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
An 11 kW motor running at 2900 rpm drives a pinion with 12 teeth, which meshes with a gear with 60 teeth. The gear pair has an efficiency of 98%. Find the torque T1 on the motor shaft.
P = 11 kWn1 = 2900 rpmN1 = 12N2 = 60η = 98%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (32 teeth) on the input shaft turning at 2900 rpm drives gear B (60 teeth). Gear C (24 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 = 2900 rpmA = 32B = 60C = 24D = 30
nout = rpm
3.Torsion of shaftsMedium
A solid brass shaft 3 m long carries a torque of 100 N·m. The angle of twist between its ends must not exceed 0.5°. Find the smallest diameter d that meets this limit.
T = 100 N·mL = 3 mG = 39 GPaφallow = 0.5°
d = mm
4.Torsion of shaftsHard
A solid steel shaft of diameter 25 mm is fixed to rigid walls at both ends A and B. A torque of 100 N·m is applied at C, 1.5 m from A and 1.2 m from B. Find the angle of rotation φC of section C, in degrees.
d = 25 mmT0 = 100 N·ma = 1.5 mb = 1.2 mG = 80 GPa
φC = °
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 90 MPa, σᵧ = −20 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 90 MPaσy = −20 MPaτxy = 30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 50 mm carries a bending moment of 300 N·m together with a torque of 1500 N·m. At the top surface, where the bending stress is tensile and largest, find the maximum in-plane shear stress τmax.
d = 50 mmM = 300 N·mT = 1500 N·m
τmax = MPa
7.Axial stress & strainMedium
A steel rod (diameter 25 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 100 mm, inner diameter 50 mm, E = 120 GPa). Both are 600 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.
P = 150 kNrod: Ø25 mm, E = 200 GPatube: Ø100/Ø50 mm, E = 120 GPaL = 600 mm
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) and a segment made of brass (diameter 30 mm, length 400 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 40 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø40 mm, 400 mm, E = 200 GPasegment 2: brass, Ø30 mm, 400 mm, E = 100 GPaP = 40 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 2.5 mE = 70 GPa (aluminium)b × h = 25 × 100 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 2 mE = 200 GPa (steel)d = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 25 mm is followed 4 cycles later by a peak of 2 mm. Find the damping ratio ζ.
x0 = 25 mmx4 = 2 mmn = 4
ζ =
12.Mechanical vibrationsHard
A 50 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 10 kN/m and the lower one k₂ = 12 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 50 kgk1 = 10 kN/mk2 = 12 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T1 = 36.22 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·2900/60 = 303.7 rad/s
T1 = P/ω₁ = 11000/303.7 = 36.22 N·m
Answer
T1 = 36.22 N·m
2.Gears & gear trains
Answer: nout = 1237 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 = 2900·32/60 = 1547 rpm = nC
Second stage
nout = 1547·24/30 = 1237 rpm
Answer
nout = 1237 rpm
total ratio = 2.344 : 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 = 100 N·m = 100000 N·mm
L = 3 m = 3000 mm
G = 39 GPa = 39000 MPa
φ = 0.5·π/180 = 0.008727 rad
J = TL/(Gφ) = 100000·3000/(39000·0.008727) = 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: φC = 1.245 °
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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).
Equilibrium and compatibility
TA + TB = 100
TA·a = TB·b
TA = 100·1.2/2.7 = 44.44 N·m
TB = 100·1.5/2.7 = 55.56 N·m
Rotation of C
Use segment AC: it is fixed at A and carries TA.
J = π·25⁴/32 = 38350 mm⁴
G = 80 GPa = 80000 MPa
φC = TA·a/(GJ) = 44444·1500/(80000·38350) = 0.02173 rad = 1.245°
Answer
φC = 1.245°
5.Mohr’s circle & stress transformation
Answer: τmax = 62.65 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 = √(((90 − −20)/2)² + 30²) = 62.65 MPa
Answer
τmax = 62.65 MPa
6.Mohr’s circle & stress transformation
Answer: τmax = 62.33 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.
Stresses at the surface
σ = 32M/(πd³) = 32·300000/(π·50³) = 24.45 MPa
τ = 16T/(πd³) = 16·1500000/(π·50³) = 61.12 MPa
Mohr’s circle
C = σ/2 = 12.22 MPa
R = √(12.22² + 61.12²) = 62.33 MPa
Answer
τmax = 62.33 MPa
7.Axial stress & strain
Answer: δ = 0.1118 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·25²/4 = 490.9 mm²
At = π(100² − 50²)/4 = 5890 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 120·10³·5890 = 8.05 × 10⁸ N
Shortening
δ = 150000·600/8.05 × 10⁸ = 0.1118 mm
Answer
δ = 0.1118 mm
8.Axial stress & strain
Answer: σ2 = 12.42 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 = 200000·1257/400 = 628319 N/mm
k₂ = 100000·706.9/400 = 176715 N/mm
Share of segment 2
F2 = 40000·176715/805033 = 8780 N
Stress
σ2 = 8780/706.9 = 12.42 MPa (compression)
Answer
σ2 = 12.42 MPa
9.Column buckling
Answer: Pcr = 57.57 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 = 100·25³/12 = 130208 mm⁴
Effective length
KL = 0.5·2500 = 1250 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·130208/1250² = 57573 N = 57.57 kN
Answer
Pcr = 57.57 kN
10.Column buckling
Answer: σcr = 6.94 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·30⁴/64 = 39761 mm⁴
Radius of gyration and slenderness
A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 2·2000 = 4000 mm
KL/r = 4000/7.5 = 533.3
Critical stress
σcr = π²E/(KL/r)² = π²·200000/533.3² = 6.94 MPa
Answer
σcr = 6.94 MPa
11.Mechanical vibrations
Answer: ζ = 0.09999
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 = (1/4)·ln(25/2) = 0.6314
Damping ratio
ζ = δ/√(4π² + δ²) = 0.6314/√(39.48 + 0.3987) = 0.09999
Answer
ζ = 0.09999
12.Mechanical vibrations
Answer: keq = 5.455 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 = 1/10 + 1/12 → keq = 10·12/(10 + 12) = 5.455 kN/m
Answer
keq = 5.455 kN/m
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