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 1440 rpm drives a worm with 2 start(s). The worm wheel has 40 teeth and the drive has an efficiency of 60%. Find the output torque T_w on the wheel shaft.

z = 2N_w = 40n = 1440 rpmP = 0.75 kWη = 60%

T_w = N·m

2.Gears & gear trainsHard

In a compound gear train, gear A (30 teeth) on the input shaft turning at 1440 rpm drives gear B (80 teeth). Gear C (24 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 = 1440 rpmA = 30B = 80C = 24D = 60

n_out = rpm

3.Torsion of shaftsMedium

A solid brass 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 = 200 N·md = 50 mmL = 3 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 30 kW at 3000 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.

P = 30 kWN = 3000 rpmτ_allow = 80 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 70 MPa, σᵧ = −30 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 = 70 MPaσ_y = −30 MPaτ_xy = 20 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 130 MPa, σᵧ = 100 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 50° counter-clockwise from the x-axis.

σ_x = 130 MPaσ_y = 100 MPaτ_xy = 30 MPaθ = 50°

σ_θ = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 20 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 40 mm, length 300 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.

segment 1: steel, Ø20 mm, 600 mm, E = 200 GPasegment 2: brass, Ø40 mm, 300 mm, E = 100 GPaδ = 0.4 mm

P = kN

8.Axial stress & strainHard

A steel segment (diameter 30 mm, length 500 mm, E = 200 GPa) and a segment made of copper (diameter 40 mm, length 500 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 100 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø30 mm, 500 mm, E = 200 GPasegment 2: copper, Ø40 mm, 500 mm, E = 120 GPaP = 100 kN

σ₂ = MPa

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 = 5 mE = 200 GPa (steel)d = 20 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 10 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 10 kNn = 2.5L = 1.5 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed 3 cycles later by a peak of 5.1 mm. Find the damping ratio ζ.

x₀ = 50 mmx₃ = 5.1 mmn = 3

ζ =

12.Mechanical vibrationsHard

A 5 kg mass hangs from spring k₃ = 25 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 100 kN/m and k₂ = 30 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 = 5 kgk₁ = 100 kN/mk₂ = 30 kN/mk₃ = 25 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: T_w = 59.68 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 = 1440·2/40 = 72 rpm

Output torque

ω_w = 2π·72/60 = 7.54 rad/s

T_w = η·P/ω_w = 0.6·750/7.54 = 59.68 N·m

Answer

T_w = 59.68 N·m

2.Gears & gear trains

Answer: n_out = 216 rpm

First stage

n_B = 1440·30/80 = 540 rpm = n_C

Second stage

n_out = 540·24/60 = 216 rpm

Answer

n_out = 216 rpm

total ratio = 6.667 : 1

3.Torsion of shafts

Answer: φ = 1.437 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 200·1000·3000/(39000·613592) = 0.02507 rad

φ = 0.02507·180/π = 1.437°

Answer

φ = 1.437°

4.Torsion of shafts

Answer: d = 18.25 mm

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s

T = P/ω = 30000/314.2 = 95.49 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·95493/(π·80)) = 18.25 mm

Answer

d = 18.25 mm

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

5.Mohr’s circle & stress transformation

Answer: θ_p = 10.9 °

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, 70 − −30) = 21.8°

Angle on the element

θ_p = 10.9°

Answer

θ_p = 10.9°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ_θ = 141.9 MPa

Transformation equation

σ_θ = 115 + 15·−0.1736 + 30·0.9848 = 141.9 MPa

Answer

σ_θ = 141.9 MPa

7.Axial stress & strain

Answer: P = 33.51 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 = 314.2 mm², A₂ = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·314.2) = 9.549 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(100000·1257) = 2.387 × 10⁻⁶ mm/N

Load

P = 0.4/1.194 × 10⁻⁵ = 33510 N = 33.51 kN

Answer

P = 33.51 kN

8.Axial stress & strain

Answer: σ₂ = 41.07 MPa

Stiffnesses

k₁ = 200000·706.9/500 = 282743 N/mm

k₂ = 120000·1257/500 = 301593 N/mm

Share of segment 2

F₂ = 100000·301593/584336 = 51613 N

Stress

σ₂ = 51613/1257 = 41.07 MPa (compression)

Answer

σ₂ = 41.07 MPa

9.Column buckling

Answer: P_cr = 2.481 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·20⁴/64 = 7854 mm⁴

Effective length

KL = 0.5·5000 = 2500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·7854/2500² = 2481 N = 2.481 kN

Answer

P_cr = 2.481 kN

10.Column buckling

Answer: d = 39.04 mm

Required critical load

P_cr = n·P = 2.5·10000 = 25000 N

KL = 2·1500 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·25000·3000²/(π³·200000))^(1/4) = 39.04 mm

Answer

d = 39.04 mm

Check: the slenderness KL/r is 307.4, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.1202

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 = (1/3)·ln(50/5.1) = 0.7609

Damping ratio

ζ = δ/√(4π² + δ²) = 0.7609/√(39.48 + 0.579) = 0.1202

Answer

ζ = 0.1202

12.Mechanical vibrations

Answer: f_n = 10.31 Hz

Equivalent stiffness

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

k₁₂ = 100 + 30 = 130 kN/m

k_eq = 130·25/(130 + 25) = 20.97 kN/m

Natural frequency

ω_n = √(20968/5) = 64.76 rad/s

f_n = 64.76/(2π) = 10.31 Hz

Answer

f_n = 10.31 Hz

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