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 125 mm driving pulley at 1500 rpm. The tension is 500 N on the tight side and 125 N on the slack side. Find the power P transmitted, in kW.
d1 = 125 mmn1 = 1500 rpmT1 = 500 NT2 = 125 N
P = kW
2.Gears & gear trainsHard
In a compound gear train, gear A (16 teeth) on the input shaft turning at 3000 rpm drives gear B (24 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D (24 teeth) on the output shaft. Find the output speed nout.
nin = 3000 rpmA = 16B = 24C = 15D = 24
nout = rpm
3.Torsion of shaftsMedium
A stepped solid steel shaft is fixed at A. Segment AB has diameter 75 mm and length 1.2 m; segment BC has diameter 30 mm and length 0.5 m. A torque of 500 N·m acts at the free end C. Find the maximum shear stress τmax in the shaft.
T = 500 N·md1 = 75 mm, L1 = 1.2 md2 = 30 mm, L2 = 0.5 m
τmax = MPa
4.Torsion of shaftsHard
A solid shaft must transmit 20 kW at 1450 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.
P = 20 kWN = 1450 rpmτallow = 60 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 70 MPa and σᵧ = −20 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 95 MPa?
σx = 70 MPaσy = −20 MPaσ1 = 95 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −60 MPa, σᵧ = 90 MPa and τₓᵧ = 60 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 45° clockwise from the x-axis.
σx = −60 MPaσy = 90 MPaτxy = 60 MPaθ = -45°
σθ = MPa
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 brass (diameter 40 mm, length 400 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.
segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: brass, Ø40 mm, 400 mm, E = 100 GPaδ = 0.2 mm
P = kN
8.Axial stress & strainHard
A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of bronze (outer diameter 80 mm, inner diameter 50 mm, E = 110 GPa, allowable stress 80 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø25 mm, E = 200 GPa, σallow = 250 MPatube: Ø80/Ø50 mm, E = 110 GPa, σallow = 80 MPa
Pmax = kN
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 = 3 mE = 70 GPa (aluminium)b × h = 25 × 100 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at both ends. It must carry 5 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 5 kNn = 3L = 4 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 500 kg mass is supported by a spring with a stiffness of 100 kN/m and a viscous damper with a damping coefficient of 2830 N·s/m. Find the damping ratio ζ.
m = 500 kgk = 100 kN/mc = 2830 N·s/m
ζ =
12.Mechanical vibrationsHard
A 40 kg mass hangs from spring k₃ = 10 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 30 kN/m and k₂ = 80 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 = 40 kgk1 = 30 kN/mk2 = 80 kN/mk3 = 10 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: P = 3.682 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 = 9.817 m/s
Power
P = (500 − 125)·9.817 = 3682 W = 3.682 kW
Answer
P = 3.682 kW
2.Gears & gear trains
Answer: nout = 1250 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 = 3000·16/24 = 2000 rpm = nC
Second stage
nout = 2000·15/24 = 1250 rpm
Answer
nout = 1250 rpm
total ratio = 2.4 : 1
3.Torsion of shafts
Answer: τmax = 94.31 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.
Stress in each segment
T = 500 N·m = 500000 N·mm
τAB = 16·500000/(π·75³) = 6.036 MPa
τBC = 16·500000/(π·30³) = 94.31 MPa
Answer
τmax = 94.31 MPa
4.Torsion of shafts
Answer: d = 22.36 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 = 20000/151.8 = 131.7 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·131714/(π·60)) = 22.36 mm
Answer
d = 22.36 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τxy = 53.62 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 = (70 + (−20))/2 = 25 MPa
R = σ1 − C = 95 − 25 = 70 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(70² − 45²) = 53.62 MPa
Answer
τxy = 53.62 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = −45 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.
Transformation equation
σθ = 15 + −75·0 + 60·−1 = −45 MPa
Answer
σθ = −45 MPa
7.Axial stress & strain
Answer: P = 20.94 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 = 1257 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·314.2) = 6.366 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 400/(100000·1257) = 3.183 × 10⁻⁶ mm/N
Load
P = 0.2/9.549 × 10⁻⁶ = 20944 N = 20.94 kN
Answer
P = 20.94 kN
8.Axial stress & strain
Answer: Pmax = 316.4 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·25²/4 = 490.9 mm²
At = π(80² − 50²)/4 = 3063 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 110·10³·3063 = 4.351 × 10⁸ N
Governing strain
ε = min(0.00125, 0.0007273) = 0.0007273 (bronze)
Largest load
Pmax = 0.0007273·4.351 × 10⁸ = 316444 N = 316.4 kN
Answer
Pmax = 316.4 kN
9.Column buckling
Answer: Pcr = 39.98 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 = 0.5·3000 = 1500 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·130208/1500² = 39981 N = 39.98 kN
Answer
Pcr = 39.98 kN
10.Column buckling
Answer: d = 28.05 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 = 3·5000 = 15000 N
KL = 0.5·4000 = 2000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·15000·2000²/(π³·200000))^(1/4) = 28.05 mm
Answer
d = 28.05 mm
Check: the slenderness KL/r is 285.2, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.2001
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·√(100000·500) = 14142 N·s/m
Damping ratio
ζ = c/cc = 2830/14142 = 0.2001
Answer
ζ = 0.2001
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 2.409 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 = 30 + 80 = 110 kN/m
keq = 110·10/(110 + 10) = 9.167 kN/m
Natural frequency
ωn = √(9167/40) = 15.14 rad/s
fn = 15.14/(2π) = 2.409 Hz
Answer
fn = 2.409 Hz
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