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​ = 150 mmd2​ = 400 mm720 rpmT1​ = 1500 NT2​ = 600 NP = ?

A belt runs on a 150 mm driving pulley at 720 rpm. The tension is 1500 N on the tight side and 600 N on the slack side. Find the power P transmitted, in kW.

d1 = 150 mmn1 = 720 rpmT1 = 1500 NT2 = 600 N
P = kW
2.Gears & gear trainsHard
Ns​ = 18nr​ = ?Np​ = 24carrier fixed

In a planetary gear set the sun gear has 18 teeth and each planet gear has 24 teeth. The planet carrier is held fixed and the sun is driven at 3000 rpm. Find the speed nr of the ring gear (magnitude).

Ns = 18Np = 24ns = 3000 rpm
nr = rpm
3.Torsion of shaftsMedium
T = 750 N·md = 40 mmL = 2.5 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 = 750 N·md = 40 mmL = 2.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
d = 60 mmT0​ = 1500 N·mACBa = 1.5 mb = 2 m

A solid aluminium shaft of diameter 60 mm is fixed to rigid walls at both ends A and B. A torque of 1500 N·m is applied at C, 1.5 m from A and 2 m from B. Find the maximum shear stress τmax in the shaft.

d = 60 mmT0 = 1500 N·ma = 1.5 mb = 2 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = −80σy​ = 70τxy​ = 30MPa

A point in a loaded part has σₓ = −80 MPa, σᵧ = 70 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = −80 MPaσy = 70 MPaτxy = 30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 130σy​ = −60τxy​ = 10MPa

A point in a thin plate is in plane stress: σₓ = 130 MPa, σᵧ = −60 MPa and τₓᵧ = 10 MPa (σz = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σx = 130 MPaσy = −60 MPaτxy = 10 MPaσY = 460 MPa
n =
7.Axial stress & strainMedium
P = ?12600 mm300 mmØ30Ø60δ = 0.5 mm

A bar is made of a steel segment (diameter 30 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 60 mm, length 300 mm, E = 70 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, Ø30 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø60 mm, 300 mm, E = 70 GPaδ = 0.5 mm
P = kN
8.Axial stress & strainHard
100 kN12300 mm500 mmØ20Ø60σ2​ = ?

A steel segment (diameter 20 mm, length 300 mm, E = 200 GPa) and a segment made of brass (diameter 60 mm, length 500 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 100 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø20 mm, 300 mm, E = 200 GPasegment 2: brass, Ø60 mm, 500 mm, E = 100 GPaP = 100 kN
σ2 = MPa
9.Column bucklingMedium
PL = 4 m20 × 60 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 = 4 mE = 70 GPa (aluminium)b × h = 20 × 60 mm
Pcr = kN
10.Column bucklingHard
PL = 4 md = 50 mmcross-section

A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.

L = 4 mE = 200 GPa (steel)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx10 mm5 cycles0.8 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 10 mm is followed 5 cycles later by a peak of 0.8 mm. Find the damping ratio ζ.

x0 = 10 mmx5 = 0.8 mmn = 5
ζ =
12.Mechanical vibrationsHard
k1​ = 40k2​ = 5k3​ = 10m = 25 kg

A 25 kg mass hangs from spring k₃ = 10 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 40 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 = 25 kgk1 = 40 kN/mk2 = 5 kN/mk3 = 10 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: P = 5.089 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 = 5.655 m/s

Power

P = (1500 − 600)·5.655 = 5089 W = 5.089 kW

Answer

P = 5.089 kW
2.Gears & gear trains

Answer: nr = 818.2 rpm

Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (ns − nc)/(nr − nc) = −Nr/Ns: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.

Ring gear teeth

Nr = Ns + 2·Np = 18 + 2·24 = 66

Willis equation with the carrier fixed

With nc = 0 the sun drives the ring through the planets, which act as idlers.

nr = −ns·Ns/Nr = −3000·18/66 = −818.2 rpm
|nr| = 818.2 rpm

Answer

nr = 818.2 rpm

The ring turns opposite to the sun.

3.Torsion of shafts

Answer: φ = 5.343 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 750·1000·2500/(80000·251327) = 0.09325 rad
φ = 0.09325·180/π = 5.343°

Answer

φ = 5.343°
4.Torsion of shafts

Answer: τmax = 20.21 MPa

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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).

Equilibrium and compatibility

TA + TB = 1500
TA·a = TB·b
TA = 1500·2/3.5 = 857.1 N·m
TB = 1500·1.5/3.5 = 642.9 N·m

Maximum shear stress

The larger of the two torques governs (same diameter).

τmax = 16·857143/(π·60³) = 20.21 MPa

Answer

τmax = 20.21 MPa
5.Mohr’s circle & stress transformation

Answer: τmax = 80.78 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 = √(((−80 − 70)/2)² + 30²) = 80.78 MPa

Answer

τmax = 80.78 MPa
6.Mohr’s circle & stress transformation

Answer: n = 2.72

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 = 35 MPa, R = 95.52 MPa
σ1 = 130.5 MPa, σ2 = −60.52 MPa, σ3 = 0

Von Mises

σvM = √(130.5² − 130.5·(−60.52) + (−60.52)²) = 169.1 MPa
n = 460/169.1 = 2.72

Answer

n = 2.72

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

7.Axial stress & strain

Answer: P = 86.81 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 = 706.9 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·706.9) = 4.244 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(70000·2827) = 1.516 × 10⁻⁶ mm/N

Load

P = 0.5/5.76 × 10⁻⁶ = 86807 N = 86.81 kN

Answer

P = 86.81 kN
8.Axial stress & strain

Answer: σ2 = 25.81 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 = 200000·314.2/300 = 209440 N/mm
k₂ = 100000·2827/500 = 565487 N/mm

Share of segment 2

F2 = 100000·565487/774926 = 72973 N

Stress

σ2 = 72973/2827 = 25.81 MPa (compression)

Answer

σ2 = 25.81 MPa
9.Column buckling

Answer: Pcr = 3.525 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 = 0.7·4000 = 2800 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·40000/2800² = 3525 N = 3.525 kN

Answer

Pcr = 3.525 kN
10.Column buckling

Answer: σcr = 19.28 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·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 1·4000 = 4000 mm
KL/r = 4000/12.5 = 320

Critical stress

σcr = π²E/(KL/r)² = π²·200000/320² = 19.28 MPa

Answer

σcr = 19.28 MPa
11.Mechanical vibrations

Answer: ζ = 0.08014

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 = (1/5)·ln(10/0.8) = 0.5051

Damping ratio

ζ = δ/√(4π² + δ²) = 0.5051/√(39.48 + 0.2552) = 0.08014

Answer

ζ = 0.08014
12.Mechanical vibrations

Answer: fn = 2.879 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 = 40 + 5 = 45 kN/m
keq = 45·10/(45 + 10) = 8.182 kN/m

Natural frequency

ωn = √(8182/25) = 18.09 rad/s
fn = 18.09/(2π) = 2.879 Hz

Answer

fn = 2.879 Hz

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