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 7.5 kW motor running at 2900 rpm drives a pinion with 16 teeth, which meshes with a gear with 32 teeth. The gear pair has an efficiency of 95%. Find the output torque T2 on the gear shaft.
P = 7.5 kWn1 = 2900 rpmN1 = 16N2 = 32η = 95%
T2 = 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 (36 teeth). Gear C (18 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 = 960 rpmA = 20B = 36C = 18D = 72
nout = rpm
3.Torsion of shaftsMedium
A hollow shaft must carry a torque of 750 N·m. Its inner diameter is to be 0.5 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter do.
T = 750 N·mdi/do = 0.5τallow = 75 MPa
do = mm
4.Torsion of shaftsHard
A solid shaft must transmit 20 kW at 3000 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.
P = 20 kWN = 3000 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 100 MPa, σᵧ = 40 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 100 MPaσy = 40 MPaτxy = 30 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 normal stress σθ on a plane whose normal is rotated 45° counter-clockwise from the x-axis.
σx = 80 MPaσy = −30 MPaτxy = 30 MPaθ = 45°
σθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the copper bar, in microstrain (µε = 10⁻⁶).
P = 45 kNd = 10 mmL = 3 mE = 117 GPa
ε = µε
8.Axial stress & strainHard
A steel segment (diameter 25 mm, length 300 mm, E = 200 GPa) and a segment made of bronze (diameter 30 mm, length 300 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 80 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: bronze, Ø30 mm, 300 mm, E = 110 GPaP = 80 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 1 mE = 200 GPa (steel)b × h = 30 × 100 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 1 mE = 70 GPa (aluminium)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed one cycle later by a peak of 23.4 mm. Find the damping ratio ζ.
x0 = 50 mmx1 = 23.4 mmn = 1
ζ =
12.Mechanical vibrationsHard
A 5 kg mass hangs from spring k₃ = 20 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 40 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 = 5 kgk1 = 40 kN/mk2 = 10 kN/mk3 = 20 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 46.92 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 = 7500/303.7 = 24.7 N·m
Output torque
T2 = 24.7·32/16·0.95 = 46.92 N·m
Answer
T2 = 46.92 N·m
2.Gears & gear trains
Answer: nout = 133.3 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 = 960·20/36 = 533.3 rpm = nC
Second stage
nout = 533.3·18/72 = 133.3 rpm
Answer
nout = 133.3 rpm
total ratio = 7.2 : 1
3.Torsion of shafts
Answer: do = 37.87 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.
Outer diameter from the allowable stress
T = 750 N·m = 750000 N·mm
1 − k⁴ = 1 − 0.5⁴ = 0.9375
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·750000/(π·75·0.9375)) = 37.87 mm
di = 0.5·37.87 = 18.94 mm
Answer
do = 37.87 mm
A solid shaft would need d = 37.07 mm; the hollow one is a little larger but much lighter.
4.Torsion of shafts
Answer: d = 16.29 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 = 20000/314.2 = 63.66 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·63662/(π·75)) = 16.29 mm
Answer
d = 16.29 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τmax = 42.43 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 = √(((100 − 40)/2)² + 30²) = 42.43 MPa
Answer
τmax = 42.43 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 55 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 = 25 + 55·0 + 30·1 = 55 MPa
Answer
σθ = 55 MPa
7.Axial stress & strain
Answer: ε = 4897 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·10²/4 = 78.54 mm²
σ = P/A = 45·1000/78.54 = 573 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 573/117000 = 0.004897 = 4897 µε
Answer
ε = 4897 µε
8.Axial stress & strain
Answer: σ2 = 50.02 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 = 200000·490.9/300 = 327249 N/mm
k₂ = 110000·706.9/300 = 259181 N/mm
Share of segment 2
F2 = 80000·259181/586431 = 35357 N
Stress
σ2 = 35357/706.9 = 50.02 MPa (compression)
Answer
σ2 = 50.02 MPa
9.Column buckling
Answer: Pcr = 111 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 = 2·1000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·225000/2000² = 111033 N = 111 kN
Answer
Pcr = 111 kN
10.Column buckling
Answer: σcr = 17.27 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·40⁴/64 = 125664 mm⁴
Radius of gyration and slenderness
A = 1257 mm²
r = √(I/A) = √(125664/1257) = 10 mm
KL = 2·1000 = 2000 mm
KL/r = 2000/10 = 200
Critical stress
σcr = π²E/(KL/r)² = π²·70000/200² = 17.27 MPa
Answer
σcr = 17.27 MPa
11.Mechanical vibrations
Answer: ζ = 0.12
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.
Logarithmic decrement
δ = (1/1)·ln(50/23.4) = 0.7593
Damping ratio
ζ = δ/√(4π² + δ²) = 0.7593/√(39.48 + 0.5765) = 0.12
Answer
ζ = 0.12
12.Mechanical vibrations
Answer: fn = 8.507 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 = 40 + 10 = 50 kN/m
keq = 50·20/(50 + 20) = 14.29 kN/m
Natural frequency
ωn = √(14286/5) = 53.45 rad/s
fn = 53.45/(2π) = 8.507 Hz
Answer
fn = 8.507 Hz
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