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

Find the polar moment of area: J = πd⁴/32 for a solid shaft.
Shear stress at the surface: τ = T·r/J, with r the outer radius.
Angle of twist: φ = TL/(GJ) in radians.
If power is given, find the torque first with P = T·ω, ω = 2πn/60.

Mohr’s circle & stress transformation

Centre: C = (σ_x + σ_y)/2
Radius: R = √(((σ_x − σ_y)/2)² + τ_xy²)
Principal stresses: σ_1,2 = C ± R
Principal direction: tan 2θ_p = 2τ_xy/(σ_x − σ_y)
Inclined plane: σ_θ = C + (σ_x − σ_y)/2·cos 2θ + τ_xy·sin 2θ

Method

Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
The principal stresses are C ± R and the largest in-plane shear stress is R.
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: ε_T = α·ΔT

Column buckling

Euler load: P_cr = π²EI/(KL)²
Effective length factor: K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress: σ_cr = P_cr/A
Slenderness: λ = KL/r, r = √(I/A)

Mechanical vibrations

Natural frequency: ω_n = √(k/m), f_n = ω_n/(2π)
From static deflection: ω_n = √(g/δ_st)
Critical damping: c_c = 2√(km) = 2mω_n
Damping ratio: ζ = c/c_c
Damped frequency: ω_d = ω_n√(1 − ζ²)
Logarithmic decrement: δ = (1/n)·ln(x₀/x_n), 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

Two parallel shafts are 200 mm apart. They must be connected by a pair of spur gears with a module of 5 mm and a speed reduction of 3 : 1. How many teeth N₂ must the gear have?

a = 200 mmm = 5 mmi = 3

N₂ =

2.Gears & gear trainsHard

In a compound gear train, gear A (18 teeth) on the input shaft turning at 1500 rpm drives gear B (30 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D (60 teeth) on the output shaft. Find the output speed n_out.

n_in = 1500 rpmA = 18B = 30C = 16D = 60

n_out = rpm

3.Torsion of shaftsMedium

A hollow brass shaft (outer diameter 30 mm, inner diameter 21 mm, length 0.5 m) is fixed at one end and twisted by a torque of 100 N·m at the other. Find the angle of twist φ in degrees.

T = 100 N·md_o = 30 mmd_i = 21 mmL = 0.5 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 15 kW at 3000 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.

P = 15 kWN = 3000 rpmτ_allow = 80 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = −70 MPa, σᵧ = −70 MPa and τₓᵧ = −40 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = −70 MPaσ_y = −70 MPaτ_xy = −40 MPaσ_z = 0

τ_abs = MPa

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 1500 mm and wall thickness 12 mm holds gas at a gauge pressure of 2.5 MPa. It is made from plate welded along a helical seam at 20° to the axis of the cylinder. Find the magnitude of the shear stress |τ_n| acting along the weld seam.

D = 1500 mmt = 12 mmp = 2.5 MPaα = 20°

|τ_n| = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 30 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 60 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 0.3 mm in total.

segment 1: steel, Ø30 mm, 600 mm, E = 200 GPasegment 2: brass, Ø60 mm, 200 mm, E = 100 GPaδ = 0.3 mm

P = kN

8.Axial stress & strainHard

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

P = 50 kNsegment 1: Ø30 mm, 2 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1.5 m, brass (E = 100 GPa)

σ_max = MPa

9.Column bucklingMedium

A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 1 mE = 70 GPa (aluminium)b × h = 30 × 80 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 140 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 140 kNn = 2L = 4 mE = 70 GPa (aluminium)

d = mm

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 15 mm is followed 2 cycles later by a peak of 2.2 mm. Find the damping ratio ζ.

x₀ = 15 mmx₂ = 2.2 mmn = 2

ζ =

12.Mechanical vibrationsHard

A 100 kg mass hangs from spring k₃ = 30 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 = 12 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency f_n of vertical vibration in Hz.

m = 100 kgk₁ = 80 kN/mk₂ = 12 kN/mk₃ = 30 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: N₂ = 60

Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: F_t = 2T/d.

Total number of teeth

N₁ + N₂ = 2a/m = 2·200/5 = 80

Split by the ratio

N₁ = 80/(1 + 3) = 20

N₂ = 3·20 = 60

Answer

N₂ = 60

2.Gears & gear trains

Answer: n_out = 240 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

n_B = 1500·18/30 = 900 rpm = n_C

Second stage

n_out = 900·16/60 = 240 rpm

Answer

n_out = 240 rpm

total ratio = 6.25 : 1

3.Torsion of shafts

Answer: φ = 1.216 °

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τ_max = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(d_o⁴ − d_i⁴)/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 − 21⁴)/32 = 60428 mm⁴

Angle of twist

T = 100 N·m = 100000 N·mm

L = 0.5 m = 500 mm

G = 39 GPa = 39000 MPa

φ = TL/(GJ) = 100000·500/(39000·60428) = 0.02122 rad

φ = 0.02122·180/π = 1.216°

Answer

φ = 1.216°

4.Torsion of shafts

Answer: d = 14.49 mm

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s

T = P/ω = 15000/314.2 = 47.75 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·47746/(π·80)) = 14.49 mm

Answer

d = 14.49 mm

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

5.Mohr’s circle & stress transformation

Answer: τ_abs = 55 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 (σ_max − σ_min)/2 over all three principal stresses; in plane stress the third one is zero.

In-plane principal stresses

C = −70 MPa, R = √(0² + (−40)²) = 40 MPa

σ₁ = −30 MPa, σ₂ = −110 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(80, 30, 110)/2 = 55 MPa

Answer

τ_abs = 55 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: |τ_n| = 25.11 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 σ_h = pr/t and axial stress σ_a = pr/(2t); these are the principal stresses in the wall.

Wall stresses

r = 1500/2 = 750 mm

σ_h = pr/t = 2.5·750/12 = 156.3 MPa

σ_a = pr/(2t) = 78.13 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

|τ_n| = (156.3 − 78.13)·sin 20°·cos 20° = 25.11 MPa

Answer

|τ_n| = 25.11 MPa

7.Axial stress & strain

Answer: P = 60.59 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, P_i = P·E_iA_i/ΣEA, and σ_i = E_i·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 = 706.9 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·706.9) = 4.244 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 200/(100000·2827) = 7.074 × 10⁻⁷ mm/N

Load

P = 0.3/4.951 × 10⁻⁶ = 60588 N = 60.59 kN

Answer

P = 60.59 kN

8.Axial stress & strain

Answer: σ_max = 248.7 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 = 70.74 MPa

σ₂ = 248.7 MPa

Largest stress

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

σ_max = 248.7 MPa

Answer

σ_max = 248.7 MPa

9.Column buckling

Answer: P_cr = 31.09 kN

Euler buckling: a slender column becomes unstable at P_cr = 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.

I_min = h·b³/12 = 80·30³/12 = 180000 mm⁴

Effective length

KL = 2·1000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·180000/2000² = 31089 N = 31.09 kN

Answer

P_cr = 31.09 kN

10.Column buckling

Answer: d = 151.6 mm

Required critical load

P_cr = n·P = 2·140000 = 280000 N

KL = 2·4000 = 8000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·280000·8000²/(π³·70000))^(1/4) = 151.6 mm

Answer

d = 151.6 mm

Check: the slenderness KL/r is 211.1, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.151

Free vibration of a mass–spring system: the undamped natural frequency is ω_n = √(k/m) in rad/s, and f_n = ω_n/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/c_c, with critical damping c_c = 2·√(k·m) = 2·m·ω_n. An underdamped system (ζ < 1) oscillates at ω_d = ω_n·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.

Logarithmic decrement

δ = (1/2)·ln(15/2.2) = 0.9598

Damping ratio

ζ = δ/√(4π² + δ²) = 0.9598/√(39.48 + 0.9212) = 0.151

Answer

ζ = 0.151

12.Mechanical vibrations

Answer: f_n = 2.394 Hz

Equivalent stiffness

Only k₁ and k₂ are in parallel. That pair is in series with k₃, so combine them as reciprocals.

k₁₂ = 80 + 12 = 92 kN/m

k_eq = 92·30/(92 + 30) = 22.62 kN/m

Natural frequency

ω_n = √(22623/100) = 15.04 rad/s

f_n = 15.04/(2π) = 2.394 Hz

Answer

f_n = 2.394 Hz

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