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​ = 30, m = 2 mm

A 22 kW motor running at 2900 rpm drives a pinion with 30 teeth and a module of 2 mm. Find the tangential force Ft on the pinion teeth.

P = 22 kWn = 2900 rpmN1 = 30m = 2 mm
Ft = N
2.Gears & gear trainsHard
d1​ = 120 mmd2​ = 400 mm2880 rpmC = 800 mmθ = ?

An open belt connects a 120 mm pulley and a 400 mm pulley whose centres are 800 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 120 mmd2 = 400 mmC = 800 mm
θ = °
3.Torsion of shaftsMedium
d = 12 mmT = 30 N·m250 mmφ = 5.41°

In a torsion test, a solid specimen of diameter 12 mm is twisted by a torque of 30 N·m. The two ends of a gauge length of 250 mm rotate 5.41° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 30 N·md = 12 mmL = 250 mmφ = 5.41°
G = GPa
4.Torsion of shaftsHard
Tdo​ = 40 mm, di​ = 20 mm

A hollow shaft transmits 50 kW at 500 rpm. Find the maximum shear stress τmax in the shaft.

P = 50 kWN = 500 rpmdo = 40 mmdi = 20 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = 70σy​ = 60τxy​ = −10MPa

A point has σₓ = 70 MPa, σᵧ = 60 MPa and τₓᵧ = −10 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σx = 70 MPaσy = 60 MPaτxy = −10 MPa
θs = °
6.Mohr’s circle & stress transformationHard
σx​ = 120σy​ = 50τxy​ = −10MPa

A point in a thin plate is in plane stress: σₓ = 120 MPa, σᵧ = 50 MPa and τₓᵧ = −10 MPa (σz = 0). The material has a yield strength of 275 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σx = 120 MPaσy = 50 MPaτxy = −10 MPaσY = 275 MPa
n =
7.Axial stress & strainMedium
Ø 16 mmSteel2 mP = 30 kN

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

P = 30 kNd = 16 mmL = 2 mE = 200 GPa
ε = µε
8.Axial stress & strainHard
Ø 30 mmCopper1 mØ 16 mmAluminium0.5 mP = 115 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.

P = 115 kNsegment 1: Ø30 mm, 1 m, copper (E = 117 GPa)segment 2: Ø16 mm, 0.5 m, aluminium (E = 70 GPa)
σmax = MPa
9.Column bucklingMedium
PL = 2.5 md = 40 mmcross-section

A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.

L = 2.5 mE = 70 GPa (aluminium)d = 40 mm
Pcr = kN
10.Column bucklingHard
PL = 4 md = 50 mmcross-section

A column of length L made of aluminium is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.

L = 4 mE = 70 GPa (aluminium)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 10 kN/mζ = 0.7m = 2 kg

A 2 kg mass on a spring with a stiffness of 10 kN/m has a viscous damper giving a damping ratio of 0.7. Find the damped natural frequency fd of its free vibration in Hz.

m = 2 kgk = 10 kN/mζ = 0.7
fd = Hz
12.Mechanical vibrationsHard
k1​ = 100k2​ = 10m = 50 kg

A 50 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 100 kN/m and the lower one k₂ = 10 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.

m = 50 kgk1 = 100 kN/mk2 = 10 kN/m
keq = kN/m

3Solutions

1.Gears & gear trains

Answer: Ft = 2415 N

Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.

Torque on the pinion

ω = 2π·2900/60 = 303.7 rad/s
T = P/ω = 22000/303.7 = 72.44 N·m

Pitch diameter

d = m·N = 2·30 = 60 mm

Tangential force

Ft = 2T/d = 2·72.44/0.06 = 2415 N

Answer

Ft = 2415 N
2.Gears & gear trains

Answer: θ = 159.8 °

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 angle

asin(0.175) = 10.08°

Angle of wrap

θ = 180 − 2·10.08 = 159.8°

Answer

θ = 159.8°
3.Torsion of shafts

Answer: G = 39.02 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 = 5.41·π/180 = 0.09442 rad

Shear modulus

T = 30 N·m = 30000 N·mm
G = TL/(Jφ) = 30000·250/(2036·0.09442) = 39018 MPa
G = 39.02 GPa

Answer

G = 39.02 GPa

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

4.Torsion of shafts

Answer: τmax = 81.06 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.

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s
T = P/ω = 50000/52.36 = 954.9 N·m

Polar second moment of area

J = π(40⁴ − 20⁴)/32 = 235619 mm⁴

Maximum shear stress

τmax = T·(do/2)/J = 954930·20/235619 = 81.06 MPa

Answer

τmax = 81.06 MPa
5.Mohr’s circle & stress transformation

Answer: θs = −76.72 °

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.

σxyCτmax​

Angle on the circle

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θs = −10/11.18 = −0.8944, sin 2θs = −5/11.18 = −0.4472
2θs = atan2(−5, −10) = −153.4°

Angle on the element

θs = −76.72°
check: θs − θp = −76.72 − (−31.72) = −45°

Answer

θs = −76.72°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: n = 2.599

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 = 85 MPa, R = 36.4 MPa
σ1 = 121.4 MPa, σ2 = 48.6 MPa, σ3 = 0

Von Mises

σvM = √(121.4² − 121.4·48.6 + 48.6²) = 105.8 MPa
n = 275/105.8 = 2.599

Answer

n = 2.599

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

7.Axial stress & strain

Answer: ε = 746 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·16²/4 = 201.1 mm²
σ = P/A = 30·1000/201.1 = 149.2 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 149.2/200000 = 0.000746 = 746 µε

Answer

ε = 746 µε
8.Axial stress & strain

Answer: σmax = 572 MPa

Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.

Same force in both segments

The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.

A1 = π·30²/4 = 706.9 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 162.7 MPa
σ2 = 572 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σmax = 572 MPa

Answer

σmax = 572 MPa
9.Column buckling

Answer: Pcr = 3.473 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 = 2·2500 = 5000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·125664/5000² = 3473 N = 3.473 kN

Answer

Pcr = 3.473 kN
10.Column buckling

Answer: σcr = 6.747 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·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 1·4000 = 4000 mm
KL/r = 4000/12.5 = 320

Critical stress

σcr = π²E/(KL/r)² = π²·70000/320² = 6.747 MPa

Answer

σcr = 6.747 MPa
11.Mechanical vibrations

Answer: fd = 8.037 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 = √(10000/2) = 70.71 rad/s
fn = 70.71/(2π) = 11.25 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 11.25·√(1 − 0.7²) = 8.037 Hz

Answer

fd = 8.037 Hz
12.Mechanical vibrations

Answer: keq = 9.091 kN/m

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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.

1/keq = 1/100 + 1/10 → keq = 100·10/(100 + 10) = 9.091 kN/m

Answer

keq = 9.091 kN/m

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