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​ = 120 mmd2​ = 200 mm2880 rpmT1​ = 1200 NT2​ = 240 NP = ?

A belt runs on a 120 mm driving pulley at 2880 rpm. The tension is 1200 N on the tight side and 240 N on the slack side. Find the power P transmitted, in kW.

d1 = 120 mmn1 = 2880 rpmT1 = 1200 NT2 = 240 N
P = kW
2.Gears & gear trainsHard
d1​ = 100 mmd2​ = 500 mm1500 rpmC = 1000 mmθ = ?

An open belt connects a 100 mm pulley and a 500 mm pulley whose centres are 1000 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 100 mmd2 = 500 mmC = 1000 mm
θ = °
3.Torsion of shaftsMedium
d = ?T = 750 N·mL = 0.8 m

A solid steel shaft 0.8 m long carries a torque of 750 N·m. The angle of twist between its ends must not exceed 1.5°. Find the smallest diameter d that meets this limit.

T = 750 N·mL = 0.8 mG = 80 GPaφallow = 1.5°
d = mm
4.Torsion of shaftsHard
Tdo​ = 50 mm, di​ = 25 mm

A hollow shaft transmits 20 kW at 1000 rpm. Find the maximum shear stress τmax in the shaft.

P = 20 kWN = 1000 rpmdo = 50 mmdi = 25 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = 30σy​ = −10τxy​ = −10MPa

A point in a thin plate is in plane stress: σₓ = 30 MPa, σᵧ = −10 MPa and τₓᵧ = −10 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).

σx = 30 MPaσy = −10 MPaτxy = −10 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
σx​ = −10σy​ = −10τxy​ = 30MPa

A point in a thin plate is in plane stress: σₓ = −10 MPa, σᵧ = −10 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 von Mises (distortion energy) criterion.

σx = −10 MPaσy = −10 MPaτxy = 30 MPaσY = 420 MPa
n =
7.Axial stress & strainMedium
P = ?12600 mm500 mmØ40Ø40δ = 0.5 mm

A bar is made of a steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 40 mm, length 500 mm, E = 110 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.

segment 1: steel, Ø40 mm, 600 mm, E = 200 GPasegment 2: bronze, Ø40 mm, 500 mm, E = 110 GPaδ = 0.5 mm
P = kN
8.Axial stress & strainHard
80 kN12600 mm400 mmØ40Ø30σ2​ = ?

A steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) and a segment made of bronze (diameter 30 mm, length 400 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 80 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø40 mm, 600 mm, E = 200 GPasegment 2: bronze, Ø30 mm, 400 mm, E = 110 GPaP = 80 kN
σ2 = MPa
9.Column bucklingMedium
PL = 4 md = 80 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 = 4 mE = 70 GPa (aluminium)d = 80 mm
Pcr = kN
10.Column bucklingHard
P = 95 kNL = 1.5 md = ?cross-section

A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 95 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 95 kNn = 3L = 1.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 80 kN/mζ = 0.45m = 150 kg

A 150 kg mass on a spring with a stiffness of 80 kN/m has a viscous damper giving a damping ratio of 0.45. Find the damped natural frequency fd of its free vibration in Hz.

m = 150 kgk = 80 kN/mζ = 0.45
fd = Hz
12.Mechanical vibrationsHard
k1​ = 50k2​ = 10m = 50 kg

A 50 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 50 kN/m and k₂ = 10 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.

m = 50 kgk1 = 50 kN/mk2 = 10 kN/m
keq = kN/m

3Solutions

1.Gears & gear trains

Answer: P = 17.37 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 = 18.1 m/s

Power

P = (1200 − 240)·18.1 = 17372 W = 17.37 kW

Answer

P = 17.37 kW
2.Gears & gear trains

Answer: θ = 156.9 °

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(0.2) = 11.54°

Angle of wrap

θ = 180 − 2·11.54 = 156.9°

Answer

θ = 156.9°
3.Torsion of shafts

Answer: d = 41.33 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.

Required polar moment

T = 750 N·m = 750000 N·mm
L = 0.8 m = 800 mm
G = 80 GPa = 80000 MPa
φ = 1.5·π/180 = 0.02618 rad
J = TL/(Gφ) = 750000·800/(80000·0.02618) = 286479 mm⁴

Diameter

d = (32J/π)1/4 = 41.33 mm

Answer

d = 41.33 mm

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

4.Torsion of shafts

Answer: τmax = 8.3 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·1000/60 = 104.7 rad/s
T = P/ω = 20000/104.7 = 191 N·m

Polar second moment of area

J = π(50⁴ − 25⁴)/32 = 575243 mm⁴

Maximum shear stress

τmax = T·(do/2)/J = 190986·25/575243 = 8.3 MPa

Answer

τmax = 8.3 MPa
5.Mohr’s circle & stress transformation

Answer: τabs = 22.36 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. In three dimensions the absolute maximum shear stress is (σmax − σmin)/2 over all three principal stresses; in plane stress the third one is zero.

στxyCσ1​σ2​

In-plane principal stresses

C = 10 MPa, R = √(20² + (−10)²) = 22.36 MPa
σ1 = 32.36 MPa, σ2 = −12.36 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(44.72, 32.36, 12.36)/2 = 22.36 MPa

Answer

τabs = 22.36 MPa

Here σ₁ and σ₂ have opposite signs, so the in-plane maximum R is also the absolute maximum.

6.Mohr’s circle & stress transformation

Answer: n = 7.937

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 = −10 MPa, R = 30 MPa
σ1 = 20 MPa, σ2 = −40 MPa, σ3 = 0

Von Mises

σvM = √(20² − 20·(−40) + (−40)²) = 52.92 MPa
n = 420/52.92 = 7.937

Answer

n = 7.937

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

7.Axial stress & strain

Answer: P = 83.27 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

A₁ = 1257 mm², A₂ = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·1257) = 2.387 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(110000·1257) = 3.617 × 10⁻⁶ mm/N

Load

P = 0.5/6.004 × 10⁻⁶ = 83271 N = 83.27 kN

Answer

P = 83.27 kN
8.Axial stress & strain

Answer: σ2 = 35.87 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 = 200000·1257/600 = 418879 N/mm
k₂ = 110000·706.9/400 = 194386 N/mm

Share of segment 2

F2 = 80000·194386/613265 = 25358 N

Stress

σ2 = 25358/706.9 = 35.87 MPa (compression)

Answer

σ2 = 35.87 MPa
9.Column buckling

Answer: Pcr = 21.7 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 2·4000 = 8000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/8000² = 21704 N = 21.7 kN

Answer

Pcr = 21.7 kN
10.Column buckling

Answer: d = 42.44 mm

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.

Required critical load

Pcr = n·P = 3·95000 = 285000 N
KL = 0.7·1500 = 1050 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·285000·1050²/(π³·200000))^(1/4) = 42.44 mm

Answer

d = 42.44 mm

Check: the slenderness KL/r is 98.97, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: fd = 3.282 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 = √(80000/150) = 23.09 rad/s
fn = 23.09/(2π) = 3.676 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 3.676·√(1 − 0.45²) = 3.282 Hz

Answer

fd = 3.282 Hz
12.Mechanical vibrations

Answer: keq = 60 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.

keq = 50 + 10 = 60 kN/m

Answer

keq = 60 kN/m

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