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​ = 20N2​ = 24

A 3 kW motor running at 1450 rpm drives a pinion with 20 teeth, which meshes with a gear with 24 teeth. The gear pair has an efficiency of 95%. Find the output torque T2 on the gear shaft.

P = 3 kWn1 = 1450 rpmN1 = 20N2 = 24η = 95%
T2 = N·m
2.Gears & gear trainsHard
A = 30B = 45C = 60D = 72

In a compound gear train, gear A (30 teeth) on the input shaft turning at 1440 rpm drives gear B (45 teeth). Gear C (60 teeth) is fixed to the same shaft as B and drives gear D (72 teeth) on the output shaft. Find the output speed nout.

nin = 1440 rpmA = 30B = 45C = 60D = 72
nout = rpm
3.Torsion of shaftsMedium
d1​ = 60 mmd2​ = 50 mmT = 750 N·mABC0.8 m1.5 m

A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 60 mm and length 0.8 m; segment BC has diameter 50 mm and length 1.5 m. A torque of 750 N·m acts at the free end C. Find the angle of twist φC of end C, in degrees.

T = 750 N·md1 = 60 mm, L1 = 0.8 md2 = 50 mm, L2 = 1.5 mG = 26 GPa
φC = °
4.Torsion of shaftsHard
d = 60 mmT0​ = 200 N·mACBa = 2.5 mb = 0.5 m

A solid brass shaft of diameter 60 mm is fixed to rigid walls at both ends A and B. A torque of 200 N·m is applied at C, 2.5 m from A and 0.5 m from B. Find the reaction torque TA at wall A.

d = 60 mmT0 = 200 N·ma = 2.5 mb = 0.5 m
TA = N·m
5.Mohr’s circle & stress transformationMedium
σx​ = 130σy​ = 60τxy​ = 60MPa

A point in a thin plate is in plane stress: σₓ = 130 MPa, σᵧ = 60 MPa and τₓᵧ = 60 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).

σx = 130 MPaσy = 60 MPaτxy = 60 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 10σy​ = 100τxy​ = −60MPa

A point in a loaded part has σₓ = 10 MPa, σᵧ = 100 MPa and τₓᵧ = −60 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 15° counter-clockwise from the x-axis.

σx = 10 MPaσy = 100 MPaτxy = −60 MPaθ = 15°
σθ = MPa
7.Axial stress & strainMedium
P = ?12600 mm300 mmØ40Ø60δ = 0.4 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 brass (diameter 60 mm, length 300 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.

segment 1: steel, Ø40 mm, 600 mm, E = 200 GPasegment 2: brass, Ø60 mm, 300 mm, E = 100 GPaδ = 0.4 mm
P = kN
8.Axial stress & strainHard
Ø 25 mmSteel2 mØ 12 mmSteel1.5 mP = 55 kN

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

P = 55 kNsegment 1: Ø25 mm, 2 m, steel (E = 200 GPa)segment 2: Ø12 mm, 1.5 m, steel (E = 200 GPa)
δ = mm
9.Column bucklingMedium
PL = 2 m30 × 50 mmcross-section

A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.

L = 2 mE = 70 GPa (aluminium)b × h = 30 × 50 mm
Pcr = kN
10.Column bucklingHard
P = 20 kNL = 2.5 md = ?cross-section

A solid round column of length L made of steel is fixed at both ends. It must carry 20 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 20 kNn = 3L = 2.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
tx30 mm4 cycles2.4 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 2.4 mm. Find the damping ratio ζ.

x0 = 30 mmx4 = 2.4 mmn = 4
ζ =
12.Mechanical vibrationsHard
k1​ = 25k2​ = 15k3​ = 25m = 40 kg

A 40 kg mass hangs from spring k₃ = 25 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 25 kN/m and k₂ = 15 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.

m = 40 kgk1 = 25 kN/mk2 = 15 kN/mk3 = 25 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: T2 = 22.52 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 = 3000/151.8 = 19.76 N·m

Output torque

T2 = 19.76·24/20·0.95 = 22.52 N·m

Answer

T2 = 22.52 N·m
2.Gears & gear trains

Answer: nout = 800 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 = 1440·30/45 = 960 rpm = nC

Second stage

nout = 960·60/72 = 800 rpm

Answer

nout = 800 rpm

total ratio = 1.8 : 1

3.Torsion of shafts

Answer: φC = 5.08 °

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·60⁴/32 = 1.272 × 10⁶ mm⁴
J2 = π·50⁴/32 = 613592 mm⁴

Twist of each segment

T = 750 N·m = 750000 N·mm
G = 26 GPa = 26000 MPa
φAB = 750000·800/(26000·1.272 × 10⁶) = 0.01814 rad
φBC = 750000·1500/(26000·613592) = 0.07052 rad

Total twist

φC = 0.01814 + 0.07052 = 0.08866 rad = 5.08°

Answer

φC = 5.08°
4.Torsion of shafts

Answer: TA = 33.33 N·m

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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).

Equilibrium and compatibility

TA + TB = 200
TA·a = TB·b
TA = 200·0.5/3 = 33.33 N·m
TB = 200·2.5/3 = 166.7 N·m

Answer

TA = 33.33 N·m
5.Mohr’s circle & stress transformation

Answer: τabs = 82.23 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 = 95 MPa, R = √(35² + 60²) = 69.46 MPa
σ1 = 164.5 MPa, σ2 = 25.54 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(138.9, 164.5, 25.54)/2 = 82.23 MPa

Answer

τabs = 82.23 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

6.Mohr’s circle & stress transformation

Answer: σθ = −13.97 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.

Transformation equation

σθ = 55 + −45·0.866 + −60·0.5 = −13.97 MPa

Answer

σθ = −13.97 MPa
7.Axial stress & strain

Answer: P = 116 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 = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·1257) = 2.387 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(100000·2827) = 1.061 × 10⁻⁶ mm/N

Load

P = 0.4/3.448 × 10⁻⁶ = 115997 N = 116 kN

Answer

P = 116 kN
8.Axial stress & strain

Answer: δ = 4.768 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·25²/4 = 490.9 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 112 MPa
σ2 = 486.3 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 112·2000/200000 = 1.12 mm
δ2 = σ2L2/E2 = 486.3·1500/200000 = 3.647 mm
δ = δ1 + δ2 = 4.768 mm

Answer

δ = 4.768 mm
9.Column buckling

Answer: Pcr = 19.43 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 = 50·30³/12 = 112500 mm⁴

Effective length

KL = 1·2000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·112500/2000² = 19431 N = 19.43 kN

Answer

Pcr = 19.43 kN
10.Column buckling

Answer: d = 31.36 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 = 3·20000 = 60000 N
KL = 0.5·2500 = 1250 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·60000·1250²/(π³·200000))^(1/4) = 31.36 mm

Answer

d = 31.36 mm

Check: the slenderness KL/r is 159.4, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.09999

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(30/2.4) = 0.6314

Damping ratio

ζ = δ/√(4π² + δ²) = 0.6314/√(39.48 + 0.3987) = 0.09999

Answer

ζ = 0.09999
12.Mechanical vibrations

Answer: fn = 3.121 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

Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.

k12 = 25 + 15 = 40 kN/m
keq = 40·25/(40 + 25) = 15.38 kN/m

Natural frequency

ωn = √(15385/40) = 19.61 rad/s
fn = 19.61/(2π) = 3.121 Hz

Answer

fn = 3.121 Hz

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