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
d1​ = 125 mmd2​ = 500 mm1500 rpmT1​ = 1000 NT2​ = 300 NP = ?

A belt runs on a 125 mm driving pulley at 1500 rpm. The tension is 1000 N on the tight side and 300 N on the slack side. Find the power P transmitted, in kW.

d1 = 125 mmn1 = 1500 rpmT1 = 1000 NT2 = 300 N
P = kW
2.Gears & gear trainsHard
d1​ = 200 mmd2​ = 250 mm720 rpmC = 600 mmT1​ = 1500 NT2​P = ?

An open belt connects a 200 mm driving pulley at 720 rpm to a 250 mm pulley; the centres are 600 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 1500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 200 mmn1 = 720 rpmd2 = 250 mmC = 600 mmμ = 0.3T1 = 1500 N
P = kW
3.Torsion of shaftsMedium
T = 750 N·md = 50 mmL = 1 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 = 750 N·md = 50 mmL = 1 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 15 kW at 1000 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 15 kWN = 1000 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = −20σy​ = −80τxy​ = −30MPa

A point in a thin plate is in plane stress: σₓ = −20 MPa, σᵧ = −80 MPa and τₓᵧ = −30 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).

σx = −20 MPaσy = −80 MPaτxy = −30 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
σx​ = −40σy​ = 100τxy​ = −50MPa

A point in a loaded part has σₓ = −40 MPa, σᵧ = 100 MPa and τₓᵧ = −50 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° clockwise from the x-axis.

σx = −40 MPaσy = 100 MPaτxy = −50 MPaθ = -30°
σθ = MPa
7.Axial stress & strainMedium
Ø 20 mmCopper3 mP = 85 kN

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

P = 85 kNd = 20 mmL = 3 mE = 117 GPa
δ = mm
8.Axial stress & strainHard
P = ?Ø25120 GPaØ100/70300 mm

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of copper (outer diameter 100 mm, inner diameter 70 mm, E = 120 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø25 mm, E = 200 GPa, σallow = 120 MPatube: Ø100/Ø70 mm, E = 120 GPa, σallow = 50 MPa
Pmax = kN
9.Column bucklingMedium
PL = 2 m40 × 100 mmcross-section

A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.

L = 2 mE = 70 GPa (aluminium)b × h = 40 × 100 mm
Pcr = kN
10.Column bucklingHard
PL = 4 md = 60 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 = 4 mE = 200 GPa (steel)d = 60 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 150 kN/mζ = 0.4m = 200 kg

A 200 kg mass on a spring with a stiffness of 150 kN/m has a viscous damper giving a damping ratio of 0.4. Find the damped natural frequency fd of its free vibration in Hz.

m = 200 kgk = 150 kN/mζ = 0.4
fd = Hz
12.Mechanical vibrationsHard
k1​ = 5k2​ = 30m = 50 kg

A 50 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 5 kN/m and the lower one k₂ = 30 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 50 kgk1 = 5 kN/mk2 = 30 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: P = 6.872 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.

Belt speed

v = 9.817 m/s

Power

P = (1000 − 300)·9.817 = 6872 W = 6.872 kW

Answer

P = 6.872 kW
2.Gears & gear trains

Answer: P = 6.791 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 = 3.058 rad

Slack-side tension

eμθ = e0.9175 = 2.503, T₂ = 1500/2.503 = 599.3 N

Power

P = (1500 − 599.3)·7.54 = 6.791 kW

Answer

P = 6.791 kW
3.Torsion of shafts

Answer: φ = 0.8754 °

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) = 750·1000·1000/(80000·613592) = 0.01528 rad
φ = 0.01528·180/π = 0.8754°

Answer

φ = 0.8754°
4.Torsion of shafts

Answer: d = 21.35 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·1000/60 = 104.7 rad/s
T = P/ω = 15000/104.7 = 143.2 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·143239/(π·75)) = 21.35 mm

Answer

d = 21.35 mm

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

5.Mohr’s circle & stress transformation

Answer: τabs = 46.21 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 = −50 MPa, R = √(30² + (−30)²) = 42.43 MPa
σ1 = −7.574 MPa, σ2 = −92.43 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(84.85, 7.574, 92.43)/2 = 46.21 MPa

Answer

τabs = 46.21 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 = 38.3 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 = 30 + −70·0.5 + −50·−0.866 = 38.3 MPa

Answer

σθ = 38.3 MPa
7.Axial stress & strain

Answer: δ = 6.938 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·20²/4 = 314.2 mm²
σ = P/A = 85·1000/314.2 = 270.6 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 270.6/117000 = 0.002313 = 2313 µε

Elongation

δ = ε·L = 0.002313·3000 mm = 6.938 mm

Answer

δ = 6.938 mm
8.Axial stress & strain

Answer: Pmax = 241.2 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

Ar = π·25²/4 = 490.9 mm²
At = π(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 120·10³·4006 = 5.788 × 10⁸ N

Governing strain

ε = min(0.0006, 0.0004167) = 0.0004167 (copper)

Largest load

Pmax = 0.0004167·5.788 × 10⁸ = 241183 N = 241.2 kN

Answer

Pmax = 241.2 kN
9.Column buckling

Answer: Pcr = 92.12 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 = 100·40³/12 = 533333 mm⁴

Effective length

KL = 1·2000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·533333/2000² = 92116 N = 92.12 kN

Answer

Pcr = 92.12 kN
10.Column buckling

Answer: σcr = 27.76 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·60⁴/64 = 636173 mm⁴

Radius of gyration and slenderness

A = 2827 mm²
r = √(I/A) = √(636173/2827) = 15 mm
KL = 1·4000 = 4000 mm
KL/r = 4000/15 = 266.7

Critical stress

σcr = π²E/(KL/r)² = π²·200000/266.7² = 27.76 MPa

Answer

σcr = 27.76 MPa
11.Mechanical vibrations

Answer: fd = 3.995 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 = √(150000/200) = 27.39 rad/s
fn = 27.39/(2π) = 4.359 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 4.359·√(1 − 0.4²) = 3.995 Hz

Answer

fd = 3.995 Hz
12.Mechanical vibrations

Answer: fn = 1.473 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 = 1/5 + 1/30 → keq = 5·30/(5 + 30) = 4.286 kN/m

Natural frequency

ωn = √(4286/50) = 9.258 rad/s
fn = 9.258/(2π) = 1.473 Hz

Answer

fn = 1.473 Hz

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