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
Nw​ = 50z = 4, 1440 rpm

A worm with 4 start(s) turns at 1440 rpm and drives a worm wheel with 50 teeth. Find the speed nw of the wheel.

z = 4Nw = 50n = 1440 rpm
nw = rpm
2.Gears & gear trainsHard
Ns​ = 24ring fixedNp​ = 24nc​ = ?

In a planetary gear set the sun gear has 24 teeth and each planet gear has 24 teeth. The ring gear is held fixed and the sun is driven at 1200 rpm. Find the speed nc of the planet carrier.

Ns = 24Np = 24ns = 1200 rpm
nc = rpm
3.Torsion of shaftsMedium
T = 500 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 = 500 N·md = 40 mmL = 3 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
do​ = ?10 kW, 750 rpm

A hollow shaft must transmit 10 kW at 750 rpm. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter do.

P = 10 kWN = 750 rpmdi/do = 0.6τallow = 40 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 110σy​ = 120τxy​ = −40MPa

A point in a loaded part has σₓ = 110 MPa, σᵧ = 120 MPa and τₓᵧ = −40 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 110 MPaσy = 120 MPaτxy = −40 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 70σy​ = 60τxy​ = 20MPa

A point in a thin plate is in plane stress: σₓ = 70 MPa, σᵧ = 60 MPa and τₓᵧ = 20 MPa (σz = 0). Find the von Mises equivalent stress σvM.

σx = 70 MPaσy = 60 MPaτxy = 20 MPa
σvM = MPa
7.Axial stress & strainMedium
50 kNØ30120 GPaØ80/60800 mmδ = ?

A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 80 mm, inner diameter 60 mm, E = 120 GPa). Both are 800 mm long and are compressed together between rigid plates by 50 kN. Find the shortening δ.

P = 50 kNrod: Ø30 mm, E = 200 GPatube: Ø80/Ø60 mm, E = 120 GPaL = 800 mm
δ = mm
8.Axial stress & strainHard
P = ?Ø30120 GPaØ120/70500 mm

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 70 mm, E = 120 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø30 mm, E = 200 GPa, σallow = 150 MPatube: Ø120/Ø70 mm, E = 120 GPa, σallow = 60 MPa
Pmax = kN
9.Column bucklingMedium
PL = 2 m30 × 100 mmcross-section

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

L = 2 mE = 200 GPa (steel)b × h = 30 × 100 mm
Pcr = kN
10.Column bucklingHard
P = 25 kNL = 3 md = ?cross-section

A solid round column of length L made of aluminium is fixed at both ends. It must carry 25 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 25 kNn = 2L = 3 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
tx12 mm2 cycles7.3 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 12 mm is followed 2 cycles later by a peak of 7.3 mm. Find the damping ratio ζ.

x0 = 12 mmx2 = 7.3 mmn = 2
ζ =
12.Mechanical vibrationsHard
k1​ = 20k2​ = 20m = 5 kg

A 5 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 20 kN/m and k₂ = 20 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.

m = 5 kgk1 = 20 kN/mk2 = 20 kN/m
keq = kN/m

3Solutions

1.Gears & gear trains

Answer: nw = 115.2 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.

Wheel speed

nw = n·z/Nw = 1440·4/50 = 115.2 rpm

Answer

nw = 115.2 rpm
2.Gears & gear trains

Answer: nc = 300 rpm

Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (ns − nc)/(nr − nc) = −Nr/Ns: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.

Ring gear teeth

Nr = Ns + 2·Np = 24 + 2·24 = 72

Willis equation with the ring fixed

Set nr = 0 and solve for the carrier speed.

(1200 − nc)/(0 − nc) = −72/24
nc = ns·Ns/(Ns + Nr) = 1200·24/96 = 300 rpm

Answer

nc = 300 rpm

The carrier turns in the same direction as the sun.

3.Torsion of shafts

Answer: φ = 4.274 °

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) = 500·1000·3000/(80000·251327) = 0.0746 rad
φ = 0.0746·180/π = 4.274°

Answer

φ = 4.274°
4.Torsion of shafts

Answer: do = 26.51 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·750/60 = 78.54 rad/s
T = P/ω = 10000/78.54 = 127.3 N·m

Outer diameter from the allowable stress

T = 127.3 N·m = 127324 N·mm
1 − k⁴ = 1 − 0.6⁴ = 0.8704
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·127324/(π·40·0.8704)) = 26.51 mm
di = 0.6·26.51 = 15.9 mm

Answer

do = 26.51 mm

A solid shaft would need d = 25.31 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: τmax = 40.31 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.

σxyCτmax​

Radius of Mohr’s circle

τmax = R = √(((110 − 120)/2)² + −40²) = 40.31 MPa

Answer

τmax = 40.31 MPa
6.Mohr’s circle & stress transformation

Answer: σvM = 74.16 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.

σxyCσ1​σ2​

Principal stresses

C = 65 MPa, R = 20.62 MPa
σ1 = 85.62 MPa, σ2 = 44.38 MPa, σ3 = 0

Von Mises

σvM = √(85.62² − 85.62·44.38 + 44.38²) = 74.16 MPa

Answer

σvM = 74.16 MPa
7.Axial stress & strain

Answer: δ = 0.0987 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(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 120·10³·2199 = 4.053 × 10⁸ N

Shortening

δ = 50000·800/4.053 × 10⁸ = 0.0987 mm

Answer

δ = 0.0987 mm
8.Axial stress & strain

Answer: Pmax = 518.4 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·30²/4 = 706.9 mm²
At = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 120·10³·7461 = 1.037 × 10⁹ N

Governing strain

ε = min(0.00075, 0.0005) = 0.0005 (copper)

Largest load

Pmax = 0.0005·1.037 × 10⁹ = 518363 N = 518.4 kN

Answer

Pmax = 518.4 kN
9.Column buckling

Answer: Pcr = 111 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 = 100·30³/12 = 225000 mm⁴

Effective length

KL = 1·2000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·225000/2000² = 111033 N = 111 kN

Answer

Pcr = 111 kN
10.Column buckling

Answer: d = 42.68 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 = 2·25000 = 50000 N
KL = 0.5·3000 = 1500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·50000·1500²/(π³·70000))^(1/4) = 42.68 mm

Answer

d = 42.68 mm

Check: the slenderness KL/r is 140.6, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.03952

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.

Logarithmic decrement

δ = (1/2)·ln(12/7.3) = 0.2485

Damping ratio

ζ = δ/√(4π² + δ²) = 0.2485/√(39.48 + 0.06176) = 0.03952

Answer

ζ = 0.03952
12.Mechanical vibrations

Answer: keq = 40 kN/m

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 = 20 + 20 = 40 kN/m

Answer

keq = 40 kN/m

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