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​ = 200 mmd2​ = 400 mm960 rpmT1​ = 800 NT2​ = 200 NP = ?

A belt runs on a 200 mm driving pulley at 960 rpm. The tension is 800 N on the tight side and 200 N on the slack side. Find the power P transmitted, in kW.

d1 = 200 mmn1 = 960 rpmT1 = 800 NT2 = 200 N
P = kW
2.Gears & gear trainsHard
A = 15B = 45C = 45D = ?

In a compound gear train, gear A (15 teeth) on the input shaft turns at 1440 rpm and drives gear B (45 teeth). Gear C (45 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 300 rpm. How many teeth ND must gear D have?

nin = 1440 rpmA = 15B = 45C = 45nout = 300 rpm
ND =
3.Torsion of shaftsMedium
do​ = 60, di​ = 30 mmT = 1500 N·mL = 2 m

A hollow brass shaft (outer diameter 60 mm, inner diameter 30 mm, length 2 m) is fixed at one end and twisted by a torque of 1500 N·m at the other. Find the angle of twist φ in degrees.

T = 1500 N·mdo = 60 mmdi = 30 mmL = 2 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
d = 60 mmT0​ = 400 N·mACBa = 1 mb = 2 m

A solid steel shaft of diameter 60 mm is fixed to rigid walls at both ends A and B. A torque of 400 N·m is applied at C, 1 m from A and 2 m from B. Find the maximum shear stress τmax in the shaft.

d = 60 mmT0 = 400 N·ma = 1 mb = 2 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = 30σy​ = 50τxy​ = 30MPa

A point in a loaded part has σₓ = 30 MPa, σᵧ = 50 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 30 MPaσy = 50 MPaτxy = 30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 150σy​ = −60τxy​ = 20MPa

A point in a thin plate is in plane stress: σₓ = 150 MPa, σᵧ = −60 MPa and τₓᵧ = 20 MPa (σz = 0). The material has a yield strength of 355 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σx = 150 MPaσy = −60 MPaτxy = 20 MPaσY = 355 MPa
n =
7.Axial stress & strainMedium
Ø 25 mmBrass1.5 mP = 85 kN

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

P = 85 kNd = 25 mmL = 1.5 mE = 100 GPa
ε = µε
8.Axial stress & strainHard
P = ?Ø3070 GPaØ120/50800 mm

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of aluminium (outer diameter 120 mm, inner diameter 50 mm, E = 70 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø30 mm, E = 200 GPa, σallow = 120 MPatube: Ø120/Ø50 mm, E = 70 GPa, σallow = 100 MPa
Pmax = kN
9.Column bucklingMedium
PL = 6 md = 30 mmcross-section

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

L = 6 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
10.Column bucklingHard
P = 65 kNL = 3 md = ?cross-section

A solid round column of length L made of steel is fixed at both ends. It must carry 65 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 65 kNn = 2.5L = 3 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 150 kN/mc = 170 N·s/mm = 5 kg

A 5 kg mass is supported by a spring with a stiffness of 150 kN/m and a viscous damper with a damping coefficient of 170 N·s/m. Find the damping ratio ζ.

m = 5 kgk = 150 kN/mc = 170 N·s/m
ζ =
12.Mechanical vibrationsHard
k1​ = 5k2​ = 30m = 25 kg

A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 5 kN/m and k₂ = 30 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 = 25 kgk1 = 5 kN/mk2 = 30 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: P = 6.032 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 = 10.05 m/s

Power

P = (800 − 200)·10.05 = 6032 W = 6.032 kW

Answer

P = 6.032 kW
2.Gears & gear trains

Answer: ND = 72

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 = 1440·15/45 = 480 rpm

Teeth on gear D

ND = C·nC/nout = 45·480/300 = 72

Answer

ND = 72
3.Torsion of shafts

Answer: φ = 3.695 °

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(60⁴ − 30⁴)/32 = 1.193 × 10⁶ mm⁴

Angle of twist

T = 1500 N·m = 1.5 × 10⁶ N·mm
L = 2 m = 2000 mm
G = 39 GPa = 39000 MPa
φ = TL/(GJ) = 1.5 × 10⁶·2000/(39000·1.193 × 10⁶) = 0.06449 rad
φ = 0.06449·180/π = 3.695°

Answer

φ = 3.695°
4.Torsion of shafts

Answer: τmax = 6.288 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. 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 = 400
TA·a = TB·b
TA = 400·2/3 = 266.7 N·m
TB = 400·1/3 = 133.3 N·m

Maximum shear stress

The larger of the two torques governs (same diameter).

τmax = 16·266667/(π·60³) = 6.288 MPa

Answer

τmax = 6.288 MPa
5.Mohr’s circle & stress transformation

Answer: τmax = 31.62 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 = √(((30 − 50)/2)² + 30²) = 31.62 MPa

Answer

τmax = 31.62 MPa
6.Mohr’s circle & stress transformation

Answer: n = 1.863

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 = 45 MPa, R = 106.9 MPa
σ1 = 151.9 MPa, σ2 = −61.89 MPa, σ3 = 0

Von Mises

σvM = √(151.9² − 151.9·(−61.89) + (−61.89)²) = 190.5 MPa
n = 355/190.5 = 1.863

Answer

n = 1.863

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

7.Axial stress & strain

Answer: ε = 1732 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·25²/4 = 490.9 mm²
σ = P/A = 85·1000/490.9 = 173.2 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 173.2/100000 = 0.001732 = 1732 µε

Answer

ε = 1732 µε
8.Axial stress & strain

Answer: Pmax = 477.4 kN

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·30²/4 = 706.9 mm²
At = π(120² − 50²)/4 = 9346 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 70·10³·9346 = 7.956 × 10⁸ N

Governing strain

ε = min(0.0006, 0.001429) = 0.0006 (steel)

Largest load

Pmax = 0.0006·7.956 × 10⁸ = 477365 N = 477.4 kN

Answer

Pmax = 477.4 kN
9.Column buckling

Answer: Pcr = 2.18 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 1·6000 = 6000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·39761/6000² = 2180 N = 2.18 kN

Answer

Pcr = 2.18 kN
10.Column buckling

Answer: d = 44.07 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 = 2.5·65000 = 162500 N
KL = 0.5·3000 = 1500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·162500·1500²/(π³·200000))^(1/4) = 44.07 mm

Answer

d = 44.07 mm

Check: the slenderness KL/r is 136.1, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.09815

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.

Critical damping

cc = 2·√(k·m) = 2·√(150000·5) = 1732 N·s/m

Damping ratio

ζ = c/cc = 170/1732 = 0.09815

Answer

ζ = 0.09815

ζ < 1, so the system is underdamped and oscillates as it decays.

12.Mechanical vibrations

Answer: fn = 5.955 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 = 5 + 30 = 35 kN/m

Natural frequency

ωn = √(35000/25) = 37.42 rad/s
fn = 37.42/(2π) = 5.955 Hz

Answer

fn = 5.955 Hz

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