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​ = 315 mm960 rpmn2​ = ?

A 150 mm pulley at 960 rpm drives a 315 mm pulley by a belt with 4 % slip. Find the speed n2 of the driven pulley.

n1 = 960 rpmd1 = 150 mmd2 = 315 mms = 4 %
n2 = rpm
2.Gears & gear trainsHard
Ns​ = 24ring fixedNp​ = 24nc​ = ?

In a planetary gear set the sun gear has 24 teeth and each planet gear has 24 teeth. The ring gear is held fixed and the sun is driven at 1200 rpm. Find the speed nc of the planet carrier.

Ns = 24Np = 24ns = 1200 rpm
nc = rpm
3.Torsion of shaftsMedium
T = 200 N·md = 20 mmL = 1 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 = 200 N·md = 20 mmL = 1 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
d = 50 mmT0​ = 1500 N·mACBa = 1 mb = 1.2 m

A solid steel shaft of diameter 50 mm is fixed to rigid walls at both ends A and B. A torque of 1500 N·m is applied at C, 1 m from A and 1.2 m from B. Find the angle of rotation φC of section C, in degrees.

d = 50 mmT0 = 1500 N·ma = 1 mb = 1.2 mG = 80 GPa
φC = °
5.Mohr’s circle & stress transformationMedium
σx​ = −70τxy​ = −60MPa

A point has σₓ = −70 MPa, σᵧ = 0 MPa and τₓᵧ = −60 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 = −70 MPaσy = 0 MPaτxy = −60 MPa
θs = °
6.Mohr’s circle & stress transformationHard
σx​ = 160σy​ = −70τxy​ = −60MPa

A point in a thin plate is in plane stress: σₓ = 160 MPa, σᵧ = −70 MPa and τₓᵧ = −60 MPa (σz = 0). Find the von Mises equivalent stress σvM.

σx = 160 MPaσy = −70 MPaτxy = −60 MPa
σvM = MPa
7.Axial stress & strainMedium
Ø 20 mmCopper1.5 mP = 80 kN

Find the elongation δ of the copper bar under the axial load.

P = 80 kNd = 20 mmL = 1.5 mE = 117 GPa
δ = mm
8.Axial stress & strainHard
30 kN12300 mm400 mmØ30Ø40σ2​ = ?

A steel segment (diameter 30 mm, length 300 mm, E = 200 GPa) and a segment made of copper (diameter 40 mm, length 400 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 30 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø30 mm, 300 mm, E = 200 GPasegment 2: copper, Ø40 mm, 400 mm, E = 120 GPaP = 30 kN
σ2 = MPa
9.Column bucklingMedium
PL = 2.5 m25 × 60 mmcross-section

A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.

L = 2.5 mE = 200 GPa (steel)b × h = 25 × 60 mm
Pcr = kN
10.Column bucklingHard
PL = 5 md = 30 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 = 5 mE = 200 GPa (steel)d = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx15 mm4 cycles7.1 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 15 mm is followed 4 cycles later by a peak of 7.1 mm. Find the damping ratio ζ.

x0 = 15 mmx4 = 7.1 mmn = 4
ζ =
12.Mechanical vibrationsHard
k1​ = 60k2​ = 30k3​ = 25m = 50 kg

A 50 kg mass hangs from spring k₃ = 25 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 60 kN/m and k₂ = 30 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 = 60 kN/mk2 = 30 kN/mk3 = 25 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: n2 = 438.9 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

960·150/315 = 457.1 rpm

With slip

n2 = 457.1·0.96 = 438.9 rpm

Answer

n2 = 438.9 rpm
2.Gears & gear trains

Answer: nc = 300 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 = 24 + 2·24 = 72

Willis equation with the ring fixed

Set nr = 0 and solve for the carrier speed.

(1200 − nc)/(0 − nc) = −72/24
nc = ns·Ns/(Ns + Nr) = 1200·24/96 = 300 rpm

Answer

nc = 300 rpm

The carrier turns in the same direction as the sun.

3.Torsion of shafts

Answer: φ = 9.119 °

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) = 200·1000·1000/(80000·15708) = 0.1592 rad
φ = 0.1592·180/π = 9.119°

Answer

φ = 9.119°
4.Torsion of shafts

Answer: φC = 0.955 °

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 = 1500·1.2/2.2 = 818.2 N·m
TB = 1500·1/2.2 = 681.8 N·m

Rotation of C

Use segment AC: it is fixed at A and carries TA.

J = π·50⁴/32 = 613592 mm⁴
G = 80 GPa = 80000 MPa
φC = TA·a/(GJ) = 818182·1000/(80000·613592) = 0.01667 rad = 0.955°

Answer

φC = 0.955°
5.Mohr’s circle & stress transformation

Answer: θs = 74.87 °

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 = −60/69.46 = −0.8638, sin 2θs = −(−35)/69.46 = 0.5039
2θs = atan2(35, −60) = 149.7°

Angle on the element

θs = 74.87°
check: θs − θp = 74.87 − (−60.13) = 135°

Answer

θs = 74.87°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σvM = 229.1 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 = 45 MPa, R = 129.7 MPa
σ1 = 174.7 MPa, σ2 = −84.71 MPa, σ3 = 0

Von Mises

σvM = √(174.7² − 174.7·(−84.71) + (−84.71)²) = 229.1 MPa

Answer

σvM = 229.1 MPa
7.Axial stress & strain

Answer: δ = 3.265 mm

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·20²/4 = 314.2 mm²
σ = P/A = 80·1000/314.2 = 254.6 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 254.6/117000 = 0.002176 = 2176 µε

Elongation

δ = ε·L = 0.002176·1500 mm = 3.265 mm

Answer

δ = 3.265 mm
8.Axial stress & strain

Answer: σ2 = 10.61 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 = 200000·706.9/300 = 471239 N/mm
k₂ = 120000·1257/400 = 376991 N/mm

Share of segment 2

F2 = 30000·376991/848230 = 13333 N

Stress

σ2 = 13333/1257 = 10.61 MPa (compression)

Answer

σ2 = 10.61 MPa
9.Column buckling

Answer: Pcr = 24.67 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 = 60·25³/12 = 78125 mm⁴

Effective length

KL = 1·2500 = 2500 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·78125/2500² = 24674 N = 24.67 kN

Answer

Pcr = 24.67 kN
10.Column buckling

Answer: σcr = 4.441 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·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 1·5000 = 5000 mm
KL/r = 5000/7.5 = 666.7

Critical stress

σcr = π²E/(KL/r)² = π²·200000/666.7² = 4.441 MPa

Answer

σcr = 4.441 MPa
11.Mechanical vibrations

Answer: ζ = 0.02975

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 = (1/4)·ln(15/7.1) = 0.187

Damping ratio

ζ = δ/√(4π² + δ²) = 0.187/√(39.48 + 0.03496) = 0.02975

Answer

ζ = 0.02975
12.Mechanical vibrations

Answer: fn = 3.148 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 = 60 + 30 = 90 kN/m
keq = 90·25/(90 + 25) = 19.57 kN/m

Natural frequency

ωn = √(19565/50) = 19.78 rad/s
fn = 19.78/(2π) = 3.148 Hz

Answer

fn = 3.148 Hz

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