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
Tw​ = ?z = 3, 2900 rpm

A 1.1 kW motor at 2900 rpm drives a worm with 3 start(s). The worm wheel has 80 teeth and the drive has an efficiency of 85%. Find the output torque Tw on the wheel shaft.

z = 3Nw = 80n = 2900 rpmP = 1.1 kWη = 85%
Tw = N·m
2.Gears & gear trainsHard
d1​ = 120 mmd2​ = 315 mm720 rpmC = 1000 mmθ = ?

An open belt connects a 120 mm pulley and a 315 mm pulley whose centres are 1000 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 120 mmd2 = 315 mmC = 1000 mm
θ = °
3.Torsion of shaftsMedium
T = 100 N·md = 20 mmL = 2.5 mφ = ?

A solid brass 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 = 100 N·md = 20 mmL = 2.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 100 kW at 1450 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 100 kWN = 1450 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 110σy​ = −70τxy​ = 40MPa

A point in a loaded part has σₓ = 110 MPa, σᵧ = −70 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 = 110 MPaσy = −70 MPaτxy = 40 MPa
θp = °
6.Mohr’s circle & stress transformationHard
σx​ = 160σy​ = −10τxy​ = −20MPa

A point in a thin plate is in plane stress: σₓ = 160 MPa, σᵧ = −10 MPa and τₓᵧ = −20 MPa (σz = 0). Find the von Mises equivalent stress σvM.

σx = 160 MPaσy = −10 MPaτxy = −20 MPa
σvM = MPa
7.Axial stress & strainMedium
P = ?12500 mm500 mmØ40Ø30δ = 0.5 mm

A bar is made of a steel segment (diameter 40 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 30 mm, length 500 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.

segment 1: steel, Ø40 mm, 500 mm, E = 200 GPasegment 2: copper, Ø30 mm, 500 mm, E = 120 GPaδ = 0.5 mm
P = kN
8.Axial stress & strainHard
P = ?Ø25100 GPaØ100/50300 mm

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of brass (outer diameter 100 mm, inner diameter 50 mm, E = 100 GPa, allowable stress 100 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 = 100 GPa, σallow = 100 MPa
Pmax = kN
9.Column bucklingMedium
PL = 2.5 md = 60 mmcross-section

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

L = 2.5 mE = 200 GPa (steel)d = 60 mm
Pcr = kN
10.Column bucklingHard
P = 15 kNL = 5 md = ?cross-section

A solid round column of length L made of steel is pinned at both ends. It must carry 15 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 15 kNn = 3L = 5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 50 kN/mc = 450 N·s/mm = 100 kg

A 100 kg mass is supported by a spring with a stiffness of 50 kN/m and a viscous damper with a damping coefficient of 450 N·s/m. Find the damping ratio ζ.

m = 100 kgk = 50 kN/mc = 450 N·s/m
ζ =
12.Mechanical vibrationsHard
k1​ = 80k2​ = 25k3​ = 40m = 20 kg

A 20 kg mass hangs from spring k₃ = 40 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 80 kN/m and k₂ = 25 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 = 20 kgk1 = 80 kN/mk2 = 25 kN/mk3 = 40 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: Tw = 82.1 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.

Wheel speed

nw = n·z/Nw = 2900·3/80 = 108.8 rpm

Output torque

ωw = 2π·108.8/60 = 11.39 rad/s
Tw = η·P/ωw = 0.85·1100/11.39 = 82.1 N·m

Answer

Tw = 82.1 N·m
2.Gears & gear trains

Answer: θ = 168.8 °

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(0.0975) = 5.595°

Angle of wrap

θ = 180 − 2·5.595 = 168.8°

Answer

θ = 168.8°
3.Torsion of shafts

Answer: φ = 23.38 °

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·20⁴/32 = 15708 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·2500/(39000·15708) = 0.4081 rad
φ = 0.4081·180/π = 23.38°

Answer

φ = 23.38°
4.Torsion of shafts

Answer: d = 35.5 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 = 100000/151.8 = 658.6 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·658572/(π·75)) = 35.5 mm

Answer

d = 35.5 mm

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

5.Mohr’s circle & stress transformation

Answer: θp = 11.98 °

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, 110 − −70) = 23.96°

Angle on the element

θp = 11.98°

Answer

θp = 11.98°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σvM = 168.8 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. 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 = 75 MPa, R = 87.32 MPa
σ1 = 162.3 MPa, σ2 = −12.32 MPa, σ3 = 0

Von Mises

σvM = √(162.3² − 162.3·(−12.32) + (−12.32)²) = 168.8 MPa

Answer

σvM = 168.8 MPa
7.Axial stress & strain

Answer: P = 63.42 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

A₁ = 1257 mm², A₂ = 706.9 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·1257) = 1.989 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(120000·706.9) = 5.895 × 10⁻⁶ mm/N

Load

P = 0.5/7.884 × 10⁻⁶ = 63419 N = 63.42 kN

Answer

P = 63.42 kN
8.Axial stress & strain

Answer: Pmax = 687.2 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 + 100·10³·5890 = 6.872 × 10⁸ N

Governing strain

ε = min(0.00125, 0.001) = 0.001 (brass)

Largest load

Pmax = 0.001·6.872 × 10⁸ = 687223 N = 687.2 kN

Answer

Pmax = 687.2 kN
9.Column buckling

Answer: Pcr = 50.23 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·60⁴/64 = 636173 mm⁴

Effective length

KL = 2·2500 = 5000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·636173/5000² = 50230 N = 50.23 kN

Answer

Pcr = 50.23 kN
10.Column buckling

Answer: d = 58.37 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 = 3·15000 = 45000 N
KL = 1·5000 = 5000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·45000·5000²/(π³·200000))^(1/4) = 58.37 mm

Answer

d = 58.37 mm

Check: the slenderness KL/r is 342.6, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.1006

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·√(50000·100) = 4472 N·s/m

Damping ratio

ζ = c/cc = 450/4472 = 0.1006

Answer

ζ = 0.1006

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

12.Mechanical vibrations

Answer: fn = 6.057 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 = 80 + 25 = 105 kN/m
keq = 105·40/(105 + 40) = 28.97 kN/m

Natural frequency

ωn = √(28966/20) = 38.06 rad/s
fn = 38.06/(2π) = 6.057 Hz

Answer

fn = 6.057 Hz

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