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​ = 200 mm720 rpmT1​ = 600 NT2​ = 150 NP = ?

A belt runs on a 150 mm driving pulley at 720 rpm. The tension is 600 N on the tight side and 150 N on the slack side. Find the power P transmitted, in kW.

d1 = 150 mmn1 = 720 rpmT1 = 600 NT2 = 150 N
P = kW
2.Gears & gear trainsHard
A = 15B = 50C = 15D = ?

In a compound gear train, gear A (15 teeth) on the input shaft turns at 1450 rpm and drives gear B (50 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 362.5 rpm. How many teeth ND must gear D have?

nin = 1450 rpmA = 15B = 50C = 15nout = 362.5 rpm
ND =
3.Torsion of shaftsMedium
do​ = ?T = 250 N·m

A hollow shaft must carry a torque of 250 N·m. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter do.

T = 250 N·mdi/do = 0.6τallow = 75 MPa
do = mm
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 15 kW at 3000 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.

P = 15 kWN = 3000 rpmτallow = 50 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = −40σy​ = 0τxy​ = 30MPa

A point in a loaded part has σₓ = −40 MPa, σᵧ = 0 MPa and τₓᵧ = 30 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σx = −40 MPaσy = 0 MPaτxy = 30 MPa
θp = °
6.Mohr’s circle & stress transformationHard
25°D = 800 mm, t = 12 mm, p = 1 MPap

A thin-walled cylindrical tank of inner diameter 800 mm and wall thickness 12 mm holds gas at a gauge pressure of 1 MPa. It is made from plate welded along a helical seam at 25° to the axis of the cylinder. Find the normal stress σn acting across the weld seam.

D = 800 mmt = 12 mmp = 1 MPaα = 25°
σn = MPa
7.Axial stress & strainMedium
Ø 16 mmBrass2.5 mP = 65 kN

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

P = 65 kNd = 16 mmL = 2.5 mE = 100 GPa
ε = µε
8.Axial stress & strainHard
20 kN12600 mm500 mmØ20Ø30σ2​ = ?

A steel segment (diameter 20 mm, length 600 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 500 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø20 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 500 mm, E = 70 GPaP = 20 kN
σ2 = MPa
9.Column bucklingMedium
PL = 2.5 md = 40 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 = 2.5 mE = 200 GPa (steel)d = 40 mm
Pcr = kN
10.Column bucklingHard
PL = 2 md = 50 mmcross-section

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

L = 2 mE = 70 GPa (aluminium)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 200 kN/mζ = 0.5m = 50 kg

A 50 kg mass on a spring with a stiffness of 200 kN/m has a viscous damper giving a damping ratio of 0.5. Find the damped natural frequency fd of its free vibration in Hz.

m = 50 kgk = 200 kN/mζ = 0.5
fd = Hz
12.Mechanical vibrationsHard
k1​ = 25k2​ = 5m = 25 kg

A 25 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 25 kN/m and the lower one k₂ = 5 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.

m = 25 kgk1 = 25 kN/mk2 = 5 kN/m
keq = kN/m

3Solutions

1.Gears & gear trains

Answer: P = 2.545 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 = 5.655 m/s

Power

P = (600 − 150)·5.655 = 2545 W = 2.545 kW

Answer

P = 2.545 kW
2.Gears & gear trains

Answer: ND = 18

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.

Intermediate shaft speed

nB = nC = 1450·15/50 = 435 rpm

Teeth on gear D

ND = C·nC/nout = 15·435/362.5 = 18

Answer

ND = 18
3.Torsion of shafts

Answer: do = 26.92 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.

Outer diameter from the allowable stress

T = 250 N·m = 250000 N·mm
1 − k⁴ = 1 − 0.6⁴ = 0.8704
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·250000/(π·75·0.8704)) = 26.92 mm
di = 0.6·26.92 = 16.15 mm

Answer

do = 26.92 mm

A solid shaft would need d = 25.7 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: d = 16.94 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 = 15000/314.2 = 47.75 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·47746/(π·50)) = 16.94 mm

Answer

d = 16.94 mm

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

5.Mohr’s circle & stress transformation

Answer: θp = 61.85 °

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(2·30, −40 − 0) = 123.7°

Angle on the element

θp = 61.85°

Answer

θp = 61.85°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σn = 30.36 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. Thin-walled cylinder: hoop stress σh = pr/t and axial stress σa = pr/(2t); these are the principal stresses in the wall.

Wall stresses

r = 800/2 = 400 mm
σh = pr/t = 1·400/12 = 33.33 MPa
σa = pr/(2t) = 16.67 MPa

Transform to the seam

Take x along the axis (σa) and y around the hoop (σh). The seam’s normal makes 90° − α with the axis.

σn = 16.67·sin²25° + 33.33·cos²25° = 30.36 MPa

Answer

σn = 30.36 MPa
7.Axial stress & strain

Answer: ε = 3233 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·16²/4 = 201.1 mm²
σ = P/A = 65·1000/201.1 = 323.3 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 323.3/100000 = 0.003233 = 3233 µε

Answer

ε = 3233 µε
8.Axial stress & strain

Answer: σ2 = 13.75 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 = 200000·314.2/600 = 104720 N/mm
k₂ = 70000·706.9/500 = 98960 N/mm

Share of segment 2

F2 = 20000·98960/203680 = 9717 N

Stress

σ2 = 9717/706.9 = 13.75 MPa (compression)

Answer

σ2 = 13.75 MPa
9.Column buckling

Answer: Pcr = 81 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·40⁴/64 = 125664 mm⁴

Effective length

KL = 0.7·2500 = 1750 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·125664/1750² = 80996 N = 81 kN

Answer

Pcr = 81 kN
10.Column buckling

Answer: σcr = 6.747 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 = 2·2000 = 4000 mm
KL/r = 4000/12.5 = 320

Critical stress

σcr = π²E/(KL/r)² = π²·70000/320² = 6.747 MPa

Answer

σcr = 6.747 MPa
11.Mechanical vibrations

Answer: fd = 8.717 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 = √(200000/50) = 63.25 rad/s
fn = 63.25/(2π) = 10.07 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 10.07·√(1 − 0.5²) = 8.717 Hz

Answer

fd = 8.717 Hz
12.Mechanical vibrations

Answer: keq = 4.167 kN/m

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 = 1/25 + 1/5 → keq = 25·5/(25 + 5) = 4.167 kN/m

Answer

keq = 4.167 kN/m

ExamProblems · examproblems.com/courses/machine-design