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 80 mm pulley at 720 rpm drives a 400 mm pulley by a belt with 3 % slip. Find the speed n₂ of the driven pulley.

n₁ = 720 rpmd₁ = 80 mmd₂ = 400 mms = 3 %

n₂ = rpm

2.Gears & gear trainsHard

In a compound gear train, gear A (15 teeth) on the input shaft turns at 2880 rpm and drives gear B (18 teeth). Gear C (20 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 960 rpm. How many teeth N_D must gear D have?

n_in = 2880 rpmA = 15B = 18C = 20n_out = 960 rpm

N_D =

3.Torsion of shaftsMedium

A hollow brass shaft (outer diameter 50 mm, inner diameter 38 mm, length 0.8 m) is fixed at one end and twisted by a torque of 400 N·m at the other. Find the angle of twist φ in degrees.

T = 400 N·md_o = 50 mmd_i = 38 mmL = 0.8 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 50 kW at 1800 rpm. Find the maximum shear stress τ_max in the shaft.

P = 50 kWN = 1800 rpmd_o = 75 mmd_i = 56 mm

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 30 MPa and σᵧ = 100 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 120 MPa?

σ_x = 30 MPaσ_y = 100 MPaσ₁ = 120 MPa

τ_xy = MPa

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −20 MPa, σᵧ = 110 MPa and τₓᵧ = 20 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 20° counter-clockwise from the x-axis.

σ_x = −20 MPaσ_y = 110 MPaτ_xy = 20 MPaθ = 20°

σ_θ = MPa

7.Axial stress & strainMedium

A steel rod (diameter 25 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 80 mm, inner diameter 50 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: Ø25 mm, E = 200 GPatube: Ø80/Ø50 mm, E = 100 GPaL = 500 mm

δ = mm

8.Axial stress & strainHard

A steel segment (diameter 20 mm, length 400 mm, E = 200 GPa) and a segment made of bronze (diameter 60 mm, length 300 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: bronze, Ø60 mm, 300 mm, E = 110 GPaP = 20 kN

σ₂ = MPa

9.Column bucklingMedium

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

L = 6 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 40 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 40 kNn = 2.5L = 4 mE = 70 GPa (aluminium)

d = mm

11.Mechanical vibrationsMedium

An 8 kg mass is supported by a spring with a stiffness of 12 kN/m and a viscous damper with a damping coefficient of 310 N·s/m. Find the damping ratio ζ.

m = 8 kgk = 12 kN/mc = 310 N·s/m

ζ =

12.Mechanical vibrationsHard

A 5 kg mass hangs from spring k₃ = 60 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 12 kN/m and k₂ = 40 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 = 12 kN/mk₂ = 40 kN/mk₃ = 60 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: n₂ = 139.7 rpm

Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = e^μθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.

Speed without slip

720·80/400 = 144 rpm

With slip

n₂ = 144·0.97 = 139.7 rpm

Answer

n₂ = 139.7 rpm

2.Gears & gear trains

Answer: N_D = 50

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.

Intermediate shaft speed

n_B = n_C = 2880·15/18 = 2400 rpm

Teeth on gear D

N_D = C·n_C/n_out = 20·2400/960 = 50

Answer

N_D = 50

3.Torsion of shafts

Answer: φ = 1.15 °

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 − 38⁴)/32 = 408885 mm⁴

Angle of twist

T = 400 N·m = 400000 N·mm

L = 0.8 m = 800 mm

G = 39 GPa = 39000 MPa

φ = TL/(GJ) = 400000·800/(39000·408885) = 0.02007 rad

φ = 0.02007·180/π = 1.15°

Answer

φ = 1.15°

4.Torsion of shafts

Answer: τ_max = 4.646 MPa

Torque

ω = 2πN/60 = 2π·1800/60 = 188.5 rad/s

T = P/ω = 50000/188.5 = 265.3 N·m

Polar second moment of area

J = π(75⁴ − 56⁴)/32 = 2.141 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 265258·37.5/2.141 × 10⁶ = 4.646 MPa

Answer

τ_max = 4.646 MPa

5.Mohr’s circle & stress transformation

Answer: τ_xy = 42.43 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.

Required radius

C = (30 + 100)/2 = 65 MPa

R = σ₁ − C = 120 − 65 = 55 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(55² − 35²) = 42.43 MPa

Answer

τ_xy = 42.43 MPa

6.Mohr’s circle & stress transformation

Answer: σ_θ = 8.063 MPa

Transformation equation

σ_θ = 45 + −65·0.766 + 20·0.6428 = 8.063 MPa

Answer

σ_θ = 8.063 MPa

7.Axial stress & strain

Answer: δ = 0.09889 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·25²/4 = 490.9 mm²

A_t = π(80² − 50²)/4 = 3063 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 100·10³·3063 = 4.045 × 10⁸ N

Shortening

δ = 80000·500/4.045 × 10⁸ = 0.09889 mm

Answer

δ = 0.09889 mm

8.Axial stress & strain

Answer: σ₂ = 6.143 MPa

Stiffnesses

k₁ = 200000·314.2/400 = 157080 N/mm

k₂ = 110000·2827/300 = 1.037 × 10⁶ N/mm

Share of segment 2

F₂ = 20000·1.037 × 10⁶/1.194 × 10⁶ = 17368 N

Stress

σ₂ = 17368/2827 = 6.143 MPa (compression)

Answer

σ₂ = 6.143 MPa

9.Column buckling

Answer: P_cr = 12.59 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 = 0.7·6000 = 4200 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·112500/4200² = 12589 N = 12.59 kN

Answer

P_cr = 12.59 kN

10.Column buckling

Answer: d = 58.6 mm

Required critical load

P_cr = n·P = 2.5·40000 = 100000 N

KL = 0.5·4000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·100000·2000²/(π³·70000))^(1/4) = 58.6 mm

Answer

d = 58.6 mm

Check: the slenderness KL/r is 136.5, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.5003

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·√(12000·8) = 619.7 N·s/m

Damping ratio

ζ = c/c_c = 310/619.7 = 0.5003

Answer

ζ = 0.5003

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

12.Mechanical vibrations

Answer: f_n = 11.88 Hz

Equivalent stiffness

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

k₁₂ = 12 + 40 = 52 kN/m

k_eq = 52·60/(52 + 60) = 27.86 kN/m

Natural frequency

ω_n = √(27857/5) = 74.64 rad/s

f_n = 74.64/(2π) = 11.88 Hz

Answer

f_n = 11.88 Hz

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