Machine Design: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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​ = 500 mm1450 rpmn2​ = ?

A 125 mm pulley at 1450 rpm drives a 500 mm pulley by a belt with 4 % slip. Find the speed n2 of the driven pulley.

n1 = 1450 rpmd1 = 125 mmd2 = 500 mms = 4 %
n2 = rpm
2.Gears & gear trainsHard
A = 12B = 20C = 32D = ?

In a compound gear train, gear A (12 teeth) on the input shaft turns at 1450 rpm and drives gear B (20 teeth). Gear C (32 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 348 rpm. How many teeth ND must gear D have?

nin = 1450 rpmA = 12B = 20C = 32nout = 348 rpm
ND =
3.Torsion of shaftsMedium
do​ = 30, di​ = 23 mmT = 100 N·mL = 1 m

A hollow brass shaft (outer diameter 30 mm, inner diameter 23 mm, length 1 m) is fixed at one end and twisted by a torque of 100 N·m at the other. Find the angle of twist φ in degrees.

T = 100 N·mdo = 30 mmdi = 23 mmL = 1 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
Tdo​ = 100 mm, di​ = 80 mm

A hollow shaft transmits 30 kW at 750 rpm. Find the maximum shear stress τmax in the shaft.

P = 30 kWN = 750 rpmdo = 100 mmdi = 80 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = −30σy​ = −70τxy​ = −10MPa

A point in a loaded part has σₓ = −30 MPa, σᵧ = −70 MPa and τₓᵧ = −10 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = −30 MPaσy = −70 MPaτxy = −10 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 70σy​ = 110τxy​ = −60MPa

A point in a thin plate is in plane stress: σₓ = 70 MPa, σᵧ = 110 MPa and τₓᵧ = −60 MPa (σz = 0). Find the von Mises equivalent stress σvM.

σx = 70 MPaσy = 110 MPaτxy = −60 MPa
σvM = MPa
7.Axial stress & strainMedium
Ø 20 mmCopper1 mP = 20 kN

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

P = 20 kNd = 20 mmL = 1 mE = 117 GPa
ε = µε
8.Axial stress & strainHard
Ø 40 mmSteel1 mØ 16 mmCopper2 mP = 120 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 120 kNsegment 1: Ø40 mm, 1 m, steel (E = 200 GPa)segment 2: Ø16 mm, 2 m, copper (E = 117 GPa)
δ = mm
9.Column bucklingMedium
PL = 1.5 md = 40 mmcross-section

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

L = 1.5 mE = 200 GPa (steel)d = 40 mm
Pcr = kN
10.Column bucklingHard
PL = 2.5 md = 30 mmcross-section

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

L = 2.5 mE = 70 GPa (aluminium)d = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx10 mm2 cycles3.6 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 10 mm is followed 2 cycles later by a peak of 3.6 mm. Find the damping ratio ζ.

x0 = 10 mmx2 = 3.6 mmn = 2
ζ =
12.Mechanical vibrationsHard
k1​ = 80k2​ = 5m = 100 kg

A 100 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 80 kN/m and the lower one k₂ = 5 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.

m = 100 kgk1 = 80 kN/mk2 = 5 kN/m
keq = kN/m

3Solutions

1.Gears & gear trains

Answer: n2 = 348 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

1450·125/500 = 362.5 rpm

With slip

n2 = 362.5·0.96 = 348 rpm

Answer

n2 = 348 rpm
2.Gears & gear trains

Answer: ND = 80

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.

Intermediate shaft speed

nB = nC = 1450·12/20 = 870 rpm

Teeth on gear D

ND = C·nC/nout = 32·870/348 = 80

Answer

ND = 80
3.Torsion of shafts

Answer: φ = 2.823 °

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(30⁴ − 23⁴)/32 = 52048 mm⁴

Angle of twist

T = 100 N·m = 100000 N·mm
L = 1 m = 1000 mm
G = 39 GPa = 39000 MPa
φ = TL/(GJ) = 100000·1000/(39000·52048) = 0.04926 rad
φ = 0.04926·180/π = 2.823°

Answer

φ = 2.823°
4.Torsion of shafts

Answer: τmax = 3.295 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·750/60 = 78.54 rad/s
T = P/ω = 30000/78.54 = 382 N·m

Polar second moment of area

J = π(100⁴ − 80⁴)/32 = 5.796 × 10⁶ mm⁴

Maximum shear stress

τmax = T·(do/2)/J = 381972·50/5.796 × 10⁶ = 3.295 MPa

Answer

τmax = 3.295 MPa
5.Mohr’s circle & stress transformation

Answer: τmax = 22.36 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.

στxyCτmax​

Radius of Mohr’s circle

τmax = R = √(((−30 − −70)/2)² + −10²) = 22.36 MPa

Answer

τmax = 22.36 MPa
6.Mohr’s circle & stress transformation

Answer: σvM = 141.8 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. 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 = 90 MPa, R = 63.25 MPa
σ1 = 153.2 MPa, σ2 = 26.75 MPa, σ3 = 0

Von Mises

σvM = √(153.2² − 153.2·26.75 + 26.75²) = 141.8 MPa

Answer

σvM = 141.8 MPa
7.Axial stress & strain

Answer: ε = 544.1 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 = 20·1000/314.2 = 63.66 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 63.66/117000 = 0.0005441 = 544.1 µε

Answer

ε = 544.1 µε
8.Axial stress & strain

Answer: δ = 10.68 mm

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·40²/4 = 1257 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 95.49 MPa
σ2 = 596.8 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 95.49·1000/200000 = 0.4775 mm
δ2 = σ2L2/E2 = 596.8·2000/117000 = 10.2 mm
δ = δ1 + δ2 = 10.68 mm

Answer

δ = 10.68 mm
9.Column buckling

Answer: Pcr = 27.56 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 = 2·1500 = 3000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·125664/3000² = 27561 N = 27.56 kN

Answer

Pcr = 27.56 kN
10.Column buckling

Answer: σcr = 24.87 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·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 0.5·2500 = 1250 mm
KL/r = 1250/7.5 = 166.7

Critical stress

σcr = π²E/(KL/r)² = π²·70000/166.7² = 24.87 MPa

Answer

σcr = 24.87 MPa
11.Mechanical vibrations

Answer: ζ = 0.08103

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 = (1/2)·ln(10/3.6) = 0.5108

Damping ratio

ζ = δ/√(4π² + δ²) = 0.5108/√(39.48 + 0.2609) = 0.08103

Answer

ζ = 0.08103
12.Mechanical vibrations

Answer: keq = 4.706 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 = 1/80 + 1/5 → keq = 80·5/(80 + 5) = 4.706 kN/m

Answer

keq = 4.706 kN/m

ExamProblems · examproblems.com/courses/machine-design