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
Two parallel shafts are 300 mm apart. They must be connected by a pair of spur gears with a module of 5 mm and a speed reduction of 4 : 1. How many teeth N2 must the gear have?
a = 300 mmm = 5 mmi = 4
N2 =
2.Gears & gear trainsHard
In a compound gear train, gear A (12 teeth) on the input shaft turns at 2880 rpm and drives gear B (60 teeth). Gear C (48 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 460.8 rpm. How many teeth ND must gear D have?
nin = 2880 rpmA = 12B = 60C = 48nout = 460.8 rpm
ND =
3.Torsion of shaftsMedium
A solid brass 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 = 50 mmL = 1.2 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid brass shaft of diameter 30 mm is fixed at A. A torque of 400 N·m acts at B, 2.5 m from A, and a torque of 250 N·m in the same direction acts at the free end C, a further 1.2 m along. Find the maximum shear stress τmax in the shaft.
d = 30 mmTB = 400 N·mTC = 250 N·mAB = 2.5 mBC = 1.2 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 100 MPa and σᵧ = −20 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 135 MPa?
σx = 100 MPaσy = −20 MPaσ1 = 135 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A thin-walled cylindrical tank of inner diameter 500 mm and wall thickness 6 mm holds gas at a gauge pressure of 3 MPa. It is made from plate welded along a helical seam at 20° to the axis of the cylinder. Find the magnitude of the shear stress |τn| acting along the weld seam.
D = 500 mmt = 6 mmp = 3 MPaα = 20°
|τn| = MPa
7.Axial stress & strainMedium
Find the elongation δ of the brass bar under the axial load.
P = 65 kNd = 12 mmL = 2 mE = 100 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 = 105 kNsegment 1: Ø40 mm, 2 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 1.5 m, aluminium (E = 70 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 5 mE = 200 GPa (steel)b × h = 50 × 80 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 2.5 mE = 200 GPa (steel)d = 60 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 10 mm is followed 3 cycles later by a peak of 1 mm. Find the damping ratio ζ.
x0 = 10 mmx3 = 1 mmn = 3
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 80 kN/m and k₂ = 40 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 10 kgk1 = 80 kN/mk2 = 40 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: N2 = 96
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.
Total number of teeth
N1 + N2 = 2a/m = 2·300/5 = 120
Split by the ratio
N1 = 120/(1 + 4) = 24
N2 = 4·24 = 96
Answer
N2 = 96
2.Gears & gear trains
Answer: ND = 60
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 = 2880·12/60 = 576 rpm
Teeth on gear D
ND = C·nC/nout = 48·576/460.8 = 60
Answer
ND = 60
3.Torsion of shafts
Answer: φ = 1.437 °
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·50⁴/32 = 613592 mm⁴
Angle of twist
φ = TL/(GJ) = 500·1000·1200/(39000·613592) = 0.02507 rad
φ = 0.02507·180/π = 1.437°
Answer
φ = 1.437°
4.Torsion of shafts
Answer: τmax = 122.6 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 = 250 N·m
TAB = TC + TB = 250 + 400 = 650 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 650 N·m
τmax = 16·650000/(π·30³) = 122.6 MPa
Answer
τmax = 122.6 MPa
5.Mohr’s circle & stress transformation
Answer: τxy = 73.65 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 = (100 + (−20))/2 = 40 MPa
R = σ1 − C = 135 − 40 = 95 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(95² − 60²) = 73.65 MPa
Answer
τxy = 73.65 MPa
6.Mohr’s circle & stress transformation
Answer: |τn| = 20.09 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. Thin-walled cylinder: hoop stress σh = pr/t and axial stress σa = pr/(2t); these are the principal stresses in the wall.
Wall stresses
r = 500/2 = 250 mm
σh = pr/t = 3·250/6 = 125 MPa
σa = pr/(2t) = 62.5 MPa
Transform to the seam
Take x along the axis (σa) and y around the hoop (σh). The seam’s normal makes 90° − α with the axis.
|τn| = (125 − 62.5)·sin 20°·cos 20° = 20.09 MPa
Answer
|τn| = 20.09 MPa
7.Axial stress & strain
Answer: δ = 11.49 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·12²/4 = 113.1 mm²
σ = P/A = 65·1000/113.1 = 574.7 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 574.7/100000 = 0.005747 = 5747 µε
Elongation
δ = ε·L = 0.005747·2000 mm = 11.49 mm
Answer
δ = 11.49 mm
8.Axial stress & strain
Answer: σmax = 928.4 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·12²/4 = 113.1 mm²
σ1 = 83.56 MPa
σ2 = 928.4 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 928.4 MPa
Answer
σmax = 928.4 MPa
9.Column buckling
Answer: Pcr = 16.45 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 = 80·50³/12 = 833333 mm⁴
Effective length
KL = 2·5000 = 10000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·833333/10000² = 16449 N = 16.45 kN
Answer
Pcr = 16.45 kN
10.Column buckling
Answer: σcr = 145 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·60⁴/64 = 636173 mm⁴
Radius of gyration and slenderness
A = 2827 mm²
r = √(I/A) = √(636173/2827) = 15 mm
KL = 0.7·2500 = 1750 mm
KL/r = 1750/15 = 116.7
Critical stress
σcr = π²E/(KL/r)² = π²·200000/116.7² = 145 MPa
Answer
σcr = 145 MPa
11.Mechanical vibrations
Answer: ζ = 0.1213
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 = (1/3)·ln(10/1) = 0.7675
Damping ratio
ζ = δ/√(4π² + δ²) = 0.7675/√(39.48 + 0.5891) = 0.1213
Answer
ζ = 0.1213
12.Mechanical vibrations
Answer: keq = 120 kN/m
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 = 80 + 40 = 120 kN/m
Answer
keq = 120 kN/m
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