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 2.2 kW motor at 1440 rpm drives a worm with 3 start(s). The worm wheel has 48 teeth and the drive has an efficiency of 75%. Find the output torque Tw on the wheel shaft.
z = 3Nw = 48n = 1440 rpmP = 2.2 kWη = 75%
Tw = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (40 teeth) on the input shaft turning at 2900 rpm drives gear B (50 teeth). Gear C (48 teeth) is fixed to the same shaft as B and drives gear D (80 teeth) on the output shaft. Find the output speed nout.
nin = 2900 rpmA = 40B = 50C = 48D = 80
nout = rpm
3.Torsion of shaftsMedium
A solid brass 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 = 750 N·md = 40 mmL = 0.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft must transmit 20 kW at 750 rpm. Its inner diameter is to be 0.8 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter do.
P = 20 kWN = 750 rpmdi/do = 0.8τallow = 75 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −40 MPa, σᵧ = 70 MPa and τₓᵧ = 10 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = −40 MPaσy = 70 MPaτxy = 10 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 160 MPa, σᵧ = 10 MPa and τₓᵧ = 60 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 20° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 160 MPaσy = 10 MPaτxy = 60 MPaθ = 20°
τθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the copper bar under the axial load.
P = 120 kNd = 25 mmL = 1 mE = 117 GPa
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 = 95 kNsegment 1: Ø40 mm, 0.5 m, copper (E = 117 GPa)segment 2: Ø12 mm, 1.5 m, steel (E = 200 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 1 mE = 200 GPa (steel)b × h = 40 × 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 105 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 105 kNn = 2.5L = 3 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 15 kg mass on a spring with a stiffness of 80 kN/m has a viscous damper giving a damping ratio of 0.5. Find the damped natural frequency fd of its free vibration in Hz.
m = 15 kgk = 80 kN/mζ = 0.5
fd = Hz
12.Mechanical vibrationsHard
A 100 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 40 kN/m and the lower one k₂ = 30 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 100 kgk1 = 40 kN/mk2 = 30 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: Tw = 175.1 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.
Wheel speed
nw = n·z/Nw = 1440·3/48 = 90 rpm
Output torque
ωw = 2π·90/60 = 9.425 rad/s
Tw = η·P/ωw = 0.75·2200/9.425 = 175.1 N·m
Answer
Tw = 175.1 N·m
2.Gears & gear trains
Answer: nout = 1392 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 = 2900·40/50 = 2320 rpm = nC
Second stage
nout = 2320·48/80 = 1392 rpm
Answer
nout = 1392 rpm
total ratio = 2.083 : 1
3.Torsion of shafts
Answer: φ = 2.192 °
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) = 750·1000·500/(39000·251327) = 0.03826 rad
φ = 0.03826·180/π = 2.192°
Answer
φ = 2.192°
4.Torsion of shafts
Answer: do = 30.82 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·750/60 = 78.54 rad/s
T = P/ω = 20000/78.54 = 254.6 N·m
Outer diameter from the allowable stress
T = 254.6 N·m = 254648 N·mm
1 − k⁴ = 1 − 0.8⁴ = 0.5904
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·254648/(π·75·0.5904)) = 30.82 mm
di = 0.8·30.82 = 24.66 mm
Answer
do = 30.82 mm
A solid shaft would need d = 25.86 mm; the hollow one is a little larger but much lighter.
5.Mohr’s circle & stress transformation
Answer: θp = 84.85 °
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
Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.
2θp = atan2(2·10, −40 − 70) = 169.7°
Angle on the element
θp = 84.85°
Answer
θp = 84.85°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: τθ = −2.246 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
τθ = −75·0.6428 + 60·0.766 = −2.246 MPa
Answer
τθ = −2.246 MPa
7.Axial stress & strain
Answer: δ = 2.089 mm
Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).
Stress
A = π·25²/4 = 490.9 mm²
σ = P/A = 120·1000/490.9 = 244.5 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 244.5/117000 = 0.002089 = 2089 µε
Elongation
δ = ε·L = 0.002089·1000 mm = 2.089 mm
Answer
δ = 2.089 mm
8.Axial stress & strain
Answer: δ = 6.623 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·40²/4 = 1257 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 75.6 MPa
σ2 = 840 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 75.6·500/117000 = 0.3231 mm
δ2 = σ2L2/E2 = 840·1500/200000 = 6.3 mm
δ = δ1 + δ2 = 6.623 mm
Answer
δ = 6.623 mm
9.Column buckling
Answer: Pcr = 157.9 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 = 60·40³/12 = 320000 mm⁴
Effective length
KL = 2·1000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·320000/2000² = 157914 N = 157.9 kN
Answer
Pcr = 157.9 kN
10.Column buckling
Answer: d = 99.38 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 = 2.5·105000 = 262500 N
KL = 2·3000 = 6000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·262500·6000²/(π³·200000))^(1/4) = 99.38 mm
Answer
d = 99.38 mm
Check: the slenderness KL/r is 241.5, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 10.07 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 = √(80000/15) = 73.03 rad/s
fn = 73.03/(2π) = 11.62 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 11.62·√(1 − 0.5²) = 10.07 Hz
Answer
fd = 10.07 Hz
12.Mechanical vibrations
Answer: keq = 17.14 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 = 1/40 + 1/30 → keq = 40·30/(40 + 30) = 17.14 kN/m
Answer
keq = 17.14 kN/m
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