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​ = 24N2​ = 60

An 11 kW motor running at 2900 rpm drives a pinion with 24 teeth, which meshes with a gear with 60 teeth. The gear pair has an efficiency of 97%. Find the output torque T2 on the gear shaft.

P = 11 kWn1 = 2900 rpmN1 = 24N2 = 60η = 97%
T2 = N·m
2.Gears & gear trainsHard
d1​ = 80 mmd2​ = 400 mm960 rpmC = 800 mmθ = ?

An open belt connects an 80 mm pulley and a 400 mm pulley whose centres are 800 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 80 mmd2 = 400 mmC = 800 mm
θ = °
3.Torsion of shaftsMedium
T = 400 N·md = 25 mmL = 3 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 = 400 N·md = 25 mmL = 3 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 15 kW at 1450 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 15 kWN = 1450 rpmτallow = 60 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = −40τxy​ = 40MPa

A point in a thin plate is in plane stress: σₓ = −40 MPa, σᵧ = 0 MPa and τₓᵧ = 40 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).

σx = −40 MPaσy = 0 MPaτxy = 40 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 150σy​ = −50τxy​ = −40MPa

A point in a loaded part has σₓ = 150 MPa, σᵧ = −50 MPa and τₓᵧ = −40 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 30° clockwise from the x-axis (positive in the same sense as τₓᵧ).

σx = 150 MPaσy = −50 MPaτxy = −40 MPaθ = -30°
τθ = MPa
7.Axial stress & strainMedium
Ø 30 mmAluminium2 mP = 45 kN

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

P = 45 kNd = 30 mmL = 2 mE = 70 GPa
δ = mm
8.Axial stress & strainHard
P = ?Ø25120 GPaØ120/60600 mm

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 60 mm, E = 120 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø25 mm, E = 200 GPa, σallow = 150 MPatube: Ø120/Ø60 mm, E = 120 GPa, σallow = 60 MPa
Pmax = kN
9.Column bucklingMedium
PL = 5 md = 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)d = 50 mm
Pcr = kN
10.Column bucklingHard
P = 75 kNL = 1.5 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 75 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 75 kNn = 2L = 1.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 4 kN/mζ = 0.5m = 5 kg

A 5 kg mass on a spring with a stiffness of 4 kN/m has a viscous damper giving a damping ratio of 0.5. Find the damped natural frequency fd of its free vibration in Hz.

m = 5 kgk = 4 kN/mζ = 0.5
fd = Hz
12.Mechanical vibrationsHard
k1​ = 60k2​ = 5k3​ = 25m = 50 kg

A 50 kg mass hangs from spring k₃ = 25 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 60 kN/m and k₂ = 5 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.

m = 50 kgk1 = 60 kN/mk2 = 5 kN/mk3 = 25 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: T2 = 87.84 N·m

Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.

Motor torque

ω₁ = 2π·2900/60 = 303.7 rad/s
T1 = P/ω₁ = 11000/303.7 = 36.22 N·m

Output torque

T2 = 36.22·60/24·0.97 = 87.84 N·m

Answer

T2 = 87.84 N·m
2.Gears & gear trains

Answer: θ = 156.9 °

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(0.2) = 11.54°

Angle of wrap

θ = 180 − 2·11.54 = 156.9°

Answer

θ = 156.9°
3.Torsion of shafts

Answer: φ = 22.41 °

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) = 400·1000·3000/(80000·38350) = 0.3911 rad
φ = 0.3911·180/π = 22.41°

Answer

φ = 22.41°
4.Torsion of shafts

Answer: d = 20.32 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·1450/60 = 151.8 rad/s
T = P/ω = 15000/151.8 = 98.79 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·98786/(π·60)) = 20.32 mm

Answer

d = 20.32 mm

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

5.Mohr’s circle & stress transformation

Answer: τabs = 44.72 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 = −20 MPa, R = √((−20)² + 40²) = 44.72 MPa
σ1 = 24.72 MPa, σ2 = −64.72 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(89.44, 24.72, 64.72)/2 = 44.72 MPa

Answer

τabs = 44.72 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: τθ = 66.6 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

τθ = −100·−0.866 + −40·0.5 = 66.6 MPa

Answer

τθ = 66.6 MPa
7.Axial stress & strain

Answer: δ = 1.819 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·30²/4 = 706.9 mm²
σ = P/A = 45·1000/706.9 = 63.66 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 63.66/70000 = 0.0009095 = 909.5 µε

Elongation

δ = ε·L = 0.0009095·2000 mm = 1.819 mm

Answer

δ = 1.819 mm
8.Axial stress & strain

Answer: Pmax = 558 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 − 60²)/4 = 8482 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 120·10³·8482 = 1.116 × 10⁹ N

Governing strain

ε = min(0.00075, 0.0005) = 0.0005 (copper)

Largest load

Pmax = 0.0005·1.116 × 10⁹ = 558025 N = 558 kN

Answer

Pmax = 558 kN
9.Column buckling

Answer: Pcr = 49.44 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·50⁴/64 = 306796 mm⁴

Effective length

KL = 0.7·5000 = 3500 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·306796/3500² = 49436 N = 49.44 kN

Answer

Pcr = 49.44 kN
10.Column buckling

Answer: d = 61.1 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 = 2·75000 = 150000 N
KL = 2·1500 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·150000·3000²/(π³·200000))^(1/4) = 61.1 mm

Answer

d = 61.1 mm

Check: the slenderness KL/r is 196.4, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: fd = 3.898 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 = √(4000/5) = 28.28 rad/s
fn = 28.28/(2π) = 4.502 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 4.502·√(1 − 0.5²) = 3.898 Hz

Answer

fd = 3.898 Hz
12.Mechanical vibrations

Answer: fn = 3.024 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

Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.

k12 = 60 + 5 = 65 kN/m
keq = 65·25/(65 + 25) = 18.06 kN/m

Natural frequency

ωn = √(18056/50) = 19 rad/s
fn = 19/(2π) = 3.024 Hz

Answer

fn = 3.024 Hz

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