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 200 mm driving pulley at 1440 rpm. The tension is 600 N on the tight side and 300 N on the slack side. Find the power P transmitted, in kW.
d1 = 200 mmn1 = 1440 rpmT1 = 600 NT2 = 300 N
P = kW
2.Gears & gear trainsHard
In a planetary gear set the sun gear has 15 teeth and each planet gear has 18 teeth. The ring gear is held fixed and the sun is driven at 900 rpm. Find the speed nc of the planet carrier.
Ns = 15Np = 18ns = 900 rpm
nc = 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 = 250 N·md = 40 mmL = 3 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 20 kW at 3000 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.
P = 20 kWN = 3000 rpmτallow = 50 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −30 MPa, σᵧ = 120 MPa and τₓᵧ = 50 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).
σx = −30 MPaσy = 120 MPaτxy = 50 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 160 MPa, σᵧ = 70 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 = 160 MPaσy = 70 MPaτxy = 30 MPaθ = 35°
τθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the steel bar, in microstrain (µε = 10⁻⁶).
P = 85 kNd = 16 mmL = 1 mE = 200 GPa
ε = µε
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 25 kNsegment 1: Ø30 mm, 2 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1.5 m, aluminium (E = 70 GPa)
δ = mm
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 = 3 mE = 200 GPa (steel)b × h = 30 × 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 = 25 mm
σcr = MPa
11.Mechanical vibrationsMedium
An 80 kg mass on a spring with a stiffness of 15 kN/m has a viscous damper giving a damping ratio of 0.6. Find the damped natural frequency fd of its free vibration in Hz.
m = 80 kgk = 15 kN/mζ = 0.6
fd = Hz
12.Mechanical vibrationsHard
A 40 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 = 100 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 40 kgk1 = 60 kN/mk2 = 100 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: P = 4.524 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 = 15.08 m/s
Power
P = (600 − 300)·15.08 = 4524 W = 4.524 kW
Answer
P = 4.524 kW
2.Gears & gear trains
Answer: nc = 204.5 rpm
Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (ns − nc)/(nr − nc) = −Nr/Ns: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.
Ring gear teeth
Nr = Ns + 2·Np = 15 + 2·18 = 51
Willis equation with the ring fixed
Set nr = 0 and solve for the carrier speed.
(900 − nc)/(0 − nc) = −51/15
nc = ns·Ns/(Ns + Nr) = 900·15/66 = 204.5 rpm
Answer
nc = 204.5 rpm
The carrier turns in the same direction as the sun.
3.Torsion of shafts
Answer: φ = 2.137 °
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) = 250·1000·3000/(80000·251327) = 0.0373 rad
φ = 0.0373·180/π = 2.137°
Answer
φ = 2.137°
4.Torsion of shafts
Answer: d = 18.65 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·3000/60 = 314.2 rad/s
T = P/ω = 20000/314.2 = 63.66 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·63662/(π·50)) = 18.65 mm
Answer
d = 18.65 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θs = 28.15 °
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.
Angle on the circle
On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.
cos 2θs = 50/90.14 = 0.5547, sin 2θs = −(−75)/90.14 = 0.8321
2θs = atan2(75, 50) = 56.31°
Angle on the element
θs = 28.15°
check: θs − θp = 28.15 − 73.15 = −45°
Answer
θs = 28.15°
Positive angles are counter-clockwise.
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: ε = 2114 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·16²/4 = 201.1 mm²
σ = P/A = 85·1000/201.1 = 422.8 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 422.8/200000 = 0.002114 = 2114 µε
Answer
ε = 2114 µε
8.Axial stress & strain
Answer: δ = 3.372 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·30²/4 = 706.9 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 35.37 MPa
σ2 = 124.3 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 35.37·2000/100000 = 0.7074 mm
δ2 = σ2L2/E2 = 124.3·1500/70000 = 2.664 mm
δ = δ1 + δ2 = 3.372 mm
Answer
δ = 3.372 mm
9.Column buckling
Answer: Pcr = 29.61 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 = 60·30³/12 = 135000 mm⁴
Effective length
KL = 1·3000 = 3000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·135000/3000² = 29609 N = 29.61 kN
Answer
Pcr = 29.61 kN
10.Column buckling
Answer: σcr = 19.28 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·25⁴/64 = 19175 mm⁴
Radius of gyration and slenderness
A = 490.9 mm²
r = √(I/A) = √(19175/490.9) = 6.25 mm
KL = 0.5·4000 = 2000 mm
KL/r = 2000/6.25 = 320
Critical stress
σcr = π²E/(KL/r)² = π²·200000/320² = 19.28 MPa
Answer
σcr = 19.28 MPa
11.Mechanical vibrations
Answer: fd = 1.743 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 = √(15000/80) = 13.69 rad/s
fn = 13.69/(2π) = 2.179 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 2.179·√(1 − 0.6²) = 1.743 Hz
Answer
fd = 1.743 Hz
12.Mechanical vibrations
Answer: keq = 37.5 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 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 + 1/100 → keq = 60·100/(60 + 100) = 37.5 kN/m
Answer
keq = 37.5 kN/m
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