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​ = 30, m = 2 mm

A 4 kW motor running at 960 rpm drives a pinion with 30 teeth and a module of 2 mm. Find the tangential force Ft on the pinion teeth.

P = 4 kWn = 960 rpmN1 = 30m = 2 mm
Ft = N
2.Gears & gear trainsHard
d1​ = 150 mmd2​ = 500 mm1450 rpmC = 600 mmT1​ = 2000 NT2​P = ?

An open belt connects a 150 mm driving pulley at 1450 rpm to a 500 mm pulley; the centres are 600 mm apart. The coefficient of friction is 0.35 and the tight-side tension may not exceed 2000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 150 mmn1 = 1450 rpmd2 = 500 mmC = 600 mmμ = 0.35T1 = 2000 N
P = kW
3.Torsion of shaftsMedium
do​ = 50, di​ = 30 mmT = 750 N·mL = 1.5 m

A hollow steel shaft (outer diameter 50 mm, inner diameter 30 mm, length 1.5 m) is fixed at one end and twisted by a torque of 750 N·m at the other. Find the angle of twist φ in degrees.

T = 750 N·mdo = 50 mmdi = 30 mmL = 1.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
d = 50 mmT0​ = 1000 N·mACBa = 3 mb = 2.5 m

A solid steel shaft of diameter 50 mm is fixed to rigid walls at both ends A and B. A torque of 1000 N·m is applied at C, 3 m from A and 2.5 m from B. Find the reaction torque TA at wall A.

d = 50 mmT0 = 1000 N·ma = 3 mb = 2.5 m
TA = N·m
5.Mohr’s circle & stress transformationMedium
σx​ = 160σy​ = −70τxy​ = 30MPa

A point in a loaded part has σₓ = 160 MPa, σᵧ = −70 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 160 MPaσy = −70 MPaτxy = 30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​τxy​top surface of the shaft

A solid shaft of diameter 50 mm carries a bending moment of 1000 N·m together with a torque of 2000 N·m. At the top surface, where the bending stress is tensile and largest, find the maximum in-plane shear stress τmax.

d = 50 mmM = 1000 N·mT = 2000 N·m
τmax = MPa
7.Axial stress & strainMedium
Ø 10 mmCopper0.5 mP = 105 kN

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

P = 105 kNd = 10 mmL = 0.5 mE = 117 GPa
ε = µε
8.Axial stress & strainHard
80 kN12500 mm300 mmØ25Ø50σ2​ = ?

A steel segment (diameter 25 mm, length 500 mm, E = 200 GPa) and a segment made of brass (diameter 50 mm, length 300 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 80 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø25 mm, 500 mm, E = 200 GPasegment 2: brass, Ø50 mm, 300 mm, E = 100 GPaP = 80 kN
σ2 = MPa
9.Column bucklingMedium
PL = 1.5 m30 × 100 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 = 1.5 mE = 200 GPa (steel)b × h = 30 × 100 mm
Pcr = kN
10.Column bucklingHard
P = 140 kNL = 2.5 md = ?cross-section

A solid round column of length L made of steel is fixed at both ends. It must carry 140 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 140 kNn = 2L = 2.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
tx50 mm5 cycles4 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed 5 cycles later by a peak of 4 mm. Find the damping ratio ζ.

x0 = 50 mmx5 = 4 mmn = 5
ζ =
12.Mechanical vibrationsHard
k1​ = 100k2​ = 20m = 10 kg

A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 100 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 = 10 kgk1 = 100 kN/mk2 = 20 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: Ft = 1326 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 = 4000/100.5 = 39.79 N·m

Pitch diameter

d = m·N = 2·30 = 60 mm

Tangential force

Ft = 2T/d = 2·39.79/0.06 = 1326 N

Answer

Ft = 1326 N
2.Gears & gear trains

Answer: P = 13.45 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.

Angle of wrap

θ = 2.55 rad

Slack-side tension

eμθ = e0.8924 = 2.441, T₂ = 2000/2.441 = 819.4 N

Power

P = (2000 − 819.4)·11.39 = 13.45 kW

Answer

P = 13.45 kW
3.Torsion of shafts

Answer: φ = 1.509 °

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 − 30⁴)/32 = 534071 mm⁴

Angle of twist

T = 750 N·m = 750000 N·mm
L = 1.5 m = 1500 mm
G = 80 GPa = 80000 MPa
φ = TL/(GJ) = 750000·1500/(80000·534071) = 0.02633 rad
φ = 0.02633·180/π = 1.509°

Answer

φ = 1.509°
4.Torsion of shafts

Answer: TA = 454.5 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 = 1000
TA·a = TB·b
TA = 1000·2.5/5.5 = 454.5 N·m
TB = 1000·3/5.5 = 545.5 N·m

Answer

TA = 454.5 N·m
5.Mohr’s circle & stress transformation

Answer: τmax = 118.8 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 = √(((160 − −70)/2)² + 30²) = 118.8 MPa

Answer

τmax = 118.8 MPa
6.Mohr’s circle & stress transformation

Answer: τmax = 91.11 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.

Stresses at the surface

σ = 32M/(πd³) = 32·1000000/(π·50³) = 81.49 MPa
τ = 16T/(πd³) = 16·2000000/(π·50³) = 81.49 MPa
στxyCτmax​

Mohr’s circle

C = σ/2 = 40.74 MPa
R = √(40.74² + 81.49²) = 91.11 MPa

Answer

τmax = 91.11 MPa
7.Axial stress & strain

Answer: ε = 11427 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·10²/4 = 78.54 mm²
σ = P/A = 105·1000/78.54 = 1337 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 1337/117000 = 0.01143 = 11427 µε

Answer

ε = 11427 µε
8.Axial stress & strain

Answer: σ2 = 31.34 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 = 200000·490.9/500 = 196350 N/mm
k₂ = 100000·1963/300 = 654498 N/mm

Share of segment 2

F2 = 80000·654498/850848 = 61538 N

Stress

σ2 = 61538/1963 = 31.34 MPa (compression)

Answer

σ2 = 31.34 MPa
9.Column buckling

Answer: Pcr = 49.35 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 = 100·30³/12 = 225000 mm⁴

Effective length

KL = 2·1500 = 3000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·225000/3000² = 49348 N = 49.35 kN

Answer

Pcr = 49.35 kN
10.Column buckling

Answer: d = 46.1 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 = 2·140000 = 280000 N
KL = 0.5·2500 = 1250 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·280000·1250²/(π³·200000))^(1/4) = 46.1 mm

Answer

d = 46.1 mm

Check: the slenderness KL/r is 108.5, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.08014

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.

Logarithmic decrement

δ = (1/5)·ln(50/4) = 0.5051

Damping ratio

ζ = δ/√(4π² + δ²) = 0.5051/√(39.48 + 0.2552) = 0.08014

Answer

ζ = 0.08014
12.Mechanical vibrations

Answer: fn = 17.43 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 = 100 + 20 = 120 kN/m

Natural frequency

ωn = √(120000/10) = 109.5 rad/s
fn = 109.5/(2π) = 17.43 Hz

Answer

fn = 17.43 Hz

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