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 2.2 kW motor running at 960 rpm drives a pinion with 45 teeth, which meshes with a gear with 48 teeth. The gear pair has an efficiency of 98%. Find the torque T1 on the motor shaft.
P = 2.2 kWn1 = 960 rpmN1 = 45N2 = 48η = 98%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (16 teeth) on the input shaft turns at 1500 rpm and drives gear B (48 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 250 rpm. How many teeth ND must gear D have?
nin = 1500 rpmA = 16B = 48C = 15nout = 250 rpm
ND =
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 = 1000 N·md = 30 mmL = 2 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid brass shaft of diameter 25 mm is fixed at A. A torque of 200 N·m acts at B, 2.5 m from A, and a torque of 100 N·m in the same direction acts at the free end C, a further 2.5 m along. Find the maximum shear stress τmax in the shaft.
d = 25 mmTB = 200 N·mTC = 100 N·mAB = 2.5 mBC = 2.5 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 120 MPa, σᵧ = 90 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 120 MPaσy = 90 MPaτxy = 30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 50 MPa, σᵧ = −70 MPa and τₓᵧ = −30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 20° counter-clockwise from the x-axis.
σx = 50 MPaσy = −70 MPaτxy = −30 MPaθ = 20°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 40 mm, length 200 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, Ø40 mm, 400 mm, E = 200 GPasegment 2: bronze, Ø40 mm, 200 mm, E = 110 GPaδ = 0.5 mm
P = kN
8.Axial stress & strainHard
A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 50 mm, E = 100 GPa, allowable stress 80 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø30 mm, E = 200 GPa, σallow = 250 MPatube: Ø120/Ø50 mm, E = 100 GPa, σallow = 80 MPa
Pmax = kN
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 = 6 mE = 200 GPa (steel)d = 80 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 65 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 65 kNn = 2.5L = 3 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 25 kg mass on a spring with a stiffness of 200 kN/m has a viscous damper giving a damping ratio of 0.45. Find the damped natural frequency fd of its free vibration in Hz.
m = 25 kgk = 200 kN/mζ = 0.45
fd = Hz
12.Mechanical vibrationsHard
A 20 kg mass hangs from spring k₃ = 30 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 80 kN/m and k₂ = 60 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 = 20 kgk1 = 80 kN/mk2 = 60 kN/mk3 = 30 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 21.88 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 = 2200/100.5 = 21.88 N·m
Answer
T1 = 21.88 N·m
2.Gears & gear trains
Answer: ND = 30
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 = 1500·16/48 = 500 rpm
Teeth on gear D
ND = C·nC/nout = 15·500/250 = 30
Answer
ND = 30
3.Torsion of shafts
Answer: φ = 18.01 °
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·30⁴/32 = 79522 mm⁴
Angle of twist
φ = TL/(GJ) = 1000·1000·2000/(80000·79522) = 0.3144 rad
φ = 0.3144·180/π = 18.01°
Answer
φ = 18.01°
4.Torsion of shafts
Answer: τmax = 97.78 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 = 100 N·m
TAB = TC + TB = 100 + 200 = 300 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 300 N·m
τmax = 16·300000/(π·25³) = 97.78 MPa
Answer
τmax = 97.78 MPa
5.Mohr’s circle & stress transformation
Answer: τmax = 33.54 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 = √(((120 − 90)/2)² + 30²) = 33.54 MPa
Answer
τmax = 33.54 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 16.68 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 = −10 + 60·0.766 + −30·0.6428 = 16.68 MPa
Answer
σθ = 16.68 MPa
7.Axial stress & strain
Answer: P = 164.6 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 = 1257 mm², A₂ = 1257 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·1257) = 1.592 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(110000·1257) = 1.447 × 10⁻⁶ mm/N
Load
P = 0.5/3.038 × 10⁻⁶ = 164560 N = 164.6 kN
Answer
P = 164.6 kN
8.Axial stress & strain
Answer: Pmax = 860.8 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
Ar = π·30²/4 = 706.9 mm²
At = π(120² − 50²)/4 = 9346 mm²
Combined axial stiffness
ΣEA = 200·10³·706.9 + 100·10³·9346 = 1.076 × 10⁹ N
Governing strain
ε = min(0.00125, 0.0008) = 0.0008 (brass)
Largest load
Pmax = 0.0008·1.076 × 10⁹ = 860796 N = 860.8 kN
Answer
Pmax = 860.8 kN
9.Column buckling
Answer: Pcr = 225 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·80⁴/64 = 2.011 × 10⁶ mm⁴
Effective length
KL = 0.7·6000 = 4200 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/4200² = 224989 N = 225 kN
Answer
Pcr = 225 kN
10.Column buckling
Answer: d = 52.15 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 = 2.5·65000 = 162500 N
KL = 0.7·3000 = 2100 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·162500·2100²/(π³·200000))^(1/4) = 52.15 mm
Answer
d = 52.15 mm
Check: the slenderness KL/r is 161.1, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 12.71 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.
Undamped natural frequency
ωn = √(200000/25) = 89.44 rad/s
fn = 89.44/(2π) = 14.24 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 14.24·√(1 − 0.45²) = 12.71 Hz
Answer
fd = 12.71 Hz
12.Mechanical vibrations
Answer: fn = 5.594 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 = 80 + 60 = 140 kN/m
keq = 140·30/(140 + 30) = 24.71 kN/m
Natural frequency
ωn = √(24706/20) = 35.15 rad/s
fn = 35.15/(2π) = 5.594 Hz
Answer
fn = 5.594 Hz
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