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
An 11 kW motor running at 1500 rpm drives a pinion with 15 teeth, which meshes with a gear with 45 teeth. The gear pair has an efficiency of 98%. Find the output torque T2 on the gear shaft.
P = 11 kWn1 = 1500 rpmN1 = 15N2 = 45η = 98%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (15 teeth) on the input shaft turning at 1440 rpm drives gear B (30 teeth). Gear C (18 teeth) is fixed to the same shaft as B and drives gear D (24 teeth) on the output shaft. Find the output speed nout.
nin = 1440 rpmA = 15B = 30C = 18D = 24
nout = rpm
3.Torsion of shaftsMedium
A solid aluminium 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 = 100 N·md = 20 mmL = 0.8 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 75 kW at 3000 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.
P = 75 kWN = 3000 rpmτallow = 60 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 90 MPa, σᵧ = −80 MPa and τₓᵧ = 60 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 90 MPaσy = −80 MPaτxy = 60 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 120 MPa, σᵧ = −50 MPa and τₓᵧ = −50 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 50° counter-clockwise from the x-axis.
σx = 120 MPaσy = −50 MPaτxy = −50 MPaθ = 50°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 25 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 30 mm, length 200 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, Ø25 mm, 500 mm, E = 200 GPasegment 2: copper, Ø30 mm, 200 mm, E = 120 GPaδ = 0.2 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 = 75 kNsegment 1: Ø25 mm, 1 m, steel (E = 200 GPa)segment 2: Ø12 mm, 2 m, brass (E = 100 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 5 mE = 70 GPa (aluminium)d = 80 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 3 mE = 200 GPa (steel)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
An 8 kg mass on a spring with a stiffness of 80 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 = 8 kgk = 80 kN/mζ = 0.45
fd = Hz
12.Mechanical vibrationsHard
A 5 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 = 30 kN/m and k₂ = 30 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 = 5 kgk1 = 30 kN/mk2 = 30 kN/mk3 = 30 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 205.9 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·1500/60 = 157.1 rad/s
T1 = P/ω₁ = 11000/157.1 = 70.03 N·m
Output torque
T2 = 70.03·45/15·0.98 = 205.9 N·m
Answer
T2 = 205.9 N·m
2.Gears & gear trains
Answer: nout = 540 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 = 1440·15/30 = 720 rpm = nC
Second stage
nout = 720·18/24 = 540 rpm
Answer
nout = 540 rpm
total ratio = 2.667 : 1
3.Torsion of shafts
Answer: φ = 11.22 °
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·20⁴/32 = 15708 mm⁴
Angle of twist
φ = TL/(GJ) = 100·1000·800/(26000·15708) = 0.1959 rad
φ = 0.1959·180/π = 11.22°
Answer
φ = 11.22°
4.Torsion of shafts
Answer: d = 27.26 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·3000/60 = 314.2 rad/s
T = P/ω = 75000/314.2 = 238.7 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·238732/(π·60)) = 27.26 mm
Answer
d = 27.26 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τmax = 104 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 = √(((90 − −80)/2)² + 60²) = 104 MPa
Answer
τmax = 104 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = −29 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 = 35 + 85·−0.1736 + −50·0.9848 = −29 MPa
Answer
σθ = −29 MPa
7.Axial stress & strain
Answer: P = 26.84 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 = 490.9 mm², A₂ = 706.9 mm²
Flexibilities
L₁/(E₁A₁) = 500/(200000·490.9) = 5.093 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(120000·706.9) = 2.358 × 10⁻⁶ mm/N
Load
P = 0.2/7.451 × 10⁻⁶ = 26843 N = 26.84 kN
Answer
P = 26.84 kN
8.Axial stress & strain
Answer: δ = 14.03 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·25²/4 = 490.9 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 152.8 MPa
σ2 = 663.1 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 152.8·1000/200000 = 0.7639 mm
δ2 = σ2L2/E2 = 663.1·2000/100000 = 13.26 mm
δ = δ1 + δ2 = 14.03 mm
Answer
δ = 14.03 mm
9.Column buckling
Answer: Pcr = 222.3 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 round section has the same I about every axis.
I = πd⁴/64 = π·80⁴/64 = 2.011 × 10⁶ mm⁴
Effective length
KL = 0.5·5000 = 2500 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/2500² = 222253 N = 222.3 kN
Answer
Pcr = 222.3 kN
10.Column buckling
Answer: σcr = 34.27 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·50⁴/64 = 306796 mm⁴
Radius of gyration and slenderness
A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 1·3000 = 3000 mm
KL/r = 3000/12.5 = 240
Critical stress
σcr = π²E/(KL/r)² = π²·200000/240² = 34.27 MPa
Answer
σcr = 34.27 MPa
11.Mechanical vibrations
Answer: fd = 14.21 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 = √(80000/8) = 100 rad/s
fn = 100/(2π) = 15.92 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 15.92·√(1 − 0.45²) = 14.21 Hz
Answer
fd = 14.21 Hz
12.Mechanical vibrations
Answer: fn = 10.07 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 = 30 + 30 = 60 kN/m
keq = 60·30/(60 + 30) = 20 kN/m
Natural frequency
ωn = √(20000/5) = 63.25 rad/s
fn = 63.25/(2π) = 10.07 Hz
Answer
fn = 10.07 Hz
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