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​ = 80 mmd2​ = 450 mm960 rpmT1​ = 400 NT2​ = 120 NP = ?

A belt runs on an 80 mm driving pulley at 960 rpm. The tension is 400 N on the tight side and 120 N on the slack side. Find the power P transmitted, in kW.

d1 = 80 mmn1 = 960 rpmT1 = 400 NT2 = 120 N
P = kW
2.Gears & gear trainsHard
Ns​ = 16ring fixedNp​ = 24nc​ = ?

In a planetary gear set the sun gear has 16 teeth and each planet gear has 24 teeth. The ring gear is held fixed and the sun is driven at 3000 rpm. Find the speed nc of the planet carrier.

Ns = 16Np = 24ns = 3000 rpm
nc = rpm
3.Torsion of shaftsMedium
d = ?T = 2000 N·mL = 1.5 m

A solid steel shaft 1.5 m long carries a torque of 2000 N·m. The angle of twist between its ends must not exceed 1.5°. Find the smallest diameter d that meets this limit.

T = 2000 N·mL = 1.5 mG = 80 GPaφallow = 1.5°
d = mm
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 30 kW at 1500 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 30 kWN = 1500 rpmτallow = 60 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 10σy​ = −80τxy​ = 10MPa

A point in a loaded part has σₓ = 10 MPa, σᵧ = −80 MPa and τₓᵧ = 10 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 10 MPaσy = −80 MPaτxy = 10 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 20σy​ = −70τxy​ = 30MPa

A point in a thin plate is in plane stress: σₓ = 20 MPa, σᵧ = −70 MPa and τₓᵧ = 30 MPa (σz = 0). Find the von Mises equivalent stress σvM.

σx = 20 MPaσy = −70 MPaτxy = 30 MPa
σvM = MPa
7.Axial stress & strainMedium
Ø 20 mmBrass0.5 mP = 50 kN

Find the elongation δ of the brass bar under the axial load.

P = 50 kNd = 20 mmL = 0.5 mE = 100 GPa
δ = mm
8.Axial stress & strainHard
P = ?Ø25100 GPaØ120/50400 mm

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 50 mm, E = 100 GPa, allowable stress 80 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø25 mm, E = 200 GPa, σallow = 120 MPatube: Ø120/Ø50 mm, E = 100 GPa, σallow = 80 MPa
Pmax = kN
9.Column bucklingMedium
PL = 1 md = 40 mmcross-section

A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.

L = 1 mE = 200 GPa (steel)d = 40 mm
Pcr = kN
10.Column bucklingHard
PL = 1 md = 30 mmcross-section

A column of length L made of steel is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.

L = 1 mE = 200 GPa (steel)d = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 40 kN/mζ = 0.3m = 15 kg

A 15 kg mass on a spring with a stiffness of 40 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 = 15 kgk = 40 kN/mζ = 0.3
fd = Hz
12.Mechanical vibrationsHard
k1​ = 12k2​ = 20k3​ = 50m = 20 kg

A 20 kg mass hangs from spring k₃ = 50 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 12 kN/m and k₂ = 20 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 = 20 kgk1 = 12 kN/mk2 = 20 kN/mk3 = 50 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: P = 1.126 kW

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.

Belt speed

v = 4.021 m/s

Power

P = (400 − 120)·4.021 = 1126 W = 1.126 kW

Answer

P = 1.126 kW
2.Gears & gear trains

Answer: nc = 600 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 = 16 + 2·24 = 64

Willis equation with the ring fixed

Set nr = 0 and solve for the carrier speed.

(3000 − nc)/(0 − nc) = −64/16
nc = ns·Ns/(Ns + Nr) = 3000·16/80 = 600 rpm

Answer

nc = 600 rpm

The carrier turns in the same direction as the sun.

3.Torsion of shafts

Answer: d = 61.8 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.

Required polar moment

T = 2000 N·m = 2 × 10⁶ N·mm
L = 1.5 m = 1500 mm
G = 80 GPa = 80000 MPa
φ = 1.5·π/180 = 0.02618 rad
J = TL/(Gφ) = 2 × 10⁶·1500/(80000·0.02618) = 1.432 × 10⁶ mm⁴

Diameter

d = (32J/π)1/4 = 61.8 mm

Answer

d = 61.8 mm

In practice you would round up to the next standard size.

4.Torsion of shafts

Answer: d = 25.31 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 = 30000/157.1 = 191 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·190986/(π·60)) = 25.31 mm

Answer

d = 25.31 mm

In practice you would round up to the next standard size.

5.Mohr’s circle & stress transformation

Answer: τmax = 46.1 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 − −80)/2)² + 10²) = 46.1 MPa

Answer

τmax = 46.1 MPa
6.Mohr’s circle & stress transformation

Answer: σvM = 96.95 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.

στxyCσ1​σ2​

Principal stresses

C = −25 MPa, R = 54.08 MPa
σ1 = 29.08 MPa, σ2 = −79.08 MPa, σ3 = 0

Von Mises

σvM = √(29.08² − 29.08·(−79.08) + (−79.08)²) = 96.95 MPa

Answer

σvM = 96.95 MPa
7.Axial stress & strain

Answer: δ = 0.7958 mm

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·20²/4 = 314.2 mm²
σ = P/A = 50·1000/314.2 = 159.2 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 159.2/100000 = 0.001592 = 1592 µε

Elongation

δ = ε·L = 0.001592·500 mm = 0.7958 mm

Answer

δ = 0.7958 mm
8.Axial stress & strain

Answer: Pmax = 619.7 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(120² − 50²)/4 = 9346 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 100·10³·9346 = 1.033 × 10⁹ N

Governing strain

ε = min(0.0006, 0.0008) = 0.0006 (steel)

Largest load

Pmax = 0.0006·1.033 × 10⁹ = 619679 N = 619.7 kN

Answer

Pmax = 619.7 kN
9.Column buckling

Answer: Pcr = 62.01 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 round section has the same I about every axis.

I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴

Effective length

KL = 2·1000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·125664/2000² = 62013 N = 62.01 kN

Answer

Pcr = 62.01 kN
10.Column buckling

Answer: σcr = 27.76 MPa

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 round section has the same I about every axis.

I = πd⁴/64 = π·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 2·1000 = 2000 mm
KL/r = 2000/7.5 = 266.7

Critical stress

σcr = π²E/(KL/r)² = π²·200000/266.7² = 27.76 MPa

Answer

σcr = 27.76 MPa
11.Mechanical vibrations

Answer: fd = 7.84 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 = √(40000/15) = 51.64 rad/s
fn = 51.64/(2π) = 8.219 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 8.219·√(1 − 0.3²) = 7.84 Hz

Answer

fd = 7.84 Hz
12.Mechanical vibrations

Answer: fn = 4.971 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 = 12 + 20 = 32 kN/m
keq = 32·50/(32 + 50) = 19.51 kN/m

Natural frequency

ωn = √(19512/20) = 31.23 rad/s
fn = 31.23/(2π) = 4.971 Hz

Answer

fn = 4.971 Hz

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