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 125 mm driving pulley at 1450 rpm. The tension is 1000 N on the tight side and 200 N on the slack side. Find the power P transmitted, in kW.

d₁ = 125 mmn₁ = 1450 rpmT₁ = 1000 NT₂ = 200 N

P = kW

2.Gears & gear trainsHard

An open belt connects a 150 mm driving pulley at 1450 rpm to a 200 mm pulley; the centres are 600 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 150 mmn₁ = 1450 rpmd₂ = 200 mmC = 600 mmμ = 0.25T₁ = 500 N

P = kW

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 = 250 N·md = 50 mmL = 1 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft must transmit 30 kW at 1500 rpm. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter d_o.

P = 30 kWN = 1500 rpmd_i/d_o = 0.6τ_allow = 75 MPa

d_o = mm

5.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = −30 MPa, σᵧ = −50 MPa and τₓᵧ = 30 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = −30 MPaσ_y = −50 MPaτ_xy = 30 MPaσ_z = 0

τ_abs = MPa

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 0 MPa, σᵧ = −20 MPa and τₓᵧ = 60 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 25° counter-clockwise from the x-axis.

σ_x = 0 MPaσ_y = −20 MPaτ_xy = 60 MPaθ = 25°

σ_θ = MPa

7.Axial stress & strainMedium

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

P = 95 kNd = 30 mmL = 0.5 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 total elongation δ.

P = 65 kNsegment 1: Ø35 mm, 2 m, steel (E = 200 GPa)segment 2: Ø12 mm, 2 m, aluminium (E = 70 GPa)

δ = mm

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 = 2.5 mE = 200 GPa (steel)d = 80 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 100 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 100 kNn = 3L = 4 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 30 mm is followed 3 cycles later by a peak of 17 mm. Find the damping ratio ζ.

x₀ = 30 mmx₃ = 17 mmn = 3

ζ =

12.Mechanical vibrationsHard

A 10 kg mass hangs from spring k₃ = 5 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 30 kN/m and k₂ = 20 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 = 10 kgk₁ = 30 kN/mk₂ = 20 kN/mk₃ = 5 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: P = 7.592 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 = 9.49 m/s

Power

P = (1000 − 200)·9.49 = 7592 W = 7.592 kW

Answer

P = 7.592 kW

2.Gears & gear trains

Answer: P = 3.043 kW

Angle of wrap

θ = 3.058 rad

Slack-side tension

e^μθ = e^0.7646 = 2.148, T₂ = 500/2.148 = 232.8 N

Power

P = (500 − 232.8)·11.39 = 3.043 kW

Answer

P = 3.043 kW

3.Torsion of shafts

Answer: φ = 0.5986 °

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) = 250·1000·1000/(39000·613592) = 0.01045 rad

φ = 0.01045·180/π = 0.5986°

Answer

φ = 0.5986°

4.Torsion of shafts

Answer: d_o = 24.61 mm

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 30000/157.1 = 191 N·m

Outer diameter from the allowable stress

T = 191 N·m = 190986 N·mm

1 − k⁴ = 1 − 0.6⁴ = 0.8704

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·190986/(π·75·0.8704)) = 24.61 mm

d_i = 0.6·24.61 = 14.76 mm

Answer

d_o = 24.61 mm

A solid shaft would need d = 23.49 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: τ_abs = 35.81 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. In three dimensions the absolute maximum shear stress is (σ_max − σ_min)/2 over all three principal stresses; in plane stress the third one is zero.

In-plane principal stresses

C = −40 MPa, R = √(10² + 30²) = 31.62 MPa

σ₁ = −8.377 MPa, σ₂ = −71.62 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(63.25, 8.377, 71.62)/2 = 35.81 MPa

Answer

τ_abs = 35.81 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

6.Mohr’s circle & stress transformation

Answer: σ_θ = 42.39 MPa

Transformation equation

σ_θ = −10 + 10·0.6428 + 60·0.766 = 42.39 MPa

Answer

σ_θ = 42.39 MPa

7.Axial stress & strain

Answer: δ = 0.96 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·30²/4 = 706.9 mm²

σ = P/A = 95·1000/706.9 = 134.4 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 134.4/70000 = 0.00192 = 1920 µε

Elongation

δ = ε·L = 0.00192·500 mm = 0.96 mm

Answer

δ = 0.96 mm

8.Axial stress & strain

Answer: δ = 17.1 mm

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·35²/4 = 962.1 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 67.56 MPa

σ₂ = 574.7 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 67.56·2000/200000 = 0.6756 mm

δ₂ = σ₂L₂/E₂ = 574.7·2000/70000 = 16.42 mm

δ = δ₁ + δ₂ = 17.1 mm

Answer

δ = 17.1 mm

9.Column buckling

Answer: P_cr = 158.8 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 2·2500 = 5000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/5000² = 158752 N = 158.8 kN

Answer

P_cr = 158.8 kN

10.Column buckling

Answer: d = 118.6 mm

Required critical load

P_cr = n·P = 3·100000 = 300000 N

KL = 2·4000 = 8000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·300000·8000²/(π³·200000))^(1/4) = 118.6 mm

Answer

d = 118.6 mm

Check: the slenderness KL/r is 269.7, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.03012

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(30/17) = 0.1893

Damping ratio

ζ = δ/√(4π² + δ²) = 0.1893/√(39.48 + 0.03585) = 0.03012

Answer

ζ = 0.03012

12.Mechanical vibrations

Answer: f_n = 3.393 Hz

Equivalent stiffness

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

k₁₂ = 30 + 20 = 50 kN/m

k_eq = 50·5/(50 + 5) = 4.545 kN/m

Natural frequency

ω_n = √(4545/10) = 21.32 rad/s

f_n = 21.32/(2π) = 3.393 Hz

Answer

f_n = 3.393 Hz

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