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 0.75 kW motor running at 960 rpm drives a pinion with 20 teeth, which meshes with a gear with 32 teeth. The gear pair has an efficiency of 95%. Find the torque T1 on the motor shaft.
P = 0.75 kWn1 = 960 rpmN1 = 20N2 = 32η = 95%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (15 teeth) on the input shaft turning at 960 rpm drives gear B (36 teeth). Gear C (20 teeth) is fixed to the same shaft as B and drives gear D (30 teeth) on the output shaft. Find the output speed nout.
nin = 960 rpmA = 15B = 36C = 20D = 30
nout = rpm
3.Torsion of shaftsMedium
A hollow shaft must carry a torque of 500 N·m. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter do.
T = 500 N·mdi/do = 0.7τallow = 40 MPa
do = mm
4.Torsion of shaftsHard
A solid brass shaft of diameter 50 mm is fixed at A. A torque of 250 N·m acts at B, 0.8 m from A, and a torque of 250 N·m in the same direction acts at the free end C, a further 0.8 m along. Find the maximum shear stress τmax in the shaft.
d = 50 mmTB = 250 N·mTC = 250 N·mAB = 0.8 mBC = 0.8 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 160 MPa, σᵧ = −10 MPa and τₓᵧ = 30 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = 160 MPaσy = −10 MPaτxy = 30 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = −30 MPa, σᵧ = 40 MPa and τₓᵧ = −30 MPa (σz = 0). The material has a yield strength of 350 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.
σx = −30 MPaσy = 40 MPaτxy = −30 MPaσY = 350 MPa
n =
7.Axial stress & strainMedium
Find the elongation δ of the aluminium bar under the axial load.
P = 70 kNd = 30 mmL = 3 mE = 70 GPa
δ = mm
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 30 kNsegment 1: Ø25 mm, 1 m, steel (E = 200 GPa)segment 2: Ø16 mm, 1.5 m, brass (E = 100 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 2.5 mE = 70 GPa (aluminium)d = 50 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is pinned at both ends. It must carry 145 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 145 kNn = 3L = 4 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed 5 cycles later by a peak of 26.7 mm. Find the damping ratio ζ.
x0 = 50 mmx5 = 26.7 mmn = 5
ζ =
12.Mechanical vibrationsHard
A 50 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 15 kN/m and k₂ = 12 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 50 kgk1 = 15 kN/mk2 = 12 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T1 = 7.46 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 = 750/100.5 = 7.46 N·m
Answer
T1 = 7.46 N·m
2.Gears & gear trains
Answer: nout = 266.7 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 = 960·15/36 = 400 rpm = nC
Second stage
nout = 400·20/30 = 266.7 rpm
Answer
nout = 266.7 rpm
total ratio = 3.6 : 1
3.Torsion of shafts
Answer: do = 43.76 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 = 500 N·m = 500000 N·mm
1 − k⁴ = 1 − 0.7⁴ = 0.7599
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·500000/(π·40·0.7599)) = 43.76 mm
di = 0.7·43.76 = 30.63 mm
Answer
do = 43.76 mm
A solid shaft would need d = 39.93 mm; the hollow one is a little larger but much lighter.
4.Torsion of shafts
Answer: τmax = 20.37 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 = 250 N·m
TAB = TC + TB = 250 + 250 = 500 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 500 N·m
τmax = 16·500000/(π·50³) = 20.37 MPa
Answer
τmax = 20.37 MPa
5.Mohr’s circle & stress transformation
Answer: θp = 9.72 °
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
Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.
2θp = atan2(2·30, 160 − −10) = 19.44°
Angle on the element
θp = 9.72°
Answer
θp = 9.72°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: n = 3.796
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 = 5 MPa, R = 46.1 MPa
σ1 = 51.1 MPa, σ2 = −41.1 MPa, σ3 = 0
Tresca
σmax − σmin = 92.2 MPa
n = 350/92.2 = 3.796
Answer
n = 3.796
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: δ = 4.244 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 = 70·1000/706.9 = 99.03 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 99.03/70000 = 0.001415 = 1415 µε
Elongation
δ = ε·L = 0.001415·3000 mm = 4.244 mm
Answer
δ = 4.244 mm
8.Axial stress & strain
Answer: δ = 2.544 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·25²/4 = 490.9 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 61.12 MPa
σ2 = 149.2 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 61.12·1000/200000 = 0.3056 mm
δ2 = σ2L2/E2 = 149.2·1500/100000 = 2.238 mm
δ = δ1 + δ2 = 2.544 mm
Answer
δ = 2.544 mm
9.Column buckling
Answer: Pcr = 8.478 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 = 2·2500 = 5000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·306796/5000² = 8478 N = 8.478 kN
Answer
Pcr = 8.478 kN
10.Column buckling
Answer: d = 92.06 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 = 3·145000 = 435000 N
KL = 1·4000 = 4000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·435000·4000²/(π³·200000))^(1/4) = 92.06 mm
Answer
d = 92.06 mm
Check: the slenderness KL/r is 173.8, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.01997
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 = (1/5)·ln(50/26.7) = 0.1255
Damping ratio
ζ = δ/√(4π² + δ²) = 0.1255/√(39.48 + 0.01574) = 0.01997
Answer
ζ = 0.01997
12.Mechanical vibrations
Answer: keq = 27 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 = 15 + 12 = 27 kN/m
Answer
keq = 27 kN/m
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