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​ = 18N2​ = 80

A 2.2 kW motor running at 1450 rpm drives a pinion with 18 teeth, which meshes with a gear with 80 teeth. The gear pair has an efficiency of 98%. Find the output torque T2 on the gear shaft.

P = 2.2 kWn1 = 1450 rpmN1 = 18N2 = 80η = 98%
T2 = N·m
2.Gears & gear trainsHard
d1​ = 160 mmd2​ = 500 mm2880 rpmC = 800 mmθ = ?

An open belt connects a 160 mm pulley and a 500 mm pulley whose centres are 800 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 160 mmd2 = 500 mmC = 800 mm
θ = °
3.Torsion of shaftsMedium
T = 100 N·md = 20 mmL = 1.5 mφ = ?

A solid brass 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 = 100 N·md = 20 mmL = 1.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
d = ?20 kW, 750 rpm

A solid steel shaft transmits 20 kW at 750 rpm. The shear stress must not exceed 75 MPa and the twist must not exceed 0.75° per m of length. Find the smallest diameter d that satisfies both limits.

P = 20 kWN = 750 rpmτallow = 75 MPaθallow = 0.75°/mG = 80 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 90τxy​ = −40MPa

A point has σₓ = 90 MPa, σᵧ = 0 MPa and τₓᵧ = −40 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 = 90 MPaσy = 0 MPaτxy = −40 MPa
θs = °
6.Mohr’s circle & stress transformationHard
σx​ = 140σy​ = −20τxy​ = −50MPa

A point in a thin plate is in plane stress: σₓ = 140 MPa, σᵧ = −20 MPa and τₓᵧ = −50 MPa (σz = 0). The material has a yield strength of 355 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σx = 140 MPaσy = −20 MPaτxy = −50 MPaσY = 355 MPa
n =
7.Axial stress & strainMedium
300 kNØ20120 GPaØ120/60600 mmδ = ?

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 60 mm, E = 120 GPa). Both are 600 mm long and are compressed together between rigid plates by 300 kN. Find the shortening δ.

P = 300 kNrod: Ø20 mm, E = 200 GPatube: Ø120/Ø60 mm, E = 120 GPaL = 600 mm
δ = mm
8.Axial stress & strainHard
P = ?Ø40120 GPaØ100/60400 mm

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of copper (outer diameter 100 mm, inner diameter 60 mm, E = 120 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø40 mm, E = 200 GPa, σallow = 160 MPatube: Ø100/Ø60 mm, E = 120 GPa, σallow = 60 MPa
Pmax = kN
9.Column bucklingMedium
PL = 4 md = 30 mmcross-section

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

L = 4 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
10.Column bucklingHard
PL = 5 md = 50 mmcross-section

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

L = 5 mE = 200 GPa (steel)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx50 mm1 cycle19.3 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed one cycle later by a peak of 19.3 mm. Find the damping ratio ζ.

x0 = 50 mmx1 = 19.3 mmn = 1
ζ =
12.Mechanical vibrationsHard
k1​ = 60k2​ = 5m = 5 kg

A 5 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 60 kN/m and k₂ = 5 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 = 5 kgk1 = 60 kN/mk2 = 5 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: T2 = 63.11 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·1450/60 = 151.8 rad/s
T1 = P/ω₁ = 2200/151.8 = 14.49 N·m

Output torque

T2 = 14.49·80/18·0.98 = 63.11 N·m

Answer

T2 = 63.11 N·m
2.Gears & gear trains

Answer: θ = 155.5 °

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 angle

asin(0.2125) = 12.27°

Angle of wrap

θ = 180 − 2·12.27 = 155.5°

Answer

θ = 155.5°
3.Torsion of shafts

Answer: φ = 14.03 °

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) = 100·1000·1500/(39000·15708) = 0.2449 rad
φ = 0.2449·180/π = 14.03°

Answer

φ = 14.03°
4.Torsion of shafts

Answer: d = 39.67 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·750/60 = 78.54 rad/s
T = P/ω = 20000/78.54 = 254.6 N·m

Diameter for strength

T = 254.6 N·m = 254648 N·mm
dτ = ∛(16·254648/(π·75)) = 25.86 mm

Diameter for stiffness

θ = 0.75·π/180/1000 = 1.309 × 10⁻⁵ rad/mm
G = 80 GPa = 80000 MPa
dφ = (32·254648/(π·80000·1.309 × 10⁻⁵))1/4 = 39.67 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(25.86, 39.67) = 39.67 mm

Answer

d = 39.67 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: θs = −65.82 °

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 = −40/60.21 = −0.6644, sin 2θs = −45/60.21 = −0.7474
2θs = atan2(−45, −40) = −131.6°

Angle on the element

θs = −65.82°
check: θs − θp = −65.82 − (−20.82) = −45°

Answer

θs = −65.82°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: n = 1.881

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 = 60 MPa, R = 94.34 MPa
σ1 = 154.3 MPa, σ2 = −34.34 MPa, σ3 = 0

Tresca

σmax − σmin = 188.7 MPa
n = 355/188.7 = 1.881

Answer

n = 1.881

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

7.Axial stress & strain

Answer: δ = 0.1666 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·20²/4 = 314.2 mm²
At = π(120² − 60²)/4 = 8482 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 120·10³·8482 = 1.081 × 10⁹ N

Shortening

δ = 300000·600/1.081 × 10⁹ = 0.1666 mm

Answer

δ = 0.1666 mm
8.Axial stress & strain

Answer: Pmax = 427.3 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·40²/4 = 1257 mm²
At = π(100² − 60²)/4 = 5027 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 120·10³·5027 = 8.545 × 10⁸ N

Governing strain

ε = min(0.0008, 0.0005) = 0.0005 (copper)

Largest load

Pmax = 0.0005·8.545 × 10⁸ = 427257 N = 427.3 kN

Answer

Pmax = 427.3 kN
9.Column buckling

Answer: Pcr = 10.01 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 0.7·4000 = 2800 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·39761/2800² = 10011 N = 10.01 kN

Answer

Pcr = 10.01 kN
10.Column buckling

Answer: σcr = 3.084 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·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 2·5000 = 10000 mm
KL/r = 10000/12.5 = 800

Critical stress

σcr = π²E/(KL/r)² = π²·200000/800² = 3.084 MPa

Answer

σcr = 3.084 MPa
11.Mechanical vibrations

Answer: ζ = 0.1498

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 = (1/1)·ln(50/19.3) = 0.9519

Damping ratio

ζ = δ/√(4π² + δ²) = 0.9519/√(39.48 + 0.9061) = 0.1498

Answer

ζ = 0.1498
12.Mechanical vibrations

Answer: fn = 18.15 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 = 60 + 5 = 65 kN/m

Natural frequency

ωn = √(65000/5) = 114 rad/s
fn = 114/(2π) = 18.15 Hz

Answer

fn = 18.15 Hz

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