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​ = 50N2​ = 60

A 7.5 kW motor running at 1440 rpm drives a pinion with 50 teeth, which meshes with a gear with 60 teeth. The gear pair has an efficiency of 96%. Find the output torque T2 on the gear shaft.

P = 7.5 kWn1 = 1440 rpmN1 = 50N2 = 60η = 96%
T2 = N·m
2.Gears & gear trainsHard
d1​ = 120 mmd2​ = 250 mm1440 rpmC = 800 mmθ = ?

An open belt connects a 120 mm pulley and a 250 mm pulley whose centres are 800 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 120 mmd2 = 250 mmC = 800 mm
θ = °
3.Torsion of shaftsMedium
T = 100 N·md = 30 mmL = 3 mφ = ?

A solid brass shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.

T = 100 N·md = 30 mmL = 3 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
Tdo​ = 75 mm, di​ = 60 mm

A hollow shaft transmits 30 kW at 500 rpm. Find the maximum shear stress τmax in the shaft.

P = 30 kWN = 500 rpmdo = 75 mmdi = 60 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = 150σy​ = −60τxy​ = 40MPa

A point in a loaded part has σₓ = 150 MPa, σᵧ = −60 MPa and τₓᵧ = 40 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σx = 150 MPaσy = −60 MPaτxy = 40 MPa
θp = °
6.Mohr’s circle & stress transformationHard
30°D = 500 mm, t = 10 mm, p = 1.5 MPap

A thin-walled cylindrical tank of inner diameter 500 mm and wall thickness 10 mm holds gas at a gauge pressure of 1.5 MPa. It is made from plate welded along a helical seam at 30° to the axis of the cylinder. Find the magnitude of the shear stress |τn| acting along the weld seam.

D = 500 mmt = 10 mmp = 1.5 MPaα = 30°
|τn| = MPa
7.Axial stress & strainMedium
200 kNØ25110 GPaØ100/60500 mmδ = ?

A steel rod (diameter 25 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 60 mm, E = 110 GPa). Both are 500 mm long and are compressed together between rigid plates by 200 kN. Find the shortening δ.

P = 200 kNrod: Ø25 mm, E = 200 GPatube: Ø100/Ø60 mm, E = 110 GPaL = 500 mm
δ = mm
8.Axial stress & strainHard
Ø 30 mmSteel1.5 mØ 16 mmSteel0.5 mP = 15 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.

P = 15 kNsegment 1: Ø30 mm, 1.5 m, steel (E = 200 GPa)segment 2: Ø16 mm, 0.5 m, steel (E = 200 GPa)
σmax = MPa
9.Column bucklingMedium
PL = 2 md = 25 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)d = 25 mm
Pcr = kN
10.Column bucklingHard
P = 85 kNL = 6 md = ?cross-section

A solid round column of length L made of steel is pinned at both ends. It must carry 85 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 85 kNn = 2.5L = 6 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 80 kN/mζ = 0.6m = 12 kg

A 12 kg mass on a spring with a stiffness of 80 kN/m has a viscous damper giving a damping ratio of 0.6. Find the damped natural frequency fd of its free vibration in Hz.

m = 12 kgk = 80 kN/mζ = 0.6
fd = Hz
12.Mechanical vibrationsHard
k1​ = 50k2​ = 60k3​ = 15m = 40 kg

A 40 kg mass hangs from spring k₃ = 15 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 50 kN/m and k₂ = 60 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 = 50 kN/mk2 = 60 kN/mk3 = 15 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: T2 = 57.3 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·1440/60 = 150.8 rad/s
T1 = P/ω₁ = 7500/150.8 = 49.74 N·m

Output torque

T2 = 49.74·60/50·0.96 = 57.3 N·m

Answer

T2 = 57.3 N·m
2.Gears & gear trains

Answer: θ = 170.7 °

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(0.08125) = 4.66°

Angle of wrap

θ = 180 − 2·4.66 = 170.7°

Answer

θ = 170.7°
3.Torsion of shafts

Answer: φ = 5.542 °

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 second moment of area

J = π·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·3000/(39000·79522) = 0.09673 rad
φ = 0.09673·180/π = 5.542°

Answer

φ = 5.542°
4.Torsion of shafts

Answer: τmax = 11.72 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·500/60 = 52.36 rad/s
T = P/ω = 30000/52.36 = 573 N·m

Polar second moment of area

J = π(75⁴ − 60⁴)/32 = 1.834 × 10⁶ mm⁴

Maximum shear stress

τmax = T·(do/2)/J = 572958·37.5/1.834 × 10⁶ = 11.72 MPa

Answer

τmax = 11.72 MPa
5.Mohr’s circle & stress transformation

Answer: θp = 10.43 °

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(2·40, 150 − −60) = 20.85°

Angle on the element

θp = 10.43°

Answer

θp = 10.43°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: |τn| = 8.119 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 = 500/2 = 250 mm
σh = pr/t = 1.5·250/10 = 37.5 MPa
σa = pr/(2t) = 18.75 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.

|τn| = (37.5 − 18.75)·sin 30°·cos 30° = 8.119 MPa

Answer

|τn| = 8.119 MPa
7.Axial stress & strain

Answer: δ = 0.1536 mm

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

Ar = π·25²/4 = 490.9 mm²
At = π(100² − 60²)/4 = 5027 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 110·10³·5027 = 6.511 × 10⁸ N

Shortening

δ = 200000·500/6.511 × 10⁸ = 0.1536 mm

Answer

δ = 0.1536 mm
8.Axial stress & strain

Answer: σmax = 74.6 MPa

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·30²/4 = 706.9 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 21.22 MPa
σ2 = 74.6 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σmax = 74.6 MPa

Answer

σmax = 74.6 MPa
9.Column buckling

Answer: Pcr = 3.312 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·25⁴/64 = 19175 mm⁴

Effective length

KL = 1·2000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·19175/2000² = 3312 N = 3.312 kN

Answer

Pcr = 3.312 kN
10.Column buckling

Answer: d = 94.26 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 = 2.5·85000 = 212500 N
KL = 1·6000 = 6000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·212500·6000²/(π³·200000))^(1/4) = 94.26 mm

Answer

d = 94.26 mm

Check: the slenderness KL/r is 254.6, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: fd = 10.4 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 = √(80000/12) = 81.65 rad/s
fn = 81.65/(2π) = 12.99 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 12.99·√(1 − 0.6²) = 10.4 Hz

Answer

fd = 10.4 Hz
12.Mechanical vibrations

Answer: fn = 2.891 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 = 50 + 60 = 110 kN/m
keq = 110·15/(110 + 15) = 13.2 kN/m

Natural frequency

ωn = √(13200/40) = 18.17 rad/s
fn = 18.17/(2π) = 2.891 Hz

Answer

fn = 2.891 Hz

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