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 worm with 1 start(s) turns at 960 rpm and drives a worm wheel with 48 teeth. Find the speed n_w of the wheel.

z = 1N_w = 48n = 960 rpm

n_w = rpm

2.Gears & gear trainsHard

In a compound gear train, gear A (16 teeth) on the input shaft turning at 960 rpm drives gear B (20 teeth). Gear C (20 teeth) is fixed to the same shaft as B and drives gear D (30 teeth) on the output shaft. Find the output speed n_out.

n_in = 960 rpmA = 16B = 20C = 20D = 30

n_out = rpm

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 = 500 N·md = 50 mmL = 1 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 50 kW at 1500 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.

P = 50 kWN = 1500 rpmτ_allow = 50 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 10 MPa and σᵧ = −40 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 60 MPa?

σ_x = 10 MPaσ_y = −40 MPaσ₁ = 60 MPa

τ_xy = MPa

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 90 MPa, σᵧ = −40 MPa and τₓᵧ = −40 MPa (σ_z = 0). Find the von Mises equivalent stress σ_vM.

σ_x = 90 MPaσ_y = −40 MPaτ_xy = −40 MPa

σ_vM = MPa

7.Axial stress & strainMedium

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

P = 65 kNd = 30 mmL = 0.5 mE = 200 GPa

δ = mm

8.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 50 kNsegment 1: Ø25 mm, 1.5 m, steel (E = 200 GPa)segment 2: Ø16 mm, 0.5 m, aluminium (E = 70 GPa)

δ = mm

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)d = 40 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 45 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 45 kNn = 2.5L = 3 mE = 70 GPa (aluminium)

d = mm

11.Mechanical vibrationsMedium

A 40 kg mass is supported by a spring with a stiffness of 20 kN/m and a viscous damper with a damping coefficient of 540 N·s/m. Find the damping ratio ζ.

m = 40 kgk = 20 kN/mc = 540 N·s/m

ζ =

12.Mechanical vibrationsHard

A 10 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 15 kN/m and the lower one k₂ = 100 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 10 kgk₁ = 15 kN/mk₂ = 100 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: n_w = 20 rpm

Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.

Wheel speed

n_w = n·z/N_w = 960·1/48 = 20 rpm

Answer

n_w = 20 rpm

2.Gears & gear trains

Answer: n_out = 512 rpm

First stage

n_B = 960·16/20 = 768 rpm = n_C

Second stage

n_out = 768·20/30 = 512 rpm

Answer

n_out = 512 rpm

total ratio = 1.875 : 1

3.Torsion of shafts

Answer: φ = 1.197 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 500·1000·1000/(39000·613592) = 0.02089 rad

φ = 0.02089·180/π = 1.197°

Answer

φ = 1.197°

4.Torsion of shafts

Answer: d = 31.89 mm

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 50000/157.1 = 318.3 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·318310/(π·50)) = 31.89 mm

Answer

d = 31.89 mm

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

5.Mohr’s circle & stress transformation

Answer: τ_xy = 70.71 MPa

Required radius

C = (10 + (−40))/2 = −15 MPa

R = σ₁ − C = 60 − (−15) = 75 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(75² − 25²) = 70.71 MPa

Answer

τ_xy = 70.71 MPa

6.Mohr’s circle & stress transformation

Answer: σ_vM = 134.5 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. 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 = 25 MPa, R = 76.32 MPa

σ₁ = 101.3 MPa, σ₂ = −51.32 MPa, σ₃ = 0

Von Mises

σ_vM = √(101.3² − 101.3·(−51.32) + (−51.32)²) = 134.5 MPa

Answer

σ_vM = 134.5 MPa

7.Axial stress & strain

Answer: δ = 0.2299 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·30²/4 = 706.9 mm²

σ = P/A = 65·1000/706.9 = 91.96 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 91.96/200000 = 0.0004598 = 459.8 µε

Elongation

δ = ε·L = 0.0004598·500 mm = 0.2299 mm

Answer

δ = 0.2299 mm

8.Axial stress & strain

Answer: δ = 2.54 mm

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·25²/4 = 490.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 101.9 MPa

σ₂ = 248.7 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 101.9·1500/200000 = 0.7639 mm

δ₂ = σ₂L₂/E₂ = 248.7·500/70000 = 1.776 mm

δ = δ₁ + δ₂ = 2.54 mm

Answer

δ = 2.54 mm

9.Column buckling

Answer: P_cr = 3.473 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 = 2·2500 = 5000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·125664/5000² = 3473 N = 3.473 kN

Answer

P_cr = 3.473 kN

10.Column buckling

Answer: d = 61.85 mm

Required critical load

P_cr = n·P = 2.5·45000 = 112500 N

KL = 0.7·3000 = 2100 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·112500·2100²/(π³·70000))^(1/4) = 61.85 mm

Answer

d = 61.85 mm

Check: the slenderness KL/r is 135.8, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.3019

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.

Critical damping

c_c = 2·√(k·m) = 2·√(20000·40) = 1789 N·s/m

Damping ratio

ζ = c/c_c = 540/1789 = 0.3019

Answer

ζ = 0.3019

ζ < 1, so the system is underdamped and oscillates as it decays.

12.Mechanical vibrations

Answer: k_eq = 13.04 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 = 1/15 + 1/100 → k_eq = 15·100/(15 + 100) = 13.04 kN/m

Answer

k_eq = 13.04 kN/m

ExamProblems · examproblems.com/courses/machine-design