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 1450 rpm drives a pinion with 32 teeth, which meshes with a gear with 40 teeth. The gear pair has an efficiency of 98%. Find the output torque T2 on the gear shaft.
P = 5.5 kWn1 = 1450 rpmN1 = 32N2 = 40η = 98%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (16 teeth) on the input shaft turning at 1450 rpm drives gear B (72 teeth). Gear C (32 teeth) is fixed to the same shaft as B and drives gear D (72 teeth) on the output shaft. Find the output speed nout.
nin = 1450 rpmA = 16B = 72C = 32D = 72
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 = 200 N·md = 30 mmL = 2.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 75 kW at 750 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.
P = 75 kWN = 750 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −40 MPa, σᵧ = −20 MPa and τₓᵧ = 40 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 = −40 MPaσy = −20 MPaτxy = 40 MPaθ = -45°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = 120 MPa, σᵧ = −70 MPa and τₓᵧ = −30 MPa (σz = 0). The material has a yield strength of 250 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.
σx = 120 MPaσy = −70 MPaτxy = −30 MPaσY = 250 MPa
n =
7.Axial stress & strainMedium
Find the axial strain ε in the steel bar, in microstrain (µε = 10⁻⁶).
P = 115 kNd = 12 mmL = 1 mE = 200 GPa
ε = µε
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 = 35 kNsegment 1: Ø35 mm, 1 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 1 m, steel (E = 200 GPa)
σmax = MPa
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 = 5 mE = 200 GPa (steel)b × h = 30 × 100 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at both ends. It must carry 15 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 15 kNn = 2.5L = 5 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 2 kg mass is supported by a spring with a stiffness of 15 kN/m and a viscous damper with a damping coefficient of 210 N·s/m. Find the damping ratio ζ.
m = 2 kgk = 15 kN/mc = 210 N·s/m
ζ =
12.Mechanical vibrationsHard
A 40 kg mass hangs from spring k₃ = 25 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 = 40 kgk1 = 100 kN/mk2 = 10 kN/mk3 = 25 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 44.37 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·1450/60 = 151.8 rad/s
T1 = P/ω₁ = 5500/151.8 = 36.22 N·m
Output torque
T2 = 36.22·40/32·0.98 = 44.37 N·m
Answer
T2 = 44.37 N·m
2.Gears & gear trains
Answer: nout = 143.2 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 = 1450·16/72 = 322.2 rpm = nC
Second stage
nout = 322.2·32/72 = 143.2 rpm
Answer
nout = 143.2 rpm
total ratio = 10.13 : 1
3.Torsion of shafts
Answer: φ = 4.503 °
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) = 200·1000·2500/(80000·79522) = 0.0786 rad
φ = 0.0786·180/π = 4.503°
Answer
φ = 4.503°
4.Torsion of shafts
Answer: d = 40.18 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·750/60 = 78.54 rad/s
T = P/ω = 75000/78.54 = 954.9 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·954930/(π·75)) = 40.18 mm
Answer
d = 40.18 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: σy′ = 10 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′ = −30 − (−10)·0 − 40·(−1) = 10 MPa
Check
The sum of the normal stresses does not change with rotation.
σx′ = −70 MPa, σx′ + σy′ = −60 = σx + σy
Answer
σy′ = 10 MPa
6.Mohr’s circle & stress transformation
Answer: n = 1.255
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. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.
Principal stresses
C = 25 MPa, R = 99.62 MPa
σ1 = 124.6 MPa, σ2 = −74.62 MPa, σ3 = 0
Tresca
σmax − σmin = 199.2 MPa
n = 250/199.2 = 1.255
Answer
n = 1.255
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: ε = 5084 µε
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 = 115·1000/113.1 = 1017 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 1017/200000 = 0.005084 = 5084 µε
Answer
ε = 5084 µε
8.Axial stress & strain
Answer: σmax = 111.4 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·35²/4 = 962.1 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 36.38 MPa
σ2 = 111.4 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 111.4 MPa
Answer
σmax = 111.4 MPa
9.Column buckling
Answer: Pcr = 17.77 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 = 100·30³/12 = 225000 mm⁴
Effective length
KL = 1·5000 = 5000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·225000/5000² = 17765 N = 17.77 kN
Answer
Pcr = 17.77 kN
10.Column buckling
Answer: d = 51.27 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 = 2.5·15000 = 37500 N
KL = 0.5·5000 = 2500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·37500·2500²/(π³·70000))^(1/4) = 51.27 mm
Answer
d = 51.27 mm
Check: the slenderness KL/r is 195, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.6062
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·√(15000·2) = 346.4 N·s/m
Damping ratio
ζ = c/cc = 210/346.4 = 0.6062
Answer
ζ = 0.6062
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 3.592 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 = 110·25/(110 + 25) = 20.37 kN/m
Natural frequency
ωn = √(20370/40) = 22.57 rad/s
fn = 22.57/(2π) = 3.592 Hz
Answer
fn = 3.592 Hz
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