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 0.75 kW motor running at 2900 rpm drives a pinion with 24 teeth, which meshes with a gear with 48 teeth. The gear pair has an efficiency of 95%. Find the output torque T2 on the gear shaft.
P = 0.75 kWn1 = 2900 rpmN1 = 24N2 = 48η = 95%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (12 teeth) on the input shaft turning at 960 rpm drives gear B (80 teeth). Gear C (20 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 = 960 rpmA = 12B = 80C = 20D = 24
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 = 1000 N·md = 40 mmL = 1.2 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid brass shaft of diameter 60 mm is fixed to rigid walls at both ends A and B. A torque of 200 N·m is applied at C, 1 m from A and 2 m from B. Find the reaction torque TA at wall A.
d = 60 mmT0 = 200 N·ma = 1 mb = 2 m
TA = N·m
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 20 MPa, σᵧ = 0 MPa and τₓᵧ = −40 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = 20 MPaσy = 0 MPaτxy = −40 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −80 MPa, σᵧ = −70 MPa and τₓᵧ = 40 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° clockwise from the x-axis.
σx = −80 MPaσy = −70 MPaτxy = 40 MPaθ = -30°
σθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the steel bar, in microstrain (µε = 10⁻⁶).
P = 30 kNd = 10 mmL = 0.5 mE = 200 GPa
ε = µε
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 100 kNsegment 1: Ø35 mm, 1 m, steel (E = 200 GPa)segment 2: Ø20 mm, 1 m, steel (E = 200 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 = 4 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 the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 5 mE = 200 GPa (steel)d = 80 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 12 kg mass on a spring with a stiffness of 20 kN/m has a viscous damper giving a damping ratio of 0.3. Find the damped natural frequency fd of its free vibration in Hz.
m = 12 kgk = 20 kN/mζ = 0.3
fd = Hz
12.Mechanical vibrationsHard
A 10 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 80 kN/m and the lower one k₂ = 30 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 10 kgk1 = 80 kN/mk2 = 30 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 4.692 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·2900/60 = 303.7 rad/s
T1 = P/ω₁ = 750/303.7 = 2.47 N·m
Output torque
T2 = 2.47·48/24·0.95 = 4.692 N·m
Answer
T2 = 4.692 N·m
2.Gears & gear trains
Answer: nout = 120 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 = 960·12/80 = 144 rpm = nC
Second stage
nout = 144·20/24 = 120 rpm
Answer
nout = 120 rpm
total ratio = 8 : 1
3.Torsion of shafts
Answer: φ = 7.015 °
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) = 1000·1000·1200/(39000·251327) = 0.1224 rad
φ = 0.1224·180/π = 7.015°
Answer
φ = 7.015°
4.Torsion of shafts
Answer: TA = 133.3 N·m
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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).
Equilibrium and compatibility
TA + TB = 200
TA·a = TB·b
TA = 200·2/3 = 133.3 N·m
TB = 200·1/3 = 66.67 N·m
Answer
TA = 133.3 N·m
5.Mohr’s circle & stress transformation
Answer: θp = −37.98 °
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·−40, 20 − 0) = −75.96°
Angle on the element
θp = −37.98°
Answer
θp = −37.98°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = −112.1 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 = −75 + −5·0.5 + 40·−0.866 = −112.1 MPa
Answer
σθ = −112.1 MPa
7.Axial stress & strain
Answer: ε = 1910 µε
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·10²/4 = 78.54 mm²
σ = P/A = 30·1000/78.54 = 382 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 382/200000 = 0.00191 = 1910 µε
Answer
ε = 1910 µε
8.Axial stress & strain
Answer: δ = 2.111 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·35²/4 = 962.1 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 103.9 MPa
σ2 = 318.3 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 103.9·1000/200000 = 0.5197 mm
δ2 = σ2L2/E2 = 318.3·1000/200000 = 1.592 mm
δ = δ1 + δ2 = 2.111 mm
Answer
δ = 2.111 mm
9.Column buckling
Answer: Pcr = 107.4 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 = 0.7·4000 = 2800 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·426667/2800² = 107424 N = 107.4 kN
Answer
Pcr = 107.4 kN
10.Column buckling
Answer: σcr = 7.896 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·80⁴/64 = 2.011 × 10⁶ mm⁴
Radius of gyration and slenderness
A = 5027 mm²
r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm
KL = 2·5000 = 10000 mm
KL/r = 10000/20 = 500
Critical stress
σcr = π²E/(KL/r)² = π²·200000/500² = 7.896 MPa
Answer
σcr = 7.896 MPa
11.Mechanical vibrations
Answer: fd = 6.198 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 = √(20000/12) = 40.82 rad/s
fn = 40.82/(2π) = 6.497 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 6.497·√(1 − 0.3²) = 6.198 Hz
Answer
fd = 6.198 Hz
12.Mechanical vibrations
Answer: fn = 7.434 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
Springs in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
1/keq = 1/80 + 1/30 → keq = 80·30/(80 + 30) = 21.82 kN/m
Natural frequency
ωn = √(21818/10) = 46.71 rad/s
fn = 46.71/(2π) = 7.434 Hz
Answer
fn = 7.434 Hz
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