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​ = 450 mm960 rpmT1​ = 1200 NT2​ = 600 NP = ?

A belt runs on an 80 mm driving pulley at 960 rpm. The tension is 1200 N on the tight side and 600 N on the slack side. Find the power P transmitted, in kW.

d1 = 80 mmn1 = 960 rpmT1 = 1200 NT2 = 600 N
P = kW
2.Gears & gear trainsHard
A = 12B = 15C = 24D = 50

A 3 kW motor at 1440 rpm drives gear A (12 teeth), which meshes with gear B (15 teeth). Gear C (24 teeth) on the same shaft as B drives gear D (50 teeth) on the output shaft. Each gear mesh has an efficiency of 97%. Find the output torque Tout.

P = 3 kWnin = 1440 rpmA = 12B = 15C = 24D = 50ηmesh = 97%
Tout = N·m
3.Torsion of shaftsMedium
d1​ = 50 mmd2​ = 30 mmT = 750 N·mABC0.5 m3 m

A stepped solid steel shaft is fixed at A. Segment AB has diameter 50 mm and length 0.5 m; segment BC has diameter 30 mm and length 3 m. A torque of 750 N·m acts at the free end C. Find the angle of twist φC of end C, in degrees.

T = 750 N·md1 = 50 mm, L1 = 0.5 md2 = 30 mm, L2 = 3 mG = 80 GPa
φC = °
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 15 kW at 1500 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.

P = 15 kWN = 1500 rpmτallow = 50 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 100σy​ = 20τxy​ = 40MPa

A point in a thin plate is in plane stress: σₓ = 100 MPa, σᵧ = 20 MPa and τₓᵧ = 40 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).

σx = 100 MPaσy = 20 MPaτxy = 40 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 40σy​ = 70τxy​ = 20MPa

A point in a thin plate is in plane stress: σₓ = 40 MPa, σᵧ = 70 MPa and τₓᵧ = 20 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 = 40 MPaσy = 70 MPaτxy = 20 MPaσY = 250 MPa
n =
7.Axial stress & strainMedium
Ø 20 mmCopper1 mP = 105 kN

Find the axial strain ε in the copper bar, in microstrain (µε = 10⁻⁶).

P = 105 kNd = 20 mmL = 1 mE = 117 GPa
ε = µε
8.Axial stress & strainHard
20 kN12300 mm300 mmØ25Ø60σ2​ = ?

A steel segment (diameter 25 mm, length 300 mm, E = 200 GPa) and a segment made of aluminium (diameter 60 mm, length 300 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: aluminium, Ø60 mm, 300 mm, E = 70 GPaP = 20 kN
σ2 = MPa
9.Column bucklingMedium
PL = 2 m30 × 50 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 = 2 mE = 70 GPa (aluminium)b × h = 30 × 50 mm
Pcr = kN
10.Column bucklingHard
P = 125 kNL = 3 md = ?cross-section

A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 125 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 125 kNn = 2L = 3 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 12 kN/mζ = 0.35m = 80 kg

An 80 kg mass on a spring with a stiffness of 12 kN/m has a viscous damper giving a damping ratio of 0.35. Find the damped natural frequency fd of its free vibration in Hz.

m = 80 kgk = 12 kN/mζ = 0.35
fd = Hz
12.Mechanical vibrationsHard
k1​ = 40k2​ = 10m = 10 kg

A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 10 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 10 kgk1 = 40 kN/mk2 = 10 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: P = 2.413 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 = 4.021 m/s

Power

P = (1200 − 600)·4.021 = 2413 W = 2.413 kW

Answer

P = 2.413 kW
2.Gears & gear trains

Answer: Tout = 48.75 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.

Output speed

nout = 1440·(12/15)·(24/50) = 553 rpm
ωout = 2π·553/60 = 57.91 rad/s

Output power and torque

Pout = 3000·0.97² = 2823 W
Tout = Pout/ωout = 2823/57.91 = 48.75 N·m

Answer

Tout = 48.75 N·m
3.Torsion of shafts

Answer: φC = 20.7 °

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 moments

J1 = π·50⁴/32 = 613592 mm⁴
J2 = π·30⁴/32 = 79522 mm⁴

Twist of each segment

T = 750 N·m = 750000 N·mm
G = 80 GPa = 80000 MPa
φAB = 750000·500/(80000·613592) = 0.007639 rad
φBC = 750000·3000/(80000·79522) = 0.3537 rad

Total twist

φC = 0.007639 + 0.3537 = 0.3613 rad = 20.7°

Answer

φC = 20.7°
4.Torsion of shafts

Answer: d = 21.35 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·1500/60 = 157.1 rad/s
T = P/ω = 15000/157.1 = 95.49 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·95493/(π·50)) = 21.35 mm

Answer

d = 21.35 mm

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

5.Mohr’s circle & stress transformation

Answer: τabs = 58.28 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. In three dimensions the absolute maximum shear stress is (σmax − σmin)/2 over all three principal stresses; in plane stress the third one is zero.

στxyCσ1​σ2​

In-plane principal stresses

C = 60 MPa, R = √(40² + 40²) = 56.57 MPa
σ1 = 116.6 MPa, σ2 = 3.431 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(113.1, 116.6, 3.431)/2 = 58.28 MPa

Answer

τabs = 58.28 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

6.Mohr’s circle & stress transformation

Answer: n = 3.125

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 = 55 MPa, R = 25 MPa
σ1 = 80 MPa, σ2 = 30 MPa, σ3 = 0

Tresca

σmax − σmin = 80 MPa
n = 250/80 = 3.125

Answer

n = 3.125

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

7.Axial stress & strain

Answer: ε = 2857 µε

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 = 105·1000/314.2 = 334.2 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 334.2/117000 = 0.002857 = 2857 µε

Answer

ε = 2857 µε
8.Axial stress & strain

Answer: σ2 = 4.728 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 = 200000·490.9/300 = 327249 N/mm
k₂ = 70000·2827/300 = 659734 N/mm

Share of segment 2

F2 = 20000·659734/986984 = 13369 N

Stress

σ2 = 13369/2827 = 4.728 MPa (compression)

Answer

σ2 = 4.728 MPa
9.Column buckling

Answer: Pcr = 39.65 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 = 50·30³/12 = 112500 mm⁴

Effective length

KL = 0.7·2000 = 1400 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·112500/1400² = 39655 N = 39.65 kN

Answer

Pcr = 39.65 kN
10.Column buckling

Answer: d = 58.08 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 = 2·125000 = 250000 N
KL = 0.7·3000 = 2100 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·250000·2100²/(π³·200000))^(1/4) = 58.08 mm

Answer

d = 58.08 mm

Check: the slenderness KL/r is 144.6, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: fd = 1.826 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.

Undamped natural frequency

ωn = √(12000/80) = 12.25 rad/s
fn = 12.25/(2π) = 1.949 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 1.949·√(1 − 0.35²) = 1.826 Hz

Answer

fd = 1.826 Hz
12.Mechanical vibrations

Answer: fn = 11.25 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

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

keq = 40 + 10 = 50 kN/m

Natural frequency

ωn = √(50000/10) = 70.71 rad/s
fn = 70.71/(2π) = 11.25 Hz

Answer

fn = 11.25 Hz

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