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 3 kW motor running at 1440 rpm drives a pinion with 24 teeth and a module of 2 mm. Find the tangential force Ft on the pinion teeth.
P = 3 kWn = 1440 rpmN1 = 24m = 2 mm
Ft = N
2.Gears & gear trainsHard
In a compound gear train, gear A (12 teeth) on the input shaft turns at 960 rpm and drives gear B (30 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 288 rpm. How many teeth ND must gear D have?
nin = 960 rpmA = 12B = 30C = 15nout = 288 rpm
ND =
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 = 200 N·md = 30 mmL = 2 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft transmits 100 kW at 300 rpm. Find the maximum shear stress τmax in the shaft.
P = 100 kWN = 300 rpmdo = 50 mmdi = 35 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 60 MPa and σᵧ = 90 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 100 MPa?
σx = 60 MPaσy = 90 MPaσ1 = 100 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 130 MPa, σᵧ = 90 MPa and τₓᵧ = −60 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 50° counter-clockwise from the x-axis.
σx = 130 MPaσy = 90 MPaτxy = −60 MPaθ = 50°
σθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the steel bar under the axial load.
P = 75 kNd = 10 mmL = 1 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 total elongation δ.
P = 20 kNsegment 1: Ø35 mm, 2 m, aluminium (E = 70 GPa)segment 2: Ø16 mm, 2 m, copper (E = 117 GPa)
δ = mm
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 mE = 70 GPa (aluminium)d = 20 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 50 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 50 kNn = 2.5L = 2 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 15 kg mass is supported by a spring with a stiffness of 5 kN/m and a viscous damper with a damping coefficient of 80 N·s/m. Find the damping ratio ζ.
m = 15 kgk = 5 kN/mc = 80 N·s/m
ζ =
12.Mechanical vibrationsHard
A 100 kg mass hangs from spring k₃ = 30 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 100 kN/m and k₂ = 12 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 = 100 kgk1 = 100 kN/mk2 = 12 kN/mk3 = 30 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: Ft = 828.9 N
Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.
Torque on the pinion
ω = 2π·1440/60 = 150.8 rad/s
T = P/ω = 3000/150.8 = 19.89 N·m
Pitch diameter
d = m·N = 2·24 = 48 mm
Tangential force
Ft = 2T/d = 2·19.89/0.048 = 828.9 N
Answer
Ft = 828.9 N
2.Gears & gear trains
Answer: ND = 20
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 = 960·12/30 = 384 rpm
Teeth on gear D
ND = C·nC/nout = 15·384/288 = 20
Answer
ND = 20
3.Torsion of shafts
Answer: φ = 7.39 °
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·30⁴/32 = 79522 mm⁴
Angle of twist
φ = TL/(GJ) = 200·1000·2000/(39000·79522) = 0.129 rad
φ = 0.129·180/π = 7.39°
Answer
φ = 7.39°
4.Torsion of shafts
Answer: τmax = 170.7 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.
Torque
ω = 2πN/60 = 2π·300/60 = 31.42 rad/s
T = P/ω = 100000/31.42 = 3183 N·m
Polar second moment of area
J = π(50⁴ − 35⁴)/32 = 466269 mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 3.183 × 10⁶·25/466269 = 170.7 MPa
Answer
τmax = 170.7 MPa
5.Mohr’s circle & stress transformation
Answer: τxy = 20 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 = (60 + 90)/2 = 75 MPa
R = σ1 − C = 100 − 75 = 25 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(25² − 15²) = 20 MPa
Answer
τxy = 20 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 47.44 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 = 110 + 20·−0.1736 + −60·0.9848 = 47.44 MPa
Answer
σθ = 47.44 MPa
7.Axial stress & strain
Answer: δ = 4.775 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·10²/4 = 78.54 mm²
σ = P/A = 75·1000/78.54 = 954.9 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 954.9/200000 = 0.004775 = 4775 µε
Elongation
δ = ε·L = 0.004775·1000 mm = 4.775 mm
Answer
δ = 4.775 mm
8.Axial stress & strain
Answer: δ = 2.294 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·16²/4 = 201.1 mm²
σ1 = 20.79 MPa
σ2 = 99.47 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 20.79·2000/70000 = 0.5939 mm
δ2 = σ2L2/E2 = 99.47·2000/117000 = 1.7 mm
δ = δ1 + δ2 = 2.294 mm
Answer
δ = 2.294 mm
9.Column buckling
Answer: Pcr = 5.426 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·20⁴/64 = 7854 mm⁴
Effective length
KL = 1·1000 = 1000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·7854/1000² = 5426 N = 5.426 kN
Answer
Pcr = 5.426 kN
10.Column buckling
Answer: d = 67.4 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 = 2.5·50000 = 125000 N
KL = 2·2000 = 4000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·125000·4000²/(π³·200000))^(1/4) = 67.4 mm
Answer
d = 67.4 mm
Check: the slenderness KL/r is 237.4, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.1461
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·√(5000·15) = 547.7 N·s/m
Damping ratio
ζ = c/cc = 80/547.7 = 0.1461
Answer
ζ = 0.1461
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 2.448 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 = 100 + 12 = 112 kN/m
keq = 112·30/(112 + 30) = 23.66 kN/m
Natural frequency
ωn = √(23662/100) = 15.38 rad/s
fn = 15.38/(2π) = 2.448 Hz
Answer
fn = 2.448 Hz
ExamProblems · examproblems.com/courses/machine-design