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 60 mm apart. They must be connected by a pair of spur gears with a module of 2 mm and a speed reduction of 3 : 1. How many teeth N2 must the gear have?

a = 60 mmm = 2 mmi = 3
N2 =
2.Gears & gear trainsHard
d1​ = 100 mmd2​ = 315 mm1450 rpmC = 800 mmT1​ = 500 NT2​P = ?

An open belt connects a 100 mm driving pulley at 1450 rpm to a 315 mm pulley; the centres are 800 mm apart. The coefficient of friction is 0.35 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 = 100 mmn1 = 1450 rpmd2 = 315 mmC = 800 mmμ = 0.35T1 = 500 N
P = kW
3.Torsion of shaftsMedium
d = ?T = 400 N·mL = 1 m

A solid aluminium shaft 1 m long carries a torque of 400 N·m. The angle of twist between its ends must not exceed 1°. Find the smallest diameter d that meets this limit.

T = 400 N·mL = 1 mG = 26 GPaφallow = 1°
d = mm
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​ = 30σy​ = 30τxy​ = −10MPa

A point in a thin plate is in plane stress: σₓ = 30 MPa, σᵧ = 30 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 = 30 MPaτxy = −10 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
σx​τxy​surface of the shaft

A solid shaft of diameter 40 mm carries an axial tensile force of 160 kN together with a torque of 2000 N·m. At the surface of the shaft, find the largest principal stress σ1.

d = 40 mmF = 160 kNT = 2000 N·m
σ1 = MPa
7.Axial stress & strainMedium
Ø 16 mmSteel2.5 mP = 65 kN

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

P = 65 kNd = 16 mmL = 2.5 mE = 200 GPa
δ = mm
8.Axial stress & strainHard
30 kN12600 mm400 mmØ40Ø30σ2​ = ?

A steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) and a segment made of bronze (diameter 30 mm, length 400 mm, E = 110 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, Ø40 mm, 600 mm, E = 200 GPasegment 2: bronze, Ø30 mm, 400 mm, E = 110 GPaP = 30 kN
σ2 = MPa
9.Column bucklingMedium
PL = 5 m25 × 50 mmcross-section

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

L = 5 mE = 200 GPa (steel)b × h = 25 × 50 mm
Pcr = kN
10.Column bucklingHard
P = 110 kNL = 2.5 md = ?cross-section

A solid round column of length L made of steel is pinned at both ends. It must carry 110 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 110 kNn = 2.5L = 2.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 4 kN/mζ = 0.4m = 40 kg

A 40 kg mass on a spring with a stiffness of 4 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 = 40 kgk = 4 kN/mζ = 0.4
fd = Hz
12.Mechanical vibrationsHard
k1​ = 12k2​ = 12m = 50 kg

A 50 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 12 kN/m and k₂ = 12 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 50 kgk1 = 12 kN/mk2 = 12 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: N2 = 45

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 = 2·60/2 = 60

Split by the ratio

N1 = 60/(1 + 3) = 15
N2 = 3·15 = 45

Answer

N2 = 45
2.Gears & gear trains

Answer: P = 2.407 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 = 2.872 rad

Slack-side tension

eμθ = e1.005 = 2.732, T₂ = 500/2.732 = 183 N

Power

P = (500 − 183)·7.592 = 2.407 kW

Answer

P = 2.407 kW
3.Torsion of shafts

Answer: d = 54.74 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.

Required polar moment

T = 400 N·m = 400000 N·mm
L = 1 m = 1000 mm
G = 26 GPa = 26000 MPa
φ = 1·π/180 = 0.01745 rad
J = TL/(Gφ) = 400000·1000/(26000·0.01745) = 881474 mm⁴

Diameter

d = (32J/π)1/4 = 54.74 mm

Answer

d = 54.74 mm

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

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 = 20 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 = 30 MPa, R = √(0² + (−10)²) = 10 MPa
σ1 = 40 MPa, σ2 = 20 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(20, 40, 20)/2 = 20 MPa

Answer

τabs = 20 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: σ1 = 235.1 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.

Stresses at the surface

σ = F/A = 160000/(π·40²/4) = 160000/1257 = 127.3 MPa
τ = 16T/(πd³) = 16·2000000/(π·40³) = 159.2 MPa
στxyCσ1​σ2​

Mohr’s circle

C = σ/2 = 63.66 MPa
R = √(63.66² + 159.2²) = 171.4 MPa
σ1 = C + R = 235.1 MPa
σ2 = C − R = −107.8 MPa

Answer

σ1 = 235.1 MPa
7.Axial stress & strain

Answer: δ = 4.041 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·16²/4 = 201.1 mm²
σ = P/A = 65·1000/201.1 = 323.3 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 323.3/200000 = 0.001616 = 1616 µε

Elongation

δ = ε·L = 0.001616·2500 mm = 4.041 mm

Answer

δ = 4.041 mm
8.Axial stress & strain

Answer: σ2 = 13.45 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 = 200000·1257/600 = 418879 N/mm
k₂ = 110000·706.9/400 = 194386 N/mm

Share of segment 2

F2 = 30000·194386/613265 = 9509 N

Stress

σ2 = 9509/706.9 = 13.45 MPa (compression)

Answer

σ2 = 13.45 MPa
9.Column buckling

Answer: Pcr = 10.49 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 = 0.7·5000 = 3500 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·65104/3500² = 10491 N = 10.49 kN

Answer

Pcr = 10.49 kN
10.Column buckling

Answer: d = 64.9 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 = 2.5·110000 = 275000 N
KL = 1·2500 = 2500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·275000·2500²/(π³·200000))^(1/4) = 64.9 mm

Answer

d = 64.9 mm

Check: the slenderness KL/r is 154.1, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: fd = 1.459 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 = √(4000/40) = 10 rad/s
fn = 10/(2π) = 1.592 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 1.592·√(1 − 0.4²) = 1.459 Hz

Answer

fd = 1.459 Hz
12.Mechanical vibrations

Answer: fn = 3.487 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.

keq = 12 + 12 = 24 kN/m

Natural frequency

ωn = √(24000/50) = 21.91 rad/s
fn = 21.91/(2π) = 3.487 Hz

Answer

fn = 3.487 Hz

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