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​ = 36N2​ = 50

A 7.5 kW motor running at 960 rpm drives a pinion with 36 teeth, which meshes with a gear with 50 teeth. The gear pair has an efficiency of 97%. Find the output torque T2 on the gear shaft.

P = 7.5 kWn1 = 960 rpmN1 = 36N2 = 50η = 97%
T2 = N·m
2.Gears & gear trainsHard
A = 36B = 40C = 25D = 40

In a compound gear train, gear A (36 teeth) on the input shaft turning at 2900 rpm drives gear B (40 teeth). Gear C (25 teeth) is fixed to the same shaft as B and drives gear D (40 teeth) on the output shaft. Find the output speed nout.

nin = 2900 rpmA = 36B = 40C = 25D = 40
nout = rpm
3.Torsion of shaftsMedium
T = 500 N·md = 50 mmL = 2.5 mφ = ?

A solid aluminium 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 = 500 N·md = 50 mmL = 2.5 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
do​ = ?50 kW, 1500 rpm

A hollow shaft must transmit 50 kW at 1500 rpm. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter do.

P = 50 kWN = 1500 rpmdi/do = 0.6τallow = 40 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 120σy​ = 40τxy​ = −60MPa

A point in a thin plate is in plane stress: σₓ = 120 MPa, σᵧ = 40 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 = 120 MPaσy = 40 MPaτxy = −60 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 130σy​ = 110τxy​ = −40MPa

A point in a thin plate is in plane stress: σₓ = 130 MPa, σᵧ = 110 MPa and τₓᵧ = −40 MPa (σz = 0). Find the von Mises equivalent stress σvM.

σx = 130 MPaσy = 110 MPaτxy = −40 MPa
σvM = MPa
7.Axial stress & strainMedium
100 kNØ25110 GPaØ120/60800 mmδ = ?

A steel rod (diameter 25 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 120 mm, inner diameter 60 mm, E = 110 GPa). Both are 800 mm long and are compressed together between rigid plates by 100 kN. Find the shortening δ.

P = 100 kNrod: Ø25 mm, E = 200 GPatube: Ø120/Ø60 mm, E = 110 GPaL = 800 mm
δ = mm
8.Axial stress & strainHard
60 kN12600 mm300 mmØ30Ø60σ2​ = ?

A steel segment (diameter 30 mm, length 600 mm, E = 200 GPa) and a segment made of copper (diameter 60 mm, length 300 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 60 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø30 mm, 600 mm, E = 200 GPasegment 2: copper, Ø60 mm, 300 mm, E = 120 GPaP = 60 kN
σ2 = MPa
9.Column bucklingMedium
PL = 2.5 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 = 2.5 mE = 70 GPa (aluminium)d = 25 mm
Pcr = kN
10.Column bucklingHard
P = 115 kNL = 2.5 md = ?cross-section

A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 115 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 115 kNn = 2L = 2.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
k = 12 kN/mζ = 0.45m = 100 kg

A 100 kg mass on a spring with a stiffness of 12 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 = 100 kgk = 12 kN/mζ = 0.45
fd = Hz
12.Mechanical vibrationsHard
k1​ = 25k2​ = 50k3​ = 5m = 100 kg

A 100 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 = 25 kN/m and k₂ = 50 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 = 100 kgk1 = 25 kN/mk2 = 50 kN/mk3 = 5 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: T2 = 100.5 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·960/60 = 100.5 rad/s
T1 = P/ω₁ = 7500/100.5 = 74.6 N·m

Output torque

T2 = 74.6·50/36·0.97 = 100.5 N·m

Answer

T2 = 100.5 N·m
2.Gears & gear trains

Answer: nout = 1631 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 = 2900·36/40 = 2610 rpm = nC

Second stage

nout = 2610·25/40 = 1631 rpm

Answer

nout = 1631 rpm

total ratio = 1.778 : 1

3.Torsion of shafts

Answer: φ = 4.489 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 500·1000·2500/(26000·613592) = 0.07835 rad
φ = 0.07835·180/π = 4.489°

Answer

φ = 4.489°
4.Torsion of shafts

Answer: do = 35.98 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·1500/60 = 157.1 rad/s
T = P/ω = 50000/157.1 = 318.3 N·m

Outer diameter from the allowable stress

T = 318.3 N·m = 318310 N·mm
1 − k⁴ = 1 − 0.6⁴ = 0.8704
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·318310/(π·40·0.8704)) = 35.98 mm
di = 0.6·35.98 = 21.59 mm

Answer

do = 35.98 mm

A solid shaft would need d = 34.35 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: τabs = 76.06 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 = 80 MPa, R = √(40² + (−60)²) = 72.11 MPa
σ1 = 152.1 MPa, σ2 = 7.889 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(144.2, 152.1, 7.889)/2 = 76.06 MPa

Answer

τabs = 76.06 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: σvM = 139.6 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. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.

σxyCσ1​σ2​

Principal stresses

C = 120 MPa, R = 41.23 MPa
σ1 = 161.2 MPa, σ2 = 78.77 MPa, σ3 = 0

Von Mises

σvM = √(161.2² − 161.2·78.77 + 78.77²) = 139.6 MPa

Answer

σvM = 139.6 MPa
7.Axial stress & strain

Answer: δ = 0.07758 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(120² − 60²)/4 = 8482 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 110·10³·8482 = 1.031 × 10⁹ N

Shortening

δ = 100000·800/1.031 × 10⁹ = 0.07758 mm

Answer

δ = 0.07758 mm
8.Axial stress & strain

Answer: σ2 = 17.56 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 = 200000·706.9/600 = 235619 N/mm
k₂ = 120000·2827/300 = 1.131 × 10⁶ N/mm

Share of segment 2

F2 = 60000·1.131 × 10⁶/1.367 × 10⁶ = 49655 N

Stress

σ2 = 49655/2827 = 17.56 MPa (compression)

Answer

σ2 = 17.56 MPa
9.Column buckling

Answer: Pcr = 2.12 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 = 1·2500 = 2500 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·19175/2500² = 2120 N = 2.12 kN

Answer

Pcr = 2.12 kN
10.Column buckling

Answer: d = 51.92 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 = 2·115000 = 230000 N
KL = 0.7·2500 = 1750 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·230000·1750²/(π³·200000))^(1/4) = 51.92 mm

Answer

d = 51.92 mm

Check: the slenderness KL/r is 134.8, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: fd = 1.557 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 = √(12000/100) = 10.95 rad/s
fn = 10.95/(2π) = 1.743 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 1.743·√(1 − 0.45²) = 1.557 Hz

Answer

fd = 1.557 Hz
12.Mechanical vibrations

Answer: fn = 1.09 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 + 50 = 75 kN/m
keq = 75·5/(75 + 5) = 4.688 kN/m

Natural frequency

ωn = √(4688/100) = 6.847 rad/s
fn = 6.847/(2π) = 1.09 Hz

Answer

fn = 1.09 Hz

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