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​ = 25, m = 2 mm

A 7.5 kW motor running at 960 rpm drives a pinion with 25 teeth and a module of 2 mm. Find the tangential force Ft on the pinion teeth.

P = 7.5 kWn = 960 rpmN1 = 25m = 2 mm
Ft = N
2.Gears & gear trainsHard
A = 72B = 80C = 40D = ?

In a compound gear train, gear A (72 teeth) on the input shaft turns at 1500 rpm and drives gear B (80 teeth). Gear C (40 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 1200 rpm. How many teeth ND must gear D have?

nin = 1500 rpmA = 72B = 80C = 40nout = 1200 rpm
ND =
3.Torsion of shaftsMedium
T = 500 N·md = 25 mmL = 1 mφ = ?

A solid brass 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 = 25 mmL = 1 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 15 kW at 750 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 15 kWN = 750 rpmτallow = 60 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = −50σy​ = 70τxy​ = 30MPa

A point has σₓ = −50 MPa, σᵧ = 70 MPa and τₓᵧ = 30 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 = −50 MPaσy = 70 MPaτxy = 30 MPa
θs = °
6.Mohr’s circle & stress transformationHard
σx​ = 10σy​ = 50τxy​ = −40MPa

A point in a thin plate is in plane stress: σₓ = 10 MPa, σᵧ = 50 MPa and τₓᵧ = −40 MPa (σz = 0). Find the von Mises equivalent stress σvM.

σx = 10 MPaσy = 50 MPaτxy = −40 MPa
σvM = MPa
7.Axial stress & strainMedium
Ø 10 mmCopper2.5 mP = 110 kN

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

P = 110 kNd = 10 mmL = 2.5 mE = 117 GPa
δ = mm
8.Axial stress & strainHard
Ø 40 mmAluminium0.5 mØ 12 mmSteel2 mP = 60 kN

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

P = 60 kNsegment 1: Ø40 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 2 m, steel (E = 200 GPa)
δ = mm
9.Column bucklingMedium
PL = 5 md = 60 mmcross-section

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

L = 5 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
10.Column bucklingHard
P = 25 kNL = 4 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 2.5 against buckling. Find the smallest diameter d.

P = 25 kNn = 2.5L = 4 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
k = 50 kN/mζ = 0.45m = 10 kg

A 10 kg mass on a spring with a stiffness of 50 kN/m has a viscous damper giving a damping ratio of 0.45. Find the damped natural frequency fd of its free vibration in Hz.

m = 10 kgk = 50 kN/mζ = 0.45
fd = Hz
12.Mechanical vibrationsHard
k1​ = 12k2​ = 100m = 50 kg

A 50 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 12 kN/m and k₂ = 100 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 = 50 kgk1 = 12 kN/mk2 = 100 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: Ft = 2984 N

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.

Torque on the pinion

ω = 2π·960/60 = 100.5 rad/s
T = P/ω = 7500/100.5 = 74.6 N·m

Pitch diameter

d = m·N = 2·25 = 50 mm

Tangential force

Ft = 2T/d = 2·74.6/0.05 = 2984 N

Answer

Ft = 2984 N
2.Gears & gear trains

Answer: ND = 45

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 = 1500·72/80 = 1350 rpm

Teeth on gear D

ND = C·nC/nout = 40·1350/1200 = 45

Answer

ND = 45
3.Torsion of shafts

Answer: φ = 19.15 °

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·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 500·1000·1000/(39000·38350) = 0.3343 rad
φ = 0.3343·180/π = 19.15°

Answer

φ = 19.15°
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·750/60 = 78.54 rad/s
T = P/ω = 15000/78.54 = 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: θs = 31.72 °

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 = 30/67.08 = 0.4472, sin 2θs = −(−60)/67.08 = 0.8944
2θs = atan2(60, 30) = 63.43°

Angle on the element

θs = 31.72°
check: θs − θp = 31.72 − 76.72 = −45°

Answer

θs = 31.72°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σvM = 83.07 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 = 30 MPa, R = 44.72 MPa
σ1 = 74.72 MPa, σ2 = −14.72 MPa, σ3 = 0

Von Mises

σvM = √(74.72² − 74.72·(−14.72) + (−14.72)²) = 83.07 MPa

Answer

σvM = 83.07 MPa
7.Axial stress & strain

Answer: δ = 29.93 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·10²/4 = 78.54 mm²
σ = P/A = 110·1000/78.54 = 1401 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 1401/117000 = 0.01197 = 11971 µε

Elongation

δ = ε·L = 0.01197·2500 mm = 29.93 mm

Answer

δ = 29.93 mm
8.Axial stress & strain

Answer: δ = 5.646 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·40²/4 = 1257 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 47.75 MPa
σ2 = 530.5 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 47.75·500/70000 = 0.341 mm
δ2 = σ2L2/E2 = 530.5·2000/200000 = 5.305 mm
δ = δ1 + δ2 = 5.646 mm

Answer

δ = 5.646 mm
9.Column buckling

Answer: Pcr = 70.32 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·60⁴/64 = 636173 mm⁴

Effective length

KL = 0.5·5000 = 2500 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·636173/2500² = 70322 N = 70.32 kN

Answer

Pcr = 70.32 kN
10.Column buckling

Answer: d = 52.11 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 = 2.5·25000 = 62500 N
KL = 0.5·4000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·62500·2000²/(π³·70000))^(1/4) = 52.11 mm

Answer

d = 52.11 mm

Check: the slenderness KL/r is 153.5, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: fd = 10.05 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 = √(50000/10) = 70.71 rad/s
fn = 70.71/(2π) = 11.25 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 11.25·√(1 − 0.45²) = 10.05 Hz

Answer

fd = 10.05 Hz
12.Mechanical vibrations

Answer: fn = 7.533 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 = 12 + 100 = 112 kN/m

Natural frequency

ωn = √(112000/50) = 47.33 rad/s
fn = 47.33/(2π) = 7.533 Hz

Answer

fn = 7.533 Hz

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