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​ = 125 mmd2​ = 450 mm720 rpmT1​ = 500 NT2​ = 200 NP = ?

A belt runs on a 125 mm driving pulley at 720 rpm. The tension is 500 N on the tight side and 200 N on the slack side. Find the power P transmitted, in kW.

d1 = 125 mmn1 = 720 rpmT1 = 500 NT2 = 200 N
P = kW
2.Gears & gear trainsHard
A = 15B = 32C = 12D = 45

In a compound gear train, gear A (15 teeth) on the input shaft turning at 1500 rpm drives gear B (32 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D (45 teeth) on the output shaft. Find the output speed nout.

nin = 1500 rpmA = 15B = 32C = 12D = 45
nout = rpm
3.Torsion of shaftsMedium
T = 750 N·md = 40 mmL = 1 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 = 750 N·md = 40 mmL = 1 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
do​ = ?50 kW, 1500 rpm

A hollow shaft must transmit 50 kW at 1500 rpm. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter do.

P = 50 kWN = 1500 rpmdi/do = 0.75τallow = 75 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 70σy​ = 100MPaτxy​ = ?

A point has σₓ = 70 MPa and σᵧ = 100 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 150 MPa?

σx = 70 MPaσy = 100 MPaσ1 = 150 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 110σy​ = −40τxy​ = 40MPa

A point in a thin plate is in plane stress: σₓ = 110 MPa, σᵧ = −40 MPa and τₓᵧ = 40 MPa (σz = 0). The material has a yield strength of 300 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σx = 110 MPaσy = −40 MPaτxy = 40 MPaσY = 300 MPa
n =
7.Axial stress & strainMedium
Ø 20 mmSteel2 mP = 70 kN

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

P = 70 kNd = 20 mmL = 2 mE = 200 GPa
ε = µε
8.Axial stress & strainHard
P = ?Ø30110 GPaØ120/60600 mm

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of bronze (outer diameter 120 mm, inner diameter 60 mm, E = 110 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø30 mm, E = 200 GPa, σallow = 120 MPatube: Ø120/Ø60 mm, E = 110 GPa, σallow = 60 MPa
Pmax = kN
9.Column bucklingMedium
PL = 3 m20 × 60 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 = 3 mE = 70 GPa (aluminium)b × h = 20 × 60 mm
Pcr = kN
10.Column bucklingHard
PL = 3 md = 40 mmcross-section

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

L = 3 mE = 200 GPa (steel)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx40 mm3 cycles4.1 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed 3 cycles later by a peak of 4.1 mm. Find the damping ratio ζ.

x0 = 40 mmx3 = 4.1 mmn = 3
ζ =
12.Mechanical vibrationsHard
k1​ = 12k2​ = 80m = 5 kg

A 5 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 12 kN/m and k₂ = 80 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 = 5 kgk1 = 12 kN/mk2 = 80 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: P = 1.414 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 = 4.712 m/s

Power

P = (500 − 200)·4.712 = 1414 W = 1.414 kW

Answer

P = 1.414 kW
2.Gears & gear trains

Answer: nout = 187.5 rpm

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.

First stage

nB = 1500·15/32 = 703.1 rpm = nC

Second stage

nout = 703.1·12/45 = 187.5 rpm

Answer

nout = 187.5 rpm

total ratio = 8 : 1

3.Torsion of shafts

Answer: φ = 4.384 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 750·1000·1000/(39000·251327) = 0.07652 rad
φ = 0.07652·180/π = 4.384°

Answer

φ = 4.384°
4.Torsion of shafts

Answer: do = 31.62 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·1500/60 = 157.1 rad/s
T = P/ω = 50000/157.1 = 318.3 N·m

Outer diameter from the allowable stress

T = 318.3 N·m = 318310 N·mm
1 − k⁴ = 1 − 0.75⁴ = 0.6836
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·318310/(π·75·0.6836)) = 31.62 mm
di = 0.75·31.62 = 23.72 mm

Answer

do = 31.62 mm

A solid shaft would need d = 27.86 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: τxy = 63.25 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.

Required radius

C = (70 + 100)/2 = 85 MPa
R = σ1 − C = 150 − 85 = 65 MPa
στxyCσ1​σ2​

Shear stress

τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(65² − 15²) = 63.25 MPa

Answer

τxy = 63.25 MPa
6.Mohr’s circle & stress transformation

Answer: n = 1.982

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 = 35 MPa, R = 85 MPa
σ1 = 120 MPa, σ2 = −50 MPa, σ3 = 0

Von Mises

σvM = √(120² − 120·(−50) + (−50)²) = 151.3 MPa
n = 300/151.3 = 1.982

Answer

n = 1.982

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

7.Axial stress & strain

Answer: ε = 1114 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·20²/4 = 314.2 mm²
σ = P/A = 70·1000/314.2 = 222.8 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 222.8/200000 = 0.001114 = 1114 µε

Answer

ε = 1114 µε
8.Axial stress & strain

Answer: Pmax = 586 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 − 60²)/4 = 8482 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 110·10³·8482 = 1.074 × 10⁹ N

Governing strain

ε = min(0.0006, 0.0005455) = 0.0005455 (bronze)

Largest load

Pmax = 0.0005455·1.074 × 10⁹ = 586050 N = 586 kN

Answer

Pmax = 586 kN
9.Column buckling

Answer: Pcr = 6.266 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 = 60·20³/12 = 40000 mm⁴

Effective length

KL = 0.7·3000 = 2100 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·40000/2100² = 6266 N = 6.266 kN

Answer

Pcr = 6.266 kN
10.Column buckling

Answer: σcr = 5.483 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·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²
r = √(I/A) = √(125664/1257) = 10 mm
KL = 2·3000 = 6000 mm
KL/r = 6000/10 = 600

Critical stress

σcr = π²E/(KL/r)² = π²·200000/600² = 5.483 MPa

Answer

σcr = 5.483 MPa
11.Mechanical vibrations

Answer: ζ = 0.12

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 = (1/3)·ln(40/4.1) = 0.7593

Damping ratio

ζ = δ/√(4π² + δ²) = 0.7593/√(39.48 + 0.5765) = 0.12

Answer

ζ = 0.12
12.Mechanical vibrations

Answer: fn = 21.59 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 = 12 + 80 = 92 kN/m

Natural frequency

ωn = √(92000/5) = 135.6 rad/s
fn = 135.6/(2π) = 21.59 Hz

Answer

fn = 21.59 Hz

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