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 3 kW motor running at 960 rpm drives a pinion with 20 teeth, which meshes with a gear with 72 teeth. Ignore losses. Find the output torque T₂ on the gear shaft.

P = 3 kWn₁ = 960 rpmN₁ = 20N₂ = 72η = 100%

T₂ = N·m

2.Gears & gear trainsHard

In a planetary gear set the sun gear has 24 teeth and each planet gear has 20 teeth. The planet carrier is held fixed and the sun is driven at 1800 rpm. Find the speed n_r of the ring gear (magnitude).

N_s = 24N_p = 20n_s = 1800 rpm

n_r = 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 = 400 N·md = 40 mmL = 2 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft must transmit 75 kW at 1000 rpm. Its inner diameter is to be 0.8 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter d_o.

P = 75 kWN = 1000 rpmd_i/d_o = 0.8τ_allow = 50 MPa

d_o = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 120 MPa, σᵧ = −80 MPa and τₓᵧ = −40 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 120 MPaσ_y = −80 MPaτ_xy = −40 MPa

τ_max = MPa

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 1000 mm and wall thickness 8 mm holds gas at a gauge pressure of 1.2 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 = 1000 mmt = 8 mmp = 1.2 MPaα = 20°

|τ_n| = MPa

7.Axial stress & strainMedium

A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 80 mm, inner diameter 60 mm, E = 100 GPa). Both are 500 mm long and are compressed together between rigid plates by 80 kN. Find the shortening δ.

P = 80 kNrod: Ø30 mm, E = 200 GPatube: Ø80/Ø60 mm, E = 100 GPaL = 500 mm

δ = mm

8.Axial stress & strainHard

A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of bronze (outer diameter 120 mm, inner diameter 70 mm, E = 110 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø20 mm, E = 200 GPa, σ_allow = 160 MPatube: Ø120/Ø70 mm, E = 110 GPa, σ_allow = 70 MPa

P_max = kN

9.Column bucklingMedium

A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 2.5 mE = 70 GPa (aluminium)b × h = 25 × 80 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of steel is fixed at both ends. It must carry 110 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 110 kNn = 2.5L = 2 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

A 25 kg mass on a spring with a stiffness of 15 kN/m has a viscous damper giving a damping ratio of 0.7. Find the damped natural frequency f_d of its free vibration in Hz.

m = 25 kgk = 15 kN/mζ = 0.7

f_d = Hz

12.Mechanical vibrationsHard

A 10 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 80 kN/m and the lower one k₂ = 12 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 10 kgk₁ = 80 kN/mk₂ = 12 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: T₂ = 107.4 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·960/60 = 100.5 rad/s

T₁ = P/ω₁ = 3000/100.5 = 29.84 N·m

Output torque

T₂ = 29.84·72/20·1 = 107.4 N·m

Answer

T₂ = 107.4 N·m

2.Gears & gear trains

Answer: n_r = 675 rpm

Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (n_s − n_c)/(n_r − n_c) = −N_r/N_s: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.

Ring gear teeth

N_r = N_s + 2·N_p = 24 + 2·20 = 64

Willis equation with the carrier fixed

With n_c = 0 the sun drives the ring through the planets, which act as idlers.

n_r = −n_s·N_s/N_r = −1800·24/64 = −675 rpm

|n_r| = 675 rpm

Answer

n_r = 675 rpm

The ring turns opposite to the sun.

3.Torsion of shafts

Answer: φ = 2.28 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·2000/(80000·251327) = 0.03979 rad

φ = 0.03979·180/π = 2.28°

Answer

φ = 2.28°

4.Torsion of shafts

Answer: d_o = 49.81 mm

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 75000/104.7 = 716.2 N·m

Outer diameter from the allowable stress

T = 716.2 N·m = 716197 N·mm

1 − k⁴ = 1 − 0.8⁴ = 0.5904

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·716197/(π·50·0.5904)) = 49.81 mm

d_i = 0.8·49.81 = 39.85 mm

Answer

d_o = 49.81 mm

A solid shaft would need d = 41.78 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: τ_max = 107.7 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 = √(((120 − −80)/2)² + −40²) = 107.7 MPa

Answer

τ_max = 107.7 MPa

6.Mohr’s circle & stress transformation

Answer: |τ_n| = 12.05 MPa

Wall stresses

r = 1000/2 = 500 mm

σ_h = pr/t = 1.2·500/8 = 75 MPa

σ_a = pr/(2t) = 37.5 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| = (75 − 37.5)·sin 20°·cos 20° = 12.05 MPa

Answer

|τ_n| = 12.05 MPa

7.Axial stress & strain

Answer: δ = 0.1107 mm

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(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 100·10³·2199 = 3.613 × 10⁸ N

Shortening

δ = 80000·500/3.613 × 10⁸ = 0.1107 mm

Answer

δ = 0.1107 mm

8.Axial stress & strain

Answer: P_max = 562.3 kN

Areas

A_r = π·20²/4 = 314.2 mm²

A_t = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 110·10³·7461 = 8.836 × 10⁸ N

Governing strain

ε = min(0.0008, 0.0006364) = 0.0006364 (bronze)

Largest load

P_max = 0.0006364·8.836 × 10⁸ = 562274 N = 562.3 kN

Answer

P_max = 562.3 kN

9.Column buckling

Answer: P_cr = 46.06 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 = 80·25³/12 = 104167 mm⁴

Effective length

KL = 0.5·2500 = 1250 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·104167/1250² = 46058 N = 46.06 kN

Answer

P_cr = 46.06 kN

10.Column buckling

Answer: d = 41.04 mm

Required critical load

P_cr = n·P = 2.5·110000 = 275000 N

KL = 0.5·2000 = 1000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·275000·1000²/(π³·200000))^(1/4) = 41.04 mm

Answer

d = 41.04 mm

Check: the slenderness KL/r is 97.45, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: f_d = 2.784 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 = √(15000/25) = 24.49 rad/s

f_n = 24.49/(2π) = 3.898 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 3.898·√(1 − 0.7²) = 2.784 Hz

Answer

f_d = 2.784 Hz

12.Mechanical vibrations

Answer: k_eq = 10.43 kN/m

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 = 1/80 + 1/12 → k_eq = 80·12/(80 + 12) = 10.43 kN/m

Answer

k_eq = 10.43 kN/m

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