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 0.75 kW motor at 1450 rpm drives a worm with 1 start(s). The worm wheel has 30 teeth and the drive has an efficiency of 65%. Find the output torque T_w on the wheel shaft.

z = 1N_w = 30n = 1450 rpmP = 0.75 kWη = 65%

T_w = N·m

2.Gears & gear trainsHard

In a compound gear train, gear A (16 teeth) on the input shaft turning at 1500 rpm drives gear B (60 teeth). Gear C (20 teeth) is fixed to the same shaft as B and drives gear D (45 teeth) on the output shaft. Find the output speed n_out.

n_in = 1500 rpmA = 16B = 60C = 20D = 45

n_out = rpm

3.Torsion of shaftsMedium

A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 60 mm and length 3 m; segment BC has diameter 30 mm and length 2.5 m. A torque of 750 N·m acts at the free end C. Find the maximum shear stress τ_max in the shaft.

T = 750 N·md₁ = 60 mm, L₁ = 3 md₂ = 30 mm, L₂ = 2.5 m

τ_max = MPa

4.Torsion of shaftsHard

A solid steel shaft of diameter 50 mm is fixed at A. A torque of 250 N·m acts at B, 1.5 m from A, and a torque of 500 N·m in the same direction acts at the free end C, a further 0.8 m along. Find the magnitude of the angle of twist φ_C of end C, in degrees.

d = 50 mmT_B = 250 N·mT_C = 500 N·mAB = 1.5 mBC = 0.8 mG = 80 GPa

φ_C = °

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 0 MPa, σᵧ = −80 MPa and τₓᵧ = 20 MPa (tension positive). Find the angle θ_p from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σ_x = 0 MPaσ_y = −80 MPaτ_xy = 20 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 90 MPa, σᵧ = 70 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 30° clockwise from the x-axis.

σ_x = 90 MPaσ_y = 70 MPaτ_xy = 30 MPaθ = -30°

σ_θ = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 30 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 60 mm, length 300 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.

segment 1: steel, Ø30 mm, 500 mm, E = 200 GPasegment 2: copper, Ø60 mm, 300 mm, E = 120 GPaδ = 0.4 mm

P = kN

8.Axial stress & strainHard

A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of bronze (outer diameter 80 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: Ø20 mm, E = 200 GPa, σ_allow = 120 MPatube: Ø80/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 both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 2.5 mE = 200 GPa (steel)d = 50 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 25 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 25 kNn = 3L = 2 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

A 5 kg mass on a spring with a stiffness of 150 kN/m has a viscous damper giving a damping ratio of 0.45. Find the damped natural frequency f_d of its free vibration in Hz.

m = 5 kgk = 150 kN/mζ = 0.45

f_d = Hz

12.Mechanical vibrationsHard

A 100 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 15 kN/m and the lower one k₂ = 40 kN/m. Ignore damping and the masses of the springs. Find the natural frequency f_n of vertical vibration in Hz.

m = 100 kgk₁ = 15 kN/mk₂ = 40 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: T_w = 96.32 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.

Wheel speed

n_w = n·z/N_w = 1450·1/30 = 48.33 rpm

Output torque

ω_w = 2π·48.33/60 = 5.061 rad/s

T_w = η·P/ω_w = 0.65·750/5.061 = 96.32 N·m

Answer

T_w = 96.32 N·m

2.Gears & gear trains

Answer: n_out = 177.8 rpm

First stage

n_B = 1500·16/60 = 400 rpm = n_C

Second stage

n_out = 400·20/45 = 177.8 rpm

Answer

n_out = 177.8 rpm

total ratio = 8.438 : 1

3.Torsion of shafts

Answer: τ_max = 141.5 MPa

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.

Stress in each segment

T = 750 N·m = 750000 N·mm

τ_AB = 16·750000/(π·60³) = 17.68 MPa

τ_BC = 16·750000/(π·30³) = 141.5 MPa

Answer

τ_max = 141.5 MPa

4.Torsion of shafts

Answer: φ_C = 1.78 °

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 500 N·m

T_AB = T_C + T_B = 500 + 250 = 750 N·m

Twist of each segment

J = π·50⁴/32 = 613592 mm⁴

G = 80 GPa = 80000 MPa

φ_AB = 750000·1500/(80000·613592) = 0.02292 rad

φ_BC = 500000·800/(80000·613592) = 0.008149 rad

Twist of end C

φ_C = 0.02292 + 0.008149 = 0.03107 rad → |φ_C| = 1.78°

Answer

φ_C = 1.78°

5.Mohr’s circle & stress transformation

Answer: θ_p = 13.28 °

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.

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θ_p = atan2(2·20, 0 − −80) = 26.57°

Angle on the element

θ_p = 13.28°

Answer

θ_p = 13.28°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ_θ = 59.02 MPa

Transformation equation

σ_θ = 80 + 10·0.5 + 30·−0.866 = 59.02 MPa

Answer

σ_θ = 59.02 MPa

7.Axial stress & strain

Answer: P = 90.48 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) = 500/(200000·706.9) = 3.537 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(120000·2827) = 8.842 × 10⁻⁷ mm/N

Load

P = 0.4/4.421 × 10⁻⁶ = 90478 N = 90.48 kN

Answer

P = 90.48 kN

8.Axial stress & strain

Answer: P_max = 239.9 kN

Areas

A_r = π·20²/4 = 314.2 mm²

A_t = π(80² − 50²)/4 = 3063 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 110·10³·3063 = 3.998 × 10⁸ N

Governing strain

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

Largest load

P_max = 0.0006·3.998 × 10⁸ = 239861 N = 239.9 kN

Answer

P_max = 239.9 kN

9.Column buckling

Answer: P_cr = 387.6 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 round section has the same I about every axis.

I = πd⁴/64 = π·50⁴/64 = 306796 mm⁴

Effective length

KL = 0.5·2500 = 1250 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·306796/1250² = 387578 N = 387.6 kN

Answer

P_cr = 387.6 kN

10.Column buckling

Answer: d = 59.32 mm

Required critical load

P_cr = n·P = 3·25000 = 75000 N

KL = 2·2000 = 4000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·75000·4000²/(π³·200000))^(1/4) = 59.32 mm

Answer

d = 59.32 mm

Check: the slenderness KL/r is 269.7, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: f_d = 24.62 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 = √(150000/5) = 173.2 rad/s

f_n = 173.2/(2π) = 27.57 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 27.57·√(1 − 0.45²) = 24.62 Hz

Answer

f_d = 24.62 Hz

12.Mechanical vibrations

Answer: f_n = 1.662 Hz

Equivalent stiffness

Springs in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k₁ + 1/k₂.

1/k_eq = 1/15 + 1/40 → k_eq = 15·40/(15 + 40) = 10.91 kN/m

Natural frequency

ω_n = √(10909/100) = 10.44 rad/s

f_n = 10.44/(2π) = 1.662 Hz

Answer

f_n = 1.662 Hz

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