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 worm with 4 start(s) turns at 1450 rpm and drives a worm wheel with 40 teeth. Find the speed nw of the wheel.
z = 4Nw = 40n = 1450 rpm
nw = rpm
2.Gears & gear trainsHard
In a compound gear train, gear A (12 teeth) on the input shaft turning at 1500 rpm drives gear B (16 teeth). Gear C (18 teeth) is fixed to the same shaft as B and drives gear D (45 teeth) on the output shaft. Find the output speed nout.
nin = 1500 rpmA = 12B = 16C = 18D = 45
nout = rpm
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 = 20 mmL = 0.8 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 20 kW at 300 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.
P = 20 kWN = 300 rpmτallow = 40 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 120 MPa, σᵧ = −20 MPa and τₓᵧ = 40 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).
σx = 120 MPaσy = −20 MPaτxy = 40 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 50 mm carries an axial tensile force of 180 kN together with a torque of 250 N·m. At the surface of the shaft, find the largest principal stress σ1.
d = 50 mmF = 180 kNT = 250 N·m
σ1 = MPa
7.Axial stress & strainMedium
A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 80 mm, inner diameter 60 mm, E = 120 GPa). Both are 400 mm long and are compressed together between rigid plates by 200 kN. Find the shortening δ.
P = 200 kNrod: Ø40 mm, E = 200 GPatube: Ø80/Ø60 mm, E = 120 GPaL = 400 mm
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 = 50 kNsegment 1: Ø25 mm, 2 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 0.5 m, brass (E = 100 GPa)
δ = mm
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 = 6 mE = 200 GPa (steel)d = 80 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 130 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 130 kNn = 2L = 1.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 250 kg mass on a spring with a stiffness of 100 kN/m has a viscous damper giving a damping ratio of 0.7. Find the damped natural frequency fd of its free vibration in Hz.
m = 250 kgk = 100 kN/mζ = 0.7
fd = Hz
12.Mechanical vibrationsHard
A 25 kg mass hangs from spring k₃ = 80 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 = 25 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 = 25 kgk1 = 30 kN/mk2 = 25 kN/mk3 = 80 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: nw = 145 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.
Wheel speed
nw = n·z/Nw = 1450·4/40 = 145 rpm
Answer
nw = 145 rpm
2.Gears & gear trains
Answer: nout = 450 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 = 1500·12/16 = 1125 rpm = nC
Second stage
nout = 1125·18/45 = 450 rpm
Answer
nout = 450 rpm
total ratio = 3.333 : 1
3.Torsion of shafts
Answer: φ = 14.59 °
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·20⁴/32 = 15708 mm⁴
Angle of twist
φ = TL/(GJ) = 400·1000·800/(80000·15708) = 0.2546 rad
φ = 0.2546·180/π = 14.59°
Answer
φ = 14.59°
4.Torsion of shafts
Answer: d = 43.28 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 = 20000/31.42 = 636.6 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·636620/(π·40)) = 43.28 mm
Answer
d = 43.28 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θs = −30.13 °
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
On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.
cos 2θs = 40/80.62 = 0.4961, sin 2θs = −70/80.62 = −0.8682
2θs = atan2(−70, 40) = −60.26°
Angle on the element
θs = −30.13°
check: θs − θp = −30.13 − 14.87 = −45°
Answer
θs = −30.13°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σ1 = 92.79 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 = F/A = 180000/(π·50²/4) = 180000/1963 = 91.67 MPa
τ = 16T/(πd³) = 16·250000/(π·50³) = 10.19 MPa
Mohr’s circle
C = σ/2 = 45.84 MPa
R = √(45.84² + 10.19²) = 46.95 MPa
σ1 = C + R = 92.79 MPa
σ2 = C − R = −1.118 MPa
Answer
σ1 = 92.79 MPa
7.Axial stress & strain
Answer: δ = 0.1553 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(80² − 60²)/4 = 2199 mm²
Combined axial stiffness
ΣEA = 200·10³·1257 + 120·10³·2199 = 5.152 × 10⁸ N
Shortening
δ = 200000·400/5.152 × 10⁸ = 0.1553 mm
Answer
δ = 0.1553 mm
8.Axial stress & strain
Answer: δ = 3.706 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·20²/4 = 314.2 mm²
σ1 = 101.9 MPa
σ2 = 159.2 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 101.9·2000/70000 = 2.91 mm
δ2 = σ2L2/E2 = 159.2·500/100000 = 0.7958 mm
δ = δ1 + δ2 = 3.706 mm
Answer
δ = 3.706 mm
9.Column buckling
Answer: Pcr = 225 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·80⁴/64 = 2.011 × 10⁶ mm⁴
Effective length
KL = 0.7·6000 = 4200 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/4200² = 224989 N = 225 kN
Answer
Pcr = 225 kN
10.Column buckling
Answer: d = 41.47 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 = 2·130000 = 260000 N
KL = 0.7·1500 = 1050 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·260000·1050²/(π³·200000))^(1/4) = 41.47 mm
Answer
d = 41.47 mm
Check: the slenderness KL/r is 101.3, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 2.273 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.
Undamped natural frequency
ωn = √(100000/250) = 20 rad/s
fn = 20/(2π) = 3.183 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 3.183·√(1 − 0.7²) = 2.273 Hz
Answer
fd = 2.273 Hz
12.Mechanical vibrations
Answer: fn = 5.747 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 + 25 = 55 kN/m
keq = 55·80/(55 + 80) = 32.59 kN/m
Natural frequency
ωn = √(32593/25) = 36.11 rad/s
fn = 36.11/(2π) = 5.747 Hz
Answer
fn = 5.747 Hz
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