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 belt runs on a 100 mm driving pulley at 960 rpm. The tension is 1500 N on the tight side and 750 N on the slack side. Find the power P transmitted, in kW.

d₁ = 100 mmn₁ = 960 rpmT₁ = 1500 NT₂ = 750 N

P = kW

2.Gears & gear trainsHard

In a compound gear train, gear A (36 teeth) on the input shaft turns at 1500 rpm and drives gear B (60 teeth). Gear C (24 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 600 rpm. How many teeth N_D must gear D have?

n_in = 1500 rpmA = 36B = 60C = 24n_out = 600 rpm

N_D =

3.Torsion of shaftsMedium

A solid aluminium 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 = 1000 N·md = 40 mmL = 1.2 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid aluminium shaft transmits 30 kW at 1450 rpm. The shear stress must not exceed 80 MPa and the twist must not exceed 0.5° per m of length. Find the smallest diameter d that satisfies both limits.

P = 30 kWN = 1450 rpmτ_allow = 80 MPaθ_allow = 0.5°/mG = 26 GPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = −80 MPa, σᵧ = −60 MPa and τₓᵧ = 30 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 = −80 MPaσ_y = −60 MPaτ_xy = 30 MPa

θ_s = °

6.Mohr’s circle & stress transformationHard

A solid shaft of diameter 50 mm carries an axial compressive force of 100 kN together with a torque of 400 N·m. At the surface of the shaft, find the largest principal stress σ₁.

d = 50 mmF = −100 kNT = 400 N·m

σ₁ = MPa

7.Axial stress & strainMedium

Find the elongation δ of the aluminium bar under the axial load.

P = 100 kNd = 25 mmL = 2.5 mE = 70 GPa

δ = mm

8.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σ_max in the bar.

P = 60 kNsegment 1: Ø30 mm, 1.5 m, copper (E = 117 GPa)segment 2: Ø20 mm, 1 m, brass (E = 100 GPa)

σ_max = MPa

9.Column bucklingMedium

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

L = 4 mE = 70 GPa (aluminium)b × h = 50 × 80 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of aluminium is pinned at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 3 mE = 70 GPa (aluminium)d = 40 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 30 mm is followed 4 cycles later by a peak of 8.5 mm. Find the damping ratio ζ.

x₀ = 30 mmx₄ = 8.5 mmn = 4

ζ =

12.Mechanical vibrationsHard

A 40 kg mass hangs from spring k₃ = 10 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 = 100 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 = 40 kgk₁ = 100 kN/mk₂ = 100 kN/mk₃ = 10 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: P = 3.77 kW

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.

Belt speed

v = 5.027 m/s

Power

P = (1500 − 750)·5.027 = 3770 W = 3.77 kW

Answer

P = 3.77 kW

2.Gears & gear trains

Answer: N_D = 36

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 = 1500·36/60 = 900 rpm

Teeth on gear D

N_D = C·n_C/n_out = 24·900/600 = 36

Answer

N_D = 36

3.Torsion of shafts

Answer: φ = 10.52 °

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) = 1000·1000·1200/(26000·251327) = 0.1836 rad

φ = 0.1836·180/π = 10.52°

Answer

φ = 10.52°

4.Torsion of shafts

Answer: d = 54.57 mm

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 30000/151.8 = 197.6 N·m

Diameter for strength

T = 197.6 N·m = 197572 N·mm

d_τ = ∛(16·197572/(π·80)) = 23.26 mm

Diameter for stiffness

θ = 0.5·π/180/1000 = 8.727 × 10⁻⁶ rad/mm

G = 26 GPa = 26000 MPa

d_φ = (32·197572/(π·26000·8.727 × 10⁻⁶))^1/4 = 54.57 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(23.26, 54.57) = 54.57 mm

Answer

d = 54.57 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: θ_s = 9.217 °

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 = 30/31.62 = 0.9487, sin 2θ_s = −(−10)/31.62 = 0.3162

2θ_s = atan2(10, 30) = 18.43°

Angle on the element

θ_s = 9.217°

check: θ_s − θ_p = 9.217 − 54.22 = −45°

Answer

θ_s = 9.217°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ₁ = 4.769 MPa

Stresses at the surface

σ = F/A = −100000/(π·50²/4) = −100000/1963 = −50.93 MPa

τ = 16T/(πd³) = 16·400000/(π·50³) = 16.3 MPa

Mohr’s circle

C = σ/2 = −25.46 MPa

R = √((−25.46)² + 16.3²) = 30.23 MPa

σ₁ = C + R = 4.769 MPa

σ₂ = C − R = −55.7 MPa

Answer

σ₁ = 4.769 MPa

7.Axial stress & strain

Answer: δ = 7.276 mm

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·25²/4 = 490.9 mm²

σ = P/A = 100·1000/490.9 = 203.7 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 203.7/70000 = 0.00291 = 2910 µε

Elongation

δ = ε·L = 0.00291·2500 mm = 7.276 mm

Answer

δ = 7.276 mm

8.Axial stress & strain

Answer: σ_max = 191 MPa

Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.

Same force in both segments

The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.

A₁ = π·30²/4 = 706.9 mm², A₂ = π·20²/4 = 314.2 mm²

σ₁ = 84.88 MPa

σ₂ = 191 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 191 MPa

Answer

σ_max = 191 MPa

9.Column buckling

Answer: P_cr = 8.996 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 = 80·50³/12 = 833333 mm⁴

Effective length

KL = 2·4000 = 8000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·833333/8000² = 8996 N = 8.996 kN

Answer

P_cr = 8.996 kN

10.Column buckling

Answer: σ_cr = 7.676 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²

r = √(I/A) = √(125664/1257) = 10 mm

KL = 1·3000 = 3000 mm

KL/r = 3000/10 = 300

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/300² = 7.676 MPa

Answer

σ_cr = 7.676 MPa

11.Mechanical vibrations

Answer: ζ = 0.05012

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 = (1/4)·ln(30/8.5) = 0.3153

Damping ratio

ζ = δ/√(4π² + δ²) = 0.3153/√(39.48 + 0.0994) = 0.05012

Answer

ζ = 0.05012

12.Mechanical vibrations

Answer: f_n = 2.456 Hz

Equivalent stiffness

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

k₁₂ = 100 + 100 = 200 kN/m

k_eq = 200·10/(200 + 10) = 9.524 kN/m

Natural frequency

ω_n = √(9524/40) = 15.43 rad/s

f_n = 15.43/(2π) = 2.456 Hz

Answer

f_n = 2.456 Hz

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