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​ = 18, m = 2 mm

A 1.5 kW motor running at 1450 rpm drives a pinion with 18 teeth and a module of 2 mm. Find the tangential force Ft on the pinion teeth.

P = 1.5 kWn = 1450 rpmN1 = 18m = 2 mm
Ft = N
2.Gears & gear trainsHard
d1​ = 120 mmd2​ = 315 mm960 rpmC = 500 mmT1​ = 1500 NT2​P = ?

An open belt connects a 120 mm driving pulley at 960 rpm to a 315 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 1500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 120 mmn1 = 960 rpmd2 = 315 mmC = 500 mmμ = 0.3T1 = 1500 N
P = kW
3.Torsion of shaftsMedium
do​ = ?T = 1000 N·m

A hollow shaft must carry a torque of 1000 N·m. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter do.

T = 1000 N·mdi/do = 0.75τallow = 40 MPa
do = mm
4.Torsion of shaftsHard
d = 60 mmTB​ = 400 N·mTC​ = 750 N·mABC0.8 m1.2 m

A solid brass shaft of diameter 60 mm is fixed at A. A torque of 400 N·m acts at B, 0.8 m from A, and a torque of 750 N·m in the opposite direction acts at the free end C, a further 1.2 m along. Find the magnitude of the angle of twist φC of end C, in degrees.

d = 60 mmTB = 400 N·mTC = 750 N·mAB = 0.8 mBC = 1.2 mG = 39 GPa
φC = °
5.Mohr’s circle & stress transformationMedium
σx​ = −20σy​ = 80τxy​ = 50MPa

A point has σₓ = −20 MPa, σᵧ = 80 MPa and τₓᵧ = 50 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 = −20 MPaσy = 80 MPaτxy = 50 MPa
θs = °
6.Mohr’s circle & stress transformationHard
σx​ = 30σy​ = 20τxy​ = −50MPa

A point in a thin plate is in plane stress: σₓ = 30 MPa, σᵧ = 20 MPa and τₓᵧ = −50 MPa (σz = 0). Find the von Mises equivalent stress σvM.

σx = 30 MPaσy = 20 MPaτxy = −50 MPa
σvM = MPa
7.Axial stress & strainMedium
Ø 20 mmSteel3 mP = 105 kN

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

P = 105 kNd = 20 mmL = 3 mE = 200 GPa
δ = mm
8.Axial stress & strainHard
Ø 30 mmSteel1 mØ 16 mmAluminium2 mP = 120 kN

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

P = 120 kNsegment 1: Ø30 mm, 1 m, steel (E = 200 GPa)segment 2: Ø16 mm, 2 m, aluminium (E = 70 GPa)
δ = mm
9.Column bucklingMedium
PL = 2 m40 × 80 mmcross-section

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

L = 2 mE = 70 GPa (aluminium)b × h = 40 × 80 mm
Pcr = kN
10.Column bucklingHard
P = 80 kNL = 3 md = ?cross-section

A solid round column of length L made of aluminium is pinned at both ends. It must carry 80 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 80 kNn = 3L = 3 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
tx40 mm4 cycles8.8 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed 4 cycles later by a peak of 8.8 mm. Find the damping ratio ζ.

x0 = 40 mmx4 = 8.8 mmn = 4
ζ =
12.Mechanical vibrationsHard
k1​ = 20k2​ = 15m = 10 kg

A 10 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 20 kN/m and the lower one k₂ = 15 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 10 kgk1 = 20 kN/mk2 = 15 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: Ft = 548.8 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·1450/60 = 151.8 rad/s
T = P/ω = 1500/151.8 = 9.879 N·m

Pitch diameter

d = m·N = 2·18 = 36 mm

Tangential force

Ft = 2T/d = 2·9.879/0.036 = 548.8 N

Answer

Ft = 548.8 N
2.Gears & gear trains

Answer: P = 5.082 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 = 2.749 rad

Slack-side tension

eμθ = e0.8247 = 2.281, T₂ = 1500/2.281 = 657.5 N

Power

P = (1500 − 657.5)·6.032 = 5.082 kW

Answer

P = 5.082 kW
3.Torsion of shafts

Answer: do = 57.11 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.

Outer diameter from the allowable stress

T = 1000 N·m = 1 × 10⁶ N·mm
1 − k⁴ = 1 − 0.75⁴ = 0.6836
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·1 × 10⁶/(π·40·0.6836)) = 57.11 mm
di = 0.75·57.11 = 42.83 mm

Answer

do = 57.11 mm

A solid shaft would need d = 50.31 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: φC = 1.362 °

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.

Internal torques

Sense of TC taken as positive.

TBC = TC = 750 N·m
TAB = TC − TB = 750 − 400 = 350 N·m

Twist of each segment

J = π·60⁴/32 = 1.272 × 10⁶ mm⁴
G = 39 GPa = 39000 MPa
φAB = 350000·800/(39000·1.272 × 10⁶) = 0.005643 rad
φBC = 750000·1200/(39000·1.272 × 10⁶) = 0.01814 rad

Twist of end C

φC = 0.005643 + 0.01814 = 0.02378 rad → |φC| = 1.362°

Answer

φC = 1.362°
5.Mohr’s circle & stress transformation

Answer: θs = 22.5 °

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 = 50/70.71 = 0.7071, sin 2θs = −(−50)/70.71 = 0.7071
2θs = atan2(50, 50) = 45°

Angle on the element

θs = 22.5°
check: θs − θp = 22.5 − 67.5 = −45°

Answer

θs = 22.5°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σvM = 90.55 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 = 50.25 MPa
σ1 = 75.25 MPa, σ2 = −25.25 MPa, σ3 = 0

Von Mises

σvM = √(75.25² − 75.25·(−25.25) + (−25.25)²) = 90.55 MPa

Answer

σvM = 90.55 MPa
7.Axial stress & strain

Answer: δ = 5.013 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 = 105·1000/314.2 = 334.2 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 334.2/200000 = 0.001671 = 1671 µε

Elongation

δ = ε·L = 0.001671·3000 mm = 5.013 mm

Answer

δ = 5.013 mm
8.Axial stress & strain

Answer: δ = 17.9 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·30²/4 = 706.9 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 169.8 MPa
σ2 = 596.8 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 169.8·1000/200000 = 0.8488 mm
δ2 = σ2L2/E2 = 596.8·2000/70000 = 17.05 mm
δ = δ1 + δ2 = 17.9 mm

Answer

δ = 17.9 mm
9.Column buckling

Answer: Pcr = 150.4 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 = 80·40³/12 = 426667 mm⁴

Effective length

KL = 0.7·2000 = 1400 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·426667/1400² = 150394 N = 150.4 kN

Answer

Pcr = 150.4 kN
10.Column buckling

Answer: d = 89.33 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 = 3·80000 = 240000 N
KL = 1·3000 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·240000·3000²/(π³·70000))^(1/4) = 89.33 mm

Answer

d = 89.33 mm

Check: the slenderness KL/r is 134.3, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.06014

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 = (1/4)·ln(40/8.8) = 0.3785

Damping ratio

ζ = δ/√(4π² + δ²) = 0.3785/√(39.48 + 0.1433) = 0.06014

Answer

ζ = 0.06014
12.Mechanical vibrations

Answer: fn = 4.66 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.

1/keq = 1/20 + 1/15 → keq = 20·15/(20 + 15) = 8.571 kN/m

Natural frequency

ωn = √(8571/10) = 29.28 rad/s
fn = 29.28/(2π) = 4.66 Hz

Answer

fn = 4.66 Hz

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