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

Two parallel shafts are 60 mm apart. They must be connected by a pair of spur gears with a module of 1.5 mm and a speed reduction of 4 : 1. How many teeth N₂ must the gear have?

a = 60 mmm = 1.5 mmi = 4

N₂ =

2.Gears & gear trainsHard

In a planetary gear set the sun gear has 15 teeth and each planet gear has 24 teeth. The ring gear is held fixed and the sun is driven at 900 rpm. Find the speed n_c of the planet carrier.

N_s = 15N_p = 24n_s = 900 rpm

n_c = rpm

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 1000 N·m. Its inner diameter is to be 0.5 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter d_o.

T = 1000 N·md_i/d_o = 0.5τ_allow = 75 MPa

d_o = mm

4.Torsion of shaftsHard

A hollow shaft transmits 75 kW at 500 rpm. Find the maximum shear stress τ_max in the shaft.

P = 75 kWN = 500 rpmd_o = 60 mmd_i = 48 mm

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = −80 MPa, σᵧ = 10 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 = −80 MPaσ_y = 10 MPaτ_xy = 30 MPaσ_z = 0

τ_abs = MPa

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 80 MPa, σᵧ = 30 MPa and τₓᵧ = −30 MPa (σ_z = 0). The material has a yield strength of 420 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = 80 MPaσ_y = 30 MPaτ_xy = −30 MPaσ_Y = 420 MPa

n =

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 40 mm, length 300 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.

segment 1: steel, Ø40 mm, 400 mm, E = 200 GPasegment 2: copper, Ø40 mm, 300 mm, E = 120 GPaδ = 0.2 mm

P = kN

8.Axial stress & strainHard

A steel segment (diameter 30 mm, length 300 mm, E = 200 GPa) and a segment made of bronze (diameter 40 mm, length 300 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 30 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø30 mm, 300 mm, E = 200 GPasegment 2: bronze, Ø40 mm, 300 mm, E = 110 GPaP = 30 kN

σ₂ = MPa

9.Column bucklingMedium

A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 1 mE = 70 GPa (aluminium)d = 40 mm

P_cr = kN

10.Column bucklingHard

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

L = 2 mE = 200 GPa (steel)d = 80 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 one cycle later by a peak of 26.5 mm. Find the damping ratio ζ.

x₀ = 30 mmx₁ = 26.5 mmn = 1

ζ =

12.Mechanical vibrationsHard

A 20 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 20 kN/m and the lower one k₂ = 40 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 20 kgk₁ = 20 kN/mk₂ = 40 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: N₂ = 64

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.

Total number of teeth

N₁ + N₂ = 2a/m = 2·60/1.5 = 80

Split by the ratio

N₁ = 80/(1 + 4) = 16

N₂ = 4·16 = 64

Answer

N₂ = 64

2.Gears & gear trains

Answer: n_c = 173.1 rpm

Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (n_s − n_c)/(n_r − n_c) = −N_r/N_s: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.

Ring gear teeth

N_r = N_s + 2·N_p = 15 + 2·24 = 63

Willis equation with the ring fixed

Set n_r = 0 and solve for the carrier speed.

(900 − n_c)/(0 − n_c) = −63/15

n_c = n_s·N_s/(N_s + N_r) = 900·15/78 = 173.1 rpm

Answer

n_c = 173.1 rpm

The carrier turns in the same direction as the sun.

3.Torsion of shafts

Answer: d_o = 41.68 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.

Outer diameter from the allowable stress

T = 1000 N·m = 1 × 10⁶ N·mm

1 − k⁴ = 1 − 0.5⁴ = 0.9375

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·1 × 10⁶/(π·75·0.9375)) = 41.68 mm

d_i = 0.5·41.68 = 20.84 mm

Answer

d_o = 41.68 mm

A solid shaft would need d = 40.8 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: τ_max = 57.2 MPa

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 75000/52.36 = 1432 N·m

Polar second moment of area

J = π(60⁴ − 48⁴)/32 = 751193 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 1.432 × 10⁶·30/751193 = 57.2 MPa

Answer

τ_max = 57.2 MPa

5.Mohr’s circle & stress transformation

Answer: τ_abs = 54.08 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 = −35 MPa, R = √((−45)² + 30²) = 54.08 MPa

σ₁ = 19.08 MPa, σ₂ = −89.08 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(108.2, 19.08, 89.08)/2 = 54.08 MPa

Answer

τ_abs = 54.08 MPa

Here σ₁ and σ₂ have opposite signs, so the in-plane maximum R is also the absolute maximum.

6.Mohr’s circle & stress transformation

Answer: n = 4.818

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. Yield criteria: von Mises predicts yield when σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = 55 MPa, R = 39.05 MPa

σ₁ = 94.05 MPa, σ₂ = 15.95 MPa, σ₃ = 0

Von Mises

σ_vM = √(94.05² − 94.05·15.95 + 15.95²) = 87.18 MPa

n = 420/87.18 = 4.818

Answer

n = 4.818

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

7.Axial stress & strain

Answer: P = 55.85 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 = 1257 mm², A₂ = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·1257) = 1.592 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(120000·1257) = 1.989 × 10⁻⁶ mm/N

Load

P = 0.2/3.581 × 10⁻⁶ = 55851 N = 55.85 kN

Answer

P = 55.85 kN

8.Axial stress & strain

Answer: σ₂ = 11.8 MPa

Stiffnesses

k₁ = 200000·706.9/300 = 471239 N/mm

k₂ = 110000·1257/300 = 460767 N/mm

Share of segment 2

F₂ = 30000·460767/932006 = 14831 N

Stress

σ₂ = 14831/1257 = 11.8 MPa (compression)

Answer

σ₂ = 11.8 MPa

9.Column buckling

Answer: P_cr = 86.82 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·40⁴/64 = 125664 mm⁴

Effective length

KL = 1·1000 = 1000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·125664/1000² = 86818 N = 86.82 kN

Answer

P_cr = 86.82 kN

10.Column buckling

Answer: σ_cr = 49.35 MPa

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

Radius of gyration and slenderness

A = 5027 mm²

r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm

KL = 2·2000 = 4000 mm

KL/r = 4000/20 = 200

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/200² = 49.35 MPa

Answer

σ_cr = 49.35 MPa

11.Mechanical vibrations

Answer: ζ = 0.01974

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 = (1/1)·ln(30/26.5) = 0.1241

Damping ratio

ζ = δ/√(4π² + δ²) = 0.1241/√(39.48 + 0.01539) = 0.01974

Answer

ζ = 0.01974

12.Mechanical vibrations

Answer: k_eq = 13.33 kN/m

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 = 1/20 + 1/40 → k_eq = 20·40/(20 + 40) = 13.33 kN/m

Answer

k_eq = 13.33 kN/m

ExamProblems · examproblems.com/courses/machine-design