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 2.2 kW motor running at 960 rpm drives a pinion with 32 teeth, which meshes with a gear with 36 teeth. The gear pair has an efficiency of 97%. Find the output torque T2 on the gear shaft.
P = 2.2 kWn1 = 960 rpmN1 = 32N2 = 36η = 97%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (16 teeth) on the input shaft turning at 960 rpm drives gear B (45 teeth). Gear C (16 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 = 960 rpmA = 16B = 45C = 16D = 50
nout = rpm
3.Torsion of shaftsMedium
A solid brass shaft 0.8 m long carries a torque of 750 N·m. The angle of twist between its ends must not exceed 1°. Find the smallest diameter d that meets this limit.
T = 750 N·mL = 0.8 mG = 39 GPaφallow = 1°
d = mm
4.Torsion of shaftsHard
A solid steel shaft of diameter 25 mm is fixed at A. A torque of 200 N·m acts at B, 3 m from A, and a torque of 750 N·m in the opposite direction acts at the free end C, a further 3 m along. Find the maximum shear stress τmax in the shaft.
d = 25 mmTB = 200 N·mTC = 750 N·mAB = 3 mBC = 3 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −80 MPa, σᵧ = 120 MPa and τₓᵧ = 30 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 = −80 MPaσy = 120 MPaτxy = 30 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A thin-walled cylindrical tank of inner diameter 1200 mm and wall thickness 5 mm holds gas at a gauge pressure of 1 MPa. It is made from plate welded along a helical seam at 30° to the axis of the cylinder. Find the normal stress σn acting across the weld seam.
D = 1200 mmt = 5 mmp = 1 MPaα = 30°
σn = MPa
7.Axial stress & strainMedium
Find the elongation δ of the copper bar under the axial load.
P = 85 kNd = 30 mmL = 1 mE = 117 GPa
δ = mm
8.Axial stress & strainHard
A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of aluminium (outer diameter 120 mm, inner diameter 50 mm, E = 70 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø20 mm, E = 200 GPa, σallow = 200 MPatube: Ø120/Ø50 mm, E = 70 GPa, σallow = 60 MPa
Pmax = kN
9.Column bucklingMedium
A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 2.5 mE = 70 GPa (aluminium)b × h = 20 × 100 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 20 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 20 kNn = 2.5L = 4 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 100 kg mass is supported by a spring with a stiffness of 12 kN/m and a viscous damper with a damping coefficient of 880 N·s/m. Find the damping ratio ζ.
m = 100 kgk = 12 kN/mc = 880 N·s/m
ζ =
12.Mechanical vibrationsHard
A 50 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 80 kN/m and k₂ = 25 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 50 kgk1 = 80 kN/mk2 = 25 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T2 = 23.88 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·960/60 = 100.5 rad/s
T1 = P/ω₁ = 2200/100.5 = 21.88 N·m
Output torque
T2 = 21.88·36/32·0.97 = 23.88 N·m
Answer
T2 = 23.88 N·m
2.Gears & gear trains
Answer: nout = 109.2 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 = 960·16/45 = 341.3 rpm = nC
Second stage
nout = 341.3·16/50 = 109.2 rpm
Answer
nout = 109.2 rpm
total ratio = 8.789 : 1
3.Torsion of shafts
Answer: d = 54.74 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.
Required polar moment
T = 750 N·m = 750000 N·mm
L = 0.8 m = 800 mm
G = 39 GPa = 39000 MPa
φ = 1·π/180 = 0.01745 rad
J = TL/(Gφ) = 750000·800/(39000·0.01745) = 881474 mm⁴
Diameter
d = (32J/π)1/4 = 54.74 mm
Answer
d = 54.74 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: τmax = 244.5 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 = 750 N·m
TAB = TC − TB = 750 − 200 = 550 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 750 N·m
τmax = 16·750000/(π·25³) = 244.5 MPa
Answer
τmax = 244.5 MPa
5.Mohr’s circle & stress transformation
Answer: θs = 36.65 °
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 = 30/104.4 = 0.2873, sin 2θs = −(−100)/104.4 = 0.9578
2θs = atan2(100, 30) = 73.3°
Angle on the element
θs = 36.65°
check: θs − θp = 36.65 − 81.65 = −45°
Answer
θs = 36.65°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σn = 105 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 = 1200/2 = 600 mm
σh = pr/t = 1·600/5 = 120 MPa
σa = pr/(2t) = 60 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.
σn = 60·sin²30° + 120·cos²30° = 105 MPa
Answer
σn = 105 MPa
7.Axial stress & strain
Answer: δ = 1.028 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·30²/4 = 706.9 mm²
σ = P/A = 85·1000/706.9 = 120.3 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 120.3/117000 = 0.001028 = 1028 µε
Elongation
δ = ε·L = 0.001028·1000 mm = 1.028 mm
Answer
δ = 1.028 mm
8.Axial stress & strain
Answer: Pmax = 614.6 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(120² − 50²)/4 = 9346 mm²
Combined axial stiffness
ΣEA = 200·10³·314.2 + 70·10³·9346 = 7.171 × 10⁸ N
Governing strain
ε = min(0.001, 0.0008571) = 0.0008571 (aluminium)
Largest load
Pmax = 0.0008571·7.171 × 10⁸ = 614630 N = 614.6 kN
Answer
Pmax = 614.6 kN
9.Column buckling
Answer: Pcr = 7.369 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 = 100·20³/12 = 66667 mm⁴
Effective length
KL = 1·2500 = 2500 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·66667/2500² = 7369 N = 7.369 kN
Answer
Pcr = 7.369 kN
10.Column buckling
Answer: d = 44.85 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 = 2.5·20000 = 50000 N
KL = 0.7·4000 = 2800 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·50000·2800²/(π³·200000))^(1/4) = 44.85 mm
Answer
d = 44.85 mm
Check: the slenderness KL/r is 249.7, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.4017
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·√(12000·100) = 2191 N·s/m
Damping ratio
ζ = c/cc = 880/2191 = 0.4017
Answer
ζ = 0.4017
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: keq = 105 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 = 80 + 25 = 105 kN/m
Answer
keq = 105 kN/m
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