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 15 kW motor running at 1500 rpm drives a pinion with 20 teeth, which meshes with a gear with 45 teeth. The gear pair has an efficiency of 96%. Find the torque T1 on the motor shaft.
P = 15 kWn1 = 1500 rpmN1 = 20N2 = 45η = 96%
T1 = N·m
2.Gears & gear trainsHard
A 1.1 kW motor at 1450 rpm drives gear A (18 teeth), which meshes with gear B (60 teeth). Gear C (18 teeth) on the same shaft as B drives gear D (80 teeth) on the output shaft. Each gear mesh has an efficiency of 96%. Find the output torque Tout.
P = 1.1 kWnin = 1450 rpmA = 18B = 60C = 18D = 80ηmesh = 96%
Tout = N·m
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 = 500 N·md = 50 mmL = 3 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft transmits 100 kW at 750 rpm. Find the maximum shear stress τmax in the shaft.
P = 100 kWN = 750 rpmdo = 100 mmdi = 80 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −70 MPa, σᵧ = 60 MPa and τₓᵧ = −10 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = −70 MPaσy = 60 MPaτxy = −10 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A thin-walled cylindrical tank of inner diameter 1500 mm and wall thickness 15 mm holds gas at a gauge pressure of 1.2 MPa. It is made from plate welded along a helical seam at 55° to the axis of the cylinder. Find the magnitude of the shear stress |τn| acting along the weld seam.
D = 1500 mmt = 15 mmp = 1.2 MPaα = 55°
|τn| = MPa
7.Axial stress & strainMedium
A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 60 mm, E = 70 GPa). Both are 800 mm long and are compressed together between rigid plates by 300 kN. Find the shortening δ.
P = 300 kNrod: Ø40 mm, E = 200 GPatube: Ø100/Ø60 mm, E = 70 GPaL = 800 mm
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 30 mm, length 300 mm, E = 200 GPa) and a segment made of copper (diameter 30 mm, length 200 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 60 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø30 mm, 300 mm, E = 200 GPasegment 2: copper, Ø30 mm, 200 mm, E = 120 GPaP = 60 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 2.5 mE = 200 GPa (steel)d = 50 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at both ends. It must carry 100 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 100 kNn = 2L = 3 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 20 kg mass on a spring with a stiffness of 8 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 = 20 kgk = 8 kN/mζ = 0.6
fd = Hz
12.Mechanical vibrationsHard
A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 12 kN/m and k₂ = 40 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 25 kgk1 = 12 kN/mk2 = 40 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 95.49 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·1500/60 = 157.1 rad/s
T1 = P/ω₁ = 15000/157.1 = 95.49 N·m
Answer
T1 = 95.49 N·m
2.Gears & gear trains
Answer: Tout = 98.91 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 = 1450·(18/60)·(18/80) = 97.88 rpm
ωout = 2π·97.88/60 = 10.25 rad/s
Output power and torque
Pout = 1100·0.96² = 1014 W
Tout = Pout/ωout = 1014/10.25 = 98.91 N·m
Answer
Tout = 98.91 N·m
3.Torsion of shafts
Answer: φ = 1.751 °
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) = 500·1000·3000/(80000·613592) = 0.03056 rad
φ = 0.03056·180/π = 1.751°
Answer
φ = 1.751°
4.Torsion of shafts
Answer: τmax = 10.98 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·750/60 = 78.54 rad/s
T = P/ω = 100000/78.54 = 1273 N·m
Polar second moment of area
J = π(100⁴ − 80⁴)/32 = 5.796 × 10⁶ mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 1.273 × 10⁶·50/5.796 × 10⁶ = 10.98 MPa
Answer
τmax = 10.98 MPa
5.Mohr’s circle & stress transformation
Answer: τmax = 65.76 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 − 60)/2)² + −10²) = 65.76 MPa
Answer
τmax = 65.76 MPa
6.Mohr’s circle & stress transformation
Answer: |τn| = 14.1 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. Thin-walled cylinder: hoop stress σh = pr/t and axial stress σa = pr/(2t); these are the principal stresses in the wall.
Wall stresses
r = 1500/2 = 750 mm
σh = pr/t = 1.2·750/15 = 60 MPa
σa = pr/(2t) = 30 MPa
Transform to the seam
Take x along the axis (σa) and y around the hoop (σh). The seam’s normal makes 90° − α with the axis.
|τn| = (60 − 30)·sin 55°·cos 55° = 14.1 MPa
Answer
|τn| = 14.1 MPa
7.Axial stress & strain
Answer: δ = 0.3979 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·40²/4 = 1257 mm²
At = π(100² − 60²)/4 = 5027 mm²
Combined axial stiffness
ΣEA = 200·10³·1257 + 70·10³·5027 = 6.032 × 10⁸ N
Shortening
δ = 300000·800/6.032 × 10⁸ = 0.3979 mm
Answer
δ = 0.3979 mm
8.Axial stress & strain
Answer: σ2 = 40.21 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 = 200000·706.9/300 = 471239 N/mm
k₂ = 120000·706.9/200 = 424115 N/mm
Share of segment 2
F2 = 60000·424115/895354 = 28421 N
Stress
σ2 = 28421/706.9 = 40.21 MPa (compression)
Answer
σ2 = 40.21 MPa
9.Column buckling
Answer: Pcr = 197.7 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 = 0.7·2500 = 1750 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·306796/1750² = 197744 N = 197.7 kN
Answer
Pcr = 197.7 kN
10.Column buckling
Answer: d = 46.42 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 = 2·100000 = 200000 N
KL = 0.5·3000 = 1500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·200000·1500²/(π³·200000))^(1/4) = 46.42 mm
Answer
d = 46.42 mm
Check: the slenderness KL/r is 129.2, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 2.546 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 = √(8000/20) = 20 rad/s
fn = 20/(2π) = 3.183 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 3.183·√(1 − 0.6²) = 2.546 Hz
Answer
fd = 2.546 Hz
12.Mechanical vibrations
Answer: fn = 7.259 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 12 + 40 = 52 kN/m
Natural frequency
ωn = √(52000/25) = 45.61 rad/s
fn = 45.61/(2π) = 7.259 Hz
Answer
fn = 7.259 Hz
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