Machine Design: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 1450 rpm drives a pinion with 18 teeth and a module of 2.5 mm. Find the tangential force F_t on the pinion teeth.

P = 3 kWn = 1450 rpmN₁ = 18m = 2.5 mm

F_t = N

2.Gears & gear trainsHard

An open belt connects a 120 mm pulley and a 250 mm pulley whose centres are 500 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d₁ = 120 mmd₂ = 250 mmC = 500 mm

θ = °

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 = 750 N·md = 40 mmL = 0.5 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid steel shaft of diameter 50 mm is fixed at A. A torque of 250 N·m acts at B, 3 m from A, and a torque of 1000 N·m in the opposite direction acts at the free end C, a further 2.5 m along. Find the magnitude of the angle of twist φ_C of end C, in degrees.

d = 50 mmT_B = 250 N·mT_C = 1000 N·mAB = 3 mBC = 2.5 mG = 80 GPa

φ_C = °

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 90 MPa and σᵧ = −20 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 150 MPa?

σ_x = 90 MPaσ_y = −20 MPaσ₁ = 150 MPa

τ_xy = MPa

6.Mohr’s circle & stress transformationHard

A solid shaft of diameter 40 mm carries an axial compressive force of 200 kN together with a torque of 400 N·m. At the surface of the shaft, find the maximum in-plane shear stress τ_max.

d = 40 mmF = −200 kNT = 400 N·m

τ_max = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 20 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 40 mm, length 500 mm, E = 110 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.25 mm in total.

segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: bronze, Ø40 mm, 500 mm, E = 110 GPaδ = 0.25 mm

P = kN

8.Axial stress & strainHard

A steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 400 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 40 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø40 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 400 mm, E = 70 GPaP = 40 kN

σ₂ = 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 = 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 aluminium is pinned at both ends. It must carry 50 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 50 kNn = 2L = 1.5 mE = 70 GPa (aluminium)

d = mm

11.Mechanical vibrationsMedium

An 8 kg mass on a spring with a stiffness of 1 kN/m has a viscous damper giving a damping ratio of 0.45. Find the damped natural frequency f_d of its free vibration in Hz.

m = 8 kgk = 1 kN/mζ = 0.45

f_d = Hz

12.Mechanical vibrationsHard

A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 12 kN/m and k₂ = 80 kN/m. The mass stays level. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 10 kgk₁ = 12 kN/mk₂ = 80 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: F_t = 878.1 N

Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: F_t = 2T/d.

Torque on the pinion

ω = 2π·1450/60 = 151.8 rad/s

T = P/ω = 3000/151.8 = 19.76 N·m

Pitch diameter

d = m·N = 2.5·18 = 45 mm

Tangential force

F_t = 2T/d = 2·19.76/0.045 = 878.1 N

Answer

F_t = 878.1 N

2.Gears & gear trains

Answer: θ = 165.1 °

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 angle

asin(0.13) = 7.47°

Angle of wrap

θ = 180 − 2·7.47 = 165.1°

Answer

θ = 165.1°

3.Torsion of shafts

Answer: φ = 2.192 °

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) = 750·1000·500/(39000·251327) = 0.03826 rad

φ = 0.03826·180/π = 2.192°

Answer

φ = 2.192°

4.Torsion of shafts

Answer: φ_C = 5.544 °

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 1000 N·m

T_AB = T_C − T_B = 1000 − 250 = 750 N·m

Twist of each segment

J = π·50⁴/32 = 613592 mm⁴

G = 80 GPa = 80000 MPa

φ_AB = 750000·3000/(80000·613592) = 0.04584 rad

φ_BC = 1 × 10⁶·2500/(80000·613592) = 0.05093 rad

Twist of end C

φ_C = 0.04584 + 0.05093 = 0.09677 rad → |φ_C| = 5.544°

Answer

φ_C = 5.544°

5.Mohr’s circle & stress transformation

Answer: τ_xy = 101 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 = (90 + (−20))/2 = 35 MPa

R = σ₁ − C = 150 − 35 = 115 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(115² − 55²) = 101 MPa

Answer

τ_xy = 101 MPa

6.Mohr’s circle & stress transformation

Answer: τ_max = 85.71 MPa

Stresses at the surface

σ = F/A = −200000/(π·40²/4) = −200000/1257 = −159.2 MPa

τ = 16T/(πd³) = 16·400000/(π·40³) = 31.83 MPa

Mohr’s circle

C = σ/2 = −79.58 MPa

R = √((−79.58)² + 31.83²) = 85.71 MPa

Answer

τ_max = 85.71 MPa

7.Axial stress & strain

Answer: P = 25.04 kN

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

A₁ = 314.2 mm², A₂ = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·314.2) = 6.366 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 500/(110000·1257) = 3.617 × 10⁻⁶ mm/N

Load

P = 0.25/9.983 × 10⁻⁶ = 25042 N = 25.04 kN

Answer

P = 25.04 kN

8.Axial stress & strain

Answer: σ₂ = 9.308 MPa

Stiffnesses

k₁ = 200000·1257/400 = 628319 N/mm

k₂ = 70000·706.9/400 = 123700 N/mm

Share of segment 2

F₂ = 40000·123700/752019 = 6580 N

Stress

σ₂ = 6580/706.9 = 9.308 MPa (compression)

Answer

σ₂ = 9.308 MPa

9.Column buckling

Answer: P_cr = 2.879 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 = 2·2500 = 5000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·104167/5000² = 2879 N = 2.879 kN

Answer

P_cr = 2.879 kN

10.Column buckling

Answer: d = 50.75 mm

Required critical load

P_cr = n·P = 2·50000 = 100000 N

KL = 1·1500 = 1500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·100000·1500²/(π³·70000))^(1/4) = 50.75 mm

Answer

d = 50.75 mm

Check: the slenderness KL/r is 118.2, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: f_d = 1.589 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 = √(1000/8) = 11.18 rad/s

f_n = 11.18/(2π) = 1.779 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 1.779·√(1 − 0.45²) = 1.589 Hz

Answer

f_d = 1.589 Hz

12.Mechanical vibrations

Answer: k_eq = 92 kN/m

Equivalent stiffness

Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k₁ + k₂.

k_eq = 12 + 80 = 92 kN/m

Answer

k_eq = 92 kN/m

ExamProblems · examproblems.com/courses/machine-design