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 1450 rpm drives a worm with 4 start(s). The worm wheel has 60 teeth and the drive has an efficiency of 60%. Find the output torque Tw on the wheel shaft.
z = 4Nw = 60n = 1450 rpmP = 0.75 kWη = 60%
Tw = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (20 teeth) on the input shaft turns at 2880 rpm and drives gear B (25 teeth). Gear C (40 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 1280 rpm. How many teeth ND must gear D have?
nin = 2880 rpmA = 20B = 25C = 40nout = 1280 rpm
ND =
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 = 1000 N·md = 40 mmL = 2.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid brass shaft of diameter 50 mm is fixed to rigid walls at both ends A and B. A torque of 250 N·m is applied at C, 0.5 m from A and 0.8 m from B. Find the reaction torque TA at wall A.
d = 50 mmT0 = 250 N·ma = 0.5 mb = 0.8 m
TA = N·m
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 30 MPa, σᵧ = −20 MPa and τₓᵧ = −50 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 = 30 MPaσy = −20 MPaτxy = −50 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = 60 MPa, σᵧ = 80 MPa and τₓᵧ = 20 MPa (σz = 0). The material has a yield strength of 420 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.
σx = 60 MPaσy = 80 MPaτxy = 20 MPaσY = 420 MPa
n =
7.Axial stress & strainMedium
Find the axial strain ε in the steel bar, in microstrain (µε = 10⁻⁶).
P = 10 kNd = 10 mmL = 1.5 mE = 200 GPa
ε = µε
8.Axial stress & strainHard
A steel segment (diameter 25 mm, length 400 mm, E = 200 GPa) and a segment made of brass (diameter 30 mm, length 400 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 50 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø25 mm, 400 mm, E = 200 GPasegment 2: brass, Ø30 mm, 400 mm, E = 100 GPaP = 50 kN
σ2 = 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 = 2 mE = 70 GPa (aluminium)b × h = 20 × 50 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is pinned at both ends. It must carry 25 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 25 kNn = 3L = 5 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed 3 cycles later by a peak of 5.1 mm. Find the damping ratio ζ.
x0 = 50 mmx3 = 5.1 mmn = 3
ζ =
12.Mechanical vibrationsHard
A 25 kg mass hangs from spring k₃ = 50 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 12 kN/m and k₂ = 80 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 = 12 kN/mk2 = 80 kN/mk3 = 50 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: Tw = 44.45 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 = 1450·4/60 = 96.67 rpm
Output torque
ωw = 2π·96.67/60 = 10.12 rad/s
Tw = η·P/ωw = 0.6·750/10.12 = 44.45 N·m
Answer
Tw = 44.45 N·m
2.Gears & gear trains
Answer: ND = 72
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.
Intermediate shaft speed
nB = nC = 2880·20/25 = 2304 rpm
Teeth on gear D
ND = C·nC/nout = 40·2304/1280 = 72
Answer
ND = 72
3.Torsion of shafts
Answer: φ = 7.124 °
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) = 1000·1000·2500/(80000·251327) = 0.1243 rad
φ = 0.1243·180/π = 7.124°
Answer
φ = 7.124°
4.Torsion of shafts
Answer: TA = 153.8 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 = 250
TA·a = TB·b
TA = 250·0.8/1.3 = 153.8 N·m
TB = 250·0.5/1.3 = 96.15 N·m
Answer
TA = 153.8 N·m
5.Mohr’s circle & stress transformation
Answer: θs = −76.72 °
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 = −50/55.9 = −0.8944, sin 2θs = −25/55.9 = −0.4472
2θs = atan2(−25, −50) = −153.4°
Angle on the element
θs = −76.72°
check: θs − θp = −76.72 − (−31.72) = −45°
Answer
θs = −76.72°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: n = 4.547
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 = 70 MPa, R = 22.36 MPa
σ1 = 92.36 MPa, σ2 = 47.64 MPa, σ3 = 0
Tresca
σmax − σmin = 92.36 MPa
n = 420/92.36 = 4.547
Answer
n = 4.547
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: ε = 636.6 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 = 10·1000/78.54 = 127.3 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 127.3/200000 = 0.0006366 = 636.6 µε
Answer
ε = 636.6 µε
8.Axial stress & strain
Answer: σ2 = 29.61 MPa
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.
Stiffnesses
k₁ = 200000·490.9/400 = 245437 N/mm
k₂ = 100000·706.9/400 = 176715 N/mm
Share of segment 2
F2 = 50000·176715/422152 = 20930 N
Stress
σ2 = 20930/706.9 = 29.61 MPa (compression)
Answer
σ2 = 29.61 MPa
9.Column buckling
Answer: Pcr = 5.757 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 = 50·20³/12 = 33333 mm⁴
Effective length
KL = 1·2000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·33333/2000² = 5757 N = 5.757 kN
Answer
Pcr = 5.757 kN
10.Column buckling
Answer: d = 86.23 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 = 3·25000 = 75000 N
KL = 1·5000 = 5000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·75000·5000²/(π³·70000))^(1/4) = 86.23 mm
Answer
d = 86.23 mm
Check: the slenderness KL/r is 231.9, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.1202
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.
Logarithmic decrement
δ = (1/3)·ln(50/5.1) = 0.7609
Damping ratio
ζ = δ/√(4π² + δ²) = 0.7609/√(39.48 + 0.579) = 0.1202
Answer
ζ = 0.1202
12.Mechanical vibrations
Answer: fn = 5.729 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 = 12 + 80 = 92 kN/m
keq = 92·50/(92 + 50) = 32.39 kN/m
Natural frequency
ωn = √(32394/25) = 36 rad/s
fn = 36/(2π) = 5.729 Hz
Answer
fn = 5.729 Hz
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