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

An 11 kW motor running at 2900 rpm drives a pinion with 30 teeth, which meshes with a gear with 80 teeth. Ignore losses. Find the torque T₁ on the motor shaft.

P = 11 kWn₁ = 2900 rpmN₁ = 30N₂ = 80η = 100%

T₁ = N·m

2.Gears & gear trainsHard

A 5.5 kW motor at 1440 rpm drives gear A (16 teeth), which meshes with gear B (80 teeth). Gear C (16 teeth) on the same shaft as B drives gear D (20 teeth) on the output shaft. Each gear mesh has an efficiency of 98%. Find the output torque T_out.

P = 5.5 kWn_in = 1440 rpmA = 16B = 80C = 16D = 20η_mesh = 98%

T_out = N·m

3.Torsion of shaftsMedium

A hollow aluminium shaft (outer diameter 75 mm, inner diameter 53 mm, length 2 m) is fixed at one end and twisted by a torque of 200 N·m at the other. Find the angle of twist φ in degrees.

T = 200 N·md_o = 75 mmd_i = 53 mmL = 2 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid brass shaft transmits 5 kW at 500 rpm. The shear stress must not exceed 75 MPa and the twist must not exceed 0.25° per m of length. Find the smallest diameter d that satisfies both limits.

P = 5 kWN = 500 rpmτ_allow = 75 MPaθ_allow = 0.25°/mG = 39 GPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 130 MPa, σᵧ = 70 MPa and τₓᵧ = 50 MPa (tension positive). Find the angle θ_s, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σ_x = 130 MPaσ_y = 70 MPaτ_xy = 50 MPa

θ_s = °

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 30 MPa, σᵧ = 50 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 40° counter-clockwise from the x-axis.

σ_x = 30 MPaσ_y = 50 MPaτ_xy = 30 MPaθ = 40°

σ_θ = MPa

7.Axial stress & strainMedium

Find the axial strain ε in the steel bar, in microstrain (µε = 10⁻⁶).

P = 30 kNd = 10 mmL = 0.5 mE = 200 GPa

ε = µε

8.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 60 mm, E = 110 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø30 mm, E = 200 GPa, σ_allow = 250 MPatube: Ø100/Ø60 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 = 20 × 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 85 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 85 kNn = 2.5L = 1 mE = 70 GPa (aluminium)

d = mm

11.Mechanical vibrationsMedium

A 500 kg mass is supported by a spring with a stiffness of 100 kN/m and a viscous damper with a damping coefficient of 1130 N·s/m. Find the damping ratio ζ.

m = 500 kgk = 100 kN/mc = 1130 N·s/m

ζ =

12.Mechanical vibrationsHard

A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 30 kN/m and k₂ = 100 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency f_n of vertical vibration in Hz.

m = 25 kgk₁ = 30 kN/mk₂ = 100 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: T₁ = 36.22 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·2900/60 = 303.7 rad/s

T₁ = P/ω₁ = 11000/303.7 = 36.22 N·m

Answer

T₁ = 36.22 N·m

2.Gears & gear trains

Answer: T_out = 218.9 N·m

Output speed

n_out = 1440·(16/80)·(16/20) = 230.4 rpm

ω_out = 2π·230.4/60 = 24.13 rad/s

Output power and torque

P_out = 5500·0.98² = 5282 W

T_out = P_out/ω_out = 5282/24.13 = 218.9 N·m

Answer

T_out = 218.9 N·m

3.Torsion of shafts

Answer: φ = 0.378 °

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(75⁴ − 53⁴)/32 = 2.332 × 10⁶ mm⁴

Angle of twist

T = 200 N·m = 200000 N·mm

L = 2 m = 2000 mm

G = 26 GPa = 26000 MPa

φ = TL/(GJ) = 200000·2000/(26000·2.332 × 10⁶) = 0.006598 rad

φ = 0.006598·180/π = 0.378°

Answer

φ = 0.378°

4.Torsion of shafts

Answer: d = 48.9 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 5000/52.36 = 95.49 N·m

Diameter for strength

T = 95.49 N·m = 95493 N·mm

d_τ = ∛(16·95493/(π·75)) = 18.65 mm

Diameter for stiffness

θ = 0.25·π/180/1000 = 4.363 × 10⁻⁶ rad/mm

G = 39 GPa = 39000 MPa

d_φ = (32·95493/(π·39000·4.363 × 10⁻⁶))^1/4 = 48.9 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(18.65, 48.9) = 48.9 mm

Answer

d = 48.9 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: θ_s = −15.48 °

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

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θ_s = 50/58.31 = 0.8575, sin 2θ_s = −30/58.31 = −0.5145

2θ_s = atan2(−30, 50) = −30.96°

Angle on the element

θ_s = −15.48°

check: θ_s − θ_p = −15.48 − 29.52 = −45°

Answer

θ_s = −15.48°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ_θ = 67.81 MPa

Transformation equation

σ_θ = 40 + −10·0.1736 + 30·0.9848 = 67.81 MPa

Answer

σ_θ = 67.81 MPa

7.Axial stress & strain

Answer: ε = 1910 µε

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·10²/4 = 78.54 mm²

σ = P/A = 30·1000/78.54 = 382 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 382/200000 = 0.00191 = 1910 µε

Answer

ε = 1910 µε

8.Axial stress & strain

Answer: P_max = 631.2 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

A_r = π·30²/4 = 706.9 mm²

A_t = π(100² − 60²)/4 = 5027 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 110·10³·5027 = 6.943 × 10⁸ N

Governing strain

ε = min(0.00125, 0.0009091) = 0.0009091 (bronze)

Largest load

P_max = 0.0009091·6.943 × 10⁸ = 631175 N = 631.2 kN

Answer

P_max = 631.2 kN

9.Column buckling

Answer: P_cr = 2.924 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 = 80·20³/12 = 53333 mm⁴

Effective length

KL = 2·3000 = 6000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·53333/6000² = 2924 N = 2.924 kN

Answer

P_cr = 2.924 kN

10.Column buckling

Answer: d = 70.76 mm

Required critical load

P_cr = n·P = 2.5·85000 = 212500 N

KL = 2·1000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·212500·2000²/(π³·70000))^(1/4) = 70.76 mm

Answer

d = 70.76 mm

Check: the slenderness KL/r is 113.1, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.0799

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.

Critical damping

c_c = 2·√(k·m) = 2·√(100000·500) = 14142 N·s/m

Damping ratio

ζ = c/c_c = 1130/14142 = 0.0799

Answer

ζ = 0.0799

ζ < 1, so the system is underdamped and oscillates as it decays.

12.Mechanical vibrations

Answer: f_n = 11.48 Hz

Equivalent stiffness

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

k_eq = 30 + 100 = 130 kN/m

Natural frequency

ω_n = √(130000/25) = 72.11 rad/s

f_n = 72.11/(2π) = 11.48 Hz

Answer

f_n = 11.48 Hz

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