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​N2​ = ?

Two parallel shafts are 210 mm apart. They must be connected by a pair of spur gears with a module of 4 mm and a speed reduction of 2.5 : 1. How many teeth N2 must the gear have?

a = 210 mmm = 4 mmi = 2.5
N2 =
2.Gears & gear trainsHard
d1​ = 150 mmd2​ = 500 mm720 rpmC = 1200 mmT1​ = 500 NT2​P = ?

An open belt connects a 150 mm driving pulley at 720 rpm to a 500 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 150 mmn1 = 720 rpmd2 = 500 mmC = 1200 mmμ = 0.3T1 = 500 N
P = kW
3.Torsion of shaftsMedium
d1​ = 60 mmd2​ = 40 mmT = 2000 N·mABC2.5 m0.5 m

A stepped solid brass shaft is fixed at A. Segment AB has diameter 60 mm and length 2.5 m; segment BC has diameter 40 mm and length 0.5 m. A torque of 2000 N·m acts at the free end C. Find the angle of twist φC of end C, in degrees.

T = 2000 N·md1 = 60 mm, L1 = 2.5 md2 = 40 mm, L2 = 0.5 mG = 39 GPa
φC = °
4.Torsion of shaftsHard
Tdo​ = 40 mm, di​ = 30 mm

A hollow shaft transmits 30 kW at 1500 rpm. Find the maximum shear stress τmax in the shaft.

P = 30 kWN = 1500 rpmdo = 40 mmdi = 30 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = −80σy​ = −10τxy​ = −10MPa

A point has σₓ = −80 MPa, σᵧ = −10 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 = −80 MPaσy = −10 MPaτxy = −10 MPa
θs = °
6.Mohr’s circle & stress transformationHard
σx​ = −40τxy​ = −50MPa

A point in a thin plate is in plane stress: σₓ = −40 MPa, σᵧ = 0 MPa and τₓᵧ = −50 MPa (σz = 0). Find the von Mises equivalent stress σvM.

σx = −40 MPaσy = 0 MPaτxy = −50 MPa
σvM = MPa
7.Axial stress & strainMedium
P = ?12500 mm500 mmØ40Ø30δ = 0.5 mm

A bar is made of a steel segment (diameter 40 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 30 mm, length 500 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.

segment 1: steel, Ø40 mm, 500 mm, E = 200 GPasegment 2: copper, Ø30 mm, 500 mm, E = 120 GPaδ = 0.5 mm
P = kN
8.Axial stress & strainHard
P = ?Ø2570 GPaØ120/70300 mm

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

rod: Ø25 mm, E = 200 GPa, σallow = 120 MPatube: Ø120/Ø70 mm, E = 70 GPa, σallow = 70 MPa
Pmax = kN
9.Column bucklingMedium
PL = 3 md = 80 mmcross-section

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

L = 3 mE = 70 GPa (aluminium)d = 80 mm
Pcr = kN
10.Column bucklingHard
P = 55 kNL = 5 md = ?cross-section

A solid round column of length L made of steel is pinned at both ends. It must carry 55 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 55 kNn = 3L = 5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 40 kN/mζ = 0.45m = 10 kg

A 10 kg mass on a spring with a stiffness of 40 kN/m has a viscous damper giving a damping ratio of 0.45. Find the damped natural frequency fd of its free vibration in Hz.

m = 10 kgk = 40 kN/mζ = 0.45
fd = Hz
12.Mechanical vibrationsHard
k1​ = 80k2​ = 60m = 25 kg

A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 80 kN/m and k₂ = 60 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.

m = 25 kgk1 = 80 kN/mk2 = 60 kN/m
keq = kN/m

3Solutions

1.Gears & gear trains

Answer: N2 = 75

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.

Total number of teeth

N1 + N2 = 2a/m = 2·210/4 = 105

Split by the ratio

N1 = 105/(1 + 2.5) = 30
N2 = 2.5·30 = 75

Answer

N2 = 75
2.Gears & gear trains

Answer: P = 1.625 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 = 2.849 rad

Slack-side tension

eμθ = e0.8547 = 2.351, T₂ = 500/2.351 = 212.7 N

Power

P = (500 − 212.7)·5.655 = 1.625 kW

Answer

P = 1.625 kW
3.Torsion of shafts

Answer: φC = 11.62 °

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 moments

J1 = π·60⁴/32 = 1.272 × 10⁶ mm⁴
J2 = π·40⁴/32 = 251327 mm⁴

Twist of each segment

T = 2000 N·m = 2 × 10⁶ N·mm
G = 39 GPa = 39000 MPa
φAB = 2 × 10⁶·2500/(39000·1.272 × 10⁶) = 0.1008 rad
φBC = 2 × 10⁶·500/(39000·251327) = 0.102 rad

Total twist

φC = 0.1008 + 0.102 = 0.2028 rad = 11.62°

Answer

φC = 11.62°
4.Torsion of shafts

Answer: τmax = 22.23 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·1500/60 = 157.1 rad/s
T = P/ω = 30000/157.1 = 191 N·m

Polar second moment of area

J = π(40⁴ − 30⁴)/32 = 171806 mm⁴

Maximum shear stress

τmax = T·(do/2)/J = 190986·20/171806 = 22.23 MPa

Answer

τmax = 22.23 MPa
5.Mohr’s circle & stress transformation

Answer: θs = 52.97 °

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​

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 = −10/36.4 = −0.2747, sin 2θs = −(−35)/36.4 = 0.9615
2θs = atan2(35, −10) = 105.9°

Angle on the element

θs = 52.97°
check: θs − θp = 52.97 − (−82.03) = 135°

Answer

θs = 52.97°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σvM = 95.39 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 = −20 MPa, R = 53.85 MPa
σ1 = 33.85 MPa, σ2 = −73.85 MPa, σ3 = 0

Von Mises

σvM = √(33.85² − 33.85·(−73.85) + (−73.85)²) = 95.39 MPa

Answer

σvM = 95.39 MPa
7.Axial stress & strain

Answer: P = 63.42 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 = 706.9 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·1257) = 1.989 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(120000·706.9) = 5.895 × 10⁻⁶ mm/N

Load

P = 0.5/7.884 × 10⁻⁶ = 63419 N = 63.42 kN

Answer

P = 63.42 kN
8.Axial stress & strain

Answer: Pmax = 372.3 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(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 70·10³·7461 = 6.205 × 10⁸ N

Governing strain

ε = min(0.0006, 0.001) = 0.0006 (steel)

Largest load

Pmax = 0.0006·6.205 × 10⁸ = 372279 N = 372.3 kN

Answer

Pmax = 372.3 kN
9.Column buckling

Answer: Pcr = 315 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 0.7·3000 = 2100 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/2100² = 314984 N = 315 kN

Answer

Pcr = 315 kN
10.Column buckling

Answer: d = 80.78 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 = 3·55000 = 165000 N
KL = 1·5000 = 5000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·165000·5000²/(π³·200000))^(1/4) = 80.78 mm

Answer

d = 80.78 mm

Check: the slenderness KL/r is 247.6, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: fd = 8.989 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 = √(40000/10) = 63.25 rad/s
fn = 63.25/(2π) = 10.07 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 10.07·√(1 − 0.45²) = 8.989 Hz

Answer

fd = 8.989 Hz
12.Mechanical vibrations

Answer: keq = 140 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.

keq = 80 + 60 = 140 kN/m

Answer

keq = 140 kN/m

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