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 1.1 kW motor running at 1500 rpm drives a pinion with 20 teeth, which meshes with a gear with 72 teeth. The gear pair has an efficiency of 98%. Find the torque T1 on the motor shaft.
P = 1.1 kWn1 = 1500 rpmN1 = 20N2 = 72η = 98%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (18 teeth) on the input shaft turns at 1440 rpm and drives gear B (45 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 204.8 rpm. How many teeth ND must gear D have?
nin = 1440 rpmA = 18B = 45C = 16nout = 204.8 rpm
ND =
3.Torsion of shaftsMedium
A solid aluminium shaft 1 m long carries a torque of 200 N·m. The angle of twist between its ends must not exceed 1.5°. Find the smallest diameter d that meets this limit.
T = 200 N·mL = 1 mG = 26 GPaφallow = 1.5°
d = mm
4.Torsion of shaftsHard
A hollow shaft must transmit 50 kW at 1450 rpm. Its inner diameter is to be 0.5 times its outer diameter, and the shear stress must not exceed 60 MPa. Find the smallest outer diameter do.
P = 50 kWN = 1450 rpmdi/do = 0.5τallow = 60 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −10 MPa and σᵧ = −20 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 40 MPa?
σx = −10 MPaσy = −20 MPaσ1 = 40 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −10 MPa, σᵧ = 40 MPa and τₓᵧ = 60 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 25° counter-clockwise from the x-axis.
σx = −10 MPaσy = 40 MPaτxy = 60 MPaθ = 25°
σθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).
P = 55 kNd = 10 mmL = 0.5 mE = 100 GPa
ε = µε
8.Axial stress & strainHard
A steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) and a segment made of bronze (diameter 40 mm, length 500 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, Ø40 mm, 400 mm, E = 200 GPasegment 2: bronze, Ø40 mm, 500 mm, E = 110 GPaP = 80 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 5 mE = 200 GPa (steel)b × h = 50 × 80 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 = 3 mE = 70 GPa (aluminium)d = 80 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 12 kg mass on a spring with a stiffness of 25 kN/m has a viscous damper giving a damping ratio of 0.5. Find the damped natural frequency fd of its free vibration in Hz.
m = 12 kgk = 25 kN/mζ = 0.5
fd = Hz
12.Mechanical vibrationsHard
A 5 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 60 kN/m and k₂ = 12 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 = 5 kgk1 = 60 kN/mk2 = 12 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 7.003 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 = 1100/157.1 = 7.003 N·m
Answer
T1 = 7.003 N·m
2.Gears & gear trains
Answer: ND = 45
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.
Intermediate shaft speed
nB = nC = 1440·18/45 = 576 rpm
Teeth on gear D
ND = C·nC/nout = 16·576/204.8 = 45
Answer
ND = 45
3.Torsion of shafts
Answer: d = 41.59 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 = 200 N·m = 200000 N·mm
L = 1 m = 1000 mm
G = 26 GPa = 26000 MPa
φ = 1.5·π/180 = 0.02618 rad
J = TL/(Gφ) = 200000·1000/(26000·0.02618) = 293825 mm⁴
Diameter
d = (32J/π)1/4 = 41.59 mm
Answer
d = 41.59 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: do = 31.01 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·1450/60 = 151.8 rad/s
T = P/ω = 50000/151.8 = 329.3 N·m
Outer diameter from the allowable stress
T = 329.3 N·m = 329286 N·mm
1 − k⁴ = 1 − 0.5⁴ = 0.9375
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·329286/(π·60·0.9375)) = 31.01 mm
di = 0.5·31.01 = 15.5 mm
Answer
do = 31.01 mm
A solid shaft would need d = 30.35 mm; the hollow one is a little larger but much lighter.
5.Mohr’s circle & stress transformation
Answer: τxy = 54.77 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 + (−20))/2 = −15 MPa
R = σ1 − C = 40 − (−15) = 55 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(55² − 5²) = 54.77 MPa
Answer
τxy = 54.77 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 44.89 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 = 15 + −25·0.6428 + 60·0.766 = 44.89 MPa
Answer
σθ = 44.89 MPa
7.Axial stress & strain
Answer: ε = 7003 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 = 55·1000/78.54 = 700.3 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 700.3/100000 = 0.007003 = 7003 µε
Answer
ε = 7003 µε
8.Axial stress & strain
Answer: σ2 = 19.45 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 = 200000·1257/400 = 628319 N/mm
k₂ = 110000·1257/500 = 276460 N/mm
Share of segment 2
F2 = 80000·276460/904779 = 24444 N
Stress
σ2 = 24444/1257 = 19.45 MPa (compression)
Answer
σ2 = 19.45 MPa
9.Column buckling
Answer: Pcr = 65.8 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 = 80·50³/12 = 833333 mm⁴
Effective length
KL = 1·5000 = 5000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·833333/5000² = 65797 N = 65.8 kN
Answer
Pcr = 65.8 kN
10.Column buckling
Answer: σcr = 7.676 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·80⁴/64 = 2.011 × 10⁶ mm⁴
Radius of gyration and slenderness
A = 5027 mm²
r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm
KL = 2·3000 = 6000 mm
KL/r = 6000/20 = 300
Critical stress
σcr = π²E/(KL/r)² = π²·70000/300² = 7.676 MPa
Answer
σcr = 7.676 MPa
11.Mechanical vibrations
Answer: fd = 6.291 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 = √(25000/12) = 45.64 rad/s
fn = 45.64/(2π) = 7.264 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 7.264·√(1 − 0.5²) = 6.291 Hz
Answer
fd = 6.291 Hz
12.Mechanical vibrations
Answer: fn = 19.1 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 = 60 + 12 = 72 kN/m
Natural frequency
ωn = √(72000/5) = 120 rad/s
fn = 120/(2π) = 19.1 Hz
Answer
fn = 19.1 Hz
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