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

A worm with 3 start(s) turns at 2900 rpm and drives a worm wheel with 40 teeth. Find the speed n_w of the wheel.

z = 3N_w = 40n = 2900 rpm

n_w = rpm

2.Gears & gear trainsHard

In a compound gear train, gear A (15 teeth) on the input shaft turning at 1450 rpm drives gear B (60 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D (30 teeth) on the output shaft. Find the output speed n_out.

n_in = 1450 rpmA = 15B = 60C = 15D = 30

n_out = rpm

3.Torsion of shaftsMedium

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 = 750 N·md = 30 mmL = 1.2 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid brass shaft of diameter 50 mm is fixed at A. A torque of 500 N·m acts at B, 1 m from A, and a torque of 1000 N·m in the opposite direction acts at the free end C, a further 1 m along. Find the magnitude of the angle of twist φ_C of end C, in degrees.

d = 50 mmT_B = 500 N·mT_C = 1000 N·mAB = 1 mBC = 1 mG = 39 GPa

φ_C = °

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −70 MPa, σᵧ = 90 MPa and τₓᵧ = −40 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = −70 MPaσ_y = 90 MPaτ_xy = −40 MPa

τ_max = MPa

6.Mohr’s circle & stress transformationHard

A solid shaft of diameter 25 mm carries an axial tensile force of 130 kN together with a torque of 400 N·m. At the surface of the shaft, find the largest principal stress σ₁.

d = 25 mmF = 130 kNT = 400 N·m

σ₁ = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 30 mm, length 200 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.

segment 1: steel, Ø40 mm, 600 mm, E = 200 GPasegment 2: copper, Ø30 mm, 200 mm, E = 120 GPaδ = 0.2 mm

P = kN

8.Axial stress & strainHard

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 50 mm, E = 110 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø25 mm, E = 200 GPa, σ_allow = 120 MPatube: Ø100/Ø50 mm, E = 110 GPa, σ_allow = 100 MPa

P_max = kN

9.Column bucklingMedium

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

L = 3 mE = 200 GPa (steel)b × h = 30 × 50 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of aluminium is fixed at both ends. It must carry 25 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 25 kNn = 3L = 2.5 mE = 70 GPa (aluminium)

d = mm

11.Mechanical vibrationsMedium

A 50 kg mass on a spring with a stiffness of 8 kN/m has a viscous damper giving a damping ratio of 0.3. Find the damped natural frequency f_d of its free vibration in Hz.

m = 50 kgk = 8 kN/mζ = 0.3

f_d = Hz

12.Mechanical vibrationsHard

A 5 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 5 kN/m and k₂ = 20 kN/m. The mass stays level. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 5 kgk₁ = 5 kN/mk₂ = 20 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: n_w = 217.5 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

n_w = n·z/N_w = 2900·3/40 = 217.5 rpm

Answer

n_w = 217.5 rpm

2.Gears & gear trains

Answer: n_out = 181.3 rpm

First stage

n_B = 1450·15/60 = 362.5 rpm = n_C

Second stage

n_out = 362.5·15/30 = 181.3 rpm

Answer

n_out = 181.3 rpm

total ratio = 8 : 1

3.Torsion of shafts

Answer: φ = 8.106 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 750·1000·1200/(80000·79522) = 0.1415 rad

φ = 0.1415·180/π = 8.106°

Answer

φ = 8.106°

4.Torsion of shafts

Answer: φ_C = 3.591 °

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 1000 N·m

T_AB = T_C − T_B = 1000 − 500 = 500 N·m

Twist of each segment

J = π·50⁴/32 = 613592 mm⁴

G = 39 GPa = 39000 MPa

φ_AB = 500000·1000/(39000·613592) = 0.02089 rad

φ_BC = 1 × 10⁶·1000/(39000·613592) = 0.04179 rad

Twist of end C

φ_C = 0.02089 + 0.04179 = 0.06268 rad → |φ_C| = 3.591°

Answer

φ_C = 3.591°

5.Mohr’s circle & stress transformation

Answer: τ_max = 89.44 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.

Radius of Mohr’s circle

τ_max = R = √(((−70 − 90)/2)² + −40²) = 89.44 MPa

Answer

τ_max = 89.44 MPa

6.Mohr’s circle & stress transformation

Answer: σ₁ = 318.2 MPa

Stresses at the surface

σ = F/A = 130000/(π·25²/4) = 130000/490.9 = 264.8 MPa

τ = 16T/(πd³) = 16·400000/(π·25³) = 130.4 MPa

Mohr’s circle

C = σ/2 = 132.4 MPa

R = √(132.4² + 130.4²) = 185.8 MPa

σ₁ = C + R = 318.2 MPa

σ₂ = C − R = −53.41 MPa

Answer

σ₁ = 318.2 MPa

7.Axial stress & strain

Answer: P = 42.15 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 = 1257 mm², A₂ = 706.9 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·1257) = 2.387 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 200/(120000·706.9) = 2.358 × 10⁻⁶ mm/N

Load

P = 0.2/4.745 × 10⁻⁶ = 42148 N = 42.15 kN

Answer

P = 42.15 kN

8.Axial stress & strain

Answer: P_max = 447.7 kN

Areas

A_r = π·25²/4 = 490.9 mm²

A_t = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 110·10³·5890 = 7.461 × 10⁸ N

Governing strain

ε = min(0.0006, 0.0009091) = 0.0006 (steel)

Largest load

P_max = 0.0006·7.461 × 10⁸ = 447677 N = 447.7 kN

Answer

P_max = 447.7 kN

9.Column buckling

Answer: P_cr = 6.169 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 = 50·30³/12 = 112500 mm⁴

Effective length

KL = 2·3000 = 6000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·112500/6000² = 6169 N = 6.169 kN

Answer

P_cr = 6.169 kN

10.Column buckling

Answer: d = 43.12 mm

Required critical load

P_cr = n·P = 3·25000 = 75000 N

KL = 0.5·2500 = 1250 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·75000·1250²/(π³·70000))^(1/4) = 43.12 mm

Answer

d = 43.12 mm

Check: the slenderness KL/r is 116, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: f_d = 1.92 Hz

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.

Undamped natural frequency

ω_n = √(8000/50) = 12.65 rad/s

f_n = 12.65/(2π) = 2.013 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 2.013·√(1 − 0.3²) = 1.92 Hz

Answer

f_d = 1.92 Hz

12.Mechanical vibrations

Answer: k_eq = 25 kN/m

Equivalent stiffness

Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k₁ + k₂.

k_eq = 5 + 20 = 25 kN/m

Answer

k_eq = 25 kN/m

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