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 belt runs on a 100 mm driving pulley at 1440 rpm. The tension is 400 N on the tight side and 100 N on the slack side. Find the power P transmitted, in kW.

d₁ = 100 mmn₁ = 1440 rpmT₁ = 400 NT₂ = 100 N

P = kW

2.Gears & gear trainsHard

An open belt connects a 160 mm pulley and a 400 mm pulley whose centres are 800 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d₁ = 160 mmd₂ = 400 mmC = 800 mm

θ = °

3.Torsion of shaftsMedium

A solid brass shaft 1 m long carries a torque of 750 N·m. The angle of twist between its ends must not exceed 2°. Find the smallest diameter d that meets this limit.

T = 750 N·mL = 1 mG = 39 GPaφ_allow = 2°

d = mm

4.Torsion of shaftsHard

A hollow shaft must transmit 20 kW at 500 rpm. Its inner diameter is to be 0.8 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter d_o.

P = 20 kWN = 500 rpmd_i/d_o = 0.8τ_allow = 40 MPa

d_o = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 110 MPa, σᵧ = 110 MPa and τₓᵧ = 50 MPa (tension positive). The element is rotated 45° clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = 110 MPaσ_y = 110 MPaτ_xy = 50 MPaθ = -45°

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 600 mm and wall thickness 6 mm holds gas at a gauge pressure of 2 MPa. It is made from plate welded along a helical seam at 55° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 600 mmt = 6 mmp = 2 MPaα = 55°

σ_n = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 30 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 40 mm, length 500 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.

segment 1: steel, Ø30 mm, 300 mm, E = 200 GPasegment 2: aluminium, Ø40 mm, 500 mm, E = 70 GPaδ = 0.3 mm

P = kN

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 = 45 kNsegment 1: Ø30 mm, 1.5 m, steel (E = 200 GPa)segment 2: Ø16 mm, 1 m, copper (E = 117 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 = 3 mE = 70 GPa (aluminium)b × h = 30 × 50 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of steel is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 1 mE = 200 GPa (steel)d = 20 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 250 kg mass on a spring with a stiffness of 80 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 = 250 kgk = 80 kN/mζ = 0.7

f_d = Hz

12.Mechanical vibrationsHard

A 5 kg mass hangs from spring k₃ = 12 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 15 kN/m and k₂ = 15 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 = 15 kN/mk₂ = 15 kN/mk₃ = 12 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: P = 2.262 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 = 7.54 m/s

Power

P = (400 − 100)·7.54 = 2262 W = 2.262 kW

Answer

P = 2.262 kW

2.Gears & gear trains

Answer: θ = 162.7 °

Belt angle

asin(0.15) = 8.627°

Angle of wrap

θ = 180 − 2·8.627 = 162.7°

Answer

θ = 162.7°

3.Torsion of shafts

Answer: d = 48.67 mm

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.

Required polar moment

T = 750 N·m = 750000 N·mm

L = 1 m = 1000 mm

G = 39 GPa = 39000 MPa

φ = 2·π/180 = 0.03491 rad

J = TL/(Gφ) = 750000·1000/(39000·0.03491) = 550921 mm⁴

Diameter

d = (32J/π)^1/4 = 48.67 mm

Answer

d = 48.67 mm

In practice you would round up to the next standard size.

4.Torsion of shafts

Answer: d_o = 43.51 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 20000/52.36 = 382 N·m

Outer diameter from the allowable stress

T = 382 N·m = 381972 N·mm

1 − k⁴ = 1 − 0.8⁴ = 0.5904

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·381972/(π·40·0.5904)) = 43.51 mm

d_i = 0.8·43.51 = 34.81 mm

Answer

d_o = 43.51 mm

A solid shaft would need d = 36.5 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: σ_y′ = 160 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.

Transformation equation for the y′ face

σ_y′ = 110 − 0·0 − 50·(−1) = 160 MPa

Check

The sum of the normal stresses does not change with rotation.

σ_x′ = 60 MPa, σ_x′ + σ_y′ = 220 = σ_x + σ_y

Answer

σ_y′ = 160 MPa

6.Mohr’s circle & stress transformation

Answer: σ_n = 66.45 MPa

Wall stresses

r = 600/2 = 300 mm

σ_h = pr/t = 2·300/6 = 100 MPa

σ_a = pr/(2t) = 50 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

σ_n = 50·sin²55° + 100·cos²55° = 66.45 MPa

Answer

σ_n = 66.45 MPa

7.Axial stress & strain

Answer: P = 38.43 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 = 706.9 mm², A₂ = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·706.9) = 2.122 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 500/(70000·1257) = 5.684 × 10⁻⁶ mm/N

Load

P = 0.3/7.806 × 10⁻⁶ = 38431 N = 38.43 kN

Answer

P = 38.43 kN

8.Axial stress & strain

Answer: σ_max = 223.8 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·16²/4 = 201.1 mm²

σ₁ = 63.66 MPa

σ₂ = 223.8 MPa

Largest stress

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

σ_max = 223.8 MPa

Answer

σ_max = 223.8 MPa

9.Column buckling

Answer: P_cr = 2.159 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 = 2·3000 = 6000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·112500/6000² = 2159 N = 2.159 kN

Answer

P_cr = 2.159 kN

10.Column buckling

Answer: σ_cr = 100.7 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·20⁴/64 = 7854 mm⁴

Radius of gyration and slenderness

A = 314.2 mm²

r = √(I/A) = √(7854/314.2) = 5 mm

KL = 0.7·1000 = 700 mm

KL/r = 700/5 = 140

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/140² = 100.7 MPa

Answer

σ_cr = 100.7 MPa

11.Mechanical vibrations

Answer: f_d = 2.033 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 = √(80000/250) = 17.89 rad/s

f_n = 17.89/(2π) = 2.847 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 2.847·√(1 − 0.7²) = 2.033 Hz

Answer

f_d = 2.033 Hz

12.Mechanical vibrations

Answer: f_n = 6.59 Hz

Equivalent stiffness

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

k₁₂ = 15 + 15 = 30 kN/m

k_eq = 30·12/(30 + 12) = 8.571 kN/m

Natural frequency

ω_n = √(8571/5) = 41.4 rad/s

f_n = 41.4/(2π) = 6.59 Hz

Answer

f_n = 6.59 Hz

ExamProblems · examproblems.com/courses/machine-design