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
N1​ = 25N2​ = 80

A 4 kW motor running at 1450 rpm drives a pinion with 25 teeth, which meshes with a gear with 80 teeth. Ignore losses. Find the torque T1 on the motor shaft.

P = 4 kWn1 = 1450 rpmN1 = 25N2 = 80η = 100%
T1 = N·m
2.Gears & gear trainsHard
d1​ = 120 mmd2​ = 500 mm1500 rpmC = 600 mmT1​ = 500 NT2​P = ?

An open belt connects a 120 mm driving pulley at 1500 rpm to a 500 mm pulley; the centres are 600 mm apart. The coefficient of friction is 0.35 and the tight-side tension may not exceed 500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 120 mmn1 = 1500 rpmd2 = 500 mmC = 600 mmμ = 0.35T1 = 500 N
P = kW
3.Torsion of shaftsMedium
T = 400 N·md = 40 mmL = 1.2 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 = 400 N·md = 40 mmL = 1.2 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 10 kW at 500 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 10 kWN = 500 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = −50σy​ = −80τxy​ = −50MPa

A point in a thin plate is in plane stress: σₓ = −50 MPa, σᵧ = −80 MPa and τₓᵧ = −50 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).

σx = −50 MPaσy = −80 MPaτxy = −50 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
σx​τxy​surface of the shaft

A solid shaft of diameter 25 mm carries an axial compressive force of 60 kN together with a torque of 400 N·m. At the surface of the shaft, find the maximum in-plane shear stress τmax.

d = 25 mmF = −60 kNT = 400 N·m
τmax = MPa
7.Axial stress & strainMedium
Ø 12 mmCopper1.5 mP = 115 kN

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

P = 115 kNd = 12 mmL = 1.5 mE = 117 GPa
ε = µε
8.Axial stress & strainHard
P = ?Ø25110 GPaØ80/60500 mm

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of bronze (outer diameter 80 mm, inner diameter 60 mm, E = 110 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø25 mm, E = 200 GPa, σallow = 160 MPatube: Ø80/Ø60 mm, E = 110 GPa, σallow = 100 MPa
Pmax = kN
9.Column bucklingMedium
PL = 3 md = 80 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 = 3 mE = 200 GPa (steel)d = 80 mm
Pcr = kN
10.Column bucklingHard
P = 80 kNL = 4 md = ?cross-section

A solid round column of length L made of steel is pinned at both ends. It must carry 80 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 80 kNn = 3L = 4 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 4 kN/mc = 60 N·s/mm = 10 kg

A 10 kg mass is supported by a spring with a stiffness of 4 kN/m and a viscous damper with a damping coefficient of 60 N·s/m. Find the damping ratio ζ.

m = 10 kgk = 4 kN/mc = 60 N·s/m
ζ =
12.Mechanical vibrationsHard
k1​ = 100k2​ = 25m = 20 kg

A 20 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 100 kN/m and k₂ = 25 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 = 20 kgk1 = 100 kN/mk2 = 25 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: T1 = 26.34 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.

Motor torque

ω₁ = 2π·1450/60 = 151.8 rad/s
T1 = P/ω₁ = 4000/151.8 = 26.34 N·m

Answer

T1 = 26.34 N·m
2.Gears & gear trains

Answer: P = 2.746 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.

Angle of wrap

θ = 2.497 rad

Slack-side tension

eμθ = e0.874 = 2.396, T₂ = 500/2.396 = 208.6 N

Power

P = (500 − 208.6)·9.425 = 2.746 kW

Answer

P = 2.746 kW
3.Torsion of shafts

Answer: φ = 2.806 °

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) = 400·1000·1200/(39000·251327) = 0.04897 rad
φ = 0.04897·180/π = 2.806°

Answer

φ = 2.806°
4.Torsion of shafts

Answer: d = 23.49 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 = 10000/52.36 = 191 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·190986/(π·75)) = 23.49 mm

Answer

d = 23.49 mm

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

5.Mohr’s circle & stress transformation

Answer: τabs = 58.6 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 = −65 MPa, R = √(15² + (−50)²) = 52.2 MPa
σ1 = −12.8 MPa, σ2 = −117.2 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(104.4, 12.8, 117.2)/2 = 58.6 MPa

Answer

τabs = 58.6 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: τmax = 144 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.

Stresses at the surface

σ = F/A = −60000/(π·25²/4) = −60000/490.9 = −122.2 MPa
τ = 16T/(πd³) = 16·400000/(π·25³) = 130.4 MPa
στxyCτmax​

Mohr’s circle

C = σ/2 = −61.12 MPa
R = √((−61.12)² + 130.4²) = 144 MPa

Answer

τmax = 144 MPa
7.Axial stress & strain

Answer: ε = 8691 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·12²/4 = 113.1 mm²
σ = P/A = 115·1000/113.1 = 1017 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 1017/117000 = 0.008691 = 8691 µε

Answer

ε = 8691 µε
8.Axial stress & strain

Answer: Pmax = 272.1 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(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 110·10³·2199 = 3.401 × 10⁸ N

Governing strain

ε = min(0.0008, 0.0009091) = 0.0008 (steel)

Largest load

Pmax = 0.0008·3.401 × 10⁸ = 272062 N = 272.1 kN

Answer

Pmax = 272.1 kN
9.Column buckling

Answer: Pcr = 441 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 1·3000 = 3000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/3000² = 440978 N = 441 kN

Answer

Pcr = 441 kN
10.Column buckling

Answer: d = 79.34 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 = 3·80000 = 240000 N
KL = 1·4000 = 4000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·240000·4000²/(π³·200000))^(1/4) = 79.34 mm

Answer

d = 79.34 mm

Check: the slenderness KL/r is 201.7, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.15

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·√(4000·10) = 400 N·s/m

Damping ratio

ζ = c/cc = 60/400 = 0.15

Answer

ζ = 0.15

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

12.Mechanical vibrations

Answer: fn = 12.58 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 = 100 + 25 = 125 kN/m

Natural frequency

ωn = √(125000/20) = 79.06 rad/s
fn = 79.06/(2π) = 12.58 Hz

Answer

fn = 12.58 Hz

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