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​ = 125 mmd2​ = 200 mm960 rpmT1​ = 1500 NT2​ = 450 NP = ?

A belt runs on a 125 mm driving pulley at 960 rpm. The tension is 1500 N on the tight side and 450 N on the slack side. Find the power P transmitted, in kW.

d1 = 125 mmn1 = 960 rpmT1 = 1500 NT2 = 450 N
P = kW
2.Gears & gear trainsHard
d1​ = 80 mmd2​ = 250 mm720 rpmC = 1200 mmT1​ = 800 NT2​P = ?

An open belt connects an 80 mm driving pulley at 720 rpm to a 250 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 800 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 80 mmn1 = 720 rpmd2 = 250 mmC = 1200 mmμ = 0.3T1 = 800 N
P = kW
3.Torsion of shaftsMedium
d = 20 mmT = 100 N·m200 mmφ = 1.87°

In a torsion test, a solid specimen of diameter 20 mm is twisted by a torque of 100 N·m. The two ends of a gauge length of 200 mm rotate 1.87° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 100 N·md = 20 mmL = 200 mmφ = 1.87°
G = GPa
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​ = 20σy​ = 10τxy​ = 20MPa

A point in a loaded part has σₓ = 20 MPa, σᵧ = 10 MPa and τₓᵧ = 20 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 20 MPaσy = 10 MPaτxy = 20 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​ = −50σy​ = −60τxy​ = −10MPa

A point in a loaded part has σₓ = −50 MPa, σᵧ = −60 MPa and τₓᵧ = −10 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 15° counter-clockwise from the x-axis.

σx = −50 MPaσy = −60 MPaτxy = −10 MPaθ = 15°
σθ = MPa
7.Axial stress & strainMedium
300 kNØ40110 GPaØ120/60400 mmδ = ?

A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 120 mm, inner diameter 60 mm, E = 110 GPa). Both are 400 mm long and are compressed together between rigid plates by 300 kN. Find the shortening δ.

P = 300 kNrod: Ø40 mm, E = 200 GPatube: Ø120/Ø60 mm, E = 110 GPaL = 400 mm
δ = mm
8.Axial stress & strainHard
30 kN12400 mm300 mmØ20Ø40σ2​ = ?

A steel segment (diameter 20 mm, length 400 mm, E = 200 GPa) and a segment made of bronze (diameter 40 mm, length 300 mm, E = 110 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, Ø20 mm, 400 mm, E = 200 GPasegment 2: bronze, Ø40 mm, 300 mm, E = 110 GPaP = 30 kN
σ2 = MPa
9.Column bucklingMedium
PL = 1.5 md = 30 mmcross-section

A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.

L = 1.5 mE = 70 GPa (aluminium)d = 30 mm
Pcr = kN
10.Column bucklingHard
PL = 3 md = 25 mmcross-section

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

L = 3 mE = 70 GPa (aluminium)d = 25 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx40 mm2 cycles5.9 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed 2 cycles later by a peak of 5.9 mm. Find the damping ratio ζ.

x0 = 40 mmx2 = 5.9 mmn = 2
ζ =
12.Mechanical vibrationsHard
k1​ = 5k2​ = 40m = 50 kg

A 50 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 5 kN/m and the lower one k₂ = 40 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 50 kgk1 = 5 kN/mk2 = 40 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: P = 6.597 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 = 6.283 m/s

Power

P = (1500 − 450)·6.283 = 6597 W = 6.597 kW

Answer

P = 6.597 kW
2.Gears & gear trains

Answer: P = 1.432 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 rad

Slack-side tension

eμθ = e0.8999 = 2.459, T₂ = 800/2.459 = 325.3 N

Power

P = (800 − 325.3)·3.016 = 1.432 kW

Answer

P = 1.432 kW
3.Torsion of shafts

Answer: G = 39.01 GPa

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 moment and angle

J = π·20⁴/32 = 15708 mm⁴
φ = 1.87·π/180 = 0.03264 rad

Shear modulus

T = 100 N·m = 100000 N·mm
G = TL/(Jφ) = 100000·200/(15708·0.03264) = 39011 MPa
G = 39.01 GPa

Answer

G = 39.01 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

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: τmax = 20.62 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.

στxyCτmax​

Radius of Mohr’s circle

τmax = R = √(((20 − 10)/2)² + 20²) = 20.62 MPa

Answer

τmax = 20.62 MPa
6.Mohr’s circle & stress transformation

Answer: σθ = −55.67 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.

Transformation equation

σθ = −55 + 5·0.866 + −10·0.5 = −55.67 MPa

Answer

σθ = −55.67 MPa
7.Axial stress & strain

Answer: δ = 0.1013 mm

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·40²/4 = 1257 mm²
At = π(120² − 60²)/4 = 8482 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 110·10³·8482 = 1.184 × 10⁹ N

Shortening

δ = 300000·400/1.184 × 10⁹ = 0.1013 mm

Answer

δ = 0.1013 mm
8.Axial stress & strain

Answer: σ2 = 17.8 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 = 200000·314.2/400 = 157080 N/mm
k₂ = 110000·1257/300 = 460767 N/mm

Share of segment 2

F2 = 30000·460767/617847 = 22373 N

Stress

σ2 = 22373/1257 = 17.8 MPa (compression)

Answer

σ2 = 17.8 MPa
9.Column buckling

Answer: Pcr = 3.052 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 2·1500 = 3000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·39761/3000² = 3052 N = 3.052 kN

Answer

Pcr = 3.052 kN
10.Column buckling

Answer: σcr = 11.99 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·25⁴/64 = 19175 mm⁴

Radius of gyration and slenderness

A = 490.9 mm²
r = √(I/A) = √(19175/490.9) = 6.25 mm
KL = 0.5·3000 = 1500 mm
KL/r = 1500/6.25 = 240

Critical stress

σcr = π²E/(KL/r)² = π²·70000/240² = 11.99 MPa

Answer

σcr = 11.99 MPa
11.Mechanical vibrations

Answer: ζ = 0.1506

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(40/5.9) = 0.957

Damping ratio

ζ = δ/√(4π² + δ²) = 0.957/√(39.48 + 0.9158) = 0.1506

Answer

ζ = 0.1506
12.Mechanical vibrations

Answer: fn = 1.501 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 = 1/5 + 1/40 → keq = 5·40/(5 + 40) = 4.444 kN/m

Natural frequency

ωn = √(4444/50) = 9.428 rad/s
fn = 9.428/(2π) = 1.501 Hz

Answer

fn = 1.501 Hz

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