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 running at 1450 rpm drives a pinion with 12 teeth, which meshes with a gear with 32 teeth. Ignore losses. Find the output torque T2 on the gear shaft.
P = 2.2 kWn1 = 1450 rpmN1 = 12N2 = 32η = 100%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (16 teeth) on the input shaft turning at 1440 rpm drives gear B (32 teeth). Gear C (48 teeth) is fixed to the same shaft as B and drives gear D (60 teeth) on the output shaft. Find the output speed nout.
nin = 1440 rpmA = 16B = 32C = 48D = 60
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 = 250 N·md = 50 mmL = 1.2 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft must transmit 30 kW at 500 rpm. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter do.
P = 30 kWN = 500 rpmdi/do = 0.75τallow = 75 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −60 MPa, σᵧ = 60 MPa and τₓᵧ = 10 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = −60 MPaσy = 60 MPaτxy = 10 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 50 MPa, σᵧ = −20 MPa and τₓᵧ = 50 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 20° clockwise from the x-axis.
σx = 50 MPaσy = −20 MPaτxy = 50 MPaθ = -20°
σθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the steel bar under the axial load.
P = 30 kNd = 20 mmL = 1.5 mE = 200 GPa
δ = mm
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.
P = 25 kNsegment 1: Ø25 mm, 2 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 1 m, copper (E = 117 GPa)
σmax = MPa
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)d = 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is pinned at both ends. It must carry 20 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 20 kNn = 2.5L = 2 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 50 kg mass on a spring with a stiffness of 50 kN/m has a viscous damper giving a damping ratio of 0.45. Find the damped natural frequency fd of its free vibration in Hz.
m = 50 kgk = 50 kN/mζ = 0.45
fd = Hz
12.Mechanical vibrationsHard
A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 100 kN/m and k₂ = 100 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 25 kgk1 = 100 kN/mk2 = 100 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T2 = 38.64 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.
Motor torque
ω₁ = 2π·1450/60 = 151.8 rad/s
T1 = P/ω₁ = 2200/151.8 = 14.49 N·m
Output torque
T2 = 14.49·32/12·1 = 38.64 N·m
Answer
T2 = 38.64 N·m
2.Gears & gear trains
Answer: nout = 576 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 = 1440·16/32 = 720 rpm = nC
Second stage
nout = 720·48/60 = 576 rpm
Answer
nout = 576 rpm
total ratio = 2.5 : 1
3.Torsion of shafts
Answer: φ = 0.3502 °
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) = 250·1000·1200/(80000·613592) = 0.006112 rad
φ = 0.006112·180/π = 0.3502°
Answer
φ = 0.3502°
4.Torsion of shafts
Answer: do = 38.47 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·500/60 = 52.36 rad/s
T = P/ω = 30000/52.36 = 573 N·m
Outer diameter from the allowable stress
T = 573 N·m = 572958 N·mm
1 − k⁴ = 1 − 0.75⁴ = 0.6836
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·572958/(π·75·0.6836)) = 38.47 mm
di = 0.75·38.47 = 28.85 mm
Answer
do = 38.47 mm
A solid shaft would need d = 33.89 mm; the hollow one is a little larger but much lighter.
5.Mohr’s circle & stress transformation
Answer: τmax = 60.83 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.
Radius of Mohr’s circle
τmax = R = √(((−60 − 60)/2)² + 10²) = 60.83 MPa
Answer
τmax = 60.83 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 9.672 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 + 35·0.766 + 50·−0.6428 = 9.672 MPa
Answer
σθ = 9.672 MPa
7.Axial stress & strain
Answer: δ = 0.7162 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·20²/4 = 314.2 mm²
σ = P/A = 30·1000/314.2 = 95.49 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 95.49/200000 = 0.0004775 = 477.5 µε
Elongation
δ = ε·L = 0.0004775·1500 mm = 0.7162 mm
Answer
δ = 0.7162 mm
8.Axial stress & strain
Answer: σmax = 79.58 MPa
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 = 50.93 MPa
σ2 = 79.58 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 79.58 MPa
Answer
σmax = 79.58 MPa
9.Column buckling
Answer: Pcr = 313.9 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 round section has the same I about every axis.
I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴
Effective length
KL = 2·1000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·636173/2000² = 313939 N = 313.9 kN
Answer
Pcr = 313.9 kN
10.Column buckling
Answer: d = 49.28 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 = 2.5·20000 = 50000 N
KL = 1·2000 = 2000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·50000·2000²/(π³·70000))^(1/4) = 49.28 mm
Answer
d = 49.28 mm
Check: the slenderness KL/r is 162.3, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 4.495 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 = √(50000/50) = 31.62 rad/s
fn = 31.62/(2π) = 5.033 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 5.033·√(1 − 0.45²) = 4.495 Hz
Answer
fd = 4.495 Hz
12.Mechanical vibrations
Answer: keq = 200 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 = 100 + 100 = 200 kN/m
Answer
keq = 200 kN/m
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