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
An 11 kW motor running at 1440 rpm drives a pinion with 25 teeth, which meshes with a gear with 32 teeth. The gear pair has an efficiency of 95%. Find the output torque T2 on the gear shaft.
P = 11 kWn1 = 1440 rpmN1 = 25N2 = 32η = 95%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (20 teeth) on the input shaft turning at 3000 rpm drives gear B (60 teeth). Gear C (32 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 = 3000 rpmA = 20B = 60C = 32D = 80
nout = rpm
3.Torsion of shaftsMedium
A hollow brass shaft (outer diameter 50 mm, inner diameter 35 mm, length 3 m) is fixed at one end and twisted by a torque of 1500 N·m at the other. Find the angle of twist φ in degrees.
T = 1500 N·mdo = 50 mmdi = 35 mmL = 3 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid brass shaft of diameter 25 mm is fixed at A. A torque of 400 N·m acts at B, 1.5 m from A, and a torque of 200 N·m in the same direction acts at the free end C, a further 0.5 m along. Find the maximum shear stress τmax in the shaft.
d = 25 mmTB = 400 N·mTC = 200 N·mAB = 1.5 mBC = 0.5 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 100 MPa and σᵧ = 30 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 155 MPa?
σx = 100 MPaσy = 30 MPaσ1 = 155 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = −40 MPa, σᵧ = −30 MPa and τₓᵧ = 10 MPa (σz = 0). The material has a yield strength of 250 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.
σx = −40 MPaσy = −30 MPaτxy = 10 MPaσY = 250 MPa
n =
7.Axial stress & strainMedium
Find the elongation δ of the steel bar under the axial load.
P = 25 kNd = 12 mmL = 0.5 mE = 200 GPa
δ = mm
8.Axial stress & strainHard
A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of bronze (outer diameter 120 mm, inner diameter 50 mm, E = 110 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø20 mm, E = 200 GPa, σallow = 160 MPatube: Ø120/Ø50 mm, E = 110 GPa, σallow = 50 MPa
Pmax = kN
9.Column bucklingMedium
A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 6 mE = 70 GPa (aluminium)b × h = 20 × 50 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 4 mE = 70 GPa (aluminium)d = 80 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 50 kg mass is supported by a spring with a stiffness of 5 kN/m and a viscous damper with a damping coefficient of 300 N·s/m. Find the damping ratio ζ.
m = 50 kgk = 5 kN/mc = 300 N·s/m
ζ =
12.Mechanical vibrationsHard
A 5 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 100 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 5 kgk1 = 40 kN/mk2 = 100 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 88.7 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·1440/60 = 150.8 rad/s
T1 = P/ω₁ = 11000/150.8 = 72.95 N·m
Output torque
T2 = 72.95·32/25·0.95 = 88.7 N·m
Answer
T2 = 88.7 N·m
2.Gears & gear trains
Answer: nout = 400 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 = 3000·20/60 = 1000 rpm = nC
Second stage
nout = 1000·32/80 = 400 rpm
Answer
nout = 400 rpm
total ratio = 7.5 : 1
3.Torsion of shafts
Answer: φ = 14.18 °
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 − 35⁴)/32 = 466269 mm⁴
Angle of twist
T = 1500 N·m = 1.5 × 10⁶ N·mm
L = 3 m = 3000 mm
G = 39 GPa = 39000 MPa
φ = TL/(GJ) = 1.5 × 10⁶·3000/(39000·466269) = 0.2475 rad
φ = 0.2475·180/π = 14.18°
Answer
φ = 14.18°
4.Torsion of shafts
Answer: τmax = 195.6 MPa
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 = 200 N·m
TAB = TC + TB = 200 + 400 = 600 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 600 N·m
τmax = 16·600000/(π·25³) = 195.6 MPa
Answer
τmax = 195.6 MPa
5.Mohr’s circle & stress transformation
Answer: τxy = 82.92 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.
Required radius
C = (100 + 30)/2 = 65 MPa
R = σ1 − C = 155 − 65 = 90 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(90² − 35²) = 82.92 MPa
Answer
τxy = 82.92 MPa
6.Mohr’s circle & stress transformation
Answer: n = 5.414
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 = −35 MPa, R = 11.18 MPa
σ1 = −23.82 MPa, σ2 = −46.18 MPa, σ3 = 0
Tresca
σmax − σmin = 46.18 MPa
n = 250/46.18 = 5.414
Answer
n = 5.414
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: δ = 0.5526 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·12²/4 = 113.1 mm²
σ = P/A = 25·1000/113.1 = 221 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 221/200000 = 0.001105 = 1105 µε
Elongation
δ = ε·L = 0.001105·500 mm = 0.5526 mm
Answer
δ = 0.5526 mm
8.Axial stress & strain
Answer: Pmax = 495.9 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
Ar = π·20²/4 = 314.2 mm²
At = π(120² − 50²)/4 = 9346 mm²
Combined axial stiffness
ΣEA = 200·10³·314.2 + 110·10³·9346 = 1.091 × 10⁹ N
Governing strain
ε = min(0.0008, 0.0004545) = 0.0004545 (bronze)
Largest load
Pmax = 0.0004545·1.091 × 10⁹ = 495872 N = 495.9 kN
Answer
Pmax = 495.9 kN
9.Column buckling
Answer: Pcr = 2.559 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 = 0.5·6000 = 3000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·33333/3000² = 2559 N = 2.559 kN
Answer
Pcr = 2.559 kN
10.Column buckling
Answer: σcr = 17.27 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 = 1·4000 = 4000 mm
KL/r = 4000/20 = 200
Critical stress
σcr = π²E/(KL/r)² = π²·70000/200² = 17.27 MPa
Answer
σcr = 17.27 MPa
11.Mechanical vibrations
Answer: ζ = 0.3
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.
Critical damping
cc = 2·√(k·m) = 2·√(5000·50) = 1000 N·s/m
Damping ratio
ζ = c/cc = 300/1000 = 0.3
Answer
ζ = 0.3
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 26.63 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 40 + 100 = 140 kN/m
Natural frequency
ωn = √(140000/5) = 167.3 rad/s
fn = 167.3/(2π) = 26.63 Hz
Answer
fn = 26.63 Hz
ExamProblems · examproblems.com/courses/machine-design