohr’s circle & stress transformation
Answer: θs = −10.9 °
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.
Angle on the circle
On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.
cos 2θs = 50/53.85 = 0.9285, sin 2θs = −20/53.85 = −0.3714
2θs = atan2(−20, 50) = −21.8°
Angle on the element
θs = −10.9°
check: θs − θp = −10.9 − 34.1 = −45°
Answer
θs = −10.9°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σ1 = 15.18 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.
Stresses at the surface
σ = F/A = −190000/(π·60²/4) = −190000/2827 = −67.2 MPa
τ = 16T/(πd³) = 16·1500000/(π·60³) = 35.37 MPa
Mohr’s circle
C = σ/2 = −33.6 MPa
R = √((−33.6)² + 35.37²) = 48.78 MPa
σ1 = C + R = 15.18 MPa
σ2 = C − R = −82.38 MPa
Answer
σ1 = 15.18 MPa
7.Axial stress & strain
Answer: δ = 3.083 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 = 85·1000/706.9 = 120.3 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 120.3/117000 = 0.001028 = 1028 µε
Elongation
δ = ε·L = 0.001028·3000 mm = 3.083 mm
Answer
δ = 3.083 mm
8.Axial stress & strain
Answer: σmax = 132.6 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 = 15.59 MPa
σ2 = 132.6 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 132.6 MPa
Answer
σmax = 132.6 MPa
9.Column buckling
Answer: Pcr = 23.32 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 = 1·2000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·135000/2000² = 23317 N = 23.32 kN
Answer
Pcr = 23.32 kN
10.Column buckling
Answer: σcr = 4.819 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 = 2·2000 = 4000 mm
KL/r = 4000/6.25 = 640
Critical stress
σcr = π²E/(KL/r)² = π²·200000/640² = 4.819 MPa
Answer
σcr = 4.819 MPa
11.Mechanical vibrations
Answer: ζ = 0.06011
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(40/27.4) = 0.3783
Damping ratio
ζ = δ/√(4π² + δ²) = 0.3783/√(39.48 + 0.1431) = 0.06011
Answer
ζ = 0.06011
12.Mechanical vibrations
Answer: fn = 2.69 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 = 10 + 40 = 50 kN/m
keq = 50·20/(50 + 20) = 14.29 kN/m
Natural frequency
ωn = √(14286/50) = 16.9 rad/s
fn = 16.9/(2π) = 2.69 Hz
Answer
fn = 2.69 Hz
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