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 18 teeth and a module of 5 mm. Find the tangential force Ft on the pinion teeth.
P = 1.1 kWn = 960 rpmN1 = 18m = 5 mm
Ft = N
2.Gears & gear trainsHard
In a compound gear train, gear A (16 teeth) on the input shaft turning at 2900 rpm drives gear B (50 teeth). Gear C (24 teeth) is fixed to the same shaft as B and drives gear D (50 teeth) on the output shaft. Find the output speed nout.
nin = 2900 rpmA = 16B = 50C = 24D = 50
nout = rpm
3.Torsion of shaftsMedium
A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 60 mm and length 0.5 m; segment BC has diameter 40 mm and length 1 m. A torque of 250 N·m acts at the free end C. Find the maximum shear stress τmax in the shaft.
T = 250 N·md1 = 60 mm, L1 = 0.5 md2 = 40 mm, L2 = 1 m
τmax = MPa
4.Torsion of shaftsHard
A solid shaft must transmit 100 kW at 1450 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.
P = 100 kWN = 1450 rpmτallow = 60 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −10 MPa, σᵧ = −80 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 = −10 MPaσy = −80 MPaτxy = −50 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 50 mm carries an axial compressive force of 160 kN together with a torque of 750 N·m. At the surface of the shaft, find the largest principal stress σ1.
d = 50 mmF = −160 kNT = 750 N·m
σ1 = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 30 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 50 mm, length 300 mm, E = 110 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.
segment 1: steel, Ø30 mm, 600 mm, E = 200 GPasegment 2: bronze, Ø50 mm, 300 mm, E = 110 GPaδ = 0.5 mm
P = kN
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 = 115 kNsegment 1: Ø35 mm, 1 m, copper (E = 117 GPa)segment 2: Ø20 mm, 1 m, brass (E = 100 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 5 mE = 70 GPa (aluminium)d = 50 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 40 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 40 kNn = 2L = 4 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 15 kg mass is supported by a spring with a stiffness of 15 kN/m and a viscous damper with a damping coefficient of 570 N·s/m. Find the damping ratio ζ.
m = 15 kgk = 15 kN/mc = 570 N·s/m
ζ =
12.Mechanical vibrationsHard
A 5 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 25 kN/m and k₂ = 5 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 5 kgk1 = 25 kN/mk2 = 5 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: Ft = 243.2 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·960/60 = 100.5 rad/s
T = P/ω = 1100/100.5 = 10.94 N·m
Pitch diameter
d = m·N = 5·18 = 90 mm
Tangential force
Ft = 2T/d = 2·10.94/0.09 = 243.2 N
Answer
Ft = 243.2 N
2.Gears & gear trains
Answer: nout = 445.4 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 = 2900·16/50 = 928 rpm = nC
Second stage
nout = 928·24/50 = 445.4 rpm
Answer
nout = 445.4 rpm
total ratio = 6.51 : 1
3.Torsion of shafts
Answer: τmax = 19.89 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.
Stress in each segment
T = 250 N·m = 250000 N·mm
τAB = 16·250000/(π·60³) = 5.895 MPa
τBC = 16·250000/(π·40³) = 19.89 MPa
Answer
τmax = 19.89 MPa
4.Torsion of shafts
Answer: d = 38.24 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 = 100000/151.8 = 658.6 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·658572/(π·60)) = 38.24 mm
Answer
d = 38.24 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θs = −72.5 °
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 = −50/61.03 = −0.8192, sin 2θs = −35/61.03 = −0.5735
2θs = atan2(−35, −50) = −145°
Angle on the element
θs = −72.5°
check: θs − θp = −72.5 − (−27.5) = −45°
Answer
θs = −72.5°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σ1 = 10.19 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 = −160000/(π·50²/4) = −160000/1963 = −81.49 MPa
τ = 16T/(πd³) = 16·750000/(π·50³) = 30.56 MPa
Mohr’s circle
C = σ/2 = −40.74 MPa
R = √((−40.74)² + 30.56²) = 50.93 MPa
σ1 = C + R = 10.19 MPa
σ2 = C − R = −91.67 MPa
Answer
σ1 = 10.19 MPa
7.Axial stress & strain
Answer: P = 88.76 kN
Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.
Areas
A₁ = 706.9 mm², A₂ = 1963 mm²
Flexibilities
L₁/(E₁A₁) = 600/(200000·706.9) = 4.244 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(110000·1963) = 1.389 × 10⁻⁶ mm/N
Load
P = 0.5/5.633 × 10⁻⁶ = 88761 N = 88.76 kN
Answer
P = 88.76 kN
8.Axial stress & strain
Answer: σmax = 366.1 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·20²/4 = 314.2 mm²
σ1 = 119.5 MPa
σ2 = 366.1 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 366.1 MPa
Answer
σmax = 366.1 MPa
9.Column buckling
Answer: Pcr = 2.12 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·50⁴/64 = 306796 mm⁴
Effective length
KL = 2·5000 = 10000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·306796/10000² = 2120 N = 2.12 kN
Answer
Pcr = 2.12 kN
10.Column buckling
Answer: d = 65.58 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 = 2·40000 = 80000 N
KL = 0.7·4000 = 2800 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·80000·2800²/(π³·70000))^(1/4) = 65.58 mm
Answer
d = 65.58 mm
Check: the slenderness KL/r is 170.8, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.6008
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·√(15000·15) = 948.7 N·s/m
Damping ratio
ζ = c/cc = 570/948.7 = 0.6008
Answer
ζ = 0.6008
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: keq = 30 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 = 25 + 5 = 30 kN/m
Answer
keq = 30 kN/m
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