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​ = 450 mm960 rpmn2​ = ?

A 125 mm pulley at 960 rpm drives a 450 mm pulley by a belt with 2 % slip. Find the speed n2 of the driven pulley.

n1 = 960 rpmd1 = 125 mmd2 = 450 mms = 2 %
n2 = rpm
2.Gears & gear trainsHard
Ns​ = 20ring fixedNp​ = 20nc​ = ?

In a planetary gear set the sun gear has 20 teeth and each planet gear has 20 teeth. The ring gear is held fixed and the sun is driven at 600 rpm. Find the speed nc of the planet carrier.

Ns = 20Np = 20ns = 600 rpm
nc = rpm
3.Torsion of shaftsMedium
do​ = 30, di​ = 18 mmT = 500 N·mL = 2.5 m

A hollow steel shaft (outer diameter 30 mm, inner diameter 18 mm, length 2.5 m) is fixed at one end and twisted by a torque of 500 N·m at the other. Find the angle of twist φ in degrees.

T = 500 N·mdo = 30 mmdi = 18 mmL = 2.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
Tdo​ = 60 mm, di​ = 30 mm

A hollow shaft transmits 50 kW at 750 rpm. Find the maximum shear stress τmax in the shaft.

P = 50 kWN = 750 rpmdo = 60 mmdi = 30 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = −10σy​ = 120τxy​ = −60MPa

A point has σₓ = −10 MPa, σᵧ = 120 MPa and τₓᵧ = −60 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σx = −10 MPaσy = 120 MPaτxy = −60 MPa
θs = °
6.Mohr’s circle & stress transformationHard
σx​ = 30σy​ = 120τxy​ = 30MPa

A point in a thin plate is in plane stress: σₓ = 30 MPa, σᵧ = 120 MPa and τₓᵧ = 30 MPa (σz = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σx = 30 MPaσy = 120 MPaτxy = 30 MPaσY = 460 MPa
n =
7.Axial stress & strainMedium
P = ?12600 mm300 mmØ40Ø40δ = 0.4 mm

A bar is made of a steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 40 mm, length 300 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.

segment 1: steel, Ø40 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø40 mm, 300 mm, E = 70 GPaδ = 0.4 mm
P = kN
8.Axial stress & strainHard
Ø 25 mmAluminium0.5 mØ 20 mmSteel1 mP = 25 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.

P = 25 kNsegment 1: Ø25 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 1 m, steel (E = 200 GPa)
σmax = MPa
9.Column bucklingMedium
PL = 5 m20 × 50 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 = 5 mE = 200 GPa (steel)b × h = 20 × 50 mm
Pcr = kN
10.Column bucklingHard
P = 45 kNL = 6 md = ?cross-section

A solid round column of length L made of steel is fixed at both ends. It must carry 45 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 45 kNn = 2L = 6 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 80 kN/mζ = 0.4m = 8 kg

An 8 kg mass on a spring with a stiffness of 80 kN/m has a viscous damper giving a damping ratio of 0.4. Find the damped natural frequency fd of its free vibration in Hz.

m = 8 kgk = 80 kN/mζ = 0.4
fd = Hz
12.Mechanical vibrationsHard
k1​ = 20k2​ = 20m = 100 kg

A 100 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. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 100 kgk1 = 20 kN/mk2 = 20 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: n2 = 261.3 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

960·125/450 = 266.7 rpm

With slip

n2 = 266.7·0.98 = 261.3 rpm

Answer

n2 = 261.3 rpm
2.Gears & gear trains

Answer: nc = 150 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 = 20 + 2·20 = 60

Willis equation with the ring fixed

Set nr = 0 and solve for the carrier speed.

(600 − nc)/(0 − nc) = −60/20
nc = ns·Ns/(Ns + Nr) = 600·20/80 = 150 rpm

Answer

nc = 150 rpm

The carrier turns in the same direction as the sun.

3.Torsion of shafts

Answer: φ = 12.93 °

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 − 18⁴)/32 = 69216 mm⁴

Angle of twist

T = 500 N·m = 500000 N·mm
L = 2.5 m = 2500 mm
G = 80 GPa = 80000 MPa
φ = TL/(GJ) = 500000·2500/(80000·69216) = 0.2257 rad
φ = 0.2257·180/π = 12.93°

Answer

φ = 12.93°
4.Torsion of shafts

Answer: τmax = 16.01 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.

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s
T = P/ω = 50000/78.54 = 636.6 N·m

Polar second moment of area

J = π(60⁴ − 30⁴)/32 = 1.193 × 10⁶ mm⁴

Maximum shear stress

τmax = T·(do/2)/J = 636620·30/1.193 × 10⁶ = 16.01 MPa

Answer

τmax = 16.01 MPa
5.Mohr’s circle & stress transformation

Answer: θs = 66.35 °

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​

Angle on the circle

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θs = −60/88.46 = −0.6783, sin 2θs = −(−65)/88.46 = 0.7348
2θs = atan2(65, −60) = 132.7°

Angle on the element

θs = 66.35°
check: θs − θp = 66.35 − (−68.65) = 135°

Answer

θs = 66.35°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: n = 3.564

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 = 75 MPa, R = 54.08 MPa
σ1 = 129.1 MPa, σ2 = 20.92 MPa, σ3 = 0

Tresca

σmax − σmin = 129.1 MPa
n = 460/129.1 = 3.564

Answer

n = 3.564

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

7.Axial stress & strain

Answer: P = 68.99 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

A₁ = 1257 mm², A₂ = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·1257) = 2.387 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(70000·1257) = 3.41 × 10⁻⁶ mm/N

Load

P = 0.4/5.798 × 10⁻⁶ = 68992 N = 68.99 kN

Answer

P = 68.99 kN
8.Axial stress & strain

Answer: σmax = 79.58 MPa

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·20²/4 = 314.2 mm²
σ1 = 50.93 MPa
σ2 = 79.58 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σmax = 79.58 MPa

Answer

σmax = 79.58 MPa
9.Column buckling

Answer: Pcr = 5.371 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 = 50·20³/12 = 33333 mm⁴

Effective length

KL = 0.7·5000 = 3500 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·33333/3500² = 5371 N = 5.371 kN

Answer

Pcr = 5.371 kN
10.Column buckling

Answer: d = 53.77 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 = 2·45000 = 90000 N
KL = 0.5·6000 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·90000·3000²/(π³·200000))^(1/4) = 53.77 mm

Answer

d = 53.77 mm

Check: the slenderness KL/r is 223.2, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: fd = 14.59 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 = √(80000/8) = 100 rad/s
fn = 100/(2π) = 15.92 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 15.92·√(1 − 0.4²) = 14.59 Hz

Answer

fd = 14.59 Hz
12.Mechanical vibrations

Answer: fn = 3.183 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 = 20 + 20 = 40 kN/m

Natural frequency

ωn = √(40000/100) = 20 rad/s
fn = 20/(2π) = 3.183 Hz

Answer

fn = 3.183 Hz

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