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 = 3 mm

A 22 kW motor running at 960 rpm drives a pinion with 30 teeth and a module of 3 mm. Find the tangential force Ft on the pinion teeth.

P = 22 kWn = 960 rpmN1 = 30m = 3 mm
Ft = N
2.Gears & gear trainsHard
d1​ = 80 mmd2​ = 315 mm720 rpmC = 1000 mmT1​ = 600 NT2​P = ?

An open belt connects an 80 mm driving pulley at 720 rpm to a 315 mm pulley; the centres are 1000 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 600 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 80 mmn1 = 720 rpmd2 = 315 mmC = 1000 mmμ = 0.3T1 = 600 N
P = kW
3.Torsion of shaftsMedium
T = 500 N·md = 30 mmL = 2.5 mφ = ?

A solid steel 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 = 500 N·md = 30 mmL = 2.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
Tdo​ = 100 mm, di​ = 50 mm

A hollow shaft transmits 100 kW at 300 rpm. Find the maximum shear stress τmax in the shaft.

P = 100 kWN = 300 rpmdo = 100 mmdi = 50 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = 150σy​ = 110τxy​ = 20MPa

A point in a thin plate is in plane stress: σₓ = 150 MPa, σᵧ = 110 MPa and τₓᵧ = 20 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).

σx = 150 MPaσy = 110 MPaτxy = 20 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 120σy​ = 50τxy​ = 20MPa

A point in a loaded part has σₓ = 120 MPa, σᵧ = 50 MPa and τₓᵧ = 20 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 20° clockwise from the x-axis.

σx = 120 MPaσy = 50 MPaτxy = 20 MPaθ = -20°
σθ = MPa
7.Axial stress & strainMedium
P = ?12400 mm400 mmØ40Ø50δ = 0.4 mm

A bar is made of a steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 50 mm, length 400 mm, E = 110 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.

segment 1: steel, Ø40 mm, 400 mm, E = 200 GPasegment 2: bronze, Ø50 mm, 400 mm, E = 110 GPaδ = 0.4 mm
P = kN
8.Axial stress & strainHard
Ø 40 mmCopper1 mØ 20 mmSteel1 mP = 115 kN

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

P = 115 kNsegment 1: Ø40 mm, 1 m, copper (E = 117 GPa)segment 2: Ø20 mm, 1 m, steel (E = 200 GPa)
δ = mm
9.Column bucklingMedium
PL = 2 m25 × 50 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 mE = 70 GPa (aluminium)b × h = 25 × 50 mm
Pcr = kN
10.Column bucklingHard
P = 5 kNL = 3 md = ?cross-section

A solid round column of length L made of steel is pinned at both ends. It must carry 5 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 5 kNn = 2L = 3 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 4 kN/mc = 80 N·s/mm = 5 kg

A 5 kg mass is supported by a spring with a stiffness of 4 kN/m and a viscous damper with a damping coefficient of 80 N·s/m. Find the damping ratio ζ.

m = 5 kgk = 4 kN/mc = 80 N·s/m
ζ =
12.Mechanical vibrationsHard
k1​ = 100k2​ = 15m = 5 kg

A 5 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 = 15 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 5 kgk1 = 100 kN/mk2 = 15 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: Ft = 4863 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·960/60 = 100.5 rad/s
T = P/ω = 22000/100.5 = 218.8 N·m

Pitch diameter

d = m·N = 3·30 = 90 mm

Tangential force

Ft = 2T/d = 2·218.8/0.09 = 4863 N

Answer

Ft = 4863 N
2.Gears & gear trains

Answer: P = 1.053 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 = 2.906 rad

Slack-side tension

eμθ = e0.8718 = 2.391, T₂ = 600/2.391 = 250.9 N

Power

P = (600 − 250.9)·3.016 = 1.053 kW

Answer

P = 1.053 kW
3.Torsion of shafts

Answer: φ = 11.26 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 500·1000·2500/(80000·79522) = 0.1965 rad
φ = 0.1965·180/π = 11.26°

Answer

φ = 11.26°
4.Torsion of shafts

Answer: τmax = 17.29 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·300/60 = 31.42 rad/s
T = P/ω = 100000/31.42 = 3183 N·m

Polar second moment of area

J = π(100⁴ − 50⁴)/32 = 9.204 × 10⁶ mm⁴

Maximum shear stress

τmax = T·(do/2)/J = 3.183 × 10⁶·50/9.204 × 10⁶ = 17.29 MPa

Answer

τmax = 17.29 MPa
5.Mohr’s circle & stress transformation

Answer: τabs = 79.14 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.

σxyCσ1​σ2​

In-plane principal stresses

C = 130 MPa, R = √(20² + 20²) = 28.28 MPa
σ1 = 158.3 MPa, σ2 = 101.7 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(56.57, 158.3, 101.7)/2 = 79.14 MPa

Answer

τabs = 79.14 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: σθ = 98.96 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.

Transformation equation

σθ = 85 + 35·0.766 + 20·−0.6428 = 98.96 MPa

Answer

σθ = 98.96 MPa
7.Axial stress & strain

Answer: P = 116.2 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

A₁ = 1257 mm², A₂ = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·1257) = 1.592 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 400/(110000·1963) = 1.852 × 10⁻⁶ mm/N

Load

P = 0.4/3.444 × 10⁻⁶ = 116160 N = 116.2 kN

Answer

P = 116.2 kN
8.Axial stress & strain

Answer: δ = 2.612 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·20²/4 = 314.2 mm²
σ1 = 91.51 MPa
σ2 = 366.1 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 91.51·1000/117000 = 0.7822 mm
δ2 = σ2L2/E2 = 366.1·1000/200000 = 1.83 mm
δ = δ1 + δ2 = 2.612 mm

Answer

δ = 2.612 mm
9.Column buckling

Answer: Pcr = 2.811 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 rectangle buckles about its weak axis, using the smaller dimension cubed.

Imin = h·b³/12 = 50·25³/12 = 65104 mm⁴

Effective length

KL = 2·2000 = 4000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·65104/4000² = 2811 N = 2.811 kN

Answer

Pcr = 2.811 kN
10.Column buckling

Answer: d = 31.04 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 = 2·5000 = 10000 N
KL = 1·3000 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·10000·3000²/(π³·200000))^(1/4) = 31.04 mm

Answer

d = 31.04 mm

Check: the slenderness KL/r is 386.5, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.2828

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·√(4000·5) = 282.8 N·s/m

Damping ratio

ζ = c/cc = 80/282.8 = 0.2828

Answer

ζ = 0.2828

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

12.Mechanical vibrations

Answer: fn = 8.129 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 = 1/100 + 1/15 → keq = 100·15/(100 + 15) = 13.04 kN/m

Natural frequency

ωn = √(13043/5) = 51.08 rad/s
fn = 51.08/(2π) = 8.129 Hz

Answer

fn = 8.129 Hz

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