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 = 4 mm

A 15 kW motor running at 960 rpm drives a pinion with 30 teeth and a module of 4 mm. Find the tangential force Ft on the pinion teeth.

P = 15 kWn = 960 rpmN1 = 30m = 4 mm
Ft = N
2.Gears & gear trainsHard
d1​ = 80 mmd2​ = 315 mm1440 rpmC = 600 mmT1​ = 600 NT2​P = ?

An open belt connects an 80 mm driving pulley at 1440 rpm to a 315 mm pulley; the centres are 600 mm apart. The coefficient of friction is 0.35 and the tight-side tension may not exceed 600 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 80 mmn1 = 1440 rpmd2 = 315 mmC = 600 mmμ = 0.35T1 = 600 N
P = kW
3.Torsion of shaftsMedium
T = 400 N·md = 50 mmL = 0.5 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 = 400 N·md = 50 mmL = 0.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
d = ?75 kW, 750 rpm

A solid brass shaft transmits 75 kW at 750 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 1° per m of length. Find the smallest diameter d that satisfies both limits.

P = 75 kWN = 750 rpmτallow = 60 MPaθallow = 1°/mG = 39 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 60σy​ = 60τxy​ = 40MPa

A point in a loaded part has σₓ = 60 MPa, σᵧ = 60 MPa and τₓᵧ = 40 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σx = 60 MPaσy = 60 MPaτxy = 40 MPa
θp = °
6.Mohr’s circle & stress transformationHard
σx​ = −70σy​ = 60τxy​ = 10MPa

A point in a thin plate is in plane stress: σₓ = −70 MPa, σᵧ = 60 MPa and τₓᵧ = 10 MPa (σz = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σx = −70 MPaσy = 60 MPaτxy = 10 MPaσY = 460 MPa
n =
7.Axial stress & strainMedium
300 kNØ40110 GPaØ100/50400 mmδ = ?

A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 50 mm, E = 110 GPa). Both are 400 mm long and are compressed together between rigid plates by 300 kN. Find the shortening δ.

P = 300 kNrod: Ø40 mm, E = 200 GPatube: Ø100/Ø50 mm, E = 110 GPaL = 400 mm
δ = mm
8.Axial stress & strainHard
Ø 25 mmSteel1.5 mØ 16 mmBrass2 mP = 45 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 = 45 kNsegment 1: Ø25 mm, 1.5 m, steel (E = 200 GPa)segment 2: Ø16 mm, 2 m, brass (E = 100 GPa)
σmax = MPa
9.Column bucklingMedium
PL = 6 m50 × 60 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 = 6 mE = 200 GPa (steel)b × h = 50 × 60 mm
Pcr = kN
10.Column bucklingHard
P = 45 kNL = 5 md = ?cross-section

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

P = 45 kNn = 2L = 5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 8 kN/mζ = 0.5m = 40 kg

A 40 kg mass on a spring with a stiffness of 8 kN/m has a viscous damper giving a damping ratio of 0.5. Find the damped natural frequency fd of its free vibration in Hz.

m = 40 kgk = 8 kN/mζ = 0.5
fd = Hz
12.Mechanical vibrationsHard
k1​ = 50k2​ = 20k3​ = 10m = 100 kg

A 100 kg mass hangs from spring k₃ = 10 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 50 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 = 100 kgk1 = 50 kN/mk2 = 20 kN/mk3 = 10 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: Ft = 2487 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 = 15000/100.5 = 149.2 N·m

Pitch diameter

d = m·N = 4·30 = 120 mm

Tangential force

Ft = 2T/d = 2·149.2/0.12 = 2487 N

Answer

Ft = 2487 N
2.Gears & gear trains

Answer: P = 2.236 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 = 2.747 rad

Slack-side tension

eμθ = e0.9616 = 2.616, T₂ = 600/2.616 = 229.4 N

Power

P = (600 − 229.4)·6.032 = 2.236 kW

Answer

P = 2.236 kW
3.Torsion of shafts

Answer: φ = 0.4789 °

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) = 400·1000·500/(39000·613592) = 0.008358 rad
φ = 0.008358·180/π = 0.4789°

Answer

φ = 0.4789°
4.Torsion of shafts

Answer: d = 61.48 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 = 75000/78.54 = 954.9 N·m

Diameter for strength

T = 954.9 N·m = 954930 N·mm
dτ = ∛(16·954930/(π·60)) = 43.28 mm

Diameter for stiffness

θ = 1·π/180/1000 = 1.745 × 10⁻⁵ rad/mm
G = 39 GPa = 39000 MPa
dφ = (32·954930/(π·39000·1.745 × 10⁻⁵))1/4 = 61.48 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(43.28, 61.48) = 61.48 mm

Answer

d = 61.48 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: θp = 45 °

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.

στxyCσ1​σ2​

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θp = atan2(2·40, 60 − 60) = 90°

Angle on the element

θp = 45°

Answer

θp = 45°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: n = 4.034

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 = −5 MPa, R = 65.76 MPa
σ1 = 60.76 MPa, σ2 = −70.76 MPa, σ3 = 0

Von Mises

σvM = √(60.76² − 60.76·(−70.76) + (−70.76)²) = 114 MPa
n = 460/114 = 4.034

Answer

n = 4.034

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

7.Axial stress & strain

Answer: δ = 0.1334 mm

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·40²/4 = 1257 mm²
At = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 110·10³·5890 = 8.993 × 10⁸ N

Shortening

δ = 300000·400/8.993 × 10⁸ = 0.1334 mm

Answer

δ = 0.1334 mm
8.Axial stress & strain

Answer: σmax = 223.8 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·16²/4 = 201.1 mm²
σ1 = 91.67 MPa
σ2 = 223.8 MPa

Largest stress

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

σmax = 223.8 MPa

Answer

σmax = 223.8 MPa
9.Column buckling

Answer: Pcr = 8.567 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 = 60·50³/12 = 625000 mm⁴

Effective length

KL = 2·6000 = 12000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·625000/12000² = 8567 N = 8.567 kN

Answer

Pcr = 8.567 kN
10.Column buckling

Answer: d = 98.17 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 = 2·5000 = 10000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·90000·10000²/(π³·200000))^(1/4) = 98.17 mm

Answer

d = 98.17 mm

Check: the slenderness KL/r is 407.4, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: fd = 1.949 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 = √(8000/40) = 14.14 rad/s
fn = 14.14/(2π) = 2.251 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 2.251·√(1 − 0.5²) = 1.949 Hz

Answer

fd = 1.949 Hz
12.Mechanical vibrations

Answer: fn = 1.489 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 = 50 + 20 = 70 kN/m
keq = 70·10/(70 + 10) = 8.75 kN/m

Natural frequency

ωn = √(8750/100) = 9.354 rad/s
fn = 9.354/(2π) = 1.489 Hz

Answer

fn = 1.489 Hz

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