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​ = 45

A 7.5 kW motor running at 1450 rpm drives a pinion with 18 teeth, which meshes with a gear with 45 teeth. The gear pair has an efficiency of 95%. Find the torque T1 on the motor shaft.

P = 7.5 kWn1 = 1450 rpmN1 = 18N2 = 45η = 95%
T1 = N·m
2.Gears & gear trainsHard
d1​ = 120 mmd2​ = 300 mm720 rpmC = 600 mmT1​ = 600 NT2​P = ?

An open belt connects a 120 mm driving pulley at 720 rpm to a 300 mm pulley; the centres are 600 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 600 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 120 mmn1 = 720 rpmd2 = 300 mmC = 600 mmμ = 0.3T1 = 600 N
P = kW
3.Torsion of shaftsMedium
do​ = 30, di​ = 21 mmT = 2000 N·mL = 0.5 m

A hollow brass shaft (outer diameter 30 mm, inner diameter 21 mm, length 0.5 m) is fixed at one end and twisted by a torque of 2000 N·m at the other. Find the angle of twist φ in degrees.

T = 2000 N·mdo = 30 mmdi = 21 mmL = 0.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 50 kW at 1450 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.

P = 50 kWN = 1450 rpmτallow = 80 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 110σy​ = −70τxy​ = 10MPa

A point in a loaded part has σₓ = 110 MPa, σᵧ = −70 MPa and τₓᵧ = 10 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σx = 110 MPaσy = −70 MPaτxy = 10 MPa
θp = °
6.Mohr’s circle & stress transformationHard
σx​ = 10σy​ = 90τxy​ = 30MPa

A point in a thin plate is in plane stress: σₓ = 10 MPa, σᵧ = 90 MPa and τₓᵧ = 30 MPa (σz = 0). The material has a yield strength of 420 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σx = 10 MPaσy = 90 MPaτxy = 30 MPaσY = 420 MPa
n =
7.Axial stress & strainMedium
Ø 16 mmAluminium1 mP = 90 kN

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

P = 90 kNd = 16 mmL = 1 mE = 70 GPa
δ = mm
8.Axial stress & strainHard
30 kN12500 mm200 mmØ40Ø40σ2​ = ?

A steel segment (diameter 40 mm, length 500 mm, E = 200 GPa) and a segment made of brass (diameter 40 mm, length 200 mm, E = 100 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, Ø40 mm, 500 mm, E = 200 GPasegment 2: brass, Ø40 mm, 200 mm, E = 100 GPaP = 30 kN
σ2 = MPa
9.Column bucklingMedium
PL = 1.5 m30 × 80 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 = 1.5 mE = 70 GPa (aluminium)b × h = 30 × 80 mm
Pcr = kN
10.Column bucklingHard
PL = 4 md = 60 mmcross-section

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

L = 4 mE = 70 GPa (aluminium)d = 60 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 15 kN/mζ = 0.4m = 80 kg

An 80 kg mass on a spring with a stiffness of 15 kN/m has a viscous damper giving a damping ratio of 0.4. Find the damped natural frequency fd of its free vibration in Hz.

m = 80 kgk = 15 kN/mζ = 0.4
fd = Hz
12.Mechanical vibrationsHard
k1​ = 80k2​ = 100m = 5 kg

A 5 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 80 kN/m and the lower one k₂ = 100 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 5 kgk1 = 80 kN/mk2 = 100 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: T1 = 49.39 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 = 7500/151.8 = 49.39 N·m

Answer

T1 = 49.39 N·m
2.Gears & gear trains

Answer: P = 1.557 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 = 2.84 rad

Slack-side tension

eμθ = e0.8521 = 2.345, T₂ = 600/2.345 = 255.9 N

Power

P = (600 − 255.9)·4.524 = 1.557 kW

Answer

P = 1.557 kW
3.Torsion of shafts

Answer: φ = 24.31 °

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(30⁴ − 21⁴)/32 = 60428 mm⁴

Angle of twist

T = 2000 N·m = 2 × 10⁶ N·mm
L = 0.5 m = 500 mm
G = 39 GPa = 39000 MPa
φ = TL/(GJ) = 2 × 10⁶·500/(39000·60428) = 0.4243 rad
φ = 0.4243·180/π = 24.31°

Answer

φ = 24.31°
4.Torsion of shafts

Answer: d = 27.57 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·1450/60 = 151.8 rad/s
T = P/ω = 50000/151.8 = 329.3 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·329286/(π·80)) = 27.57 mm

Answer

d = 27.57 mm

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

5.Mohr’s circle & stress transformation

Answer: θp = 3.17 °

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.

στxyCσ1​σ2​

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θp = atan2(2·10, 110 − −70) = 6.34°

Angle on the element

θp = 3.17°

Answer

θp = 3.17°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: n = 4.2

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 = 50 MPa, R = 50 MPa
σ1 = 100 MPa, σ2 = 0 MPa, σ3 = 0

Tresca

σmax − σmin = 100 MPa
n = 420/100 = 4.2

Answer

n = 4.2

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

7.Axial stress & strain

Answer: δ = 6.395 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·16²/4 = 201.1 mm²
σ = P/A = 90·1000/201.1 = 447.6 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 447.6/70000 = 0.006395 = 6395 µε

Elongation

δ = ε·L = 0.006395·1000 mm = 6.395 mm

Answer

δ = 6.395 mm
8.Axial stress & strain

Answer: σ2 = 13.26 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 = 200000·1257/500 = 502655 N/mm
k₂ = 100000·1257/200 = 628319 N/mm

Share of segment 2

F2 = 30000·628319/1.131 × 10⁶ = 16667 N

Stress

σ2 = 16667/1257 = 13.26 MPa (compression)

Answer

σ2 = 13.26 MPa
9.Column buckling

Answer: Pcr = 112.8 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 = 80·30³/12 = 180000 mm⁴

Effective length

KL = 0.7·1500 = 1050 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·180000/1050² = 112795 N = 112.8 kN

Answer

Pcr = 112.8 kN
10.Column buckling

Answer: σcr = 19.83 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·60⁴/64 = 636173 mm⁴

Radius of gyration and slenderness

A = 2827 mm²
r = √(I/A) = √(636173/2827) = 15 mm
KL = 0.7·4000 = 2800 mm
KL/r = 2800/15 = 186.7

Critical stress

σcr = π²E/(KL/r)² = π²·70000/186.7² = 19.83 MPa

Answer

σcr = 19.83 MPa
11.Mechanical vibrations

Answer: fd = 1.997 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 = √(15000/80) = 13.69 rad/s
fn = 13.69/(2π) = 2.179 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 2.179·√(1 − 0.4²) = 1.997 Hz

Answer

fd = 1.997 Hz
12.Mechanical vibrations

Answer: fn = 15.01 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 = 1/80 + 1/100 → keq = 80·100/(80 + 100) = 44.44 kN/m

Natural frequency

ωn = √(44444/5) = 94.28 rad/s
fn = 94.28/(2π) = 15.01 Hz

Answer

fn = 15.01 Hz

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