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
Nw​ = 45z = 4, 1450 rpm

A worm with 4 start(s) turns at 1450 rpm and drives a worm wheel with 45 teeth. Find the speed nw of the wheel.

z = 4Nw = 45n = 1450 rpm
nw = rpm
2.Gears & gear trainsHard
d1​ = 200 mmd2​ = 250 mm720 rpmC = 1200 mmT1​ = 500 NT2​P = ?

An open belt connects a 200 mm driving pulley at 720 rpm to a 250 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.35 and the tight-side tension may not exceed 500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 200 mmn1 = 720 rpmd2 = 250 mmC = 1200 mmμ = 0.35T1 = 500 N
P = kW
3.Torsion of shaftsMedium
T = 100 N·md = 25 mmL = 1 mφ = ?

A solid steel 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 = 25 mmL = 1 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
d = ?15 kW, 300 rpm

A solid steel shaft transmits 15 kW at 300 rpm. The shear stress must not exceed 50 MPa and the twist must not exceed 0.5° per m of length. Find the smallest diameter d that satisfies both limits.

P = 15 kWN = 300 rpmτallow = 50 MPaθallow = 0.5°/mG = 80 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 110σy​ = 80τxy​ = −10MPa

A point in a loaded part has σₓ = 110 MPa, σᵧ = 80 MPa and τₓᵧ = −10 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 110 MPaσy = 80 MPaτxy = −10 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 30σy​ = −60τxy​ = 30MPa

A point in a loaded part has σₓ = 30 MPa, σᵧ = −60 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 25° counter-clockwise from the x-axis.

σx = 30 MPaσy = −60 MPaτxy = 30 MPaθ = 25°
σθ = MPa
7.Axial stress & strainMedium
P = ?12500 mm200 mmØ25Ø30δ = 0.4 mm

A bar is made of a steel segment (diameter 25 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 30 mm, length 200 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, Ø25 mm, 500 mm, E = 200 GPasegment 2: brass, Ø30 mm, 200 mm, E = 100 GPaδ = 0.4 mm
P = kN
8.Axial stress & strainHard
20 kN12400 mm400 mmØ40Ø30σ2​ = ?

A steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 400 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, Ø40 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 400 mm, E = 70 GPaP = 20 kN
σ2 = MPa
9.Column bucklingMedium
PL = 3 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 = 3 mE = 200 GPa (steel)d = 40 mm
Pcr = kN
10.Column bucklingHard
PL = 3 md = 40 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 = 3 mE = 70 GPa (aluminium)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx15 mm5 cycles0.6 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 15 mm is followed 5 cycles later by a peak of 0.6 mm. Find the damping ratio ζ.

x0 = 15 mmx5 = 0.6 mmn = 5
ζ =
12.Mechanical vibrationsHard
k1​ = 5k2​ = 12k3​ = 60m = 50 kg

A 50 kg mass hangs from spring k₃ = 60 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 5 kN/m and k₂ = 12 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 = 50 kgk1 = 5 kN/mk2 = 12 kN/mk3 = 60 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: nw = 128.9 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.

Wheel speed

nw = n·z/Nw = 1450·4/45 = 128.9 rpm

Answer

nw = 128.9 rpm
2.Gears & gear trains

Answer: P = 2.496 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.

Angle of wrap

θ = 3.1 rad

Slack-side tension

eμθ = e1.085 = 2.959, T₂ = 500/2.959 = 169 N

Power

P = (500 − 169)·7.54 = 2.496 kW

Answer

P = 2.496 kW
3.Torsion of shafts

Answer: φ = 1.868 °

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·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·1000/(80000·38350) = 0.03259 rad
φ = 0.03259·180/π = 1.868°

Answer

φ = 1.868°
4.Torsion of shafts

Answer: d = 51.37 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·300/60 = 31.42 rad/s
T = P/ω = 15000/31.42 = 477.5 N·m

Diameter for strength

T = 477.5 N·m = 477465 N·mm
dτ = ∛(16·477465/(π·50)) = 36.5 mm

Diameter for stiffness

θ = 0.5·π/180/1000 = 8.727 × 10⁻⁶ rad/mm
G = 80 GPa = 80000 MPa
dφ = (32·477465/(π·80000·8.727 × 10⁻⁶))1/4 = 51.37 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(36.5, 51.37) = 51.37 mm

Answer

d = 51.37 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: τmax = 18.03 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.

σxyCτmax​

Radius of Mohr’s circle

τmax = R = √(((110 − 80)/2)² + −10²) = 18.03 MPa

Answer

τmax = 18.03 MPa
6.Mohr’s circle & stress transformation

Answer: σθ = 36.91 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 = −15 + 45·0.6428 + 30·0.766 = 36.91 MPa

Answer

σθ = 36.91 MPa
7.Axial stress & strain

Answer: P = 50.49 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 = 706.9 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·490.9) = 5.093 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(100000·706.9) = 2.829 × 10⁻⁶ mm/N

Load

P = 0.4/7.922 × 10⁻⁶ = 50490 N = 50.49 kN

Answer

P = 50.49 kN
8.Axial stress & strain

Answer: σ2 = 4.654 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 = 200000·1257/400 = 628319 N/mm
k₂ = 70000·706.9/400 = 123700 N/mm

Share of segment 2

F2 = 20000·123700/752019 = 3290 N

Stress

σ2 = 3290/706.9 = 4.654 MPa (compression)

Answer

σ2 = 4.654 MPa
9.Column buckling

Answer: Pcr = 56.25 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 = 0.7·3000 = 2100 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·125664/2100² = 56247 N = 56.25 kN

Answer

Pcr = 56.25 kN
10.Column buckling

Answer: σcr = 1.919 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·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²
r = √(I/A) = √(125664/1257) = 10 mm
KL = 2·3000 = 6000 mm
KL/r = 6000/10 = 600

Critical stress

σcr = π²E/(KL/r)² = π²·70000/600² = 1.919 MPa

Answer

σcr = 1.919 MPa
11.Mechanical vibrations

Answer: ζ = 0.1019

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 = (1/5)·ln(15/0.6) = 0.6438

Damping ratio

ζ = δ/√(4π² + δ²) = 0.6438/√(39.48 + 0.4144) = 0.1019

Answer

ζ = 0.1019
12.Mechanical vibrations

Answer: fn = 2.591 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 = 5 + 12 = 17 kN/m
keq = 17·60/(17 + 60) = 13.25 kN/m

Natural frequency

ωn = √(13247/50) = 16.28 rad/s
fn = 16.28/(2π) = 2.591 Hz

Answer

fn = 2.591 Hz

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