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​ = 160 mmd2​ = 400 mm1500 rpmn2​ = ?

A 160 mm pulley at 1500 rpm drives a 400 mm pulley by a belt with 2 % slip. Find the speed n2 of the driven pulley.

n1 = 1500 rpmd1 = 160 mmd2 = 400 mms = 2 %
n2 = rpm
2.Gears & gear trainsHard
A = 16B = 18C = 40D = 48

In a compound gear train, gear A (16 teeth) on the input shaft turning at 1450 rpm drives gear B (18 teeth). Gear C (40 teeth) is fixed to the same shaft as B and drives gear D (48 teeth) on the output shaft. Find the output speed nout.

nin = 1450 rpmA = 16B = 18C = 40D = 48
nout = rpm
3.Torsion of shaftsMedium
T = 500 N·md = 50 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 = 500 N·md = 50 mmL = 3 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
do​ = ?5 kW, 750 rpm

A hollow shaft must transmit 5 kW at 750 rpm. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter do.

P = 5 kWN = 750 rpmdi/do = 0.7τallow = 40 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
σx​ = −30σy​ = 20τxy​ = 10MPa

A point in a loaded part has σₓ = −30 MPa, σᵧ = 20 MPa and τₓᵧ = 10 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = −30 MPaσy = 20 MPaτxy = 10 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 150σy​ = 40τxy​ = −50MPa

A point in a loaded part has σₓ = 150 MPa, σᵧ = 40 MPa and τₓᵧ = −50 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 15° counter-clockwise from the x-axis.

σx = 150 MPaσy = 40 MPaτxy = −50 MPaθ = 15°
σθ = MPa
7.Axial stress & strainMedium
150 kNØ40120 GPaØ120/50400 mmδ = ?

A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 50 mm, E = 120 GPa). Both are 400 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.

P = 150 kNrod: Ø40 mm, E = 200 GPatube: Ø120/Ø50 mm, E = 120 GPaL = 400 mm
δ = mm
8.Axial stress & strainHard
P = ?Ø25110 GPaØ80/50800 mm

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of bronze (outer diameter 80 mm, inner diameter 50 mm, E = 110 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø25 mm, E = 200 GPa, σallow = 200 MPatube: Ø80/Ø50 mm, E = 110 GPa, σallow = 60 MPa
Pmax = kN
9.Column bucklingMedium
PL = 5 m20 × 60 mmcross-section

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

L = 5 mE = 200 GPa (steel)b × h = 20 × 60 mm
Pcr = kN
10.Column bucklingHard
PL = 2 md = 40 mmcross-section

A column of length L made of aluminium is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.

L = 2 mE = 70 GPa (aluminium)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx15 mm4 cycles5.5 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 15 mm is followed 4 cycles later by a peak of 5.5 mm. Find the damping ratio ζ.

x0 = 15 mmx4 = 5.5 mmn = 4
ζ =
12.Mechanical vibrationsHard
k1​ = 20k2​ = 25m = 25 kg

A 25 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 20 kN/m and the lower one k₂ = 25 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.

m = 25 kgk1 = 20 kN/mk2 = 25 kN/m
keq = kN/m

3Solutions

1.Gears & gear trains

Answer: n2 = 588 rpm

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.

Speed without slip

1500·160/400 = 600 rpm

With slip

n2 = 600·0.98 = 588 rpm

Answer

n2 = 588 rpm
2.Gears & gear trains

Answer: nout = 1074 rpm

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.

First stage

nB = 1450·16/18 = 1289 rpm = nC

Second stage

nout = 1289·40/48 = 1074 rpm

Answer

nout = 1074 rpm

total ratio = 1.35 : 1

3.Torsion of shafts

Answer: φ = 1.751 °

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) = 500·1000·3000/(80000·613592) = 0.03056 rad
φ = 0.03056·180/π = 1.751°

Answer

φ = 1.751°
4.Torsion of shafts

Answer: do = 22.01 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·750/60 = 78.54 rad/s
T = P/ω = 5000/78.54 = 63.66 N·m

Outer diameter from the allowable stress

T = 63.66 N·m = 63662 N·mm
1 − k⁴ = 1 − 0.7⁴ = 0.7599
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·63662/(π·40·0.7599)) = 22.01 mm
di = 0.7·22.01 = 15.41 mm

Answer

do = 22.01 mm

A solid shaft would need d = 20.09 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: τmax = 26.93 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.

στxyCτmax​

Radius of Mohr’s circle

τmax = R = √(((−30 − 20)/2)² + 10²) = 26.93 MPa

Answer

τmax = 26.93 MPa
6.Mohr’s circle & stress transformation

Answer: σθ = 117.6 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

σθ = 95 + 55·0.866 + −50·0.5 = 117.6 MPa

Answer

σθ = 117.6 MPa
7.Axial stress & strain

Answer: δ = 0.0437 mm

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·40²/4 = 1257 mm²
At = π(120² − 50²)/4 = 9346 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 120·10³·9346 = 1.373 × 10⁹ N

Shortening

δ = 150000·400/1.373 × 10⁹ = 0.0437 mm

Answer

δ = 0.0437 mm
8.Axial stress & strain

Answer: Pmax = 237.3 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(80² − 50²)/4 = 3063 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 110·10³·3063 = 4.351 × 10⁸ N

Governing strain

ε = min(0.001, 0.0005455) = 0.0005455 (bronze)

Largest load

Pmax = 0.0005455·4.351 × 10⁸ = 237333 N = 237.3 kN

Answer

Pmax = 237.3 kN
9.Column buckling

Answer: Pcr = 3.158 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 = 60·20³/12 = 40000 mm⁴

Effective length

KL = 1·5000 = 5000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·40000/5000² = 3158 N = 3.158 kN

Answer

Pcr = 3.158 kN
10.Column buckling

Answer: σcr = 4.318 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·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²
r = √(I/A) = √(125664/1257) = 10 mm
KL = 2·2000 = 4000 mm
KL/r = 4000/10 = 400

Critical stress

σcr = π²E/(KL/r)² = π²·70000/400² = 4.318 MPa

Answer

σcr = 4.318 MPa
11.Mechanical vibrations

Answer: ζ = 0.03989

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.

Logarithmic decrement

δ = (1/4)·ln(15/5.5) = 0.2508

Damping ratio

ζ = δ/√(4π² + δ²) = 0.2508/√(39.48 + 0.06291) = 0.03989

Answer

ζ = 0.03989
12.Mechanical vibrations

Answer: keq = 11.11 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.

1/keq = 1/20 + 1/25 → keq = 20·25/(20 + 25) = 11.11 kN/m

Answer

keq = 11.11 kN/m

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