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​ = 24N2​ = 40

A 15 kW motor running at 960 rpm drives a pinion with 24 teeth, which meshes with a gear with 40 teeth. The gear pair has an efficiency of 95%. Find the torque T1 on the motor shaft.

P = 15 kWn1 = 960 rpmN1 = 24N2 = 40η = 95%
T1 = N·m
2.Gears & gear trainsHard
d1​ = 100 mmd2​ = 200 mm1440 rpmC = 1200 mmθ = ?

An open belt connects a 100 mm pulley and a 200 mm pulley whose centres are 1200 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 100 mmd2 = 200 mmC = 1200 mm
θ = °
3.Torsion of shaftsMedium
T = 200 N·md = 50 mmL = 1.5 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 = 200 N·md = 50 mmL = 1.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
d = 30 mmTB​ = 1000 N·mTC​ = 400 N·mABC1.5 m1.2 m

A solid brass shaft of diameter 30 mm is fixed at A. A torque of 1000 N·m acts at B, 1.5 m from A, and a torque of 400 N·m in the same direction acts at the free end C, a further 1.2 m along. Find the maximum shear stress τmax in the shaft.

d = 30 mmTB = 1000 N·mTC = 400 N·mAB = 1.5 mBC = 1.2 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = −30σy​ = −80τxy​ = −10MPa

A point in a thin plate is in plane stress: σₓ = −30 MPa, σᵧ = −80 MPa and τₓᵧ = −10 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).

σx = −30 MPaσy = −80 MPaτxy = −10 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 110σy​ = 120τxy​ = 60MPa

A point in a loaded part has σₓ = 110 MPa, σᵧ = 120 MPa and τₓᵧ = 60 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 40° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σx = 110 MPaσy = 120 MPaτxy = 60 MPaθ = 40°
τθ = MPa
7.Axial stress & strainMedium
Ø 25 mmBrass3 mP = 40 kN

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

P = 40 kNd = 25 mmL = 3 mE = 100 GPa
δ = mm
8.Axial stress & strainHard
30 kN12300 mm200 mmØ30Ø30σ2​ = ?

A steel segment (diameter 30 mm, length 300 mm, E = 200 GPa) and a segment made of copper (diameter 30 mm, length 200 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 30 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø30 mm, 300 mm, E = 200 GPasegment 2: copper, Ø30 mm, 200 mm, E = 120 GPaP = 30 kN
σ2 = MPa
9.Column bucklingMedium
PL = 2 md = 30 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 = 2 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
10.Column bucklingHard
PL = 3 md = 30 mmcross-section

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

L = 3 mE = 200 GPa (steel)d = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 15 kN/mζ = 0.3m = 12 kg

A 12 kg mass on a spring with a stiffness of 15 kN/m has a viscous damper giving a damping ratio of 0.3. Find the damped natural frequency fd of its free vibration in Hz.

m = 12 kgk = 15 kN/mζ = 0.3
fd = Hz
12.Mechanical vibrationsHard
k1​ = 40k2​ = 80m = 50 kg

A 50 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 40 kN/m and the lower one k₂ = 80 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 50 kgk1 = 40 kN/mk2 = 80 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: T1 = 149.2 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·960/60 = 100.5 rad/s
T1 = P/ω₁ = 15000/100.5 = 149.2 N·m

Answer

T1 = 149.2 N·m
2.Gears & gear trains

Answer: θ = 175.2 °

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(0.04167) = 2.388°

Angle of wrap

θ = 180 − 2·2.388 = 175.2°

Answer

θ = 175.2°
3.Torsion of shafts

Answer: φ = 0.7183 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 200·1000·1500/(39000·613592) = 0.01254 rad
φ = 0.01254·180/π = 0.7183°

Answer

φ = 0.7183°
4.Torsion of shafts

Answer: τmax = 264.1 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.

Internal torques

Sense of TC taken as positive.

TBC = TC = 400 N·m
TAB = TC + TB = 400 + 1000 = 1400 N·m

Maximum shear stress

The segment with the largest internal torque has the largest stress (same diameter).

|T|max = 1400 N·m
τmax = 16·1.4 × 10⁶/(π·30³) = 264.1 MPa

Answer

τmax = 264.1 MPa
5.Mohr’s circle & stress transformation

Answer: τabs = 40.96 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 = −55 MPa, R = √(25² + (−10)²) = 26.93 MPa
σ1 = −28.07 MPa, σ2 = −81.93 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(53.85, 28.07, 81.93)/2 = 40.96 MPa

Answer

τabs = 40.96 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: τθ = 15.34 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.

Transformation equation

τθ = −−5·0.9848 + 60·0.1736 = 15.34 MPa

Answer

τθ = 15.34 MPa
7.Axial stress & strain

Answer: δ = 2.445 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·25²/4 = 490.9 mm²
σ = P/A = 40·1000/490.9 = 81.49 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 81.49/100000 = 0.0008149 = 814.9 µε

Elongation

δ = ε·L = 0.0008149·3000 mm = 2.445 mm

Answer

δ = 2.445 mm
8.Axial stress & strain

Answer: σ2 = 20.1 MPa

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.

Stiffnesses

k₁ = 200000·706.9/300 = 471239 N/mm
k₂ = 120000·706.9/200 = 424115 N/mm

Share of segment 2

F2 = 30000·424115/895354 = 14211 N

Stress

σ2 = 14211/706.9 = 20.1 MPa (compression)

Answer

σ2 = 20.1 MPa
9.Column buckling

Answer: Pcr = 4.905 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 2·2000 = 4000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·39761/4000² = 4905 N = 4.905 kN

Answer

Pcr = 4.905 kN
10.Column buckling

Answer: σcr = 12.34 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 = 1·3000 = 3000 mm
KL/r = 3000/7.5 = 400

Critical stress

σcr = π²E/(KL/r)² = π²·200000/400² = 12.34 MPa

Answer

σcr = 12.34 MPa
11.Mechanical vibrations

Answer: fd = 5.368 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 = √(15000/12) = 35.36 rad/s
fn = 35.36/(2π) = 5.627 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 5.627·√(1 − 0.3²) = 5.368 Hz

Answer

fd = 5.368 Hz
12.Mechanical vibrations

Answer: fn = 3.676 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.

1/keq = 1/40 + 1/80 → keq = 40·80/(40 + 80) = 26.67 kN/m

Natural frequency

ωn = √(26667/50) = 23.09 rad/s
fn = 23.09/(2π) = 3.676 Hz

Answer

fn = 3.676 Hz

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