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 720 rpm. The tension is 600 N on the tight side and 240 N on the slack side. Find the power P transmitted, in kW.
d1 = 80 mmn1 = 720 rpmT1 = 600 NT2 = 240 N
P = kW
2.Gears & gear trainsHard
In a compound gear train, gear A (50 teeth) on the input shaft turning at 1450 rpm drives gear B (60 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D (50 teeth) on the output shaft. Find the output speed nout.
nin = 1450 rpmA = 50B = 60C = 15D = 50
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 = 750 N·md = 40 mmL = 1.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 75 kW at 1450 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.
P = 75 kWN = 1450 rpmτallow = 80 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 30 MPa, σᵧ = −20 MPa and τₓᵧ = −50 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 30 MPaσy = −20 MPaτxy = −50 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = 140 MPa, σᵧ = 120 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 = 140 MPaσy = 120 MPaτxy = −60 MPaσY = 460 MPa
n =
7.Axial stress & strainMedium
Find the elongation δ of the steel bar under the axial load.
P = 15 kNd = 20 mmL = 0.5 mE = 200 GPa
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 20 mm, length 500 mm, E = 200 GPa) and a segment made of copper (diameter 40 mm, length 300 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 30 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø20 mm, 500 mm, E = 200 GPasegment 2: copper, Ø40 mm, 300 mm, E = 120 GPaP = 30 kN
σ2 = MPa
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 = 4 mE = 200 GPa (steel)b × h = 25 × 100 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 15 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 15 kNn = 3L = 6 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 12 mm is followed 3 cycles later by a peak of 5.6 mm. Find the damping ratio ζ.
x0 = 12 mmx3 = 5.6 mmn = 3
ζ =
12.Mechanical vibrationsHard
A 40 kg mass hangs from spring k₃ = 15 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 20 kN/m and k₂ = 20 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 = 40 kgk1 = 20 kN/mk2 = 20 kN/mk3 = 15 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: P = 1.086 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 = 3.016 m/s
Power
P = (600 − 240)·3.016 = 1086 W = 1.086 kW
Answer
P = 1.086 kW
2.Gears & gear trains
Answer: nout = 362.5 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 = 1450·50/60 = 1208 rpm = nC
Second stage
nout = 1208·15/50 = 362.5 rpm
Answer
nout = 362.5 rpm
total ratio = 4 : 1
3.Torsion of shafts
Answer: φ = 3.206 °
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·40⁴/32 = 251327 mm⁴
Angle of twist
φ = TL/(GJ) = 750·1000·1500/(80000·251327) = 0.05595 rad
φ = 0.05595·180/π = 3.206°
Answer
φ = 3.206°
4.Torsion of shafts
Answer: d = 31.56 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 = 75000/151.8 = 493.9 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·493929/(π·80)) = 31.56 mm
Answer
d = 31.56 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τmax = 55.9 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 = √(((30 − −20)/2)² + −50²) = 55.9 MPa
Answer
τmax = 55.9 MPa
6.Mohr’s circle & stress transformation
Answer: n = 2.749
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 = 130 MPa, R = 60.83 MPa
σ1 = 190.8 MPa, σ2 = 69.17 MPa, σ3 = 0
Von Mises
σvM = √(190.8² − 190.8·69.17 + 69.17²) = 167.3 MPa
n = 460/167.3 = 2.749
Answer
n = 2.749
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: δ = 0.1194 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·20²/4 = 314.2 mm²
σ = P/A = 15·1000/314.2 = 47.75 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 47.75/200000 = 0.0002387 = 238.7 µε
Elongation
δ = ε·L = 0.0002387·500 mm = 0.1194 mm
Answer
δ = 0.1194 mm
8.Axial stress & strain
Answer: σ2 = 19.1 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 = 200000·314.2/500 = 125664 N/mm
k₂ = 120000·1257/300 = 502655 N/mm
Share of segment 2
F2 = 30000·502655/628319 = 24000 N
Stress
σ2 = 24000/1257 = 19.1 MPa (compression)
Answer
σ2 = 19.1 MPa
9.Column buckling
Answer: Pcr = 4.016 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 = 100·25³/12 = 130208 mm⁴
Effective length
KL = 2·4000 = 8000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·130208/8000² = 4016 N = 4.016 kN
Answer
Pcr = 4.016 kN
10.Column buckling
Answer: d = 69.56 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 = 3·15000 = 45000 N
KL = 0.7·6000 = 4200 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·45000·4200²/(π³·70000))^(1/4) = 69.56 mm
Answer
d = 69.56 mm
Check: the slenderness KL/r is 241.5, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.0404
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 = (1/3)·ln(12/5.6) = 0.254
Damping ratio
ζ = δ/√(4π² + δ²) = 0.254/√(39.48 + 0.06454) = 0.0404
Answer
ζ = 0.0404
12.Mechanical vibrations
Answer: fn = 2.628 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 = 20 + 20 = 40 kN/m
keq = 40·15/(40 + 15) = 10.91 kN/m
Natural frequency
ωn = √(10909/40) = 16.51 rad/s
fn = 16.51/(2π) = 2.628 Hz
Answer
fn = 2.628 Hz
ExamProblems · examproblems.com/courses/machine-design