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​ = 150 mmd2​ = 450 mm2880 rpmn2​ = ?

A 150 mm pulley at 2880 rpm drives a 450 mm pulley by a belt with 2.5 % slip. Find the speed n2 of the driven pulley.

n1 = 2880 rpmd1 = 150 mmd2 = 450 mms = 2.5 %
n2 = rpm
2.Gears & gear trainsHard
A = 25B = 45C = 16D = 60

A 5.5 kW motor at 960 rpm drives gear A (25 teeth), which meshes with gear B (45 teeth). Gear C (16 teeth) on the same shaft as B drives gear D (60 teeth) on the output shaft. Each gear mesh has an efficiency of 96%. Find the output torque Tout.

P = 5.5 kWnin = 960 rpmA = 25B = 45C = 16D = 60ηmesh = 96%
Tout = N·m
3.Torsion of shaftsMedium
T = 200 N·md = 50 mmL = 2.5 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 = 200 N·md = 50 mmL = 2.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
d = 25 mmT0​ = 2000 N·mACBa = 3 mb = 0.5 m

A solid brass shaft of diameter 25 mm is fixed to rigid walls at both ends A and B. A torque of 2000 N·m is applied at C, 3 m from A and 0.5 m from B. Find the reaction torque TA at wall A.

d = 25 mmT0 = 2000 N·ma = 3 mb = 0.5 m
TA = N·m
5.Mohr’s circle & stress transformationMedium
σx​ = 60σy​ = 120τxy​ = 30MPa

A point in a loaded part has σₓ = 60 MPa, σᵧ = 120 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 60 MPaσy = 120 MPaτxy = 30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 150σy​ = −50τxy​ = −60MPa

A point in a thin plate is in plane stress: σₓ = 150 MPa, σᵧ = −50 MPa and τₓᵧ = −60 MPa (σz = 0). Find the von Mises equivalent stress σvM.

σx = 150 MPaσy = −50 MPaτxy = −60 MPa
σvM = MPa
7.Axial stress & strainMedium
Ø 25 mmCopper3 mP = 100 kN

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

P = 100 kNd = 25 mmL = 3 mE = 117 GPa
ε = µε
8.Axial stress & strainHard
60 kN12400 mm200 mmØ30Ø50σ2​ = ?

A steel segment (diameter 30 mm, length 400 mm, E = 200 GPa) and a segment made of copper (diameter 50 mm, length 200 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 60 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø30 mm, 400 mm, E = 200 GPasegment 2: copper, Ø50 mm, 200 mm, E = 120 GPaP = 60 kN
σ2 = MPa
9.Column bucklingMedium
PL = 1.5 md = 60 mmcross-section

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

L = 1.5 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
10.Column bucklingHard
P = 50 kNL = 2 md = ?cross-section

A solid round column of length L made of steel is pinned at both ends. It must carry 50 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 50 kNn = 2.5L = 2 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 2 kN/mζ = 0.7m = 10 kg

A 10 kg mass on a spring with a stiffness of 2 kN/m has a viscous damper giving a damping ratio of 0.7. Find the damped natural frequency fd of its free vibration in Hz.

m = 10 kgk = 2 kN/mζ = 0.7
fd = Hz
12.Mechanical vibrationsHard
k1​ = 25k2​ = 10m = 5 kg

A 5 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 25 kN/m and k₂ = 10 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 = 5 kgk1 = 25 kN/mk2 = 10 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: n2 = 936 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

2880·150/450 = 960 rpm

With slip

n2 = 960·0.975 = 936 rpm

Answer

n2 = 936 rpm
2.Gears & gear trains

Answer: Tout = 340.3 N·m

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.

Output speed

nout = 960·(25/45)·(16/60) = 142.2 rpm
ωout = 2π·142.2/60 = 14.89 rad/s

Output power and torque

Pout = 5500·0.96² = 5069 W
Tout = Pout/ωout = 5069/14.89 = 340.3 N·m

Answer

Tout = 340.3 N·m
3.Torsion of shafts

Answer: φ = 0.5836 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 200·1000·2500/(80000·613592) = 0.01019 rad
φ = 0.01019·180/π = 0.5836°

Answer

φ = 0.5836°
4.Torsion of shafts

Answer: TA = 285.7 N·m

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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).

Equilibrium and compatibility

TA + TB = 2000
TA·a = TB·b
TA = 2000·0.5/3.5 = 285.7 N·m
TB = 2000·3/3.5 = 1714 N·m

Answer

TA = 285.7 N·m
5.Mohr’s circle & stress transformation

Answer: τmax = 42.43 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 = √(((60 − 120)/2)² + 30²) = 42.43 MPa

Answer

τmax = 42.43 MPa
6.Mohr’s circle & stress transformation

Answer: σvM = 208.1 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. 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 = 50 MPa, R = 116.6 MPa
σ1 = 166.6 MPa, σ2 = −66.62 MPa, σ3 = 0

Von Mises

σvM = √(166.6² − 166.6·(−66.62) + (−66.62)²) = 208.1 MPa

Answer

σvM = 208.1 MPa
7.Axial stress & strain

Answer: ε = 1741 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 = 100·1000/490.9 = 203.7 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 203.7/117000 = 0.001741 = 1741 µε

Answer

ε = 1741 µε
8.Axial stress & strain

Answer: σ2 = 23.51 MPa

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.

Stiffnesses

k₁ = 200000·706.9/400 = 353429 N/mm
k₂ = 120000·1963/200 = 1.178 × 10⁶ N/mm

Share of segment 2

F2 = 60000·1.178 × 10⁶/1.532 × 10⁶ = 46154 N

Stress

σ2 = 46154/1963 = 23.51 MPa (compression)

Answer

σ2 = 23.51 MPa
9.Column buckling

Answer: Pcr = 195.3 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 = 1·1500 = 1500 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·636173/1500² = 195340 N = 195.3 kN

Answer

Pcr = 195.3 kN
10.Column buckling

Answer: d = 47.66 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 = 2.5·50000 = 125000 N
KL = 1·2000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·125000·2000²/(π³·200000))^(1/4) = 47.66 mm

Answer

d = 47.66 mm

Check: the slenderness KL/r is 167.9, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: fd = 1.607 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 = √(2000/10) = 14.14 rad/s
fn = 14.14/(2π) = 2.251 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 2.251·√(1 − 0.7²) = 1.607 Hz

Answer

fd = 1.607 Hz
12.Mechanical vibrations

Answer: fn = 13.32 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 = 25 + 10 = 35 kN/m

Natural frequency

ωn = √(35000/5) = 83.67 rad/s
fn = 83.67/(2π) = 13.32 Hz

Answer

fn = 13.32 Hz

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