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 150 mm driving pulley at 2880 rpm. The tension is 1200 N on the tight side and 600 N on the slack side. Find the power P transmitted, in kW.
d1 = 150 mmn1 = 2880 rpmT1 = 1200 NT2 = 600 N
P = kW
2.Gears & gear trainsHard
A 5.5 kW motor at 2900 rpm drives gear A (15 teeth), which meshes with gear B (20 teeth). Gear C (12 teeth) on the same shaft as B drives gear D (18 teeth) on the output shaft. Each gear mesh has an efficiency of 95%. Find the output torque Tout.
P = 5.5 kWnin = 2900 rpmA = 15B = 20C = 12D = 18ηmesh = 95%
Tout = N·m
3.Torsion of shaftsMedium
A hollow shaft must carry a torque of 2000 N·m. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter do.
T = 2000 N·mdi/do = 0.75τallow = 40 MPa
do = mm
4.Torsion of shaftsHard
A solid brass shaft of diameter 40 mm is fixed at A. A torque of 400 N·m acts at B, 1.5 m from A, and a torque of 500 N·m in the same direction acts at the free end C, a further 0.5 m along. Find the maximum shear stress τmax in the shaft.
d = 40 mmTB = 400 N·mTC = 500 N·mAB = 1.5 mBC = 0.5 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 110 MPa, σᵧ = 110 MPa and τₓᵧ = −20 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 110 MPaσy = 110 MPaτxy = −20 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A thin-walled cylindrical tank of inner diameter 1200 mm and wall thickness 8 mm holds gas at a gauge pressure of 0.5 MPa. It is made from plate welded along a helical seam at 20° to the axis of the cylinder. Find the normal stress σn acting across the weld seam.
D = 1200 mmt = 8 mmp = 0.5 MPaα = 20°
σn = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the steel bar, in microstrain (µε = 10⁻⁶).
P = 30 kNd = 25 mmL = 3 mE = 200 GPa
ε = µε
8.Axial stress & strainHard
A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 50 mm, E = 100 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø20 mm, E = 200 GPa, σallow = 160 MPatube: Ø120/Ø50 mm, E = 100 GPa, σallow = 60 MPa
Pmax = kN
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 85 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 85 kNn = 2.5L = 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 30 mm is followed 2 cycles later by a peak of 20.6 mm. Find the damping ratio ζ.
x0 = 30 mmx2 = 20.6 mmn = 2
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from spring k₃ = 25 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 80 kN/m and k₂ = 40 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 = 10 kgk1 = 80 kN/mk2 = 40 kN/mk3 = 25 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: P = 13.57 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 = 22.62 m/s
Power
P = (1200 − 600)·22.62 = 13572 W = 13.57 kW
Answer
P = 13.57 kW
2.Gears & gear trains
Answer: Tout = 32.69 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 = 2900·(15/20)·(12/18) = 1450 rpm
ωout = 2π·1450/60 = 151.8 rad/s
Output power and torque
Pout = 5500·0.95² = 4964 W
Tout = Pout/ωout = 4964/151.8 = 32.69 N·m
Answer
Tout = 32.69 N·m
3.Torsion of shafts
Answer: do = 71.95 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.
Outer diameter from the allowable stress
T = 2000 N·m = 2 × 10⁶ N·mm
1 − k⁴ = 1 − 0.75⁴ = 0.6836
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·2 × 10⁶/(π·40·0.6836)) = 71.95 mm
di = 0.75·71.95 = 53.96 mm
Answer
do = 71.95 mm
A solid shaft would need d = 63.38 mm; the hollow one is a little larger but much lighter.
4.Torsion of shafts
Answer: τmax = 71.62 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 = 500 N·m
TAB = TC + TB = 500 + 400 = 900 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 900 N·m
τmax = 16·900000/(π·40³) = 71.62 MPa
Answer
τmax = 71.62 MPa
5.Mohr’s circle & stress transformation
Answer: τmax = 20 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 = √(((110 − 110)/2)² + −20²) = 20 MPa
Answer
τmax = 20 MPa
6.Mohr’s circle & stress transformation
Answer: σn = 35.31 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 = 0.5·600/8 = 37.5 MPa
σa = pr/(2t) = 18.75 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 = 18.75·sin²20° + 37.5·cos²20° = 35.31 MPa
Answer
σn = 35.31 MPa
7.Axial stress & strain
Answer: ε = 305.6 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·25²/4 = 490.9 mm²
σ = P/A = 30·1000/490.9 = 61.12 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 61.12/200000 = 0.0003056 = 305.6 µε
Answer
ε = 305.6 µε
8.Axial stress & strain
Answer: Pmax = 598.5 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 + 100·10³·9346 = 9.975 × 10⁸ N
Governing strain
ε = min(0.0008, 0.0006) = 0.0006 (brass)
Largest load
Pmax = 0.0006·9.975 × 10⁸ = 598473 N = 598.5 kN
Answer
Pmax = 598.5 kN
9.Column buckling
Answer: Pcr = 224.2 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·60⁴/64 = 636173 mm⁴
Effective length
KL = 0.7·2000 = 1400 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·636173/1400² = 224242 N = 224.2 kN
Answer
Pcr = 224.2 kN
10.Column buckling
Answer: d = 173.3 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 = 2.5·85000 = 212500 N
KL = 2·6000 = 12000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·212500·12000²/(π³·70000))^(1/4) = 173.3 mm
Answer
d = 173.3 mm
Check: the slenderness KL/r is 277, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.0299
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 = (1/2)·ln(30/20.6) = 0.188
Damping ratio
ζ = δ/√(4π² + δ²) = 0.188/√(39.48 + 0.03533) = 0.0299
Answer
ζ = 0.0299
12.Mechanical vibrations
Answer: fn = 7.239 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 = 80 + 40 = 120 kN/m
keq = 120·25/(120 + 25) = 20.69 kN/m
Natural frequency
ωn = √(20690/10) = 45.49 rad/s
fn = 45.49/(2π) = 7.239 Hz
Answer
fn = 7.239 Hz
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