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​ = 120 mmd2​ = 500 mm1440 rpmT1​ = 1000 NT2​ = 500 NP = ?

A belt runs on a 120 mm driving pulley at 1440 rpm. The tension is 1000 N on the tight side and 500 N on the slack side. Find the power P transmitted, in kW.

d1 = 120 mmn1 = 1440 rpmT1 = 1000 NT2 = 500 N
P = kW
2.Gears & gear trainsHard
d1​ = 160 mmd2​ = 400 mm960 rpmC = 600 mmT1​ = 2000 NT2​P = ?

An open belt connects a 160 mm driving pulley at 960 rpm to a 400 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 = 160 mmn1 = 960 rpmd2 = 400 mmC = 600 mmμ = 0.35T1 = 2000 N
P = kW
3.Torsion of shaftsMedium
do​ = ?T = 2000 N·m

A hollow shaft must carry a torque of 2000 N·m. Its inner diameter is to be 0.8 times its outer diameter, and the shear stress must not exceed 80 MPa. Find the smallest outer diameter do.

T = 2000 N·mdi/do = 0.8τallow = 80 MPa
do = mm
4.Torsion of shaftsHard
Tdo​ = 75 mm, di​ = 60 mm

A hollow shaft transmits 75 kW at 3000 rpm. Find the maximum shear stress τmax in the shaft.

P = 75 kWN = 3000 rpmdo = 75 mmdi = 60 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = 70σy​ = 20τxy​ = 40MPa

A point in a loaded part has σₓ = 70 MPa, σᵧ = 20 MPa and τₓᵧ = 40 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 70 MPaσy = 20 MPaτxy = 40 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​τxy​surface of the shaft

A solid shaft of diameter 25 mm carries an axial tensile force of 90 kN together with a torque of 400 N·m. At the surface of the shaft, find the largest principal stress σ1.

d = 25 mmF = 90 kNT = 400 N·m
σ1 = MPa
7.Axial stress & strainMedium
Ø 20 mmSteel3 mP = 115 kN

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

P = 115 kNd = 20 mmL = 3 mE = 200 GPa
δ = mm
8.Axial stress & strainHard
50 kN12600 mm200 mmØ20Ø50σ2​ = ?

A steel segment (diameter 20 mm, length 600 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 50 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø20 mm, 600 mm, E = 200 GPasegment 2: copper, Ø50 mm, 200 mm, E = 120 GPaP = 50 kN
σ2 = MPa
9.Column bucklingMedium
PL = 5 md = 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 = 5 mE = 70 GPa (aluminium)d = 80 mm
Pcr = kN
10.Column bucklingHard
P = 15 kNL = 4 md = ?cross-section

A solid round column of length L made of steel is fixed at both ends. It must carry 15 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 15 kNn = 2L = 4 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 20 kN/mζ = 0.35m = 5 kg

A 5 kg mass on a spring with a stiffness of 20 kN/m has a viscous damper giving a damping ratio of 0.35. Find the damped natural frequency fd of its free vibration in Hz.

m = 5 kgk = 20 kN/mζ = 0.35
fd = Hz
12.Mechanical vibrationsHard
k1​ = 50k2​ = 30m = 5 kg

A 5 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 50 kN/m and the lower one k₂ = 30 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 5 kgk1 = 50 kN/mk2 = 30 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: P = 4.524 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.

Belt speed

v = 9.048 m/s

Power

P = (1000 − 500)·9.048 = 4524 W = 4.524 kW

Answer

P = 4.524 kW
2.Gears & gear trains

Answer: P = 9.918 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 = 2.739 rad

Slack-side tension

eμθ = e0.9586 = 2.608, T₂ = 2000/2.608 = 766.9 N

Power

P = (2000 − 766.9)·8.042 = 9.918 kW

Answer

P = 9.918 kW
3.Torsion of shafts

Answer: do = 59.97 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 = 2000 N·m = 2 × 10⁶ N·mm
1 − k⁴ = 1 − 0.8⁴ = 0.5904
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·2 × 10⁶/(π·80·0.5904)) = 59.97 mm
di = 0.8·59.97 = 47.97 mm

Answer

do = 59.97 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: τmax = 4.881 MPa

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·3000/60 = 314.2 rad/s
T = P/ω = 75000/314.2 = 238.7 N·m

Polar second moment of area

J = π(75⁴ − 60⁴)/32 = 1.834 × 10⁶ mm⁴

Maximum shear stress

τmax = T·(do/2)/J = 238732·37.5/1.834 × 10⁶ = 4.881 MPa

Answer

τmax = 4.881 MPa
5.Mohr’s circle & stress transformation

Answer: τmax = 47.17 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 = √(((70 − 20)/2)² + 40²) = 47.17 MPa

Answer

τmax = 47.17 MPa
6.Mohr’s circle & stress transformation

Answer: σ1 = 251.1 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.

Stresses at the surface

σ = F/A = 90000/(π·25²/4) = 90000/490.9 = 183.3 MPa
τ = 16T/(πd³) = 16·400000/(π·25³) = 130.4 MPa
στxyCσ1​σ2​

Mohr’s circle

C = σ/2 = 91.67 MPa
R = √(91.67² + 130.4²) = 159.4 MPa
σ1 = C + R = 251.1 MPa
σ2 = C − R = −67.71 MPa

Answer

σ1 = 251.1 MPa
7.Axial stress & strain

Answer: δ = 5.491 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 = 115·1000/314.2 = 366.1 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 366.1/200000 = 0.00183 = 1830 µε

Elongation

δ = ε·L = 0.00183·3000 mm = 5.491 mm

Answer

δ = 5.491 mm
8.Axial stress & strain

Answer: σ2 = 23.39 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 = 200000·314.2/600 = 104720 N/mm
k₂ = 120000·1963/200 = 1.178 × 10⁶ N/mm

Share of segment 2

F2 = 50000·1.178 × 10⁶/1.283 × 10⁶ = 45918 N

Stress

σ2 = 45918/1963 = 23.39 MPa (compression)

Answer

σ2 = 23.39 MPa
9.Column buckling

Answer: Pcr = 113.4 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 round section has the same I about every axis.

I = πd⁴/64 = π·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 0.7·5000 = 3500 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/3500² = 113394 N = 113.4 kN

Answer

Pcr = 113.4 kN
10.Column buckling

Answer: d = 33.36 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 = 2·15000 = 30000 N
KL = 0.5·4000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·30000·2000²/(π³·200000))^(1/4) = 33.36 mm

Answer

d = 33.36 mm

Check: the slenderness KL/r is 239.8, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: fd = 9.429 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 = √(20000/5) = 63.25 rad/s
fn = 63.25/(2π) = 10.07 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 10.07·√(1 − 0.35²) = 9.429 Hz

Answer

fd = 9.429 Hz
12.Mechanical vibrations

Answer: fn = 9.746 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 = 1/50 + 1/30 → keq = 50·30/(50 + 30) = 18.75 kN/m

Natural frequency

ωn = √(18750/5) = 61.24 rad/s
fn = 61.24/(2π) = 9.746 Hz

Answer

fn = 9.746 Hz

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