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 2.2 kW motor running at 960 rpm drives a pinion with 48 teeth, which meshes with a gear with 50 teeth. The gear pair has an efficiency of 98%. Find the output torque T2 on the gear shaft.
P = 2.2 kWn1 = 960 rpmN1 = 48N2 = 50η = 98%
T2 = N·m
2.Gears & gear trainsHard
A 3 kW motor at 2900 rpm drives gear A (15 teeth), which meshes with gear B (80 teeth). Gear C (15 teeth) on the same shaft as B drives gear D (24 teeth) on the output shaft. Each gear mesh has an efficiency of 96%. Find the output torque Tout.
P = 3 kWnin = 2900 rpmA = 15B = 80C = 15D = 24ηmesh = 96%
Tout = N·m
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 = 250 N·md = 50 mmL = 1.2 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A solid steel shaft of diameter 60 mm is fixed at A. A torque of 750 N·m acts at B, 3 m from A, and a torque of 750 N·m in the same direction acts at the free end C, a further 1.2 m along. Find the magnitude of the angle of twist φC of end C, in degrees.
d = 60 mmTB = 750 N·mTC = 750 N·mAB = 3 mBC = 1.2 mG = 80 GPa
φC = °
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −70 MPa, σᵧ = 70 MPa and τₓᵧ = −40 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = −70 MPaσy = 70 MPaτxy = −40 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −20 MPa, σᵧ = 60 MPa and τₓᵧ = −50 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 35° counter-clockwise from the x-axis.
σx = −20 MPaσy = 60 MPaτxy = −50 MPaθ = 35°
σθ = MPa
7.Axial stress & strainMedium
A steel rod (diameter 25 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 50 mm, E = 110 GPa). Both are 800 mm long and are compressed together between rigid plates by 100 kN. Find the shortening δ.
P = 100 kNrod: Ø25 mm, E = 200 GPatube: Ø100/Ø50 mm, E = 110 GPaL = 800 mm
δ = mm
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 30 kNsegment 1: Ø25 mm, 0.5 m, steel (E = 200 GPa)segment 2: Ø20 mm, 0.5 m, brass (E = 100 GPa)
δ = mm
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 = 1.5 mE = 200 GPa (steel)b × h = 30 × 80 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 4 mE = 200 GPa (steel)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 2 kg mass is supported by a spring with a stiffness of 25 kN/m and a viscous damper with a damping coefficient of 270 N·s/m. Find the damping ratio ζ.
m = 2 kgk = 25 kN/mc = 270 N·s/m
ζ =
12.Mechanical vibrationsHard
A 20 kg mass hangs from spring k₃ = 25 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 = 40 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 = 20 kgk1 = 40 kN/mk2 = 40 kN/mk3 = 25 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 22.34 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 = 2200/100.5 = 21.88 N·m
Output torque
T2 = 21.88·50/48·0.98 = 22.34 N·m
Answer
T2 = 22.34 N·m
2.Gears & gear trains
Answer: Tout = 77.69 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.
Output speed
nout = 2900·(15/80)·(15/24) = 339.8 rpm
ωout = 2π·339.8/60 = 35.59 rad/s
Output power and torque
Pout = 3000·0.96² = 2765 W
Tout = Pout/ωout = 2765/35.59 = 77.69 N·m
Answer
Tout = 77.69 N·m
3.Torsion of shafts
Answer: φ = 1.077 °
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·50⁴/32 = 613592 mm⁴
Angle of twist
φ = TL/(GJ) = 250·1000·1200/(26000·613592) = 0.0188 rad
φ = 0.0188·180/π = 1.077°
Answer
φ = 1.077°
4.Torsion of shafts
Answer: φC = 3.04 °
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.
Internal torques
Sense of TC taken as positive.
TBC = TC = 750 N·m
TAB = TC + TB = 750 + 750 = 1500 N·m
Twist of each segment
J = π·60⁴/32 = 1.272 × 10⁶ mm⁴
G = 80 GPa = 80000 MPa
φAB = 1.5 × 10⁶·3000/(80000·1.272 × 10⁶) = 0.04421 rad
φBC = 750000·1200/(80000·1.272 × 10⁶) = 0.008842 rad
Twist of end C
φC = 0.04421 + 0.008842 = 0.05305 rad → |φC| = 3.04°
Answer
φC = 3.04°
5.Mohr’s circle & stress transformation
Answer: τmax = 80.62 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 = √(((−70 − 70)/2)² + −40²) = 80.62 MPa
Answer
τmax = 80.62 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = −40.67 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 = 20 + −40·0.342 + −50·0.9397 = −40.67 MPa
Answer
σθ = −40.67 MPa
7.Axial stress & strain
Answer: δ = 0.1072 mm
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
Ar = π·25²/4 = 490.9 mm²
At = π(100² − 50²)/4 = 5890 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 110·10³·5890 = 7.461 × 10⁸ N
Shortening
δ = 100000·800/7.461 × 10⁸ = 0.1072 mm
Answer
δ = 0.1072 mm
8.Axial stress & strain
Answer: δ = 0.6303 mm
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·25²/4 = 490.9 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 61.12 MPa
σ2 = 95.49 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 61.12·500/200000 = 0.1528 mm
δ2 = σ2L2/E2 = 95.49·500/100000 = 0.4775 mm
δ = δ1 + δ2 = 0.6303 mm
Answer
δ = 0.6303 mm
9.Column buckling
Answer: Pcr = 39.48 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 = 80·30³/12 = 180000 mm⁴
Effective length
KL = 2·1500 = 3000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·180000/3000² = 39478 N = 39.48 kN
Answer
Pcr = 39.48 kN
10.Column buckling
Answer: σcr = 3.084 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 = 2·4000 = 8000 mm
KL/r = 8000/10 = 800
Critical stress
σcr = π²E/(KL/r)² = π²·200000/800² = 3.084 MPa
Answer
σcr = 3.084 MPa
11.Mechanical vibrations
Answer: ζ = 0.6037
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·√(25000·2) = 447.2 N·s/m
Damping ratio
ζ = c/cc = 270/447.2 = 0.6037
Answer
ζ = 0.6037
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 4.912 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 + 40 = 80 kN/m
keq = 80·25/(80 + 25) = 19.05 kN/m
Natural frequency
ωn = √(19048/20) = 30.86 rad/s
fn = 30.86/(2π) = 4.912 Hz
Answer
fn = 4.912 Hz
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