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​ = 80 mmd2​ = 315 mm1440 rpmn2​ = ?

An 80 mm pulley at 1440 rpm drives a 315 mm pulley by a belt with 4 % slip. Find the speed n2 of the driven pulley.

n1 = 1440 rpmd1 = 80 mmd2 = 315 mms = 4 %
n2 = rpm
2.Gears & gear trainsHard
A = 15B = 24C = 20D = 50

A 15 kW motor at 2900 rpm drives gear A (15 teeth), which meshes with gear B (24 teeth). Gear C (20 teeth) on the same shaft as B drives gear D (50 teeth) on the output shaft. Each gear mesh has an efficiency of 98%. Find the output torque Tout.

P = 15 kWnin = 2900 rpmA = 15B = 24C = 20D = 50ηmesh = 98%
Tout = N·m
3.Torsion of shaftsMedium
d = ?T = 1500 N·mL = 0.8 m

A solid aluminium shaft 0.8 m long carries a torque of 1500 N·m. The angle of twist between its ends must not exceed 1.5°. Find the smallest diameter d that meets this limit.

T = 1500 N·mL = 0.8 mG = 26 GPaφallow = 1.5°
d = mm
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 0.8 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter do.

P = 50 kWN = 1500 rpmdi/do = 0.8τallow = 50 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
σy​ = −30MPaτxy​ = ?

A point has σₓ = 0 MPa and σᵧ = −30 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 40 MPa?

σx = 0 MPaσy = −30 MPaσ1 = 40 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 70σy​ = −10τxy​ = 20MPa

A point in a thin plate is in plane stress: σₓ = 70 MPa, σᵧ = −10 MPa and τₓᵧ = 20 MPa (σz = 0). The material has a yield strength of 250 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σx = 70 MPaσy = −10 MPaτxy = 20 MPaσY = 250 MPa
n =
7.Axial stress & strainMedium
Ø 10 mmSteel1 mP = 75 kN

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

P = 75 kNd = 10 mmL = 1 mE = 200 GPa
δ = mm
8.Axial stress & strainHard
P = ?Ø25100 GPaØ120/50800 mm

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 50 mm, E = 100 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø25 mm, E = 200 GPa, σallow = 150 MPatube: Ø120/Ø50 mm, E = 100 GPa, σallow = 50 MPa
Pmax = kN
9.Column bucklingMedium
PL = 6 md = 40 mmcross-section

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

L = 6 mE = 70 GPa (aluminium)d = 40 mm
Pcr = kN
10.Column bucklingHard
P = 145 kNL = 4 md = ?cross-section

A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 145 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 145 kNn = 2.5L = 4 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
k = 10 kN/mζ = 0.4m = 8 kg

An 8 kg mass on a spring with a stiffness of 10 kN/m has a viscous damper giving a damping ratio of 0.4. Find the damped natural frequency fd of its free vibration in Hz.

m = 8 kgk = 10 kN/mζ = 0.4
fd = Hz
12.Mechanical vibrationsHard
k1​ = 40k2​ = 15m = 25 kg

A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 15 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 = 40 kN/mk2 = 15 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: n2 = 351.1 rpm

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.

Speed without slip

1440·80/315 = 365.7 rpm

With slip

n2 = 365.7·0.96 = 351.1 rpm

Answer

n2 = 351.1 rpm
2.Gears & gear trains

Answer: Tout = 189.7 N·m

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.

Output speed

nout = 2900·(15/24)·(20/50) = 725 rpm
ωout = 2π·725/60 = 75.92 rad/s

Output power and torque

Pout = 15000·0.98² = 14406 W
Tout = Pout/ωout = 14406/75.92 = 189.7 N·m

Answer

Tout = 189.7 N·m
3.Torsion of shafts

Answer: d = 65.1 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.

Required polar moment

T = 1500 N·m = 1.5 × 10⁶ N·mm
L = 0.8 m = 800 mm
G = 26 GPa = 26000 MPa
φ = 1.5·π/180 = 0.02618 rad
J = TL/(Gφ) = 1.5 × 10⁶·800/(26000·0.02618) = 1.763 × 10⁶ mm⁴

Diameter

d = (32J/π)1/4 = 65.1 mm

Answer

d = 65.1 mm

In practice you would round up to the next standard size.

4.Torsion of shafts

Answer: do = 38.01 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 − 0.8⁴ = 0.5904
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·318310/(π·50·0.5904)) = 38.01 mm
di = 0.8·38.01 = 30.41 mm

Answer

do = 38.01 mm

A solid shaft would need d = 31.89 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: τxy = 52.92 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 = (0 + (−30))/2 = −15 MPa
R = σ1 − C = 40 − (−15) = 55 MPa
στxyCσ1​σ2​

Shear stress

τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(55² − 15²) = 52.92 MPa

Answer

τxy = 52.92 MPa
6.Mohr’s circle & stress transformation

Answer: n = 3.01

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 = 30 MPa, R = 44.72 MPa
σ1 = 74.72 MPa, σ2 = −14.72 MPa, σ3 = 0

Von Mises

σvM = √(74.72² − 74.72·(−14.72) + (−14.72)²) = 83.07 MPa
n = 250/83.07 = 3.01

Answer

n = 3.01

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

7.Axial stress & strain

Answer: δ = 4.775 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·10²/4 = 78.54 mm²
σ = P/A = 75·1000/78.54 = 954.9 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 954.9/200000 = 0.004775 = 4775 µε

Elongation

δ = ε·L = 0.004775·1000 mm = 4.775 mm

Answer

δ = 4.775 mm
8.Axial stress & strain

Answer: Pmax = 516.4 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·25²/4 = 490.9 mm²
At = π(120² − 50²)/4 = 9346 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 100·10³·9346 = 1.033 × 10⁹ N

Governing strain

ε = min(0.00075, 0.0005) = 0.0005 (brass)

Largest load

Pmax = 0.0005·1.033 × 10⁹ = 516399 N = 516.4 kN

Answer

Pmax = 516.4 kN
9.Column buckling

Answer: Pcr = 9.646 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·40⁴/64 = 125664 mm⁴

Effective length

KL = 0.5·6000 = 3000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·125664/3000² = 9646 N = 9.646 kN

Answer

Pcr = 9.646 kN
10.Column buckling

Answer: d = 161.7 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 = 2.5·145000 = 362500 N
KL = 2·4000 = 8000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·362500·8000²/(π³·70000))^(1/4) = 161.7 mm

Answer

d = 161.7 mm

Check: the slenderness KL/r is 197.9, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: fd = 5.157 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 = √(10000/8) = 35.36 rad/s
fn = 35.36/(2π) = 5.627 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 5.627·√(1 − 0.4²) = 5.157 Hz

Answer

fd = 5.157 Hz
12.Mechanical vibrations

Answer: fn = 7.465 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 = 40 + 15 = 55 kN/m

Natural frequency

ωn = √(55000/25) = 46.9 rad/s
fn = 46.9/(2π) = 7.465 Hz

Answer

fn = 7.465 Hz

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