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​ = 200 mm2880 rpmT1​ = 1000 NT2​ = 300 NP = ?

A belt runs on a 120 mm driving pulley at 2880 rpm. The tension is 1000 N on the tight side and 300 N on the slack side. Find the power P transmitted, in kW.

d1 = 120 mmn1 = 2880 rpmT1 = 1000 NT2 = 300 N
P = kW
2.Gears & gear trainsHard
A = 18B = 36C = 12D = 40

In a compound gear train, gear A (18 teeth) on the input shaft turning at 3000 rpm drives gear B (36 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D (40 teeth) on the output shaft. Find the output speed nout.

nin = 3000 rpmA = 18B = 36C = 12D = 40
nout = rpm
3.Torsion of shaftsMedium
d = 12 mmT = 40 N·m250 mmφ = 7.22°

In a torsion test, a solid specimen of diameter 12 mm is twisted by a torque of 40 N·m. The two ends of a gauge length of 250 mm rotate 7.22° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 40 N·md = 12 mmL = 250 mmφ = 7.22°
G = GPa
4.Torsion of shaftsHard
d = ?50 kW, 500 rpm

A solid brass shaft transmits 50 kW at 500 rpm. The shear stress must not exceed 40 MPa and the twist must not exceed 0.5° per m of length. Find the smallest diameter d that satisfies both limits.

P = 50 kWN = 500 rpmτallow = 40 MPaθallow = 0.5°/mG = 39 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 80σy​ = −50τxy​ = 30MPa

A point in a thin plate is in plane stress: σₓ = 80 MPa, σᵧ = −50 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 = 80 MPaσy = −50 MPaτxy = 30 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 70σy​ = −40τxy​ = −50MPa

A point in a thin plate is in plane stress: σₓ = 70 MPa, σᵧ = −40 MPa and τₓᵧ = −50 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 = 70 MPaσy = −40 MPaτxy = −50 MPaσY = 355 MPa
n =
7.Axial stress & strainMedium
80 kNØ20120 GPaØ100/70300 mmδ = ?

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 100 mm, inner diameter 70 mm, E = 120 GPa). Both are 300 mm long and are compressed together between rigid plates by 80 kN. Find the shortening δ.

P = 80 kNrod: Ø20 mm, E = 200 GPatube: Ø100/Ø70 mm, E = 120 GPaL = 300 mm
δ = mm
8.Axial stress & strainHard
P = ?Ø40120 GPaØ120/50800 mm

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 50 mm, E = 120 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø40 mm, E = 200 GPa, σallow = 250 MPatube: Ø120/Ø50 mm, E = 120 GPa, σallow = 60 MPa
Pmax = kN
9.Column bucklingMedium
PL = 4 md = 60 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 = 4 mE = 200 GPa (steel)d = 60 mm
Pcr = kN
10.Column bucklingHard
P = 35 kNL = 2.5 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 35 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 35 kNn = 3L = 2.5 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
k = 15 kN/mc = 60 N·s/mm = 25 kg

A 25 kg mass is supported by a spring with a stiffness of 15 kN/m and a viscous damper with a damping coefficient of 60 N·s/m. Find the damping ratio ζ.

m = 25 kgk = 15 kN/mc = 60 N·s/m
ζ =
12.Mechanical vibrationsHard
k1​ = 10k2​ = 40k3​ = 5m = 5 kg

A 5 kg mass hangs from spring k₃ = 5 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 10 kN/m and k₂ = 40 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.

m = 5 kgk1 = 10 kN/mk2 = 40 kN/mk3 = 5 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: P = 12.67 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 = 18.1 m/s

Power

P = (1000 − 300)·18.1 = 12667 W = 12.67 kW

Answer

P = 12.67 kW
2.Gears & gear trains

Answer: nout = 450 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 = 3000·18/36 = 1500 rpm = nC

Second stage

nout = 1500·12/40 = 450 rpm

Answer

nout = 450 rpm

total ratio = 6.667 : 1

3.Torsion of shafts

Answer: G = 38.98 GPa

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 moment and angle

J = π·12⁴/32 = 2036 mm⁴
φ = 7.22·π/180 = 0.126 rad

Shear modulus

T = 40 N·m = 40000 N·mm
G = TL/(Jφ) = 40000·250/(2036·0.126) = 38982 MPa
G = 38.98 GPa

Answer

G = 38.98 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

4.Torsion of shafts

Answer: d = 73.12 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·500/60 = 52.36 rad/s
T = P/ω = 50000/52.36 = 954.9 N·m

Diameter for strength

T = 954.9 N·m = 954930 N·mm
dτ = ∛(16·954930/(π·40)) = 49.54 mm

Diameter for stiffness

θ = 0.5·π/180/1000 = 8.727 × 10⁻⁶ rad/mm
G = 39 GPa = 39000 MPa
dφ = (32·954930/(π·39000·8.727 × 10⁻⁶))1/4 = 73.12 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(49.54, 73.12) = 73.12 mm

Answer

d = 73.12 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: τabs = 71.59 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 = 15 MPa, R = √(65² + 30²) = 71.59 MPa
σ1 = 86.59 MPa, σ2 = −56.59 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(143.2, 86.59, 56.59)/2 = 71.59 MPa

Answer

τabs = 71.59 MPa

Here σ₁ and σ₂ have opposite signs, so the in-plane maximum R is also the absolute maximum.

6.Mohr’s circle & stress transformation

Answer: n = 2.739

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 = 15 MPa, R = 74.33 MPa
σ1 = 89.33 MPa, σ2 = −59.33 MPa, σ3 = 0

Von Mises

σvM = √(89.33² − 89.33·(−59.33) + (−59.33)²) = 129.6 MPa
n = 355/129.6 = 2.739

Answer

n = 2.739

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

7.Axial stress & strain

Answer: δ = 0.04416 mm

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·20²/4 = 314.2 mm²
At = π(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 120·10³·4006 = 5.435 × 10⁸ N

Shortening

δ = 80000·300/5.435 × 10⁸ = 0.04416 mm

Answer

δ = 0.04416 mm
8.Axial stress & strain

Answer: Pmax = 686.4 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·40²/4 = 1257 mm²
At = π(120² − 50²)/4 = 9346 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 120·10³·9346 = 1.373 × 10⁹ N

Governing strain

ε = min(0.00125, 0.0005) = 0.0005 (copper)

Largest load

Pmax = 0.0005·1.373 × 10⁹ = 686438 N = 686.4 kN

Answer

Pmax = 686.4 kN
9.Column buckling

Answer: Pcr = 313.9 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·60⁴/64 = 636173 mm⁴

Effective length

KL = 0.5·4000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·636173/2000² = 313939 N = 313.9 kN

Answer

Pcr = 313.9 kN
10.Column buckling

Answer: d = 93.8 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 = 3·35000 = 105000 N
KL = 2·2500 = 5000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·105000·5000²/(π³·70000))^(1/4) = 93.8 mm

Answer

d = 93.8 mm

Check: the slenderness KL/r is 213.2, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.04899

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·√(15000·25) = 1225 N·s/m

Damping ratio

ζ = c/cc = 60/1225 = 0.04899

Answer

ζ = 0.04899

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

12.Mechanical vibrations

Answer: fn = 4.799 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

Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.

k12 = 10 + 40 = 50 kN/m
keq = 50·5/(50 + 5) = 4.545 kN/m

Natural frequency

ωn = √(4545/5) = 30.15 rad/s
fn = 30.15/(2π) = 4.799 Hz

Answer

fn = 4.799 Hz

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