Machine Design: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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​ = 150 mmd2​ = 400 mm720 rpmn2​ = ?

A 150 mm pulley at 720 rpm drives a 400 mm pulley by a belt with 4 % slip. Find the speed n2 of the driven pulley.

n1 = 720 rpmd1 = 150 mmd2 = 400 mms = 4 %
n2 = rpm
2.Gears & gear trainsHard
A = 72B = 80C = 15D = 18

In a compound gear train, gear A (72 teeth) on the input shaft turning at 3000 rpm drives gear B (80 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D (18 teeth) on the output shaft. Find the output speed nout.

nin = 3000 rpmA = 72B = 80C = 15D = 18
nout = rpm
3.Torsion of shaftsMedium
d1​ = 75 mmd2​ = 30 mmT = 250 N·mABC1 m1.2 m

A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 75 mm and length 1 m; segment BC has diameter 30 mm and length 1.2 m. A torque of 250 N·m acts at the free end C. Find the angle of twist φC of end C, in degrees.

T = 250 N·md1 = 75 mm, L1 = 1 md2 = 30 mm, L2 = 1.2 mG = 26 GPa
φC = °
4.Torsion of shaftsHard
Tdo​ = 50 mm, di​ = 38 mm

A hollow shaft transmits 30 kW at 1450 rpm. Find the maximum shear stress τmax in the shaft.

P = 30 kWN = 1450 rpmdo = 50 mmdi = 38 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = 30σy​ = −50MPaτxy​ = ?

A point has σₓ = 30 MPa and σᵧ = −50 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 80 MPa?

σx = 30 MPaσy = −50 MPaσ1 = 80 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 100σy​ = −10τxy​ = 30MPa

A point in a thin plate is in plane stress: σₓ = 100 MPa, σᵧ = −10 MPa and τₓᵧ = 30 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 = 100 MPaσy = −10 MPaτxy = 30 MPaσY = 250 MPa
n =
7.Axial stress & strainMedium
Ø 25 mmSteel2 mP = 30 kN

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

P = 30 kNd = 25 mmL = 2 mE = 200 GPa
ε = µε
8.Axial stress & strainHard
Ø 35 mmBrass1.5 mØ 20 mmBrass2 mP = 85 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 85 kNsegment 1: Ø35 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø20 mm, 2 m, brass (E = 100 GPa)
δ = mm
9.Column bucklingMedium
PL = 2.5 md = 30 mmcross-section

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

L = 2.5 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
10.Column bucklingHard
PL = 2 md = 30 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 = 2 mE = 70 GPa (aluminium)d = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx30 mm4 cycles4 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 30 mm is followed 4 cycles later by a peak of 4 mm. Find the damping ratio ζ.

x0 = 30 mmx4 = 4 mmn = 4
ζ =
12.Mechanical vibrationsHard
k1​ = 30k2​ = 20m = 100 kg

A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 30 kN/m and k₂ = 20 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 = 100 kgk1 = 30 kN/mk2 = 20 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: n2 = 259.2 rpm

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.

Speed without slip

720·150/400 = 270 rpm

With slip

n2 = 270·0.96 = 259.2 rpm

Answer

n2 = 259.2 rpm
2.Gears & gear trains

Answer: nout = 2250 rpm

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.

First stage

nB = 3000·72/80 = 2700 rpm = nC

Second stage

nout = 2700·15/18 = 2250 rpm

Answer

nout = 2250 rpm

total ratio = 1.333 : 1

3.Torsion of shafts

Answer: φC = 8.491 °

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·75⁴/32 = 3.106 × 10⁶ mm⁴
J2 = π·30⁴/32 = 79522 mm⁴

Twist of each segment

T = 250 N·m = 250000 N·mm
G = 26 GPa = 26000 MPa
φAB = 250000·1000/(26000·3.106 × 10⁶) = 0.003095 rad
φBC = 250000·1200/(26000·79522) = 0.1451 rad

Total twist

φC = 0.003095 + 0.1451 = 0.1482 rad = 8.491°

Answer

φC = 8.491°
4.Torsion of shafts

Answer: τmax = 12.08 MPa

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 = 30000/151.8 = 197.6 N·m

Polar second moment of area

J = π(50⁴ − 38⁴)/32 = 408885 mm⁴

Maximum shear stress

τmax = T·(do/2)/J = 197572·25/408885 = 12.08 MPa

Answer

τmax = 12.08 MPa
5.Mohr’s circle & stress transformation

Answer: τxy = 80.62 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.

Required radius

C = (30 + (−50))/2 = −10 MPa
R = σ1 − C = 80 − (−10) = 90 MPa
στxyCσ1​σ2​

Shear stress

τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(90² − 40²) = 80.62 MPa

Answer

τxy = 80.62 MPa
6.Mohr’s circle & stress transformation

Answer: n = 1.995

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 = 45 MPa, R = 62.65 MPa
σ1 = 107.6 MPa, σ2 = −17.65 MPa, σ3 = 0

Tresca

σmax − σmin = 125.3 MPa
n = 250/125.3 = 1.995

Answer

n = 1.995

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

7.Axial stress & strain

Answer: ε = 305.6 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·25²/4 = 490.9 mm²
σ = P/A = 30·1000/490.9 = 61.12 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 61.12/200000 = 0.0003056 = 305.6 µε

Answer

ε = 305.6 µε
8.Axial stress & strain

Answer: δ = 6.736 mm

Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.

Same force in both segments

The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.

A1 = π·35²/4 = 962.1 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 88.35 MPa
σ2 = 270.6 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 88.35·1500/100000 = 1.325 mm
δ2 = σ2L2/E2 = 270.6·2000/100000 = 5.411 mm
δ = δ1 + δ2 = 6.736 mm

Answer

δ = 6.736 mm
9.Column buckling

Answer: Pcr = 3.139 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 = 2·2500 = 5000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·39761/5000² = 3139 N = 3.139 kN

Answer

Pcr = 3.139 kN
10.Column buckling

Answer: σcr = 38.86 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·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 0.5·2000 = 1000 mm
KL/r = 1000/7.5 = 133.3

Critical stress

σcr = π²E/(KL/r)² = π²·70000/133.3² = 38.86 MPa

Answer

σcr = 38.86 MPa
11.Mechanical vibrations

Answer: ζ = 0.07991

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(30/4) = 0.5037

Damping ratio

ζ = δ/√(4π² + δ²) = 0.5037/√(39.48 + 0.2537) = 0.07991

Answer

ζ = 0.07991
12.Mechanical vibrations

Answer: fn = 3.559 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 = 30 + 20 = 50 kN/m

Natural frequency

ωn = √(50000/100) = 22.36 rad/s
fn = 22.36/(2π) = 3.559 Hz

Answer

fn = 3.559 Hz

ExamProblems · examproblems.com/courses/machine-design