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​ = 100 mmd2​ = 300 mm2880 rpmT1​ = 1000 NT2​ = 250 NP = ?

A belt runs on a 100 mm driving pulley at 2880 rpm. The tension is 1000 N on the tight side and 250 N on the slack side. Find the power P transmitted, in kW.

d1 = 100 mmn1 = 2880 rpmT1 = 1000 NT2 = 250 N
P = kW
2.Gears & gear trainsHard
A = 12B = 40C = 24D = ?

In a compound gear train, gear A (12 teeth) on the input shaft turns at 960 rpm and drives gear B (40 teeth). Gear C (24 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 216 rpm. How many teeth ND must gear D have?

nin = 960 rpmA = 12B = 40C = 24nout = 216 rpm
ND =
3.Torsion of shaftsMedium
T = 400 N·md = 30 mmL = 0.8 mφ = ?

A solid aluminium 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 = 30 mmL = 0.8 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
d = ?10 kW, 750 rpm

A solid aluminium shaft transmits 10 kW at 750 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 0.75° per m of length. Find the smallest diameter d that satisfies both limits.

P = 10 kWN = 750 rpmτallow = 60 MPaθallow = 0.75°/mG = 26 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 150σy​ = 70τxy​ = 30MPa

A point in a thin plate is in plane stress: σₓ = 150 MPa, σᵧ = 70 MPa and τₓᵧ = 30 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).

σx = 150 MPaσy = 70 MPaτxy = 30 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 100σy​ = −20τxy​ = −40MPa

A point in a thin plate is in plane stress: σₓ = 100 MPa, σᵧ = −20 MPa and τₓᵧ = −40 MPa (σz = 0). Find the von Mises equivalent stress σvM.

σx = 100 MPaσy = −20 MPaτxy = −40 MPa
σvM = MPa
7.Axial stress & strainMedium
Ø 10 mmSteel2.5 mP = 10 kN

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

P = 10 kNd = 10 mmL = 2.5 mE = 200 GPa
ε = µε
8.Axial stress & strainHard
20 kN12600 mm500 mmØ20Ø30σ2​ = ?

A steel segment (diameter 20 mm, length 600 mm, E = 200 GPa) and a segment made of copper (diameter 30 mm, length 500 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 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, Ø30 mm, 500 mm, E = 120 GPaP = 20 kN
σ2 = MPa
9.Column bucklingMedium
PL = 3 m20 × 40 mmcross-section

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

L = 3 mE = 200 GPa (steel)b × h = 20 × 40 mm
Pcr = kN
10.Column bucklingHard
PL = 6 md = 50 mmcross-section

A column of length L made of aluminium is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.

L = 6 mE = 70 GPa (aluminium)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 80 kN/mζ = 0.5m = 25 kg

A 25 kg mass on a spring with a stiffness of 80 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 = 25 kgk = 80 kN/mζ = 0.5
fd = Hz
12.Mechanical vibrationsHard
k1​ = 30k2​ = 80m = 100 kg

A 100 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 30 kN/m and the lower one k₂ = 80 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.

m = 100 kgk1 = 30 kN/mk2 = 80 kN/m
keq = kN/m

3Solutions

1.Gears & gear trains

Answer: P = 11.31 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 = 15.08 m/s

Power

P = (1000 − 250)·15.08 = 11310 W = 11.31 kW

Answer

P = 11.31 kW
2.Gears & gear trains

Answer: ND = 32

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.

Intermediate shaft speed

nB = nC = 960·12/40 = 288 rpm

Teeth on gear D

ND = C·nC/nout = 24·288/216 = 32

Answer

ND = 32
3.Torsion of shafts

Answer: φ = 8.868 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·800/(26000·79522) = 0.1548 rad
φ = 0.1548·180/π = 8.868°

Answer

φ = 8.868°
4.Torsion of shafts

Answer: d = 44.18 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 = 10000/78.54 = 127.3 N·m

Diameter for strength

T = 127.3 N·m = 127324 N·mm
dτ = ∛(16·127324/(π·60)) = 22.11 mm

Diameter for stiffness

θ = 0.75·π/180/1000 = 1.309 × 10⁻⁵ rad/mm
G = 26 GPa = 26000 MPa
dφ = (32·127324/(π·26000·1.309 × 10⁻⁵))1/4 = 44.18 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(22.11, 44.18) = 44.18 mm

Answer

d = 44.18 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: τabs = 80 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. In three dimensions the absolute maximum shear stress is (σmax − σmin)/2 over all three principal stresses; in plane stress the third one is zero.

στxyCσ1​σ2​

In-plane principal stresses

C = 110 MPa, R = √(40² + 30²) = 50 MPa
σ1 = 160 MPa, σ2 = 60 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(100, 160, 60)/2 = 80 MPa

Answer

τabs = 80 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

6.Mohr’s circle & stress transformation

Answer: σvM = 131.1 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 = 40 MPa, R = 72.11 MPa
σ1 = 112.1 MPa, σ2 = −32.11 MPa, σ3 = 0

Von Mises

σvM = √(112.1² − 112.1·(−32.11) + (−32.11)²) = 131.1 MPa

Answer

σvM = 131.1 MPa
7.Axial stress & strain

Answer: ε = 636.6 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 = 10·1000/78.54 = 127.3 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 127.3/200000 = 0.0006366 = 636.6 µε

Answer

ε = 636.6 µε
8.Axial stress & strain

Answer: σ2 = 17.49 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 = 120000·706.9/500 = 169646 N/mm

Share of segment 2

F2 = 20000·169646/274366 = 12366 N

Stress

σ2 = 12366/706.9 = 17.49 MPa (compression)

Answer

σ2 = 17.49 MPa
9.Column buckling

Answer: Pcr = 23.39 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 = 40·20³/12 = 26667 mm⁴

Effective length

KL = 0.5·3000 = 1500 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·26667/1500² = 23395 N = 23.39 kN

Answer

Pcr = 23.39 kN
10.Column buckling

Answer: σcr = 6.12 MPa

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·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 0.7·6000 = 4200 mm
KL/r = 4200/12.5 = 336

Critical stress

σcr = π²E/(KL/r)² = π²·70000/336² = 6.12 MPa

Answer

σcr = 6.12 MPa
11.Mechanical vibrations

Answer: fd = 7.797 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 = √(80000/25) = 56.57 rad/s
fn = 56.57/(2π) = 9.003 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 9.003·√(1 − 0.5²) = 7.797 Hz

Answer

fd = 7.797 Hz
12.Mechanical vibrations

Answer: keq = 21.82 kN/m

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 = 1/30 + 1/80 → keq = 30·80/(30 + 80) = 21.82 kN/m

Answer

keq = 21.82 kN/m

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