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 belt runs on a 160 mm driving pulley at 1500 rpm. The tension is 1500 N on the tight side and 375 N on the slack side. Find the power P transmitted, in kW.
d1 = 160 mmn1 = 1500 rpmT1 = 1500 NT2 = 375 N
P = kW
2.Gears & gear trainsHard
In a compound gear train, gear A (16 teeth) on the input shaft turning at 1500 rpm drives gear B (72 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D (32 teeth) on the output shaft. Find the output speed nout.
nin = 1500 rpmA = 16B = 72C = 16D = 32
nout = rpm
3.Torsion of shaftsMedium
A solid brass 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 = 100 N·md = 25 mmL = 0.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid aluminium shaft of diameter 40 mm is fixed at A. A torque of 750 N·m acts at B, 0.8 m from A, and a torque of 1500 N·m in the same direction acts at the free end C, a further 1 m along. Find the maximum shear stress τmax in the shaft.
d = 40 mmTB = 750 N·mTC = 1500 N·mAB = 0.8 mBC = 1 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −20 MPa and σᵧ = −50 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 35 MPa?
σx = −20 MPaσy = −50 MPaσ1 = 35 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 70 MPa, σᵧ = −20 MPa and τₓᵧ = 30 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 35° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 70 MPaσy = −20 MPaτxy = 30 MPaθ = 35°
τθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the aluminium bar under the axial load.
P = 80 kNd = 25 mmL = 1 mE = 70 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 = 80 kNsegment 1: Ø40 mm, 1 m, copper (E = 117 GPa)segment 2: Ø16 mm, 1.5 m, steel (E = 200 GPa)
σmax = 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 = 4 mE = 200 GPa (steel)b × h = 20 × 60 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 4 mE = 200 GPa (steel)d = 60 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 5 kg mass is supported by a spring with a stiffness of 10 kN/m and a viscous damper with a damping coefficient of 130 N·s/m. Find the damping ratio ζ.
m = 5 kgk = 10 kN/mc = 130 N·s/m
ζ =
12.Mechanical vibrationsHard
A 100 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 60 kN/m and the lower one k₂ = 20 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 100 kgk1 = 60 kN/mk2 = 20 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: P = 14.14 kW
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Belt speed
v = 12.57 m/s
Power
P = (1500 − 375)·12.57 = 14137 W = 14.14 kW
Answer
P = 14.14 kW
2.Gears & gear trains
Answer: nout = 166.7 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 = 1500·16/72 = 333.3 rpm = nC
Second stage
nout = 333.3·16/32 = 166.7 rpm
Answer
nout = 166.7 rpm
total ratio = 9 : 1
3.Torsion of shafts
Answer: φ = 1.915 °
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·25⁴/32 = 38350 mm⁴
Angle of twist
φ = TL/(GJ) = 100·1000·500/(39000·38350) = 0.03343 rad
φ = 0.03343·180/π = 1.915°
Answer
φ = 1.915°
4.Torsion of shafts
Answer: τmax = 179 MPa
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 = 1500 N·m
TAB = TC + TB = 1500 + 750 = 2250 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 2250 N·m
τmax = 16·2.25 × 10⁶/(π·40³) = 179 MPa
Answer
τmax = 179 MPa
5.Mohr’s circle & stress transformation
Answer: τxy = 68.37 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 = (−20 + (−50))/2 = −35 MPa
R = σ1 − C = 35 − (−35) = 70 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(70² − 15²) = 68.37 MPa
Answer
τxy = 68.37 MPa
6.Mohr’s circle & stress transformation
Answer: τθ = −32.03 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 = −45·0.9397 + 30·0.342 = −32.03 MPa
Answer
τθ = −32.03 MPa
7.Axial stress & strain
Answer: δ = 2.328 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·25²/4 = 490.9 mm²
σ = P/A = 80·1000/490.9 = 163 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 163/70000 = 0.002328 = 2328 µε
Elongation
δ = ε·L = 0.002328·1000 mm = 2.328 mm
Answer
δ = 2.328 mm
8.Axial stress & strain
Answer: σmax = 397.9 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·40²/4 = 1257 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 63.66 MPa
σ2 = 397.9 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 397.9 MPa
Answer
σmax = 397.9 MPa
9.Column buckling
Answer: Pcr = 4.935 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 = 60·20³/12 = 40000 mm⁴
Effective length
KL = 1·4000 = 4000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·40000/4000² = 4935 N = 4.935 kN
Answer
Pcr = 4.935 kN
10.Column buckling
Answer: σcr = 111 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·60⁴/64 = 636173 mm⁴
Radius of gyration and slenderness
A = 2827 mm²
r = √(I/A) = √(636173/2827) = 15 mm
KL = 0.5·4000 = 2000 mm
KL/r = 2000/15 = 133.3
Critical stress
σcr = π²E/(KL/r)² = π²·200000/133.3² = 111 MPa
Answer
σcr = 111 MPa
11.Mechanical vibrations
Answer: ζ = 0.2907
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·√(10000·5) = 447.2 N·s/m
Damping ratio
ζ = c/cc = 130/447.2 = 0.2907
Answer
ζ = 0.2907
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 1.949 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 = 1/60 + 1/20 → keq = 60·20/(60 + 20) = 15 kN/m
Natural frequency
ωn = √(15000/100) = 12.25 rad/s
fn = 12.25/(2π) = 1.949 Hz
Answer
fn = 1.949 Hz
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