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 15 kW motor running at 2900 rpm drives a pinion with 32 teeth, which meshes with a gear with 72 teeth. The gear pair has an efficiency of 95%. Find the output torque T2 on the gear shaft.
P = 15 kWn1 = 2900 rpmN1 = 32N2 = 72η = 95%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (18 teeth) on the input shaft turns at 2880 rpm and drives gear B (36 teeth). Gear C (30 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 864 rpm. How many teeth ND must gear D have?
nin = 2880 rpmA = 18B = 36C = 30nout = 864 rpm
ND =
3.Torsion of shaftsMedium
A solid steel shaft 2.5 m long carries a torque of 100 N·m. The angle of twist between its ends must not exceed 1°. Find the smallest diameter d that meets this limit.
T = 100 N·mL = 2.5 mG = 80 GPaφallow = 1°
d = mm
4.Torsion of shaftsHard
A solid brass shaft transmits 20 kW at 300 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 0.25° per m of length. Find the smallest diameter d that satisfies both limits.
P = 20 kWN = 300 rpmτallow = 60 MPaθallow = 0.25°/mG = 39 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 110 MPa, σᵧ = −30 MPa and τₓᵧ = 40 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 = 110 MPaσy = −30 MPaτxy = 40 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 90 MPa, σᵧ = 0 MPa and τₓᵧ = −50 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 35° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 90 MPaσy = 0 MPaτxy = −50 MPaθ = 35°
τθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 30 mm, length 300 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.
segment 1: steel, Ø40 mm, 600 mm, E = 200 GPasegment 2: copper, Ø30 mm, 300 mm, E = 120 GPaδ = 0.2 mm
P = kN
8.Axial stress & strainHard
A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of copper (outer diameter 100 mm, inner diameter 60 mm, E = 120 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø25 mm, E = 200 GPa, σallow = 150 MPatube: Ø100/Ø60 mm, E = 120 GPa, σallow = 100 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 = 5 mE = 200 GPa (steel)b × h = 50 × 80 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium 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 = 70 GPa (aluminium)d = 25 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 15 kg mass on a spring with a stiffness of 25 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 = 15 kgk = 25 kN/mζ = 0.45
fd = Hz
12.Mechanical vibrationsHard
A 10 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 = 100 kN/m and k₂ = 10 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 = 10 kgk1 = 100 kN/mk2 = 10 kN/mk3 = 30 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 105.6 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·2900/60 = 303.7 rad/s
T1 = P/ω₁ = 15000/303.7 = 49.39 N·m
Output torque
T2 = 49.39·72/32·0.95 = 105.6 N·m
Answer
T2 = 105.6 N·m
2.Gears & gear trains
Answer: ND = 50
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 = 2880·18/36 = 1440 rpm
Teeth on gear D
ND = C·nC/nout = 30·1440/864 = 50
Answer
ND = 50
3.Torsion of shafts
Answer: d = 36.75 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 = 100 N·m = 100000 N·mm
L = 2.5 m = 2500 mm
G = 80 GPa = 80000 MPa
φ = 1·π/180 = 0.01745 rad
J = TL/(Gφ) = 100000·2500/(80000·0.01745) = 179049 mm⁴
Diameter
d = (32J/π)1/4 = 36.75 mm
Answer
d = 36.75 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: d = 78.57 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·300/60 = 31.42 rad/s
T = P/ω = 20000/31.42 = 636.6 N·m
Diameter for strength
T = 636.6 N·m = 636620 N·mm
dτ = ∛(16·636620/(π·60)) = 37.81 mm
Diameter for stiffness
θ = 0.25·π/180/1000 = 4.363 × 10⁻⁶ rad/mm
G = 39 GPa = 39000 MPa
dφ = (32·636620/(π·39000·4.363 × 10⁻⁶))1/4 = 78.57 mm
Governing diameter
Both limits must hold, so the larger diameter governs.
d = max(37.81, 78.57) = 78.57 mm
Answer
d = 78.57 mm
Here stiffness governs.
5.Mohr’s circle & stress transformation
Answer: θs = −30.13 °
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 = 40/80.62 = 0.4961, sin 2θs = −70/80.62 = −0.8682
2θs = atan2(−70, 40) = −60.26°
Angle on the element
θs = −30.13°
check: θs − θp = −30.13 − 14.87 = −45°
Answer
θs = −30.13°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: τθ = −59.39 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
τθ = −45·0.9397 + −50·0.342 = −59.39 MPa
Answer
τθ = −59.39 MPa
7.Axial stress & strain
Answer: P = 33.76 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 = 1257 mm², A₂ = 706.9 mm²
Flexibilities
L₁/(E₁A₁) = 600/(200000·1257) = 2.387 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(120000·706.9) = 3.537 × 10⁻⁶ mm/N
Load
P = 0.2/5.924 × 10⁻⁶ = 33760 N = 33.76 kN
Answer
P = 33.76 kN
8.Axial stress & strain
Answer: Pmax = 526 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·25²/4 = 490.9 mm²
At = π(100² − 60²)/4 = 5027 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 120·10³·5027 = 7.014 × 10⁸ N
Governing strain
ε = min(0.00075, 0.0008333) = 0.00075 (steel)
Largest load
Pmax = 0.00075·7.014 × 10⁸ = 526020 N = 526 kN
Answer
Pmax = 526 kN
9.Column buckling
Answer: Pcr = 134.3 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 = 0.7·5000 = 3500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·833333/3500² = 134280 N = 134.3 kN
Answer
Pcr = 134.3 kN
10.Column buckling
Answer: σcr = 8.812 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 = 0.7·2500 = 1750 mm
KL/r = 1750/6.25 = 280
Critical stress
σcr = π²E/(KL/r)² = π²·70000/280² = 8.812 MPa
Answer
σcr = 8.812 MPa
11.Mechanical vibrations
Answer: fd = 5.802 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 = √(25000/15) = 40.82 rad/s
fn = 40.82/(2π) = 6.497 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 6.497·√(1 − 0.45²) = 5.802 Hz
Answer
fd = 5.802 Hz
12.Mechanical vibrations
Answer: fn = 7.727 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 = 100 + 10 = 110 kN/m
keq = 110·30/(110 + 30) = 23.57 kN/m
Natural frequency
ωn = √(23571/10) = 48.55 rad/s
fn = 48.55/(2π) = 7.727 Hz
Answer
fn = 7.727 Hz
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