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 5.5 kW motor running at 1440 rpm drives a pinion with 15 teeth, which meshes with a gear with 50 teeth. The gear pair has an efficiency of 95%. Find the torque T1 on the motor shaft.
P = 5.5 kWn1 = 1440 rpmN1 = 15N2 = 50η = 95%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (12 teeth) on the input shaft turning at 1500 rpm drives gear B (24 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D (16 teeth) on the output shaft. Find the output speed nout.
nin = 1500 rpmA = 12B = 24C = 12D = 16
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 = 750 N·md = 25 mmL = 0.8 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 50 kW at 1500 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.
P = 50 kWN = 1500 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 60 MPa, σᵧ = −50 MPa and τₓᵧ = −10 MPa (tension positive). The element is rotated 45° clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.
σx = 60 MPaσy = −50 MPaτxy = −10 MPaθ = -45°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 160 MPa, σᵧ = −40 MPa and τₓᵧ = −60 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 40° counter-clockwise from the x-axis.
σx = 160 MPaσy = −40 MPaτxy = −60 MPaθ = 40°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 20 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 60 mm, length 300 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.25 mm in total.
segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: brass, Ø60 mm, 300 mm, E = 100 GPaδ = 0.25 mm
P = kN
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 20 kNsegment 1: Ø35 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø16 mm, 1 m, copper (E = 117 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 4 mE = 200 GPa (steel)d = 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 125 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 125 kNn = 2L = 1.5 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 12 kg mass on a spring with a stiffness of 8 kN/m has a viscous damper giving a damping ratio of 0.5. Find the damped natural frequency fd of its free vibration in Hz.
m = 12 kgk = 8 kN/mζ = 0.5
fd = Hz
12.Mechanical vibrationsHard
A 5 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 12 kN/m and the lower one k₂ = 40 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 5 kgk1 = 12 kN/mk2 = 40 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 36.47 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·1440/60 = 150.8 rad/s
T1 = P/ω₁ = 5500/150.8 = 36.47 N·m
Answer
T1 = 36.47 N·m
2.Gears & gear trains
Answer: nout = 562.5 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 = 1500·12/24 = 750 rpm = nC
Second stage
nout = 750·12/16 = 562.5 rpm
Answer
nout = 562.5 rpm
total ratio = 2.667 : 1
3.Torsion of shafts
Answer: φ = 11.21 °
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·25⁴/32 = 38350 mm⁴
Angle of twist
φ = TL/(GJ) = 750·1000·800/(80000·38350) = 0.1956 rad
φ = 0.1956·180/π = 11.21°
Answer
φ = 11.21°
4.Torsion of shafts
Answer: d = 27.86 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·1500/60 = 157.1 rad/s
T = P/ω = 50000/157.1 = 318.3 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·318310/(π·75)) = 27.86 mm
Answer
d = 27.86 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: σy′ = −5 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 for the y′ face
σy′ = 5 − 55·0 − (−10)·(−1) = −5 MPa
Check
The sum of the normal stresses does not change with rotation.
σx′ = 15 MPa, σx′ + σy′ = 10 = σx + σy
Answer
σy′ = −5 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 18.28 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 = 60 + 100·0.1736 + −60·0.9848 = 18.28 MPa
Answer
σθ = 18.28 MPa
7.Axial stress & strain
Answer: P = 33.66 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 = 314.2 mm², A₂ = 2827 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·314.2) = 6.366 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(100000·2827) = 1.061 × 10⁻⁶ mm/N
Load
P = 0.25/7.427 × 10⁻⁶ = 33660 N = 33.66 kN
Answer
P = 33.66 kN
8.Axial stress & strain
Answer: δ = 0.9987 mm
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·16²/4 = 201.1 mm²
σ1 = 20.79 MPa
σ2 = 99.47 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 20.79·500/70000 = 0.1485 mm
δ2 = σ2L2/E2 = 99.47·1000/117000 = 0.8502 mm
δ = δ1 + δ2 = 0.9987 mm
Answer
δ = 0.9987 mm
9.Column buckling
Answer: Pcr = 78.48 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·60⁴/64 = 636173 mm⁴
Effective length
KL = 1·4000 = 4000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·636173/4000² = 78485 N = 78.48 kN
Answer
Pcr = 78.48 kN
10.Column buckling
Answer: d = 90.25 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 = 2·125000 = 250000 N
KL = 2·1500 = 3000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·250000·3000²/(π³·70000))^(1/4) = 90.25 mm
Answer
d = 90.25 mm
Check: the slenderness KL/r is 133, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 3.559 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 = √(8000/12) = 25.82 rad/s
fn = 25.82/(2π) = 4.109 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 4.109·√(1 − 0.5²) = 3.559 Hz
Answer
fd = 3.559 Hz
12.Mechanical vibrations
Answer: fn = 6.838 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
1/keq = 1/12 + 1/40 → keq = 12·40/(12 + 40) = 9.231 kN/m
Natural frequency
ωn = √(9231/5) = 42.97 rad/s
fn = 42.97/(2π) = 6.838 Hz
Answer
fn = 6.838 Hz
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