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​ = 100 mmd2​ = 400 mm960 rpmT1​ = 800 NT2​ = 320 NP = ?

A belt runs on a 100 mm driving pulley at 960 rpm. The tension is 800 N on the tight side and 320 N on the slack side. Find the power P transmitted, in kW.

d1 = 100 mmn1 = 960 rpmT1 = 800 NT2 = 320 N
P = kW
2.Gears & gear trainsHard
A = 16B = 80C = 36D = 48

In a compound gear train, gear A (16 teeth) on the input shaft turning at 2900 rpm drives gear B (80 teeth). Gear C (36 teeth) is fixed to the same shaft as B and drives gear D (48 teeth) on the output shaft. Find the output speed nout.

nin = 2900 rpmA = 16B = 80C = 36D = 48
nout = rpm
3.Torsion of shaftsMedium
d1​ = 50 mmd2​ = 25 mmT = 100 N·mABC1 m3 m

A stepped solid brass shaft is fixed at A. Segment AB has diameter 50 mm and length 1 m; segment BC has diameter 25 mm and length 3 m. A torque of 100 N·m acts at the free end C. Find the maximum shear stress τmax in the shaft.

T = 100 N·md1 = 50 mm, L1 = 1 md2 = 25 mm, L2 = 3 m
τmax = MPa
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 10 kW at 1450 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 10 kWN = 1450 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 100σy​ = 40τxy​ = −30MPa

A point in a thin plate is in plane stress: σₓ = 100 MPa, σᵧ = 40 MPa and τₓᵧ = −30 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).

σx = 100 MPaσy = 40 MPaτxy = −30 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 40σy​ = −10τxy​ = −10MPa

A point in a loaded part has σₓ = 40 MPa, σᵧ = −10 MPa and τₓᵧ = −10 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 25° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σx = 40 MPaσy = −10 MPaτxy = −10 MPaθ = 25°
τθ = MPa
7.Axial stress & strainMedium
P = ?12300 mm500 mmØ25Ø60δ = 0.5 mm

A bar is made of a steel segment (diameter 25 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 60 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, Ø25 mm, 300 mm, E = 200 GPasegment 2: bronze, Ø60 mm, 500 mm, E = 110 GPaδ = 0.5 mm
P = kN
8.Axial stress & strainHard
30 kN12600 mm400 mmØ40Ø60σ2​ = ?

A steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) and a segment made of copper (diameter 60 mm, length 400 mm, E = 120 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, 600 mm, E = 200 GPasegment 2: copper, Ø60 mm, 400 mm, E = 120 GPaP = 30 kN
σ2 = MPa
9.Column bucklingMedium
PL = 6 md = 80 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 = 6 mE = 200 GPa (steel)d = 80 mm
Pcr = kN
10.Column bucklingHard
PL = 3 md = 50 mmcross-section

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

L = 3 mE = 200 GPa (steel)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 100 kN/mc = 180 N·s/mm = 8 kg

An 8 kg mass is supported by a spring with a stiffness of 100 kN/m and a viscous damper with a damping coefficient of 180 N·s/m. Find the damping ratio ζ.

m = 8 kgk = 100 kN/mc = 180 N·s/m
ζ =
12.Mechanical vibrationsHard
k1​ = 12k2​ = 50m = 100 kg

A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 12 kN/m and k₂ = 50 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 = 12 kN/mk2 = 50 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: P = 2.413 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 = 5.027 m/s

Power

P = (800 − 320)·5.027 = 2413 W = 2.413 kW

Answer

P = 2.413 kW
2.Gears & gear trains

Answer: nout = 435 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 = 2900·16/80 = 580 rpm = nC

Second stage

nout = 580·36/48 = 435 rpm

Answer

nout = 435 rpm

total ratio = 6.667 : 1

3.Torsion of shafts

Answer: τmax = 32.59 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.

Stress in each segment

T = 100 N·m = 100000 N·mm
τAB = 16·100000/(π·50³) = 4.074 MPa
τBC = 16·100000/(π·25³) = 32.59 MPa

Answer

τmax = 32.59 MPa
4.Torsion of shafts

Answer: d = 16.48 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 = 10000/151.8 = 65.86 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·65857/(π·75)) = 16.48 mm

Answer

d = 16.48 mm

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

5.Mohr’s circle & stress transformation

Answer: τabs = 56.21 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 = 70 MPa, R = √(30² + (−30)²) = 42.43 MPa
σ1 = 112.4 MPa, σ2 = 27.57 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(84.85, 112.4, 27.57)/2 = 56.21 MPa

Answer

τabs = 56.21 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: τθ = −25.58 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

τθ = −25·0.766 + −10·0.6428 = −25.58 MPa

Answer

τθ = −25.58 MPa
7.Axial stress & strain

Answer: P = 107.2 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 = 490.9 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·490.9) = 3.056 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(110000·2827) = 1.608 × 10⁻⁶ mm/N

Load

P = 0.5/4.663 × 10⁻⁶ = 107218 N = 107.2 kN

Answer

P = 107.2 kN
8.Axial stress & strain

Answer: σ2 = 7.103 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 = 120000·2827/400 = 848230 N/mm

Share of segment 2

F2 = 30000·848230/1.267 × 10⁶ = 20083 N

Stress

σ2 = 20083/2827 = 7.103 MPa (compression)

Answer

σ2 = 7.103 MPa
9.Column buckling

Answer: Pcr = 27.56 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 = 2·6000 = 12000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/12000² = 27561 N = 27.56 kN

Answer

Pcr = 27.56 kN
10.Column buckling

Answer: σcr = 8.567 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 = 2·3000 = 6000 mm
KL/r = 6000/12.5 = 480

Critical stress

σcr = π²E/(KL/r)² = π²·200000/480² = 8.567 MPa

Answer

σcr = 8.567 MPa
11.Mechanical vibrations

Answer: ζ = 0.1006

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.

Critical damping

cc = 2·√(k·m) = 2·√(100000·8) = 1789 N·s/m

Damping ratio

ζ = c/cc = 180/1789 = 0.1006

Answer

ζ = 0.1006

ζ < 1, so the system is underdamped and oscillates as it decays.

12.Mechanical vibrations

Answer: fn = 3.963 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 = 12 + 50 = 62 kN/m

Natural frequency

ωn = √(62000/100) = 24.9 rad/s
fn = 24.9/(2π) = 3.963 Hz

Answer

fn = 3.963 Hz

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