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​ = 315 mm1500 rpmT1​ = 1500 NT2​ = 750 NP = ?

A belt runs on a 160 mm driving pulley at 1500 rpm. The tension is 1500 N on the tight side and 750 N on the slack side. Find the power P transmitted, in kW.

d1 = 160 mmn1 = 1500 rpmT1 = 1500 NT2 = 750 N
P = kW
2.Gears & gear trainsHard
d1​ = 125 mmd2​ = 315 mm2880 rpmC = 800 mmT1​ = 1200 NT2​P = ?

An open belt connects a 125 mm driving pulley at 2880 rpm to a 315 mm pulley; the centres are 800 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 1200 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 125 mmn1 = 2880 rpmd2 = 315 mmC = 800 mmμ = 0.25T1 = 1200 N
P = kW
3.Torsion of shaftsMedium
d = ?T = 250 N·mL = 2.5 m

A solid aluminium shaft 2.5 m long carries a torque of 250 N·m. The angle of twist between its ends must not exceed 2°. Find the smallest diameter d that meets this limit.

T = 250 N·mL = 2.5 mG = 26 GPaφallow = 2°
d = mm
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 30 kW at 500 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 30 kWN = 500 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = −50σy​ = 60τxy​ = −40MPa

A point in a loaded part has σₓ = −50 MPa, σᵧ = 60 MPa and τₓᵧ = −40 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σx = −50 MPaσy = 60 MPaτxy = −40 MPa
θp = °
6.Mohr’s circle & stress transformationHard
σx​ = 140σy​ = 110τxy​ = 10MPa

A point in a thin plate is in plane stress: σₓ = 140 MPa, σᵧ = 110 MPa and τₓᵧ = 10 MPa (σz = 0). The material has a yield strength of 250 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σx = 140 MPaσy = 110 MPaτxy = 10 MPaσY = 250 MPa
n =
7.Axial stress & strainMedium
50 kNØ30110 GPaØ120/70600 mmδ = ?

A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 120 mm, inner diameter 70 mm, E = 110 GPa). Both are 600 mm long and are compressed together between rigid plates by 50 kN. Find the shortening δ.

P = 50 kNrod: Ø30 mm, E = 200 GPatube: Ø120/Ø70 mm, E = 110 GPaL = 600 mm
δ = mm
8.Axial stress & strainHard
P = ?Ø25110 GPaØ100/50300 mm

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 50 mm, E = 110 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø25 mm, E = 200 GPa, σallow = 250 MPatube: Ø100/Ø50 mm, E = 110 GPa, σallow = 50 MPa
Pmax = kN
9.Column bucklingMedium
PL = 6 m25 × 50 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 = 6 mE = 70 GPa (aluminium)b × h = 25 × 50 mm
Pcr = kN
10.Column bucklingHard
P = 20 kNL = 2.5 md = ?cross-section

A solid round column of length L made of aluminium is pinned at both ends. It must carry 20 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 20 kNn = 2.5L = 2.5 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
tx50 mm3 cycles11 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed 3 cycles later by a peak of 11 mm. Find the damping ratio ζ.

x0 = 50 mmx3 = 11 mmn = 3
ζ =
12.Mechanical vibrationsHard
k1​ = 10k2​ = 5m = 25 kg

A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 10 kN/m and k₂ = 5 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 = 25 kgk1 = 10 kN/mk2 = 5 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: P = 9.425 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.

Belt speed

v = 12.57 m/s

Power

P = (1500 − 750)·12.57 = 9425 W = 9.425 kW

Answer

P = 9.425 kW
2.Gears & gear trains

Answer: P = 11.67 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 = 2.904 rad

Slack-side tension

eμθ = e0.7259 = 2.067, T₂ = 1200/2.067 = 580.7 N

Power

P = (1200 − 580.7)·18.85 = 11.67 kW

Answer

P = 11.67 kW
3.Torsion of shafts

Answer: d = 51.46 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 = 250 N·m = 250000 N·mm
L = 2.5 m = 2500 mm
G = 26 GPa = 26000 MPa
φ = 2·π/180 = 0.03491 rad
J = TL/(Gφ) = 250000·2500/(26000·0.03491) = 688651 mm⁴

Diameter

d = (32J/π)1/4 = 51.46 mm

Answer

d = 51.46 mm

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

4.Torsion of shafts

Answer: d = 33.89 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·500/60 = 52.36 rad/s
T = P/ω = 30000/52.36 = 573 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·572958/(π·75)) = 33.89 mm

Answer

d = 33.89 mm

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

5.Mohr’s circle & stress transformation

Answer: θp = −71.99 °

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.

στxyCσ1​σ2​

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θp = atan2(2·−40, −50 − 60) = −144°

Angle on the element

θp = −71.99°

Answer

θp = −71.99°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: n = 1.748

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.

σxyCσ1​σ2​

Principal stresses

C = 125 MPa, R = 18.03 MPa
σ1 = 143 MPa, σ2 = 107 MPa, σ3 = 0

Tresca

σmax − σmin = 143 MPa
n = 250/143 = 1.748

Answer

n = 1.748

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

7.Axial stress & strain

Answer: δ = 0.03118 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·30²/4 = 706.9 mm²
At = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 110·10³·7461 = 9.621 × 10⁸ N

Shortening

δ = 50000·600/9.621 × 10⁸ = 0.03118 mm

Answer

δ = 0.03118 mm
8.Axial stress & strain

Answer: Pmax = 339.1 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(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 110·10³·5890 = 7.461 × 10⁸ N

Governing strain

ε = min(0.00125, 0.0004545) = 0.0004545 (bronze)

Largest load

Pmax = 0.0004545·7.461 × 10⁸ = 339149 N = 339.1 kN

Answer

Pmax = 339.1 kN
9.Column buckling

Answer: Pcr = 2.55 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 = 0.7·6000 = 4200 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·65104/4200² = 2550 N = 2.55 kN

Answer

Pcr = 2.55 kN
10.Column buckling

Answer: d = 55.1 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 = 2.5·20000 = 50000 N
KL = 1·2500 = 2500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·50000·2500²/(π³·70000))^(1/4) = 55.1 mm

Answer

d = 55.1 mm

Check: the slenderness KL/r is 181.5, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.08007

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 = (1/3)·ln(50/11) = 0.5047

Damping ratio

ζ = δ/√(4π² + δ²) = 0.5047/√(39.48 + 0.2547) = 0.08007

Answer

ζ = 0.08007
12.Mechanical vibrations

Answer: fn = 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.

Equivalent stiffness

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

keq = 10 + 5 = 15 kN/m

Natural frequency

ωn = √(15000/25) = 24.49 rad/s
fn = 24.49/(2π) = 3.898 Hz

Answer

fn = 3.898 Hz

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