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 4 kW motor running at 2900 rpm drives a pinion with 36 teeth, which meshes with a gear with 80 teeth. The gear pair has an efficiency of 95%. Find the torque T1 on the motor shaft.
P = 4 kWn1 = 2900 rpmN1 = 36N2 = 80η = 95%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (20 teeth) on the input shaft turning at 960 rpm drives gear B (25 teeth). Gear C (32 teeth) is fixed to the same shaft as B and drives gear D (60 teeth) on the output shaft. Find the output speed nout.
nin = 960 rpmA = 20B = 25C = 32D = 60
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 = 200 N·md = 40 mmL = 2 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 20 kW at 1800 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.
P = 20 kWN = 1800 rpmτallow = 50 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 120 MPa, σᵧ = 10 MPa and τₓᵧ = −50 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 120 MPaσy = 10 MPaτxy = −50 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 80 MPa, σᵧ = 30 MPa and τₓᵧ = 30 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 45° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 80 MPaσy = 30 MPaτxy = 30 MPaθ = 45°
τθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the copper bar under the axial load.
P = 25 kNd = 12 mmL = 1 mE = 117 GPa
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 20 mm, length 400 mm, E = 200 GPa) and a segment made of aluminium (diameter 50 mm, length 400 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 40 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø50 mm, 400 mm, E = 70 GPaP = 40 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 3 mE = 200 GPa (steel)b × h = 40 × 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 45 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 45 kNn = 3L = 1 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 25 kg mass is supported by a spring with a stiffness of 4 kN/m and a viscous damper with a damping coefficient of 320 N·s/m. Find the damping ratio ζ.
m = 25 kgk = 4 kN/mc = 320 N·s/m
ζ =
12.Mechanical vibrationsHard
A 5 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 15 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 = 5 kgk1 = 15 kN/mk2 = 60 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 13.17 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 = 4000/303.7 = 13.17 N·m
Answer
T1 = 13.17 N·m
2.Gears & gear trains
Answer: nout = 409.6 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 = 960·20/25 = 768 rpm = nC
Second stage
nout = 768·32/60 = 409.6 rpm
Answer
nout = 409.6 rpm
total ratio = 2.344 : 1
3.Torsion of shafts
Answer: φ = 3.507 °
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·40⁴/32 = 251327 mm⁴
Angle of twist
φ = TL/(GJ) = 200·1000·2000/(26000·251327) = 0.06121 rad
φ = 0.06121·180/π = 3.507°
Answer
φ = 3.507°
4.Torsion of shafts
Answer: d = 22.11 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·1800/60 = 188.5 rad/s
T = P/ω = 20000/188.5 = 106.1 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·106103/(π·50)) = 22.11 mm
Answer
d = 22.11 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τmax = 74.33 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 = √(((120 − 10)/2)² + −50²) = 74.33 MPa
Answer
τmax = 74.33 MPa
6.Mohr’s circle & stress transformation
Answer: τθ = −25 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 = −25·1 + 30·0 = −25 MPa
Answer
τθ = −25 MPa
7.Axial stress & strain
Answer: δ = 1.889 mm
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 = 25·1000/113.1 = 221 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 221/117000 = 0.001889 = 1889 µε
Elongation
δ = ε·L = 0.001889·1000 mm = 1.889 mm
Answer
δ = 1.889 mm
8.Axial stress & strain
Answer: σ2 = 13.98 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 = 200000·314.2/400 = 157080 N/mm
k₂ = 70000·1963/400 = 343612 N/mm
Share of segment 2
F2 = 40000·343612/500691 = 27451 N
Stress
σ2 = 27451/1963 = 13.98 MPa (compression)
Answer
σ2 = 13.98 MPa
9.Column buckling
Answer: Pcr = 280.7 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 = 60·40³/12 = 320000 mm⁴
Effective length
KL = 0.5·3000 = 1500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·320000/1500² = 280735 N = 280.7 kN
Answer
Pcr = 280.7 kN
10.Column buckling
Answer: d = 48.59 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 = 3·45000 = 135000 N
KL = 2·1000 = 2000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·135000·2000²/(π³·200000))^(1/4) = 48.59 mm
Answer
d = 48.59 mm
Check: the slenderness KL/r is 164.7, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.506
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·√(4000·25) = 632.5 N·s/m
Damping ratio
ζ = c/cc = 320/632.5 = 0.506
Answer
ζ = 0.506
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 7.797 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 = 1/15 + 1/60 → keq = 15·60/(15 + 60) = 12 kN/m
Natural frequency
ωn = √(12000/5) = 48.99 rad/s
fn = 48.99/(2π) = 7.797 Hz
Answer
fn = 7.797 Hz
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