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 1.1 kW motor running at 1440 rpm drives a pinion with 12 teeth, which meshes with a gear with 80 teeth. The gear pair has an efficiency of 95%. Find the output torque T2 on the gear shaft.
P = 1.1 kWn1 = 1440 rpmN1 = 12N2 = 80η = 95%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (48 teeth) on the input shaft turning at 1450 rpm drives gear B (80 teeth). Gear C (20 teeth) is fixed to the same shaft as B and drives gear D (25 teeth) on the output shaft. Find the output speed nout.
nin = 1450 rpmA = 48B = 80C = 20D = 25
nout = rpm
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 = 100 N·md = 30 mmL = 3 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft must transmit 10 kW at 1500 rpm. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter do.
P = 10 kWN = 1500 rpmdi/do = 0.75τallow = 50 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 50 MPa and σᵧ = −30 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 85 MPa?
σx = 50 MPaσy = −30 MPaσ1 = 85 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −50 MPa, σᵧ = 20 MPa and τₓᵧ = 40 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 45° counter-clockwise from the x-axis.
σx = −50 MPaσy = 20 MPaτxy = 40 MPaθ = 45°
σθ = MPa
7.Axial stress & strainMedium
A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 100 mm, inner diameter 50 mm, E = 100 GPa). Both are 800 mm long and are compressed together between rigid plates by 200 kN. Find the shortening δ.
P = 200 kNrod: Ø30 mm, E = 200 GPatube: Ø100/Ø50 mm, E = 100 GPaL = 800 mm
δ = mm
8.Axial stress & strainHard
A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 70 mm, E = 110 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø25 mm, E = 200 GPa, σallow = 120 MPatube: Ø100/Ø70 mm, E = 110 GPa, σallow = 60 MPa
Pmax = kN
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 = 1 mE = 200 GPa (steel)b × h = 25 × 50 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 85 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 85 kNn = 2L = 3 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 25 kg mass on a spring with a stiffness of 4 kN/m has a viscous damper giving a damping ratio of 0.3. Find the damped natural frequency fd of its free vibration in Hz.
m = 25 kgk = 4 kN/mζ = 0.3
fd = Hz
12.Mechanical vibrationsHard
A 10 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 100 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 = 10 kgk1 = 100 kN/mk2 = 5 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 46.2 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 = 1100/150.8 = 7.295 N·m
Output torque
T2 = 7.295·80/12·0.95 = 46.2 N·m
Answer
T2 = 46.2 N·m
2.Gears & gear trains
Answer: nout = 696 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 = 1450·48/80 = 870 rpm = nC
Second stage
nout = 870·20/25 = 696 rpm
Answer
nout = 696 rpm
total ratio = 2.083 : 1
3.Torsion of shafts
Answer: φ = 2.702 °
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) = 100·1000·3000/(80000·79522) = 0.04716 rad
φ = 0.04716·180/π = 2.702°
Answer
φ = 2.702°
4.Torsion of shafts
Answer: do = 21.17 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 = 10000/157.1 = 63.66 N·m
Outer diameter from the allowable stress
T = 63.66 N·m = 63662 N·mm
1 − k⁴ = 1 − 0.75⁴ = 0.6836
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·63662/(π·50·0.6836)) = 21.17 mm
di = 0.75·21.17 = 15.88 mm
Answer
do = 21.17 mm
A solid shaft would need d = 18.65 mm; the hollow one is a little larger but much lighter.
5.Mohr’s circle & stress transformation
Answer: τxy = 63.44 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.
Required radius
C = (50 + (−30))/2 = 10 MPa
R = σ1 − C = 85 − 10 = 75 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(75² − 40²) = 63.44 MPa
Answer
τxy = 63.44 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 25 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 = −15 + −35·0 + 40·1 = 25 MPa
Answer
σθ = 25 MPa
7.Axial stress & strain
Answer: δ = 0.2191 mm
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·30²/4 = 706.9 mm²
At = π(100² − 50²)/4 = 5890 mm²
Combined axial stiffness
ΣEA = 200·10³·706.9 + 100·10³·5890 = 7.304 × 10⁸ N
Shortening
δ = 200000·800/7.304 × 10⁸ = 0.2191 mm
Answer
δ = 0.2191 mm
8.Axial stress & strain
Answer: Pmax = 293.9 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·25²/4 = 490.9 mm²
At = π(100² − 70²)/4 = 4006 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 110·10³·4006 = 5.388 × 10⁸ N
Governing strain
ε = min(0.0006, 0.0005455) = 0.0005455 (bronze)
Largest load
Pmax = 0.0005455·5.388 × 10⁸ = 293882 N = 293.9 kN
Answer
Pmax = 293.9 kN
9.Column buckling
Answer: Pcr = 262.3 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 = 50·25³/12 = 65104 mm⁴
Effective length
KL = 0.7·1000 = 700 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·65104/700² = 262266 N = 262.3 kN
Answer
Pcr = 262.3 kN
10.Column buckling
Answer: d = 115.9 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 = 2·85000 = 170000 N
KL = 2·3000 = 6000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·170000·6000²/(π³·70000))^(1/4) = 115.9 mm
Answer
d = 115.9 mm
Check: the slenderness KL/r is 207.1, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 1.92 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 = √(4000/25) = 12.65 rad/s
fn = 12.65/(2π) = 2.013 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 2.013·√(1 − 0.3²) = 1.92 Hz
Answer
fd = 1.92 Hz
12.Mechanical vibrations
Answer: fn = 3.473 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 = 1/100 + 1/5 → keq = 100·5/(100 + 5) = 4.762 kN/m
Natural frequency
ωn = √(4762/10) = 21.82 rad/s
fn = 21.82/(2π) = 3.473 Hz
Answer
fn = 3.473 Hz
ExamProblems · examproblems.com/courses/machine-design