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 15 kW motor running at 960 rpm drives a pinion with 25 teeth and a module of 5 mm. Find the tangential force Ft on the pinion teeth.
P = 15 kWn = 960 rpmN1 = 25m = 5 mm
Ft = N
2.Gears & gear trainsHard
A 4 kW motor at 2900 rpm drives gear A (25 teeth), which meshes with gear B (45 teeth). Gear C (30 teeth) on the same shaft as B drives gear D (50 teeth) on the output shaft. Each gear mesh has an efficiency of 95%. Find the output torque Tout.
P = 4 kWnin = 2900 rpmA = 25B = 45C = 30D = 50ηmesh = 95%
Tout = N·m
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 = 1000 N·md = 40 mmL = 2 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft must transmit 15 kW at 300 rpm. Its inner diameter is to be 0.5 times its outer diameter, and the shear stress must not exceed 60 MPa. Find the smallest outer diameter do.
P = 15 kWN = 300 rpmdi/do = 0.5τallow = 60 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 0 MPa and σᵧ = 30 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 45 MPa?
σx = 0 MPaσy = 30 MPaσ1 = 45 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −70 MPa, σᵧ = 50 MPa and τₓᵧ = −40 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 35° counter-clockwise from the x-axis.
σx = −70 MPaσy = 50 MPaτxy = −40 MPaθ = 35°
σθ = MPa
7.Axial stress & strainMedium
A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 60 mm, E = 110 GPa). Both are 300 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.
P = 150 kNrod: Ø40 mm, E = 200 GPatube: Ø100/Ø60 mm, E = 110 GPaL = 300 mm
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 20 mm, length 500 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 500 mm, E = 70 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, Ø20 mm, 500 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 500 mm, E = 70 GPaP = 40 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 5 mE = 200 GPa (steel)b × h = 40 × 80 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 2.5 mE = 200 GPa (steel)d = 20 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 30 mm is followed one cycle later by a peak of 21.9 mm. Find the damping ratio ζ.
x0 = 30 mmx1 = 21.9 mmn = 1
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from spring k₃ = 20 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 5 kN/m and k₂ = 12 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 = 5 kN/mk2 = 12 kN/mk3 = 20 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: Ft = 2387 N
Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.
Torque on the pinion
ω = 2π·960/60 = 100.5 rad/s
T = P/ω = 15000/100.5 = 149.2 N·m
Pitch diameter
d = m·N = 5·25 = 125 mm
Tangential force
Ft = 2T/d = 2·149.2/0.125 = 2387 N
Answer
Ft = 2387 N
2.Gears & gear trains
Answer: Tout = 35.66 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·(25/45)·(30/50) = 966.7 rpm
ωout = 2π·966.7/60 = 101.2 rad/s
Output power and torque
Pout = 4000·0.95² = 3610 W
Tout = Pout/ωout = 3610/101.2 = 35.66 N·m
Answer
Tout = 35.66 N·m
3.Torsion of shafts
Answer: φ = 5.699 °
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·40⁴/32 = 251327 mm⁴
Angle of twist
φ = TL/(GJ) = 1000·1000·2000/(80000·251327) = 0.09947 rad
φ = 0.09947·180/π = 5.699°
Answer
φ = 5.699°
4.Torsion of shafts
Answer: do = 35.1 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·300/60 = 31.42 rad/s
T = P/ω = 15000/31.42 = 477.5 N·m
Outer diameter from the allowable stress
T = 477.5 N·m = 477465 N·mm
1 − k⁴ = 1 − 0.5⁴ = 0.9375
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·477465/(π·60·0.9375)) = 35.1 mm
di = 0.5·35.1 = 17.55 mm
Answer
do = 35.1 mm
A solid shaft would need d = 34.35 mm; the hollow one is a little larger but much lighter.
5.Mohr’s circle & stress transformation
Answer: τxy = 25.98 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 = (0 + 30)/2 = 15 MPa
R = σ1 − C = 45 − 15 = 30 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(30² − 15²) = 25.98 MPa
Answer
τxy = 25.98 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = −68.11 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 = −10 + −60·0.342 + −40·0.9397 = −68.11 MPa
Answer
σθ = −68.11 MPa
7.Axial stress & strain
Answer: δ = 0.05595 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·40²/4 = 1257 mm²
At = π(100² − 60²)/4 = 5027 mm²
Combined axial stiffness
ΣEA = 200·10³·1257 + 110·10³·5027 = 8.042 × 10⁸ N
Shortening
δ = 150000·300/8.042 × 10⁸ = 0.05595 mm
Answer
δ = 0.05595 mm
8.Axial stress & strain
Answer: σ2 = 24.93 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 = 200000·314.2/500 = 125664 N/mm
k₂ = 70000·706.9/500 = 98960 N/mm
Share of segment 2
F2 = 40000·98960/224624 = 17622 N
Stress
σ2 = 17622/706.9 = 24.93 MPa (compression)
Answer
σ2 = 24.93 MPa
9.Column buckling
Answer: Pcr = 68.75 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 = 80·40³/12 = 426667 mm⁴
Effective length
KL = 0.7·5000 = 3500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·426667/3500² = 68752 N = 68.75 kN
Answer
Pcr = 68.75 kN
10.Column buckling
Answer: σcr = 31.58 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·20⁴/64 = 7854 mm⁴
Radius of gyration and slenderness
A = 314.2 mm²
r = √(I/A) = √(7854/314.2) = 5 mm
KL = 0.5·2500 = 1250 mm
KL/r = 1250/5 = 250
Critical stress
σcr = π²E/(KL/r)² = π²·200000/250² = 31.58 MPa
Answer
σcr = 31.58 MPa
11.Mechanical vibrations
Answer: ζ = 0.05003
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 = (1/1)·ln(30/21.9) = 0.3147
Damping ratio
ζ = δ/√(4π² + δ²) = 0.3147/√(39.48 + 0.09904) = 0.05003
Answer
ζ = 0.05003
12.Mechanical vibrations
Answer: fn = 4.825 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 = 5 + 12 = 17 kN/m
keq = 17·20/(17 + 20) = 9.189 kN/m
Natural frequency
ωn = √(9189/10) = 30.31 rad/s
fn = 30.31/(2π) = 4.825 Hz
Answer
fn = 4.825 Hz
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