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 2900 rpm drives a pinion with 32 teeth, which meshes with a gear with 48 teeth. The gear pair has an efficiency of 95%. Find the torque T1 on the motor shaft.
P = 3 kWn1 = 2900 rpmN1 = 32N2 = 48η = 95%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (32 teeth) on the input shaft turning at 1450 rpm drives gear B (40 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D (32 teeth) on the output shaft. Find the output speed nout.
nin = 1450 rpmA = 32B = 40C = 12D = 32
nout = rpm
3.Torsion of shaftsMedium
In a torsion test, a solid specimen of diameter 15 mm is twisted by a torque of 20 N·m. The two ends of a gauge length of 200 mm rotate 1.77° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 20 N·md = 15 mmL = 200 mmφ = 1.77°
G = GPa
4.Torsion of shaftsHard
A solid shaft must transmit 20 kW at 3000 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.
P = 20 kWN = 3000 rpmτallow = 60 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 40 MPa, σᵧ = −70 MPa and τₓᵧ = −20 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 40 MPaσy = −70 MPaτxy = −20 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = 30 MPa, σᵧ = −20 MPa and τₓᵧ = −60 MPa (σz = 0). The material has a yield strength of 300 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.
σx = 30 MPaσy = −20 MPaτxy = −60 MPaσY = 300 MPa
n =
7.Axial stress & strainMedium
Find the elongation δ of the brass bar under the axial load.
P = 85 kNd = 20 mmL = 2 mE = 100 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 = 65 kNsegment 1: Ø40 mm, 2 m, copper (E = 117 GPa)segment 2: Ø16 mm, 2 m, brass (E = 100 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 1.5 mE = 70 GPa (aluminium)d = 25 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 100 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 100 kNn = 2.5L = 2 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 10 kg mass on a spring with a stiffness of 2 kN/m has a viscous damper giving a damping ratio of 0.7. Find the damped natural frequency fd of its free vibration in Hz.
m = 10 kgk = 2 kN/mζ = 0.7
fd = Hz
12.Mechanical vibrationsHard
A 5 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 50 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 = 50 kN/mk2 = 100 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 9.879 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·2900/60 = 303.7 rad/s
T1 = P/ω₁ = 3000/303.7 = 9.879 N·m
Answer
T1 = 9.879 N·m
2.Gears & gear trains
Answer: nout = 435 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 = 1450·32/40 = 1160 rpm = nC
Second stage
nout = 1160·12/32 = 435 rpm
Answer
nout = 435 rpm
total ratio = 3.333 : 1
3.Torsion of shafts
Answer: G = 26.05 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 = 1.77·π/180 = 0.03089 rad
Shear modulus
T = 20 N·m = 20000 N·mm
G = TL/(Jφ) = 20000·200/(4970·0.03089) = 26052 MPa
G = 26.05 GPa
Answer
G = 26.05 GPa
Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.
4.Torsion of shafts
Answer: d = 17.55 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·3000/60 = 314.2 rad/s
T = P/ω = 20000/314.2 = 63.66 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·63662/(π·60)) = 17.55 mm
Answer
d = 17.55 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τmax = 58.52 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.
Radius of Mohr’s circle
τmax = R = √(((40 − −70)/2)² + −20²) = 58.52 MPa
Answer
τmax = 58.52 MPa
6.Mohr’s circle & stress transformation
Answer: n = 2.662
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 = 5 MPa, R = 65 MPa
σ1 = 70 MPa, σ2 = −60 MPa, σ3 = 0
Von Mises
σvM = √(70² − 70·(−60) + (−60)²) = 112.7 MPa
n = 300/112.7 = 2.662
Answer
n = 2.662
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: δ = 5.411 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·20²/4 = 314.2 mm²
σ = P/A = 85·1000/314.2 = 270.6 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 270.6/100000 = 0.002706 = 2706 µε
Elongation
δ = ε·L = 0.002706·2000 mm = 5.411 mm
Answer
δ = 5.411 mm
8.Axial stress & strain
Answer: δ = 7.35 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·40²/4 = 1257 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 51.73 MPa
σ2 = 323.3 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 51.73·2000/117000 = 0.8842 mm
δ2 = σ2L2/E2 = 323.3·2000/100000 = 6.466 mm
δ = δ1 + δ2 = 7.35 mm
Answer
δ = 7.35 mm
9.Column buckling
Answer: Pcr = 12.02 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·25⁴/64 = 19175 mm⁴
Effective length
KL = 0.7·1500 = 1050 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·19175/1050² = 12016 N = 12.02 kN
Answer
Pcr = 12.02 kN
10.Column buckling
Answer: d = 80.16 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 = 2.5·100000 = 250000 N
KL = 2·2000 = 4000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·250000·4000²/(π³·200000))^(1/4) = 80.16 mm
Answer
d = 80.16 mm
Check: the slenderness KL/r is 199.6, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 1.607 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 = √(2000/10) = 14.14 rad/s
fn = 14.14/(2π) = 2.251 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 2.251·√(1 − 0.7²) = 1.607 Hz
Answer
fd = 1.607 Hz
12.Mechanical vibrations
Answer: fn = 27.57 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 = 50 + 100 = 150 kN/m
Natural frequency
ωn = √(150000/5) = 173.2 rad/s
fn = 173.2/(2π) = 27.57 Hz
Answer
fn = 27.57 Hz
ExamProblems · examproblems.com/courses/machine-design