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 150 mm pulley at 720 rpm drives a 450 mm pulley by a belt with 1.5 % slip. Find the speed n2 of the driven pulley.
n1 = 720 rpmd1 = 150 mmd2 = 450 mms = 1.5 %
n2 = rpm
2.Gears & gear trainsHard
An open belt connects a 125 mm driving pulley at 720 rpm to a 400 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 800 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
d1 = 125 mmn1 = 720 rpmd2 = 400 mmC = 500 mmμ = 0.25T1 = 800 N
P = kW
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 = 400 N·md = 50 mmL = 2 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 100 kW at 1450 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.
P = 100 kWN = 1450 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 110 MPa and σᵧ = −30 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 170 MPa?
σx = 110 MPaσy = −30 MPaσ1 = 170 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = 40 MPa, σᵧ = 30 MPa and τₓᵧ = 30 MPa (σz = 0). The material has a yield strength of 355 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.
σx = 40 MPaσy = 30 MPaτxy = 30 MPaσY = 355 MPa
n =
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 20 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 60 mm, length 300 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.
segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: copper, Ø60 mm, 300 mm, E = 120 GPaδ = 0.3 mm
P = kN
8.Axial stress & strainHard
A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 70 mm, E = 110 GPa, allowable stress 80 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø20 mm, E = 200 GPa, σallow = 160 MPatube: Ø100/Ø70 mm, E = 110 GPa, σallow = 80 MPa
Pmax = kN
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 = 1.5 mE = 70 GPa (aluminium)b × h = 25 × 80 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 3 mE = 70 GPa (aluminium)d = 80 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 250 kg mass is supported by a spring with a stiffness of 100 kN/m and a viscous damper with a damping coefficient of 1500 N·s/m. Find the damping ratio ζ.
m = 250 kgk = 100 kN/mc = 1500 N·s/m
ζ =
12.Mechanical vibrationsHard
A 5 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 50 kN/m and k₂ = 60 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 5 kgk1 = 50 kN/mk2 = 60 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: n2 = 236.4 rpm
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.
Speed without slip
720·150/450 = 240 rpm
With slip
n2 = 240·0.985 = 236.4 rpm
Answer
n2 = 236.4 rpm
2.Gears & gear trains
Answer: P = 1.794 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.
Angle of wrap
θ = 2.584 rad
Slack-side tension
eμθ = e0.6461 = 1.908, T₂ = 800/1.908 = 419.3 N
Power
P = (800 − 419.3)·4.712 = 1.794 kW
Answer
P = 1.794 kW
3.Torsion of shafts
Answer: φ = 0.9338 °
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) = 400·1000·2000/(80000·613592) = 0.0163 rad
φ = 0.0163·180/π = 0.9338°
Answer
φ = 0.9338°
4.Torsion of shafts
Answer: d = 35.5 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.
Torque
ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s
T = P/ω = 100000/151.8 = 658.6 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·658572/(π·75)) = 35.5 mm
Answer
d = 35.5 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τxy = 109.5 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.
Required radius
C = (110 + (−30))/2 = 40 MPa
R = σ1 − C = 170 − 40 = 130 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(130² − 70²) = 109.5 MPa
Answer
τxy = 109.5 MPa
6.Mohr’s circle & stress transformation
Answer: n = 5.613
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 = 35 MPa, R = 30.41 MPa
σ1 = 65.41 MPa, σ2 = 4.586 MPa, σ3 = 0
Von Mises
σvM = √(65.41² − 65.41·4.586 + 4.586²) = 63.25 MPa
n = 355/63.25 = 5.613
Answer
n = 5.613
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: P = 41.38 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
A₁ = 314.2 mm², A₂ = 2827 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·314.2) = 6.366 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(120000·2827) = 8.842 × 10⁻⁷ mm/N
Load
P = 0.3/7.25 × 10⁻⁶ = 41377 N = 41.38 kN
Answer
P = 41.38 kN
8.Axial stress & strain
Answer: Pmax = 366.1 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(100² − 70²)/4 = 4006 mm²
Combined axial stiffness
ΣEA = 200·10³·314.2 + 110·10³·4006 = 5.034 × 10⁸ N
Governing strain
ε = min(0.0008, 0.0007273) = 0.0007273 (bronze)
Largest load
Pmax = 0.0007273·5.034 × 10⁸ = 366138 N = 366.1 kN
Answer
Pmax = 366.1 kN
9.Column buckling
Answer: Pcr = 7.996 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 = 80·25³/12 = 104167 mm⁴
Effective length
KL = 2·1500 = 3000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·104167/3000² = 7996 N = 7.996 kN
Answer
Pcr = 7.996 kN
10.Column buckling
Answer: σcr = 62.66 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·80⁴/64 = 2.011 × 10⁶ mm⁴
Radius of gyration and slenderness
A = 5027 mm²
r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm
KL = 0.7·3000 = 2100 mm
KL/r = 2100/20 = 105
Critical stress
σcr = π²E/(KL/r)² = π²·70000/105² = 62.66 MPa
Answer
σcr = 62.66 MPa
11.Mechanical vibrations
Answer: ζ = 0.15
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·√(100000·250) = 10000 N·s/m
Damping ratio
ζ = c/cc = 1500/10000 = 0.15
Answer
ζ = 0.15
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: keq = 110 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 = 50 + 60 = 110 kN/m
Answer
keq = 110 kN/m
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