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​ = 48N2​ = 72

An 11 kW motor running at 960 rpm drives a pinion with 48 teeth, which meshes with a gear with 72 teeth. The gear pair has an efficiency of 95%. Find the torque T1 on the motor shaft.

P = 11 kWn1 = 960 rpmN1 = 48N2 = 72η = 95%
T1 = N·m
2.Gears & gear trainsHard
d1​ = 150 mmd2​ = 250 mm1440 rpmC = 1200 mmT1​ = 2000 NT2​P = ?

An open belt connects a 150 mm driving pulley at 1440 rpm to a 250 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 2000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 150 mmn1 = 1440 rpmd2 = 250 mmC = 1200 mmμ = 0.25T1 = 2000 N
P = kW
3.Torsion of shaftsMedium
T = 500 N·md = 25 mmL = 1.5 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 = 500 N·md = 25 mmL = 1.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 20 kW at 3000 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 20 kWN = 3000 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 110σy​ = 30τxy​ = 10MPa

A point in a thin plate is in plane stress: σₓ = 110 MPa, σᵧ = 30 MPa and τₓᵧ = 10 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).

σx = 110 MPaσy = 30 MPaτxy = 10 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
σx​ = −10σy​ = −60τxy​ = 20MPa

A point in a thin plate is in plane stress: σₓ = −10 MPa, σᵧ = −60 MPa and τₓᵧ = 20 MPa (σz = 0). The material has a yield strength of 420 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σx = −10 MPaσy = −60 MPaτxy = 20 MPaσY = 420 MPa
n =
7.Axial stress & strainMedium
Ø 25 mmAluminium3 mP = 30 kN

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

P = 30 kNd = 25 mmL = 3 mE = 70 GPa
δ = mm
8.Axial stress & strainHard
P = ?Ø30120 GPaØ120/50400 mm

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 50 mm, E = 120 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø30 mm, E = 200 GPa, σallow = 150 MPatube: Ø120/Ø50 mm, E = 120 GPa, σallow = 50 MPa
Pmax = kN
9.Column bucklingMedium
PL = 2 md = 80 mmcross-section

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

L = 2 mE = 70 GPa (aluminium)d = 80 mm
Pcr = kN
10.Column bucklingHard
P = 90 kNL = 1.5 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 90 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 90 kNn = 2L = 1.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
tx40 mm4 cycles14.6 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed 4 cycles later by a peak of 14.6 mm. Find the damping ratio ζ.

x0 = 40 mmx4 = 14.6 mmn = 4
ζ =
12.Mechanical vibrationsHard
k1​ = 12k2​ = 60m = 25 kg

A 25 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 12 kN/m and the lower one k₂ = 60 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 25 kgk1 = 12 kN/mk2 = 60 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: T1 = 109.4 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.

Motor torque

ω₁ = 2π·960/60 = 100.5 rad/s
T1 = P/ω₁ = 11000/100.5 = 109.4 N·m

Answer

T1 = 109.4 N·m
2.Gears & gear trains

Answer: P = 12.09 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.

Angle of wrap

θ = 3.058 rad

Slack-side tension

eμθ = e0.7646 = 2.148, T₂ = 2000/2.148 = 931.1 N

Power

P = (2000 − 931.1)·11.31 = 12.09 kW

Answer

P = 12.09 kW
3.Torsion of shafts

Answer: φ = 14.01 °

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·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 500·1000·1500/(80000·38350) = 0.2445 rad
φ = 0.2445·180/π = 14.01°

Answer

φ = 14.01°
4.Torsion of shafts

Answer: d = 16.29 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·3000/60 = 314.2 rad/s
T = P/ω = 20000/314.2 = 63.66 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·63662/(π·75)) = 16.29 mm

Answer

d = 16.29 mm

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

5.Mohr’s circle & stress transformation

Answer: τabs = 55.62 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 = 70 MPa, R = √(40² + 10²) = 41.23 MPa
σ1 = 111.2 MPa, σ2 = 28.77 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(82.46, 111.2, 28.77)/2 = 55.62 MPa

Answer

τabs = 55.62 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 = 6.405

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 = 32.02 MPa
σ1 = −2.984 MPa, σ2 = −67.02 MPa, σ3 = 0

Von Mises

σvM = √((−2.984)² − (−2.984)·(−67.02) + (−67.02)²) = 65.57 MPa
n = 420/65.57 = 6.405

Answer

n = 6.405

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

7.Axial stress & strain

Answer: δ = 2.619 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·25²/4 = 490.9 mm²
σ = P/A = 30·1000/490.9 = 61.12 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 61.12/70000 = 0.0008731 = 873.1 µε

Elongation

δ = ε·L = 0.0008731·3000 mm = 2.619 mm

Answer

δ = 2.619 mm
8.Axial stress & strain

Answer: Pmax = 526.2 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·30²/4 = 706.9 mm²
At = π(120² − 50²)/4 = 9346 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 120·10³·9346 = 1.263 × 10⁹ N

Governing strain

ε = min(0.00075, 0.0004167) = 0.0004167 (copper)

Largest load

Pmax = 0.0004167·1.263 × 10⁹ = 526217 N = 526.2 kN

Answer

Pmax = 526.2 kN
9.Column buckling

Answer: Pcr = 347.3 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 = 1·2000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/2000² = 347270 N = 347.3 kN

Answer

Pcr = 347.3 kN
10.Column buckling

Answer: d = 37.83 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 = 2·90000 = 180000 N
KL = 0.7·1500 = 1050 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·180000·1050²/(π³·200000))^(1/4) = 37.83 mm

Answer

d = 37.83 mm

Check: the slenderness KL/r is 111, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.04007

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(40/14.6) = 0.252

Damping ratio

ζ = δ/√(4π² + δ²) = 0.252/√(39.48 + 0.06349) = 0.04007

Answer

ζ = 0.04007
12.Mechanical vibrations

Answer: fn = 3.183 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.

1/keq = 1/12 + 1/60 → keq = 12·60/(12 + 60) = 10 kN/m

Natural frequency

ωn = √(10000/25) = 20 rad/s
fn = 20/(2π) = 3.183 Hz

Answer

fn = 3.183 Hz

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