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 150 mm apart. They must be connected by a pair of spur gears with a module of 5 mm and a speed reduction of 3 : 1. How many teeth N₂ must the gear have?

a = 150 mmm = 5 mmi = 3

N₂ =

2.Gears & gear trainsHard

An open belt connects a 100 mm driving pulley at 2880 rpm to a 400 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 1000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 100 mmn₁ = 2880 rpmd₂ = 400 mmC = 1200 mmμ = 0.3T₁ = 1000 N

P = kW

3.Torsion of shaftsMedium

A solid brass shaft 2.5 m long carries a torque of 1500 N·m. The angle of twist between its ends must not exceed 1.5°. Find the smallest diameter d that meets this limit.

T = 1500 N·mL = 2.5 mG = 39 GPaφ_allow = 1.5°

d = mm

4.Torsion of shaftsHard

A solid shaft must transmit 50 kW at 750 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.

P = 50 kWN = 750 rpmτ_allow = 80 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 90 MPa, σᵧ = −80 MPa and τₓᵧ = −20 MPa (tension positive). The element is rotated 25° counter-clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = 90 MPaσ_y = −80 MPaτ_xy = −20 MPaθ = 25°

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 1500 mm and wall thickness 15 mm holds gas at a gauge pressure of 1.5 MPa. It is made from plate welded along a helical seam at 20° to the axis of the cylinder. Find the magnitude of the shear stress |τ_n| acting along the weld seam.

D = 1500 mmt = 15 mmp = 1.5 MPaα = 20°

|τ_n| = MPa

7.Axial stress & strainMedium

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

P = 35 kNd = 16 mmL = 2.5 mE = 70 GPa

δ = mm

8.Axial stress & strainHard

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 50 mm, E = 110 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø40 mm, E = 200 GPa, σ_allow = 120 MPatube: Ø100/Ø50 mm, E = 110 GPa, σ_allow = 100 MPa

P_max = kN

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 = 1.5 mE = 200 GPa (steel)d = 20 mm

P_cr = kN

10.Column bucklingHard

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

L = 3 mE = 200 GPa (steel)d = 40 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 15 kg mass on a spring with a stiffness of 80 kN/m has a viscous damper giving a damping ratio of 0.4. Find the damped natural frequency f_d of its free vibration in Hz.

m = 15 kgk = 80 kN/mζ = 0.4

f_d = Hz

12.Mechanical vibrationsHard

A 40 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 60 kN/m and k₂ = 20 kN/m. The mass stays level. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 40 kgk₁ = 60 kN/mk₂ = 20 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: N₂ = 45

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 = 2·150/5 = 60

Split by the ratio

N₁ = 60/(1 + 3) = 15

N₂ = 3·15 = 45

Answer

N₂ = 45

2.Gears & gear trains

Answer: P = 8.745 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.

Angle of wrap

θ = 2.891 rad

Slack-side tension

e^μθ = e^0.8673 = 2.38, T₂ = 1000/2.38 = 420.1 N

Power

P = (1000 − 420.1)·15.08 = 8.745 kW

Answer

P = 8.745 kW

3.Torsion of shafts

Answer: d = 78.21 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 = 1500 N·m = 1.5 × 10⁶ N·mm

L = 2.5 m = 2500 mm

G = 39 GPa = 39000 MPa

φ = 1.5·π/180 = 0.02618 rad

J = TL/(Gφ) = 1.5 × 10⁶·2500/(39000·0.02618) = 3.673 × 10⁶ mm⁴

Diameter

d = (32J/π)^1/4 = 78.21 mm

Answer

d = 78.21 mm

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

4.Torsion of shafts

Answer: d = 34.35 mm

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 50000/78.54 = 636.6 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·636620/(π·80)) = 34.35 mm

Answer

d = 34.35 mm

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

5.Mohr’s circle & stress transformation

Answer: σ_y′ = −34.32 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′ = 5 − 85·0.6428 − (−20)·0.766 = −34.32 MPa

Check

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

σ_x′ = 44.32 MPa, σ_x′ + σ_y′ = 10 = σ_x + σ_y

Answer

σ_y′ = −34.32 MPa

6.Mohr’s circle & stress transformation

Answer: |τ_n| = 12.05 MPa

Wall stresses

r = 1500/2 = 750 mm

σ_h = pr/t = 1.5·750/15 = 75 MPa

σ_a = pr/(2t) = 37.5 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.

|τ_n| = (75 − 37.5)·sin 20°·cos 20° = 12.05 MPa

Answer

|τ_n| = 12.05 MPa

7.Axial stress & strain

Answer: δ = 6.217 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·16²/4 = 201.1 mm²

σ = P/A = 35·1000/201.1 = 174.1 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 174.1/70000 = 0.002487 = 2487 µε

Elongation

δ = ε·L = 0.002487·2500 mm = 6.217 mm

Answer

δ = 6.217 mm

8.Axial stress & strain

Answer: P_max = 539.6 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

A_r = π·40²/4 = 1257 mm²

A_t = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 110·10³·5890 = 8.993 × 10⁸ N

Governing strain

ε = min(0.0006, 0.0009091) = 0.0006 (steel)

Largest load

P_max = 0.0006·8.993 × 10⁸ = 539569 N = 539.6 kN

Answer

P_max = 539.6 kN

9.Column buckling

Answer: P_cr = 14.06 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·20⁴/64 = 7854 mm⁴

Effective length

KL = 0.7·1500 = 1050 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·7854/1050² = 14062 N = 14.06 kN

Answer

P_cr = 14.06 kN

10.Column buckling

Answer: σ_cr = 21.93 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²

r = √(I/A) = √(125664/1257) = 10 mm

KL = 1·3000 = 3000 mm

KL/r = 3000/10 = 300

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/300² = 21.93 MPa

Answer

σ_cr = 21.93 MPa

11.Mechanical vibrations

Answer: f_d = 10.65 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 = √(80000/15) = 73.03 rad/s

f_n = 73.03/(2π) = 11.62 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 11.62·√(1 − 0.4²) = 10.65 Hz

Answer

f_d = 10.65 Hz

12.Mechanical vibrations

Answer: k_eq = 80 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 = 60 + 20 = 80 kN/m

Answer

k_eq = 80 kN/m

ExamProblems · examproblems.com/courses/machine-design