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 15 kW motor running at 1440 rpm drives a pinion with 15 teeth, which meshes with a gear with 50 teeth. The gear pair has an efficiency of 96%. Find the torque T1 on the motor shaft.
P = 15 kWn1 = 1440 rpmN1 = 15N2 = 50η = 96%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (36 teeth) on the input shaft turning at 2900 rpm drives gear B (45 teeth). Gear C (45 teeth) is fixed to the same shaft as B and drives gear D (48 teeth) on the output shaft. Find the output speed nout.
nin = 2900 rpmA = 36B = 45C = 45D = 48
nout = rpm
3.Torsion of shaftsMedium
A solid aluminium 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 = 400 N·md = 50 mmL = 1.5 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft must transmit 50 kW at 1500 rpm. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 60 MPa. Find the smallest outer diameter do.
P = 50 kWN = 1500 rpmdi/do = 0.7τallow = 60 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −10 MPa, σᵧ = 30 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 = −10 MPaσy = 30 MPaτxy = −50 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = 50 MPa, σᵧ = 90 MPa and τₓᵧ = 60 MPa (σz = 0). Find the von Mises equivalent stress σvM.
σx = 50 MPaσy = 90 MPaτxy = 60 MPa
σvM = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the aluminium bar, in microstrain (µε = 10⁻⁶).
P = 105 kNd = 25 mmL = 1.5 mE = 70 GPa
ε = µε
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 = 100 kNsegment 1: Ø30 mm, 2 m, steel (E = 200 GPa)segment 2: Ø12 mm, 0.5 m, copper (E = 117 GPa)
σmax = MPa
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 = 5 mE = 200 GPa (steel)b × h = 30 × 40 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 15 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 15 kNn = 2.5L = 4 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 5 kg mass is supported by a spring with a stiffness of 50 kN/m and a viscous damper with a damping coefficient of 50 N·s/m. Find the damping ratio ζ.
m = 5 kgk = 50 kN/mc = 50 N·s/m
ζ =
12.Mechanical vibrationsHard
A 40 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 15 kN/m and the lower one k₂ = 5 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 40 kgk1 = 15 kN/mk2 = 5 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 99.47 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 = 15000/150.8 = 99.47 N·m
Answer
T1 = 99.47 N·m
2.Gears & gear trains
Answer: nout = 2175 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 = 2900·36/45 = 2320 rpm = nC
Second stage
nout = 2320·45/48 = 2175 rpm
Answer
nout = 2175 rpm
total ratio = 1.333 : 1
3.Torsion of shafts
Answer: φ = 2.155 °
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·50⁴/32 = 613592 mm⁴
Angle of twist
φ = TL/(GJ) = 400·1000·1500/(26000·613592) = 0.03761 rad
φ = 0.03761·180/π = 2.155°
Answer
φ = 2.155°
4.Torsion of shafts
Answer: do = 32.88 mm
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·1500/60 = 157.1 rad/s
T = P/ω = 50000/157.1 = 318.3 N·m
Outer diameter from the allowable stress
T = 318.3 N·m = 318310 N·mm
1 − k⁴ = 1 − 0.7⁴ = 0.7599
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·318310/(π·60·0.7599)) = 32.88 mm
di = 0.7·32.88 = 23.02 mm
Answer
do = 32.88 mm
A solid shaft would need d = 30.01 mm; the hollow one is a little larger but much lighter.
5.Mohr’s circle & stress transformation
Answer: θs = 79.1 °
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 = −50/53.85 = −0.9285, sin 2θs = −(−20)/53.85 = 0.3714
2θs = atan2(20, −50) = 158.2°
Angle on the element
θs = 79.1°
check: θs − θp = 79.1 − (−55.9) = 135°
Answer
θs = 79.1°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σvM = 130 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 = 70 MPa, R = 63.25 MPa
σ1 = 133.2 MPa, σ2 = 6.754 MPa, σ3 = 0
Von Mises
σvM = √(133.2² − 133.2·6.754 + 6.754²) = 130 MPa
Answer
σvM = 130 MPa
7.Axial stress & strain
Answer: ε = 3056 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·25²/4 = 490.9 mm²
σ = P/A = 105·1000/490.9 = 213.9 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 213.9/70000 = 0.003056 = 3056 µε
Answer
ε = 3056 µε
8.Axial stress & strain
Answer: σmax = 884.2 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 = 141.5 MPa
σ2 = 884.2 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 884.2 MPa
Answer
σmax = 884.2 MPa
9.Column buckling
Answer: Pcr = 14.5 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 = 40·30³/12 = 90000 mm⁴
Effective length
KL = 0.7·5000 = 3500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·90000/3500² = 14502 N = 14.5 kN
Answer
Pcr = 14.5 kN
10.Column buckling
Answer: d = 41.74 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 = 2.5·15000 = 37500 N
KL = 0.7·4000 = 2800 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·37500·2800²/(π³·200000))^(1/4) = 41.74 mm
Answer
d = 41.74 mm
Check: the slenderness KL/r is 268.4, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.05
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·√(50000·5) = 1000 N·s/m
Damping ratio
ζ = c/cc = 50/1000 = 0.05
Answer
ζ = 0.05
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 1.541 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 = 1/15 + 1/5 → keq = 15·5/(15 + 5) = 3.75 kN/m
Natural frequency
ωn = √(3750/40) = 9.682 rad/s
fn = 9.682/(2π) = 1.541 Hz
Answer
fn = 1.541 Hz
ExamProblems · examproblems.com/courses/machine-design