Machine Design: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

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1Formulas and methods

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

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

  1. Find the polar moment of area: J = πd⁴/32 for a solid shaft.
  2. Shear stress at the surface: τ = T·r/J, with r the outer radius.
  3. Angle of twist: φ = TL/(GJ) in radians.
  4. 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

  1. Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
  2. Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
  3. The principal stresses are C ± R and the largest in-plane shear stress is R.
  4. 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
N1​N2​ = ?

Two parallel shafts are 105 mm apart. They must be connected by a pair of spur gears with a module of 2 mm and a speed reduction of 2.5 : 1. How many teeth N2 must the gear have?

a = 105 mmm = 2 mmi = 2.5
N2 =
2.Gears & gear trainsHard
A = 45B = 50C = 12D = ?

In a compound gear train, gear A (45 teeth) on the input shaft turns at 1500 rpm and drives gear B (50 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 202.5 rpm. How many teeth ND must gear D have?

nin = 1500 rpmA = 45B = 50C = 12nout = 202.5 rpm
ND =
3.Torsion of shaftsMedium
T = 100 N·md = 40 mmL = 3 mφ = ?

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 = 40 mmL = 3 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
d = ?50 kW, 1500 rpm

A solid steel shaft transmits 50 kW at 1500 rpm. The shear stress must not exceed 40 MPa and the twist must not exceed 0.75° per m of length. Find the smallest diameter d that satisfies both limits.

P = 50 kWN = 1500 rpmτallow = 40 MPaθallow = 0.75°/mG = 80 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = −10σy​ = 120τxy​ = −50MPa

A point in a loaded part has σₓ = −10 MPa, σᵧ = 120 MPa and τₓᵧ = −50 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = −10 MPaσy = 120 MPaτxy = −50 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​ = −80σy​ = 120τxy​ = 10MPa

A point in a thin plate is in plane stress: σₓ = −80 MPa, σᵧ = 120 MPa and τₓᵧ = 10 MPa (σz = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σx = −80 MPaσy = 120 MPaτxy = 10 MPaσY = 460 MPa
n =
7.Axial stress & strainMedium
Ø 16 mmSteel2 mP = 120 kN

Find the axial strain ε in the steel bar, in microstrain (µε = 10⁻⁶).

P = 120 kNd = 16 mmL = 2 mE = 200 GPa
ε = µε
8.Axial stress & strainHard
Ø 25 mmCopper2 mØ 16 mmAluminium1.5 mP = 15 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 15 kNsegment 1: Ø25 mm, 2 m, copper (E = 117 GPa)segment 2: Ø16 mm, 1.5 m, aluminium (E = 70 GPa)
δ = mm
9.Column bucklingMedium
PL = 5 m50 × 80 mmcross-section

A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.

L = 5 mE = 200 GPa (steel)b × h = 50 × 80 mm
Pcr = kN
10.Column bucklingHard
P = 95 kNL = 2.5 md = ?cross-section

A solid round column of length L made of steel is pinned at both ends. It must carry 95 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 95 kNn = 2.5L = 2.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 5 kN/mζ = 0.3m = 40 kg

A 40 kg mass on a spring with a stiffness of 5 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 = 40 kgk = 5 kN/mζ = 0.3
fd = Hz
12.Mechanical vibrationsHard
k1​ = 25k2​ = 50k3​ = 15m = 40 kg

A 40 kg mass hangs from spring k₃ = 15 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 25 kN/m and k₂ = 50 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 = 40 kgk1 = 25 kN/mk2 = 50 kN/mk3 = 15 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: N2 = 75

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.

Total number of teeth

N1 + N2 = 2a/m = 2·105/2 = 105

Split by the ratio

N1 = 105/(1 + 2.5) = 30
N2 = 2.5·30 = 75

Answer

N2 = 75
2.Gears & gear trains

Answer: ND = 80

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 = 1500·45/50 = 1350 rpm

Teeth on gear D

ND = C·nC/nout = 12·1350/202.5 = 80

Answer

ND = 80
3.Torsion of shafts

Answer: φ = 0.8549 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·3000/(80000·251327) = 0.01492 rad
φ = 0.01492·180/π = 0.8549°

Answer

φ = 0.8549°
4.Torsion of shafts

Answer: d = 41.95 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

Diameter for strength

T = 318.3 N·m = 318310 N·mm
dτ = ∛(16·318310/(π·40)) = 34.35 mm

Diameter for stiffness

θ = 0.75·π/180/1000 = 1.309 × 10⁻⁵ rad/mm
G = 80 GPa = 80000 MPa
dφ = (32·318310/(π·80000·1.309 × 10⁻⁵))1/4 = 41.95 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(34.35, 41.95) = 41.95 mm

Answer

d = 41.95 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: τmax = 82.01 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.

στxyCτmax​

Radius of Mohr’s circle

τmax = R = √(((−10 − 120)/2)² + −50²) = 82.01 MPa

Answer

τmax = 82.01 MPa
6.Mohr’s circle & stress transformation

Answer: n = 2.625

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.

στxyCσ1​σ2​

Principal stresses

C = 20 MPa, R = 100.5 MPa
σ1 = 120.5 MPa, σ2 = −80.5 MPa, σ3 = 0

Von Mises

σvM = √(120.5² − 120.5·(−80.5) + (−80.5)²) = 175.2 MPa
n = 460/175.2 = 2.625

Answer

n = 2.625

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

7.Axial stress & strain

Answer: ε = 2984 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·16²/4 = 201.1 mm²
σ = P/A = 120·1000/201.1 = 596.8 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 596.8/200000 = 0.002984 = 2984 µε

Answer

ε = 2984 µε
8.Axial stress & strain

Answer: δ = 2.121 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·25²/4 = 490.9 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 30.56 MPa
σ2 = 74.6 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 30.56·2000/117000 = 0.5224 mm
δ2 = σ2L2/E2 = 74.6·1500/70000 = 1.599 mm
δ = δ1 + δ2 = 2.121 mm

Answer

δ = 2.121 mm
9.Column buckling

Answer: Pcr = 263.2 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 = 80·50³/12 = 833333 mm⁴

Effective length

KL = 0.5·5000 = 2500 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·833333/2500² = 263189 N = 263.2 kN

Answer

Pcr = 263.2 kN
10.Column buckling

Answer: d = 62.56 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 = 2.5·95000 = 237500 N
KL = 1·2500 = 2500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·237500·2500²/(π³·200000))^(1/4) = 62.56 mm

Answer

d = 62.56 mm

Check: the slenderness KL/r is 159.8, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: fd = 1.697 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 = √(5000/40) = 11.18 rad/s
fn = 11.18/(2π) = 1.779 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 1.779·√(1 − 0.3²) = 1.697 Hz

Answer

fd = 1.697 Hz
12.Mechanical vibrations

Answer: fn = 2.813 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 = 25 + 50 = 75 kN/m
keq = 75·15/(75 + 15) = 12.5 kN/m

Natural frequency

ωn = √(12500/40) = 17.68 rad/s
fn = 17.68/(2π) = 2.813 Hz

Answer

fn = 2.813 Hz

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