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​ = 80 mmd2​ = 400 mm1440 rpmn2​ = ?

An 80 mm pulley at 1440 rpm drives a 400 mm pulley by a belt with 2 % slip. Find the speed n2 of the driven pulley.

n1 = 1440 rpmd1 = 80 mmd2 = 400 mms = 2 %
n2 = rpm
2.Gears & gear trainsHard
A = 12B = 45C = 16D = ?

In a compound gear train, gear A (12 teeth) on the input shaft turns at 2880 rpm and drives gear B (45 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 409.6 rpm. How many teeth ND must gear D have?

nin = 2880 rpmA = 12B = 45C = 16nout = 409.6 rpm
ND =
3.Torsion of shaftsMedium
T = 250 N·md = 50 mmL = 1.2 mφ = ?

A solid aluminium 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 = 250 N·md = 50 mmL = 1.2 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
d = 30 mmT0​ = 2000 N·mACBa = 0.8 mb = 2.5 m

A solid steel shaft of diameter 30 mm is fixed to rigid walls at both ends A and B. A torque of 2000 N·m is applied at C, 0.8 m from A and 2.5 m from B. Find the reaction torque TA at wall A.

d = 30 mmT0 = 2000 N·ma = 0.8 mb = 2.5 m
TA = N·m
5.Mohr’s circle & stress transformationMedium
σx​ = 20σy​ = 70τxy​ = 50MPa

A point in a loaded part has σₓ = 20 MPa, σᵧ = 70 MPa and τₓᵧ = 50 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σx = 20 MPaσy = 70 MPaτxy = 50 MPa
θp = °
6.Mohr’s circle & stress transformationHard
σx​ = 50σy​ = −70τxy​ = −50MPa

A point in a thin plate is in plane stress: σₓ = 50 MPa, σᵧ = −70 MPa and τₓᵧ = −50 MPa (σz = 0). The material has a yield strength of 350 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σx = 50 MPaσy = −70 MPaτxy = −50 MPaσY = 350 MPa
n =
7.Axial stress & strainMedium
P = ?12300 mm200 mmØ20Ø30δ = 0.5 mm

A bar is made of a steel segment (diameter 20 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 30 mm, length 200 mm, E = 110 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, Ø20 mm, 300 mm, E = 200 GPasegment 2: bronze, Ø30 mm, 200 mm, E = 110 GPaδ = 0.5 mm
P = kN
8.Axial stress & strainHard
P = ?Ø20100 GPaØ120/70600 mm

A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 70 mm, E = 100 GPa, allowable stress 80 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø20 mm, E = 200 GPa, σallow = 120 MPatube: Ø120/Ø70 mm, E = 100 GPa, σallow = 80 MPa
Pmax = kN
9.Column bucklingMedium
PL = 4 md = 80 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 = 4 mE = 200 GPa (steel)d = 80 mm
Pcr = kN
10.Column bucklingHard
PL = 6 md = 80 mmcross-section

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

L = 6 mE = 70 GPa (aluminium)d = 80 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 4 kN/mc = 30 N·s/mm = 20 kg

A 20 kg mass is supported by a spring with a stiffness of 4 kN/m and a viscous damper with a damping coefficient of 30 N·s/m. Find the damping ratio ζ.

m = 20 kgk = 4 kN/mc = 30 N·s/m
ζ =
12.Mechanical vibrationsHard
k1​ = 100k2​ = 5m = 10 kg

A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 100 kN/m and k₂ = 5 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.

m = 10 kgk1 = 100 kN/mk2 = 5 kN/m
keq = kN/m

3Solutions

1.Gears & gear trains

Answer: n2 = 282.2 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

1440·80/400 = 288 rpm

With slip

n2 = 288·0.98 = 282.2 rpm

Answer

n2 = 282.2 rpm
2.Gears & gear trains

Answer: ND = 30

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.

Intermediate shaft speed

nB = nC = 2880·12/45 = 768 rpm

Teeth on gear D

ND = C·nC/nout = 16·768/409.6 = 30

Answer

ND = 30
3.Torsion of shafts

Answer: φ = 1.077 °

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) = 250·1000·1200/(26000·613592) = 0.0188 rad
φ = 0.0188·180/π = 1.077°

Answer

φ = 1.077°
4.Torsion of shafts

Answer: TA = 1515 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 = 2000·2.5/3.3 = 1515 N·m
TB = 2000·0.8/3.3 = 484.8 N·m

Answer

TA = 1515 N·m
5.Mohr’s circle & stress transformation

Answer: θp = 58.28 °

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(2·50, 20 − 70) = 116.6°

Angle on the element

θp = 58.28°

Answer

θp = 58.28°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: n = 2.241

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 = −10 MPa, R = 78.1 MPa
σ1 = 68.1 MPa, σ2 = −88.1 MPa, σ3 = 0

Tresca

σmax − σmin = 156.2 MPa
n = 350/156.2 = 2.241

Answer

n = 2.241

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

7.Axial stress & strain

Answer: P = 68.06 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 = 314.2 mm², A₂ = 706.9 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·314.2) = 4.775 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(110000·706.9) = 2.572 × 10⁻⁶ mm/N

Load

P = 0.5/7.347 × 10⁻⁶ = 68056 N = 68.06 kN

Answer

P = 68.06 kN
8.Axial stress & strain

Answer: Pmax = 485.4 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·20²/4 = 314.2 mm²
At = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 100·10³·7461 = 8.09 × 10⁸ N

Governing strain

ε = min(0.0006, 0.0008) = 0.0006 (steel)

Largest load

Pmax = 0.0006·8.09 × 10⁸ = 485376 N = 485.4 kN

Answer

Pmax = 485.4 kN
9.Column buckling

Answer: Pcr = 62.01 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 2·4000 = 8000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/8000² = 62013 N = 62.01 kN

Answer

Pcr = 62.01 kN
10.Column buckling

Answer: σcr = 15.67 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Radius of gyration and slenderness

A = 5027 mm²
r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm
KL = 0.7·6000 = 4200 mm
KL/r = 4200/20 = 210

Critical stress

σcr = π²E/(KL/r)² = π²·70000/210² = 15.67 MPa

Answer

σcr = 15.67 MPa
11.Mechanical vibrations

Answer: ζ = 0.05303

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·√(4000·20) = 565.7 N·s/m

Damping ratio

ζ = c/cc = 30/565.7 = 0.05303

Answer

ζ = 0.05303

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

12.Mechanical vibrations

Answer: keq = 105 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.

keq = 100 + 5 = 105 kN/m

Answer

keq = 105 kN/m

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