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
d1​ = 125 mmd2​ = 250 mm960 rpmn2​ = ?

A 125 mm pulley at 960 rpm drives a 250 mm pulley by a belt with 4 % slip. Find the speed n2 of the driven pulley.

n1 = 960 rpmd1 = 125 mmd2 = 250 mms = 4 %
n2 = rpm
2.Gears & gear trainsHard
A = 45B = 48C = 15D = 30

In a compound gear train, gear A (45 teeth) on the input shaft turning at 960 rpm drives gear B (48 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D (30 teeth) on the output shaft. Find the output speed nout.

nin = 960 rpmA = 45B = 48C = 15D = 30
nout = rpm
3.Torsion of shaftsMedium
do​ = 60, di​ = 30 mmT = 1500 N·mL = 0.5 m

A hollow steel shaft (outer diameter 60 mm, inner diameter 30 mm, length 0.5 m) is fixed at one end and twisted by a torque of 1500 N·m at the other. Find the angle of twist φ in degrees.

T = 1500 N·mdo = 60 mmdi = 30 mmL = 0.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
d = 25 mmTB​ = 750 N·mTC​ = 250 N·mABC3 m0.5 m

A solid aluminium shaft of diameter 25 mm is fixed at A. A torque of 750 N·m acts at B, 3 m from A, and a torque of 250 N·m in the opposite direction acts at the free end C, a further 0.5 m along. Find the maximum shear stress τmax in the shaft.

d = 25 mmTB = 750 N·mTC = 250 N·mAB = 3 mBC = 0.5 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = 50σy​ = 40τxy​ = −60MPa

A point has σₓ = 50 MPa, σᵧ = 40 MPa and τₓᵧ = −60 MPa (tension positive). The element is rotated 45° clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σx = 50 MPaσy = 40 MPaτxy = −60 MPaθ = -45°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
σx​ = −10σy​ = −10τxy​ = 30MPa

A point in a thin plate is in plane stress: σₓ = −10 MPa, σᵧ = −10 MPa and τₓᵧ = 30 MPa (σz = 0). The material has a yield strength of 250 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σx = −10 MPaσy = −10 MPaτxy = 30 MPaσY = 250 MPa
n =
7.Axial stress & strainMedium
Ø 25 mmBrass2.5 mP = 115 kN

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

P = 115 kNd = 25 mmL = 2.5 mE = 100 GPa
ε = µε
8.Axial stress & strainHard
Ø 35 mmCopper1.5 mØ 12 mmAluminium1.5 mP = 85 kN

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

P = 85 kNsegment 1: Ø35 mm, 1.5 m, copper (E = 117 GPa)segment 2: Ø12 mm, 1.5 m, aluminium (E = 70 GPa)
δ = mm
9.Column bucklingMedium
PL = 1.5 m40 × 80 mmcross-section

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 = 1.5 mE = 200 GPa (steel)b × h = 40 × 80 mm
Pcr = kN
10.Column bucklingHard
P = 145 kNL = 6 md = ?cross-section

A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 145 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 145 kNn = 3L = 6 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 80 kN/mc = 760 N·s/mm = 5 kg

A 5 kg mass is supported by a spring with a stiffness of 80 kN/m and a viscous damper with a damping coefficient of 760 N·s/m. Find the damping ratio ζ.

m = 5 kgk = 80 kN/mc = 760 N·s/m
ζ =
12.Mechanical vibrationsHard
k1​ = 15k2​ = 80m = 10 kg

A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 15 kN/m and k₂ = 80 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 = 10 kgk1 = 15 kN/mk2 = 80 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: n2 = 460.8 rpm

Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.

Speed without slip

960·125/250 = 480 rpm

With slip

n2 = 480·0.96 = 460.8 rpm

Answer

n2 = 460.8 rpm
2.Gears & gear trains

Answer: nout = 450 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 = 960·45/48 = 900 rpm = nC

Second stage

nout = 900·15/30 = 450 rpm

Answer

nout = 450 rpm

total ratio = 2.133 : 1

3.Torsion of shafts

Answer: φ = 0.4503 °

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(60⁴ − 30⁴)/32 = 1.193 × 10⁶ mm⁴

Angle of twist

T = 1500 N·m = 1.5 × 10⁶ N·mm
L = 0.5 m = 500 mm
G = 80 GPa = 80000 MPa
φ = TL/(GJ) = 1.5 × 10⁶·500/(80000·1.193 × 10⁶) = 0.00786 rad
φ = 0.00786·180/π = 0.4503°

Answer

φ = 0.4503°
4.Torsion of shafts

Answer: τmax = 163 MPa

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.

Internal torques

Sense of TC taken as positive.

TBC = TC = 250 N·m
TAB = TC − TB = 250 − 750 = −500 N·m

Maximum shear stress

The segment with the largest internal torque has the largest stress (same diameter).

|T|max = 500 N·m
τmax = 16·500000/(π·25³) = 163 MPa

Answer

τmax = 163 MPa
5.Mohr’s circle & stress transformation

Answer: σy′ = −15 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 for the y′ face

σy′ = 45 − 5·0 − (−60)·(−1) = −15 MPa

Check

The sum of the normal stresses does not change with rotation.

σx′ = 105 MPa, σx′ + σy′ = 90 = σx + σy

Answer

σy′ = −15 MPa
6.Mohr’s circle & stress transformation

Answer: n = 4.167

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 = −10 MPa, R = 30 MPa
σ1 = 20 MPa, σ2 = −40 MPa, σ3 = 0

Tresca

σmax − σmin = 60 MPa
n = 250/60 = 4.167

Answer

n = 4.167

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

7.Axial stress & strain

Answer: ε = 2343 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 = 115·1000/490.9 = 234.3 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 234.3/100000 = 0.002343 = 2343 µε

Answer

ε = 2343 µε
8.Axial stress & strain

Answer: δ = 17.24 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·35²/4 = 962.1 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 88.35 MPa
σ2 = 751.6 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 88.35·1500/117000 = 1.133 mm
δ2 = σ2L2/E2 = 751.6·1500/70000 = 16.1 mm
δ = δ1 + δ2 = 17.24 mm

Answer

δ = 17.24 mm
9.Column buckling

Answer: Pcr = 763.9 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 = 80·40³/12 = 426667 mm⁴

Effective length

KL = 0.7·1500 = 1050 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·426667/1050² = 763906 N = 763.9 kN

Answer

Pcr = 763.9 kN
10.Column buckling

Answer: d = 94.33 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 = 3·145000 = 435000 N
KL = 0.7·6000 = 4200 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·435000·4200²/(π³·200000))^(1/4) = 94.33 mm

Answer

d = 94.33 mm

Check: the slenderness KL/r is 178.1, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.6008

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·√(80000·5) = 1265 N·s/m

Damping ratio

ζ = c/cc = 760/1265 = 0.6008

Answer

ζ = 0.6008

ζ < 1, so the system is underdamped and oscillates as it decays.

12.Mechanical vibrations

Answer: fn = 15.51 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 = 15 + 80 = 95 kN/m

Natural frequency

ωn = √(95000/10) = 97.47 rad/s
fn = 97.47/(2π) = 15.51 Hz

Answer

fn = 15.51 Hz

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