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​ = 16, m = 2.5 mm

A 2.2 kW motor running at 2900 rpm drives a pinion with 16 teeth and a module of 2.5 mm. Find the tangential force Ft on the pinion teeth.

P = 2.2 kWn = 2900 rpmN1 = 16m = 2.5 mm
Ft = N
2.Gears & gear trainsHard
A = 25B = 80C = 16D = 45

In a compound gear train, gear A (25 teeth) on the input shaft turning at 960 rpm drives gear B (80 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D (45 teeth) on the output shaft. Find the output speed nout.

nin = 960 rpmA = 25B = 80C = 16D = 45
nout = rpm
3.Torsion of shaftsMedium
T = 750 N·md = 50 mmL = 0.5 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 = 50 mmL = 0.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
d = 40 mmT0​ = 2000 N·mACBa = 2.5 mb = 0.5 m

A solid aluminium shaft of diameter 40 mm is fixed to rigid walls at both ends A and B. A torque of 2000 N·m is applied at C, 2.5 m from A and 0.5 m from B. Find the reaction torque TA at wall A.

d = 40 mmT0 = 2000 N·ma = 2.5 mb = 0.5 m
TA = N·m
5.Mohr’s circle & stress transformationMedium
σx​ = −80σy​ = −10τxy​ = 50MPa

A point has σₓ = −80 MPa, σᵧ = −10 MPa and τₓᵧ = 50 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σx = −80 MPaσy = −10 MPaτxy = 50 MPa
θs = °
6.Mohr’s circle & stress transformationHard
σx​ = 60σy​ = −50τxy​ = 40MPa

A point in a thin plate is in plane stress: σₓ = 60 MPa, σᵧ = −50 MPa and τₓᵧ = 40 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 = 60 MPaσy = −50 MPaτxy = 40 MPaσY = 250 MPa
n =
7.Axial stress & strainMedium
P = ?12300 mm200 mmØ40Ø40δ = 0.4 mm

A bar is made of a steel segment (diameter 40 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 40 mm, length 200 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.

segment 1: steel, Ø40 mm, 300 mm, E = 200 GPasegment 2: brass, Ø40 mm, 200 mm, E = 100 GPaδ = 0.4 mm
P = kN
8.Axial stress & strainHard
Ø 35 mmCopper1.5 mØ 12 mmBrass1.5 mP = 40 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 40 kNsegment 1: Ø35 mm, 1.5 m, copper (E = 117 GPa)segment 2: Ø12 mm, 1.5 m, brass (E = 100 GPa)
δ = mm
9.Column bucklingMedium
PL = 2.5 md = 25 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 = 2.5 mE = 200 GPa (steel)d = 25 mm
Pcr = kN
10.Column bucklingHard
P = 50 kNL = 2.5 md = ?cross-section

A solid round column of length L made of aluminium is fixed at both ends. It must carry 50 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 50 kNn = 3L = 2.5 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
tx30 mm2 cycles10.9 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 30 mm is followed 2 cycles later by a peak of 10.9 mm. Find the damping ratio ζ.

x0 = 30 mmx2 = 10.9 mmn = 2
ζ =
12.Mechanical vibrationsHard
k1​ = 15k2​ = 25k3​ = 50m = 50 kg

A 50 kg mass hangs from spring k₃ = 50 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 15 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 = 50 kgk1 = 15 kN/mk2 = 25 kN/mk3 = 50 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: Ft = 362.2 N

Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.

Torque on the pinion

ω = 2π·2900/60 = 303.7 rad/s
T = P/ω = 2200/303.7 = 7.244 N·m

Pitch diameter

d = m·N = 2.5·16 = 40 mm

Tangential force

Ft = 2T/d = 2·7.244/0.04 = 362.2 N

Answer

Ft = 362.2 N
2.Gears & gear trains

Answer: nout = 106.7 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 = 960·25/80 = 300 rpm = nC

Second stage

nout = 300·16/45 = 106.7 rpm

Answer

nout = 106.7 rpm

total ratio = 9 : 1

3.Torsion of shafts

Answer: φ = 0.4377 °

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) = 750·1000·500/(80000·613592) = 0.007639 rad
φ = 0.007639·180/π = 0.4377°

Answer

φ = 0.4377°
4.Torsion of shafts

Answer: TA = 333.3 N·m

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 = 2000
TA·a = TB·b
TA = 2000·0.5/3 = 333.3 N·m
TB = 2000·2.5/3 = 1667 N·m

Answer

TA = 333.3 N·m
5.Mohr’s circle & stress transformation

Answer: θs = 17.5 °

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​

Angle on the circle

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θs = 50/61.03 = 0.8192, sin 2θs = −(−35)/61.03 = 0.5735
2θs = atan2(35, 50) = 34.99°

Angle on the element

θs = 17.5°
check: θs − θp = 17.5 − 62.5 = −45°

Answer

θs = 17.5°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: n = 1.838

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 = 5 MPa, R = 68.01 MPa
σ1 = 73.01 MPa, σ2 = −63.01 MPa, σ3 = 0

Tresca

σmax − σmin = 136 MPa
n = 250/136 = 1.838

Answer

n = 1.838

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

7.Axial stress & strain

Answer: P = 143.6 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 = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·1257) = 1.194 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(100000·1257) = 1.592 × 10⁻⁶ mm/N

Load

P = 0.4/2.785 × 10⁻⁶ = 143616 N = 143.6 kN

Answer

P = 143.6 kN
8.Axial stress & strain

Answer: δ = 5.838 mm

Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.

Same force in both segments

The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.

A1 = π·35²/4 = 962.1 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 41.58 MPa
σ2 = 353.7 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 41.58·1500/117000 = 0.533 mm
δ2 = σ2L2/E2 = 353.7·1500/100000 = 5.305 mm
δ = δ1 + δ2 = 5.838 mm

Answer

δ = 5.838 mm
9.Column buckling

Answer: Pcr = 12.36 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·25⁴/64 = 19175 mm⁴

Effective length

KL = 0.7·2500 = 1750 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·19175/1750² = 12359 N = 12.36 kN

Answer

Pcr = 12.36 kN
10.Column buckling

Answer: d = 51.27 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 = 3·50000 = 150000 N
KL = 0.5·2500 = 1250 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·150000·1250²/(π³·70000))^(1/4) = 51.27 mm

Answer

d = 51.27 mm

Check: the slenderness KL/r is 97.52, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.08031

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 = (1/2)·ln(30/10.9) = 0.5062

Damping ratio

ζ = δ/√(4π² + δ²) = 0.5062/√(39.48 + 0.2563) = 0.08031

Answer

ζ = 0.08031
12.Mechanical vibrations

Answer: fn = 3.355 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 = 15 + 25 = 40 kN/m
keq = 40·50/(40 + 50) = 22.22 kN/m

Natural frequency

ωn = √(22222/50) = 21.08 rad/s
fn = 21.08/(2π) = 3.355 Hz

Answer

fn = 3.355 Hz

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