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 100 mm pulley at 720 rpm drives a 315 mm pulley by a belt with 2 % slip. Find the speed n₂ of the driven pulley.

n₁ = 720 rpmd₁ = 100 mmd₂ = 315 mms = 2 %

n₂ = rpm

2.Gears & gear trainsHard

An open belt connects a 200 mm driving pulley at 720 rpm to a 250 mm pulley; the centres are 1000 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 2000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 200 mmn₁ = 720 rpmd₂ = 250 mmC = 1000 mmμ = 0.3T₁ = 2000 N

P = kW

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 = 750 N·md = 40 mmL = 1 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft must transmit 15 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 = 15 kWN = 1500 rpmd_i/d_o = 0.6τ_allow = 75 MPa

d_o = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 100 MPa, σᵧ = −70 MPa and τₓᵧ = 30 MPa (tension positive). Find the angle θ_p from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σ_x = 100 MPaσ_y = −70 MPaτ_xy = 30 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −80 MPa, σᵧ = 30 MPa and τₓᵧ = −30 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 60° counter-clockwise from the x-axis.

σ_x = −80 MPaσ_y = 30 MPaτ_xy = −30 MPaθ = 60°

σ_θ = MPa

7.Axial stress & strainMedium

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

P = 60 kNd = 12 mmL = 2 mE = 100 GPa

ε = µε

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 = 115 kNsegment 1: Ø35 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 2 m, aluminium (E = 70 GPa)

σ_max = 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 = 3 mE = 200 GPa (steel)b × h = 25 × 100 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 130 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 130 kNn = 2.5L = 4 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

A 500 kg mass on a spring with a stiffness of 50 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 = 500 kgk = 50 kN/mζ = 0.7

f_d = Hz

12.Mechanical vibrationsHard

A 50 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 60 kN/m and the lower one k₂ = 30 kN/m. Ignore damping and the masses of the springs. Find the natural frequency f_n of vertical vibration in Hz.

m = 50 kgk₁ = 60 kN/mk₂ = 30 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: n₂ = 224 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

720·100/315 = 228.6 rpm

With slip

n₂ = 228.6·0.98 = 224 rpm

Answer

n₂ = 224 rpm

2.Gears & gear trains

Answer: P = 9.115 kW

Angle of wrap

θ = 3.092 rad

Slack-side tension

e^μθ = e^0.9275 = 2.528, T₂ = 2000/2.528 = 791.1 N

Power

P = (2000 − 791.1)·7.54 = 9.115 kW

Answer

P = 9.115 kW

3.Torsion of shafts

Answer: φ = 2.137 °

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) = 750·1000·1000/(80000·251327) = 0.0373 rad

φ = 0.0373·180/π = 2.137°

Answer

φ = 2.137°

4.Torsion of shafts

Answer: d_o = 19.53 mm

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 15000/157.1 = 95.49 N·m

Outer diameter from the allowable stress

T = 95.49 N·m = 95493 N·mm

1 − k⁴ = 1 − 0.6⁴ = 0.8704

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·95493/(π·75·0.8704)) = 19.53 mm

d_i = 0.6·19.53 = 11.72 mm

Answer

d_o = 19.53 mm

A solid shaft would need d = 18.65 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: θ_p = 9.72 °

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

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θ_p = atan2(2·30, 100 − −70) = 19.44°

Angle on the element

θ_p = 9.72°

Answer

θ_p = 9.72°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ_θ = −23.48 MPa

Transformation equation

σ_θ = −25 + −55·−0.5 + −30·0.866 = −23.48 MPa

Answer

σ_θ = −23.48 MPa

7.Axial stress & strain

Answer: ε = 5305 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·12²/4 = 113.1 mm²

σ = P/A = 60·1000/113.1 = 530.5 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 530.5/100000 = 0.005305 = 5305 µε

Answer

ε = 5305 µε

8.Axial stress & strain

Answer: σ_max = 1017 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·35²/4 = 962.1 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 119.5 MPa

σ₂ = 1017 MPa

Largest stress

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

σ_max = 1017 MPa

Answer

σ_max = 1017 MPa

9.Column buckling

Answer: P_cr = 58.28 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 = 100·25³/12 = 130208 mm⁴

Effective length

KL = 0.7·3000 = 2100 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·130208/2100² = 58281 N = 58.28 kN

Answer

P_cr = 58.28 kN

10.Column buckling

Answer: d = 71.61 mm

Required critical load

P_cr = n·P = 2.5·130000 = 325000 N

KL = 0.7·4000 = 2800 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·325000·2800²/(π³·200000))^(1/4) = 71.61 mm

Answer

d = 71.61 mm

Check: the slenderness KL/r is 156.4, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: f_d = 1.137 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 = √(50000/500) = 10 rad/s

f_n = 10/(2π) = 1.592 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 1.592·√(1 − 0.7²) = 1.137 Hz

Answer

f_d = 1.137 Hz

12.Mechanical vibrations

Answer: f_n = 3.183 Hz

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 = 1/60 + 1/30 → k_eq = 60·30/(60 + 30) = 20 kN/m

Natural frequency

ω_n = √(20000/50) = 20 rad/s

f_n = 20/(2π) = 3.183 Hz

Answer

f_n = 3.183 Hz

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