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 0.75 kW motor running at 2900 rpm drives a pinion with 18 teeth, which meshes with a gear with 25 teeth. The gear pair has an efficiency of 96%. Find the torque T1 on the motor shaft.
P = 0.75 kWn1 = 2900 rpmN1 = 18N2 = 25η = 96%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (18 teeth) on the input shaft turning at 1440 rpm drives gear B (20 teeth). Gear C (30 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 = 1440 rpmA = 18B = 20C = 30D = 72
nout = rpm
3.Torsion of shaftsMedium
A solid aluminium shaft 1 m long carries a torque of 250 N·m. The angle of twist between its ends must not exceed 2°. Find the smallest diameter d that meets this limit.
T = 250 N·mL = 1 mG = 26 GPaφallow = 2°
d = mm
4.Torsion of shaftsHard
A hollow shaft must transmit 10 kW at 500 rpm. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter do.
P = 10 kWN = 500 rpmdi/do = 0.6τallow = 40 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 100 MPa, σᵧ = 120 MPa and τₓᵧ = −40 MPa (tension positive). The element is rotated 20° clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.
σx = 100 MPaσy = 120 MPaτxy = −40 MPaθ = -20°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 110 MPa, σᵧ = 80 MPa and τₓᵧ = −20 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 60° counter-clockwise from the x-axis.
σx = 110 MPaσy = 80 MPaτxy = −20 MPaθ = 60°
σθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the steel bar under the axial load.
P = 90 kNd = 12 mmL = 3 mE = 200 GPa
δ = mm
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 = 25 kNsegment 1: Ø35 mm, 0.5 m, copper (E = 117 GPa)segment 2: Ø20 mm, 1.5 m, steel (E = 200 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 6 mE = 70 GPa (aluminium)d = 40 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is pinned at both ends. It must carry 150 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 150 kNn = 3L = 2 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 5 kg mass is supported by a spring with a stiffness of 4 kN/m and a viscous damper with a damping coefficient of 140 N·s/m. Find the damping ratio ζ.
m = 5 kgk = 4 kN/mc = 140 N·s/m
ζ =
12.Mechanical vibrationsHard
A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 5 kN/m and k₂ = 10 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 25 kgk1 = 5 kN/mk2 = 10 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 2.47 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 = 750/303.7 = 2.47 N·m
Answer
T1 = 2.47 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 = 1440·18/20 = 1296 rpm = nC
Second stage
nout = 1296·30/72 = 540 rpm
Answer
nout = 540 rpm
total ratio = 2.667 : 1
3.Torsion of shafts
Answer: d = 40.93 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.
Required polar moment
T = 250 N·m = 250000 N·mm
L = 1 m = 1000 mm
G = 26 GPa = 26000 MPa
φ = 2·π/180 = 0.03491 rad
J = TL/(Gφ) = 250000·1000/(26000·0.03491) = 275460 mm⁴
Diameter
d = (32J/π)1/4 = 40.93 mm
Answer
d = 40.93 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: do = 30.34 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·500/60 = 52.36 rad/s
T = P/ω = 10000/52.36 = 191 N·m
Outer diameter from the allowable stress
T = 191 N·m = 190986 N·mm
1 − k⁴ = 1 − 0.6⁴ = 0.8704
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·190986/(π·40·0.8704)) = 30.34 mm
di = 0.6·30.34 = 18.21 mm
Answer
do = 30.34 mm
A solid shaft would need d = 28.97 mm; the hollow one is a little larger but much lighter.
5.Mohr’s circle & stress transformation
Answer: σy′ = 91.95 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′ = 110 − (−10)·0.766 − (−40)·(−0.6428) = 91.95 MPa
Check
The sum of the normal stresses does not change with rotation.
σx′ = 128.1 MPa, σx′ + σy′ = 220 = σx + σy
Answer
σy′ = 91.95 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 70.18 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 = 95 + 15·−0.5 + −20·0.866 = 70.18 MPa
Answer
σθ = 70.18 MPa
7.Axial stress & strain
Answer: δ = 11.94 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 = 90·1000/113.1 = 795.8 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 795.8/200000 = 0.003979 = 3979 µε
Elongation
δ = ε·L = 0.003979·3000 mm = 11.94 mm
Answer
δ = 11.94 mm
8.Axial stress & strain
Answer: σmax = 79.58 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 = 25.98 MPa
σ2 = 79.58 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 79.58 MPa
Answer
σmax = 79.58 MPa
9.Column buckling
Answer: Pcr = 2.412 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·40⁴/64 = 125664 mm⁴
Effective length
KL = 1·6000 = 6000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·125664/6000² = 2412 N = 2.412 kN
Answer
Pcr = 2.412 kN
10.Column buckling
Answer: d = 85.35 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 = 3·150000 = 450000 N
KL = 1·2000 = 2000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·450000·2000²/(π³·70000))^(1/4) = 85.35 mm
Answer
d = 85.35 mm
Check: the slenderness KL/r is 93.73, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.495
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·√(4000·5) = 282.8 N·s/m
Damping ratio
ζ = c/cc = 140/282.8 = 0.495
Answer
ζ = 0.495
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 3.898 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 5 + 10 = 15 kN/m
Natural frequency
ωn = √(15000/25) = 24.49 rad/s
fn = 24.49/(2π) = 3.898 Hz
Answer
fn = 3.898 Hz
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