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​ = 200 mmd2​ = 300 mm1440 rpmT1​ = 1500 NT2​ = 450 NP = ?

A belt runs on a 200 mm driving pulley at 1440 rpm. The tension is 1500 N on the tight side and 450 N on the slack side. Find the power P transmitted, in kW.

d1 = 200 mmn1 = 1440 rpmT1 = 1500 NT2 = 450 N
P = kW
2.Gears & gear trainsHard
d1​ = 120 mmd2​ = 250 mm2880 rpmC = 500 mmT1​ = 500 NT2​P = ?

An open belt connects a 120 mm driving pulley at 2880 rpm to a 250 mm pulley; the centres are 500 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 = 120 mmn1 = 2880 rpmd2 = 250 mmC = 500 mmμ = 0.35T1 = 500 N
P = kW
3.Torsion of shaftsMedium
d1​ = 50 mmd2​ = 40 mmT = 1000 N·mABC2.5 m1 m

A stepped solid steel shaft is fixed at A. Segment AB has diameter 50 mm and length 2.5 m; segment BC has diameter 40 mm and length 1 m. A torque of 1000 N·m acts at the free end C. Find the maximum shear stress τmax in the shaft.

T = 1000 N·md1 = 50 mm, L1 = 2.5 md2 = 40 mm, L2 = 1 m
τmax = MPa
4.Torsion of shaftsHard
do​ = ?100 kW, 750 rpm

A hollow shaft must transmit 100 kW at 750 rpm. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 60 MPa. Find the smallest outer diameter do.

P = 100 kWN = 750 rpmdi/do = 0.75τallow = 60 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
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 maximum in-plane shear stress τmax.

σx = 30 MPaσy = 60 MPaτxy = 30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​ = −50σy​ = 80τxy​ = −20MPa

A point in a loaded part has σₓ = −50 MPa, σᵧ = 80 MPa and τₓᵧ = −20 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 15° counter-clockwise from the x-axis.

σx = −50 MPaσy = 80 MPaτxy = −20 MPaθ = 15°
σθ = MPa
7.Axial stress & strainMedium
100 kNØ40110 GPaØ80/60500 mmδ = ?

A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 80 mm, inner diameter 60 mm, E = 110 GPa). Both are 500 mm long and are compressed together between rigid plates by 100 kN. Find the shortening δ.

P = 100 kNrod: Ø40 mm, E = 200 GPatube: Ø80/Ø60 mm, E = 110 GPaL = 500 mm
δ = mm
8.Axial stress & strainHard
Ø 25 mmSteel1.5 mØ 12 mmCopper1 mP = 100 kN

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

P = 100 kNsegment 1: Ø25 mm, 1.5 m, steel (E = 200 GPa)segment 2: Ø12 mm, 1 m, copper (E = 117 GPa)
δ = mm
9.Column bucklingMedium
PL = 4 m25 × 40 mmcross-section

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

L = 4 mE = 200 GPa (steel)b × h = 25 × 40 mm
Pcr = kN
10.Column bucklingHard
PL = 4 md = 40 mmcross-section

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

L = 4 mE = 70 GPa (aluminium)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 50 kN/mζ = 0.45m = 2 kg

A 2 kg mass on a spring with a stiffness of 50 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 = 2 kgk = 50 kN/mζ = 0.45
fd = Hz
12.Mechanical vibrationsHard
k1​ = 80k2​ = 60k3​ = 5m = 50 kg

A 50 kg mass hangs from spring k₃ = 5 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 80 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 = 50 kgk1 = 80 kN/mk2 = 60 kN/mk3 = 5 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: P = 15.83 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 = 15.08 m/s

Power

P = (1500 − 450)·15.08 = 15834 W = 15.83 kW

Answer

P = 15.83 kW
2.Gears & gear trains

Answer: P = 5.747 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 = 2.881 rad

Slack-side tension

eμθ = e1.008 = 2.741, T₂ = 500/2.741 = 182.4 N

Power

P = (500 − 182.4)·18.1 = 5.747 kW

Answer

P = 5.747 kW
3.Torsion of shafts

Answer: τmax = 79.58 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 = 1000 N·m = 1 × 10⁶ N·mm
τAB = 16·1 × 10⁶/(π·50³) = 40.74 MPa
τBC = 16·1 × 10⁶/(π·40³) = 79.58 MPa

Answer

τmax = 79.58 MPa
4.Torsion of shafts

Answer: do = 54.07 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·750/60 = 78.54 rad/s
T = P/ω = 100000/78.54 = 1273 N·m

Outer diameter from the allowable stress

T = 1273 N·m = 1.273 × 10⁶ N·mm
1 − k⁴ = 1 − 0.75⁴ = 0.6836
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·1.273 × 10⁶/(π·60·0.6836)) = 54.07 mm
di = 0.75·54.07 = 40.55 mm

Answer

do = 54.07 mm

A solid shaft would need d = 47.63 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: τmax = 33.54 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.

στxyCτmax​

Radius of Mohr’s circle

τmax = R = √(((30 − 60)/2)² + 30²) = 33.54 MPa

Answer

τmax = 33.54 MPa
6.Mohr’s circle & stress transformation

Answer: σθ = −51.29 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 + −65·0.866 + −20·0.5 = −51.29 MPa

Answer

σθ = −51.29 MPa
7.Axial stress & strain

Answer: δ = 0.1014 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·40²/4 = 1257 mm²
At = π(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 110·10³·2199 = 4.932 × 10⁸ N

Shortening

δ = 100000·500/4.932 × 10⁸ = 0.1014 mm

Answer

δ = 0.1014 mm
8.Axial stress & strain

Answer: δ = 9.085 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 = 203.7 MPa
σ2 = 884.2 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 203.7·1500/200000 = 1.528 mm
δ2 = σ2L2/E2 = 884.2·1000/117000 = 7.557 mm
δ = δ1 + δ2 = 9.085 mm

Answer

δ = 9.085 mm
9.Column buckling

Answer: Pcr = 25.7 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 rectangle buckles about its weak axis, using the smaller dimension cubed.

Imin = h·b³/12 = 40·25³/12 = 52083 mm⁴

Effective length

KL = 0.5·4000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·52083/2000² = 25702 N = 25.7 kN

Answer

Pcr = 25.7 kN
10.Column buckling

Answer: σcr = 8.812 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 = 0.7·4000 = 2800 mm
KL/r = 2800/10 = 280

Critical stress

σcr = π²E/(KL/r)² = π²·70000/280² = 8.812 MPa

Answer

σcr = 8.812 MPa
11.Mechanical vibrations

Answer: fd = 22.47 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 = √(50000/2) = 158.1 rad/s
fn = 158.1/(2π) = 25.16 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 25.16·√(1 − 0.45²) = 22.47 Hz

Answer

fd = 22.47 Hz
12.Mechanical vibrations

Answer: fn = 1.564 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 = 80 + 60 = 140 kN/m
keq = 140·5/(140 + 5) = 4.828 kN/m

Natural frequency

ωn = √(4828/50) = 9.826 rad/s
fn = 9.826/(2π) = 1.564 Hz

Answer

fn = 1.564 Hz

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