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​ = 25, m = 5 mm

A 7.5 kW motor running at 2900 rpm drives a pinion with 25 teeth and a module of 5 mm. Find the tangential force Ft on the pinion teeth.

P = 7.5 kWn = 2900 rpmN1 = 25m = 5 mm
Ft = N
2.Gears & gear trainsHard
d1​ = 120 mmd2​ = 400 mm2880 rpmC = 1000 mmθ = ?

An open belt connects a 120 mm pulley and a 400 mm pulley whose centres are 1000 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 120 mmd2 = 400 mmC = 1000 mm
θ = °
3.Torsion of shaftsMedium
T = 750 N·md = 25 mmL = 1.5 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 = 750 N·md = 25 mmL = 1.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
Tdo​ = 80 mm, di​ = 56 mm

A hollow shaft transmits 100 kW at 1450 rpm. Find the maximum shear stress τmax in the shaft.

P = 100 kWN = 1450 rpmdo = 80 mmdi = 56 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = −40σy​ = 90τxy​ = −30MPa

A point in a thin plate is in plane stress: σₓ = −40 MPa, σᵧ = 90 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 = −40 MPaσy = 90 MPaτxy = −30 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
σx​ = −80σy​ = 10τxy​ = 10MPa

A point in a thin plate is in plane stress: σₓ = −80 MPa, σᵧ = 10 MPa and τₓᵧ = 10 MPa (σz = 0). The material has a yield strength of 355 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σx = −80 MPaσy = 10 MPaτxy = 10 MPaσY = 355 MPa
n =
7.Axial stress & strainMedium
Ø 20 mmSteel1 mP = 50 kN

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

P = 50 kNd = 20 mmL = 1 mE = 200 GPa
δ = mm
8.Axial stress & strainHard
80 kN12400 mm400 mmØ25Ø40σ2​ = ?

A steel segment (diameter 25 mm, length 400 mm, E = 200 GPa) and a segment made of copper (diameter 40 mm, length 400 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 80 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø25 mm, 400 mm, E = 200 GPasegment 2: copper, Ø40 mm, 400 mm, E = 120 GPaP = 80 kN
σ2 = MPa
9.Column bucklingMedium
PL = 3 md = 25 mmcross-section

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

L = 3 mE = 70 GPa (aluminium)d = 25 mm
Pcr = kN
10.Column bucklingHard
P = 25 kNL = 2.5 md = ?cross-section

A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 25 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 25 kNn = 3L = 2.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 12 kN/mc = 120 N·s/mm = 5 kg

A 5 kg mass is supported by a spring with a stiffness of 12 kN/m and a viscous damper with a damping coefficient of 120 N·s/m. Find the damping ratio ζ.

m = 5 kgk = 12 kN/mc = 120 N·s/m
ζ =
12.Mechanical vibrationsHard
k1​ = 80k2​ = 20m = 20 kg

A 20 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 = 20 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 20 kgk1 = 80 kN/mk2 = 20 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: Ft = 395.1 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·2900/60 = 303.7 rad/s
T = P/ω = 7500/303.7 = 24.7 N·m

Pitch diameter

d = m·N = 5·25 = 125 mm

Tangential force

Ft = 2T/d = 2·24.7/0.125 = 395.1 N

Answer

Ft = 395.1 N
2.Gears & gear trains

Answer: θ = 163.9 °

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(0.14) = 8.048°

Angle of wrap

θ = 180 − 2·8.048 = 163.9°

Answer

θ = 163.9°
3.Torsion of shafts

Answer: φ = 21.01 °

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·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 750·1000·1500/(80000·38350) = 0.3667 rad
φ = 0.3667·180/π = 21.01°

Answer

φ = 21.01°
4.Torsion of shafts

Answer: τmax = 8.621 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·1450/60 = 151.8 rad/s
T = P/ω = 100000/151.8 = 658.6 N·m

Polar second moment of area

J = π(80⁴ − 56⁴)/32 = 3.056 × 10⁶ mm⁴

Maximum shear stress

τmax = T·(do/2)/J = 658572·40/3.056 × 10⁶ = 8.621 MPa

Answer

τmax = 8.621 MPa
5.Mohr’s circle & stress transformation

Answer: τabs = 71.59 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 = 25 MPa, R = √((−65)² + (−30)²) = 71.59 MPa
σ1 = 96.59 MPa, σ2 = −46.59 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(143.2, 96.59, 46.59)/2 = 71.59 MPa

Answer

τabs = 71.59 MPa

Here σ₁ and σ₂ have opposite signs, so the in-plane maximum R is also the absolute maximum.

6.Mohr’s circle & stress transformation

Answer: n = 4.072

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 = −35 MPa, R = 46.1 MPa
σ1 = 11.1 MPa, σ2 = −81.1 MPa, σ3 = 0

Von Mises

σvM = √(11.1² − 11.1·(−81.1) + (−81.1)²) = 87.18 MPa
n = 355/87.18 = 4.072

Answer

n = 4.072

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

7.Axial stress & strain

Answer: δ = 0.7958 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 = 50·1000/314.2 = 159.2 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 159.2/200000 = 0.0007958 = 795.8 µε

Elongation

δ = ε·L = 0.0007958·1000 mm = 0.7958 mm

Answer

δ = 0.7958 mm
8.Axial stress & strain

Answer: σ2 = 38.56 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 = 200000·490.9/400 = 245437 N/mm
k₂ = 120000·1257/400 = 376991 N/mm

Share of segment 2

F2 = 80000·376991/622428 = 48454 N

Stress

σ2 = 48454/1257 = 38.56 MPa (compression)

Answer

σ2 = 38.56 MPa
9.Column buckling

Answer: Pcr = 5.888 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·25⁴/64 = 19175 mm⁴

Effective length

KL = 0.5·3000 = 1500 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·19175/1500² = 5888 N = 5.888 kN

Answer

Pcr = 5.888 kN
10.Column buckling

Answer: d = 66.32 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 = 3·25000 = 75000 N
KL = 2·2500 = 5000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·75000·5000²/(π³·200000))^(1/4) = 66.32 mm

Answer

d = 66.32 mm

Check: the slenderness KL/r is 301.5, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.2449

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·√(12000·5) = 489.9 N·s/m

Damping ratio

ζ = c/cc = 120/489.9 = 0.2449

Answer

ζ = 0.2449

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

12.Mechanical vibrations

Answer: fn = 4.502 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 = 1/80 + 1/20 → keq = 80·20/(80 + 20) = 16 kN/m

Natural frequency

ωn = √(16000/20) = 28.28 rad/s
fn = 28.28/(2π) = 4.502 Hz

Answer

fn = 4.502 Hz

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