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 at 2900 rpm drives a worm with 2 start(s). The worm wheel has 40 teeth and the drive has an efficiency of 55%. Find the output torque Tw on the wheel shaft.
z = 2Nw = 40n = 2900 rpmP = 0.75 kWη = 55%
Tw = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (16 teeth) on the input shaft turning at 1500 rpm drives gear B (45 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D (15 teeth) on the output shaft. Find the output speed nout.
nin = 1500 rpmA = 16B = 45C = 12D = 15
nout = rpm
3.Torsion of shaftsMedium
In a torsion test, a solid specimen of diameter 15 mm is twisted by a torque of 50 N·m. The two ends of a gauge length of 250 mm rotate 1.8° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 50 N·md = 15 mmL = 250 mmφ = 1.8°
G = GPa
4.Torsion of shaftsHard
A solid steel shaft of diameter 50 mm is fixed at A. A torque of 200 N·m acts at B, 0.5 m from A, and a torque of 250 N·m in the opposite direction acts at the free end C, a further 3 m along. Find the magnitude of the angle of twist φC of end C, in degrees.
d = 50 mmTB = 200 N·mTC = 250 N·mAB = 0.5 mBC = 3 mG = 80 GPa
φC = °
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −40 MPa, σᵧ = 0 MPa and τₓᵧ = 30 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 = −40 MPaσy = 0 MPaτxy = 30 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = 0 MPa, σᵧ = −80 MPa and τₓᵧ = −20 MPa (σz = 0). Find the von Mises equivalent stress σvM.
σx = 0 MPaσy = −80 MPaτxy = −20 MPa
σvM = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 25 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 60 mm, length 300 mm, E = 110 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.
segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: bronze, Ø60 mm, 300 mm, E = 110 GPaδ = 0.5 mm
P = kN
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 = 35 kNsegment 1: Ø30 mm, 0.5 m, steel (E = 200 GPa)segment 2: Ø12 mm, 2 m, aluminium (E = 70 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 1.5 mE = 70 GPa (aluminium)b × h = 40 × 50 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 = 6 mE = 200 GPa (steel)d = 80 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 15 kg mass on a spring with a stiffness of 150 kN/m has a viscous damper giving a damping ratio of 0.4. Find the damped natural frequency fd of its free vibration in Hz.
m = 15 kgk = 150 kN/mζ = 0.4
fd = Hz
12.Mechanical vibrationsHard
A 5 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 60 kN/m and the lower one k₂ = 100 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 5 kgk1 = 60 kN/mk2 = 100 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: Tw = 27.17 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 = 2900·2/40 = 145 rpm
Output torque
ωw = 2π·145/60 = 15.18 rad/s
Tw = η·P/ωw = 0.55·750/15.18 = 27.17 N·m
Answer
Tw = 27.17 N·m
2.Gears & gear trains
Answer: nout = 426.7 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 = 1500·16/45 = 533.3 rpm = nC
Second stage
nout = 533.3·12/15 = 426.7 rpm
Answer
nout = 426.7 rpm
total ratio = 3.516 : 1
3.Torsion of shafts
Answer: G = 80.06 GPa
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 moment and angle
J = π·15⁴/32 = 4970 mm⁴
φ = 1.8·π/180 = 0.03142 rad
Shear modulus
T = 50 N·m = 50000 N·mm
G = TL/(Jφ) = 50000·250/(4970·0.03142) = 80056 MPa
G = 80.06 GPa
Answer
G = 80.06 GPa
Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.
4.Torsion of shafts
Answer: φC = 0.9046 °
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.
Internal torques
Sense of TC taken as positive.
TBC = TC = 250 N·m
TAB = TC − TB = 250 − 200 = 50 N·m
Twist of each segment
J = π·50⁴/32 = 613592 mm⁴
G = 80 GPa = 80000 MPa
φAB = 50000·500/(80000·613592) = 0.0005093 rad
φBC = 250000·3000/(80000·613592) = 0.01528 rad
Twist of end C
φC = 0.0005093 + 0.01528 = 0.01579 rad → |φC| = 0.9046°
Answer
φC = 0.9046°
5.Mohr’s circle & stress transformation
Answer: θs = 16.85 °
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
On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.
cos 2θs = 30/36.06 = 0.8321, sin 2θs = −(−20)/36.06 = 0.5547
2θs = atan2(20, 30) = 33.69°
Angle on the element
θs = 16.85°
check: θs − θp = 16.85 − 61.85 = −45°
Answer
θs = 16.85°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σvM = 87.18 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. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.
Principal stresses
C = −40 MPa, R = 44.72 MPa
σ1 = 4.721 MPa, σ2 = −84.72 MPa, σ3 = 0
Von Mises
σvM = √(4.721² − 4.721·(−84.72) + (−84.72)²) = 87.18 MPa
Answer
σvM = 87.18 MPa
7.Axial stress & strain
Answer: P = 124.4 kN
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
A₁ = 490.9 mm², A₂ = 2827 mm²
Flexibilities
L₁/(E₁A₁) = 300/(200000·490.9) = 3.056 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(110000·2827) = 9.646 × 10⁻⁷ mm/N
Load
P = 0.5/4.02 × 10⁻⁶ = 124367 N = 124.4 kN
Answer
P = 124.4 kN
8.Axial stress & strain
Answer: σmax = 309.5 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·30²/4 = 706.9 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 49.51 MPa
σ2 = 309.5 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 309.5 MPa
Answer
σmax = 309.5 MPa
9.Column buckling
Answer: Pcr = 81.88 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 = 50·40³/12 = 266667 mm⁴
Effective length
KL = 1·1500 = 1500 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·266667/1500² = 81881 N = 81.88 kN
Answer
Pcr = 81.88 kN
10.Column buckling
Answer: σcr = 87.73 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 = 0.5·6000 = 3000 mm
KL/r = 3000/20 = 150
Critical stress
σcr = π²E/(KL/r)² = π²·200000/150² = 87.73 MPa
Answer
σcr = 87.73 MPa
11.Mechanical vibrations
Answer: fd = 14.59 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 = √(150000/15) = 100 rad/s
fn = 100/(2π) = 15.92 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 15.92·√(1 − 0.4²) = 14.59 Hz
Answer
fd = 14.59 Hz
12.Mechanical vibrations
Answer: keq = 37.5 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 = 1/60 + 1/100 → keq = 60·100/(60 + 100) = 37.5 kN/m
Answer
keq = 37.5 kN/m
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