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 an 80 mm driving pulley at 1500 rpm. The tension is 800 N on the tight side and 400 N on the slack side. Find the power P transmitted, in kW.
d1 = 80 mmn1 = 1500 rpmT1 = 800 NT2 = 400 N
P = kW
2.Gears & gear trainsHard
In a compound gear train, gear A (15 teeth) on the input shaft turning at 1440 rpm drives gear B (40 teeth). Gear C (25 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 = 1440 rpmA = 15B = 40C = 25D = 32
nout = rpm
3.Torsion of shaftsMedium
A solid steel 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 = 200 N·md = 50 mmL = 3 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft transmits 100 kW at 1500 rpm. Find the maximum shear stress τmax in the shaft.
P = 100 kWN = 1500 rpmdo = 50 mmdi = 35 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −20 MPa, σᵧ = −60 MPa and τₓᵧ = 60 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = −20 MPaσy = −60 MPaτxy = 60 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = −30 MPa, σᵧ = −20 MPa and τₓᵧ = 60 MPa (σz = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.
σx = −30 MPaσy = −20 MPaτxy = 60 MPaσY = 460 MPa
n =
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 40 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 50 mm, length 300 mm, E = 110 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.25 mm in total.
segment 1: steel, Ø40 mm, 300 mm, E = 200 GPasegment 2: bronze, Ø50 mm, 300 mm, E = 110 GPaδ = 0.25 mm
P = kN
8.Axial stress & strainHard
A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of copper (outer diameter 100 mm, inner diameter 50 mm, E = 120 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø20 mm, E = 200 GPa, σallow = 150 MPatube: Ø100/Ø50 mm, E = 120 GPa, σallow = 70 MPa
Pmax = kN
9.Column bucklingMedium
A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 5 mE = 200 GPa (steel)d = 50 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is pinned at both ends. It must carry 65 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 65 kNn = 2L = 3 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 30 mm is followed one cycle later by a peak of 16 mm. Find the damping ratio ζ.
x0 = 30 mmx1 = 16 mmn = 1
ζ =
12.Mechanical vibrationsHard
A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 30 kN/m and k₂ = 60 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 25 kgk1 = 30 kN/mk2 = 60 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: P = 2.513 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 = 6.283 m/s
Power
P = (800 − 400)·6.283 = 2513 W = 2.513 kW
Answer
P = 2.513 kW
2.Gears & gear trains
Answer: nout = 421.9 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 = 1440·15/40 = 540 rpm = nC
Second stage
nout = 540·25/32 = 421.9 rpm
Answer
nout = 421.9 rpm
total ratio = 3.413 : 1
3.Torsion of shafts
Answer: φ = 0.7003 °
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) = 200·1000·3000/(80000·613592) = 0.01222 rad
φ = 0.01222·180/π = 0.7003°
Answer
φ = 0.7003°
4.Torsion of shafts
Answer: τmax = 34.13 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.
Torque
ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s
T = P/ω = 100000/157.1 = 636.6 N·m
Polar second moment of area
J = π(50⁴ − 35⁴)/32 = 466269 mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 636620·25/466269 = 34.13 MPa
Answer
τmax = 34.13 MPa
5.Mohr’s circle & stress transformation
Answer: τmax = 63.25 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 = √(((−20 − −60)/2)² + 60²) = 63.25 MPa
Answer
τmax = 63.25 MPa
6.Mohr’s circle & stress transformation
Answer: n = 4.29
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. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.
Principal stresses
C = −25 MPa, R = 60.21 MPa
σ1 = 35.21 MPa, σ2 = −85.21 MPa, σ3 = 0
Von Mises
σvM = √(35.21² − 35.21·(−85.21) + (−85.21)²) = 107.2 MPa
n = 460/107.2 = 4.29
Answer
n = 4.29
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: P = 96.8 kN
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
A₁ = 1257 mm², A₂ = 1963 mm²
Flexibilities
L₁/(E₁A₁) = 300/(200000·1257) = 1.194 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(110000·1963) = 1.389 × 10⁻⁶ mm/N
Load
P = 0.25/2.583 × 10⁻⁶ = 96800 N = 96.8 kN
Answer
P = 96.8 kN
8.Axial stress & strain
Answer: Pmax = 449 kN
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·20²/4 = 314.2 mm²
At = π(100² − 50²)/4 = 5890 mm²
Combined axial stiffness
ΣEA = 200·10³·314.2 + 120·10³·5890 = 7.697 × 10⁸ N
Governing strain
ε = min(0.00075, 0.0005833) = 0.0005833 (copper)
Largest load
Pmax = 0.0005833·7.697 × 10⁸ = 448986 N = 449 kN
Answer
Pmax = 449 kN
9.Column buckling
Answer: Pcr = 96.89 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·50⁴/64 = 306796 mm⁴
Effective length
KL = 0.5·5000 = 2500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·306796/2500² = 96895 N = 96.89 kN
Answer
Pcr = 96.89 kN
10.Column buckling
Answer: d = 76.64 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 = 2·65000 = 130000 N
KL = 1·3000 = 3000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·130000·3000²/(π³·70000))^(1/4) = 76.64 mm
Answer
d = 76.64 mm
Check: the slenderness KL/r is 156.6, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.09955
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(30/16) = 0.6286
Damping ratio
ζ = δ/√(4π² + δ²) = 0.6286/√(39.48 + 0.3951) = 0.09955
Answer
ζ = 0.09955
12.Mechanical vibrations
Answer: keq = 90 kN/m
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 = 30 + 60 = 90 kN/m
Answer
keq = 90 kN/m
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