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 3 kW motor running at 960 rpm drives a pinion with 48 teeth, which meshes with a gear with 60 teeth. The gear pair has an efficiency of 98%. Find the output torque T2 on the gear shaft.
P = 3 kWn1 = 960 rpmN1 = 48N2 = 60η = 98%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (24 teeth) on the input shaft turning at 3000 rpm drives gear B (45 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 = 3000 rpmA = 24B = 45C = 12D = 16
nout = rpm
3.Torsion of shaftsMedium
In a torsion test, a solid specimen of diameter 12 mm is twisted by a torque of 50 N·m. The two ends of a gauge length of 200 mm rotate 10.82° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 50 N·md = 12 mmL = 200 mmφ = 10.82°
G = GPa
4.Torsion of shaftsHard
A hollow shaft transmits 100 kW at 1450 rpm. Find the maximum shear stress τmax in the shaft.
P = 100 kWN = 1450 rpmdo = 50 mmdi = 38 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 10 MPa and σᵧ = 50 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 95 MPa?
σx = 10 MPaσy = 50 MPaσ1 = 95 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 10 MPa, σᵧ = 110 MPa and τₓᵧ = −20 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° clockwise from the x-axis.
σx = 10 MPaσy = 110 MPaτxy = −20 MPaθ = -30°
σθ = 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 aluminium (diameter 60 mm, length 500 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.
segment 1: steel, Ø40 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø60 mm, 500 mm, E = 70 GPaδ = 0.4 mm
P = kN
8.Axial stress & strainHard
A steel segment (diameter 25 mm, length 600 mm, E = 200 GPa) and a segment made of copper (diameter 50 mm, length 400 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø25 mm, 600 mm, E = 200 GPasegment 2: copper, Ø50 mm, 400 mm, E = 120 GPaP = 20 kN
σ2 = MPa
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 = 4 mE = 200 GPa (steel)b × h = 20 × 50 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 = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 250 kg mass is supported by a spring with a stiffness of 100 kN/m and a viscous damper with a damping coefficient of 4000 N·s/m. Find the damping ratio ζ.
m = 250 kgk = 100 kN/mc = 4000 N·s/m
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 80 kN/m and the lower one k₂ = 60 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 10 kgk1 = 80 kN/mk2 = 60 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 36.56 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 = 3000/100.5 = 29.84 N·m
Output torque
T2 = 29.84·60/48·0.98 = 36.56 N·m
Answer
T2 = 36.56 N·m
2.Gears & gear trains
Answer: nout = 1200 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 = 3000·24/45 = 1600 rpm = nC
Second stage
nout = 1600·12/16 = 1200 rpm
Answer
nout = 1200 rpm
total ratio = 2.5 : 1
3.Torsion of shafts
Answer: G = 26.01 GPa
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 moment and angle
J = π·12⁴/32 = 2036 mm⁴
φ = 10.82·π/180 = 0.1888 rad
Shear modulus
T = 50 N·m = 50000 N·mm
G = TL/(Jφ) = 50000·200/(2036·0.1888) = 26012 MPa
G = 26.01 GPa
Answer
G = 26.01 GPa
Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.
4.Torsion of shafts
Answer: τmax = 40.27 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.
Torque
ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s
T = P/ω = 100000/151.8 = 658.6 N·m
Polar second moment of area
J = π(50⁴ − 38⁴)/32 = 408885 mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 658572·25/408885 = 40.27 MPa
Answer
τmax = 40.27 MPa
5.Mohr’s circle & stress transformation
Answer: τxy = 61.85 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.
Required radius
C = (10 + 50)/2 = 30 MPa
R = σ1 − C = 95 − 30 = 65 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(65² − 20²) = 61.85 MPa
Answer
τxy = 61.85 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 52.32 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 + −50·0.5 + −20·−0.866 = 52.32 MPa
Answer
σθ = 52.32 MPa
7.Axial stress & strain
Answer: P = 81.41 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 = 2827 mm²
Flexibilities
L₁/(E₁A₁) = 600/(200000·1257) = 2.387 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(70000·2827) = 2.526 × 10⁻⁶ mm/N
Load
P = 0.4/4.914 × 10⁻⁶ = 81407 N = 81.41 kN
Answer
P = 81.41 kN
8.Axial stress & strain
Answer: σ2 = 7.972 MPa
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.
Stiffnesses
k₁ = 200000·490.9/600 = 163625 N/mm
k₂ = 120000·1963/400 = 589049 N/mm
Share of segment 2
F2 = 20000·589049/752673 = 15652 N
Stress
σ2 = 15652/1963 = 7.972 MPa (compression)
Answer
σ2 = 7.972 MPa
9.Column buckling
Answer: Pcr = 8.393 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 = 50·20³/12 = 33333 mm⁴
Effective length
KL = 0.7·4000 = 2800 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·33333/2800² = 8393 N = 8.393 kN
Answer
Pcr = 8.393 kN
10.Column buckling
Answer: σcr = 12.69 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·30⁴/64 = 39761 mm⁴
Radius of gyration and slenderness
A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 0.7·2500 = 1750 mm
KL/r = 1750/7.5 = 233.3
Critical stress
σcr = π²E/(KL/r)² = π²·70000/233.3² = 12.69 MPa
Answer
σcr = 12.69 MPa
11.Mechanical vibrations
Answer: ζ = 0.4
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·√(100000·250) = 10000 N·s/m
Damping ratio
ζ = c/cc = 4000/10000 = 0.4
Answer
ζ = 0.4
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 9.319 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 = 1/80 + 1/60 → keq = 80·60/(80 + 60) = 34.29 kN/m
Natural frequency
ωn = √(34286/10) = 58.55 rad/s
fn = 58.55/(2π) = 9.319 Hz
Answer
fn = 9.319 Hz
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