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 0.75 kW motor running at 2900 rpm drives a pinion with 20 teeth, which meshes with a gear with 30 teeth. The gear pair has an efficiency of 95%. Find the output torque T₂ on the gear shaft.

P = 0.75 kWn₁ = 2900 rpmN₁ = 20N₂ = 30η = 95%

T₂ = N·m

2.Gears & gear trainsHard

In a compound gear train, gear A (36 teeth) on the input shaft turns at 2880 rpm and drives gear B (72 teeth). Gear C (18 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 720 rpm. How many teeth N_D must gear D have?

n_in = 2880 rpmA = 36B = 72C = 18n_out = 720 rpm

N_D =

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 20 mm is twisted by a torque of 50 N·m. The two ends of a gauge length of 400 mm rotate 0.91° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 50 N·md = 20 mmL = 400 mmφ = 0.91°

G = GPa

4.Torsion of shaftsHard

A solid brass shaft of diameter 40 mm is fixed to rigid walls at both ends A and B. A torque of 200 N·m is applied at C, 0.8 m from A and 2 m from B. Find the angle of rotation φ_C of section C, in degrees.

d = 40 mmT₀ = 200 N·ma = 0.8 mb = 2 mG = 39 GPa

φ_C = °

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 100 MPa, σᵧ = −60 MPa and τₓᵧ = 30 MPa (tension positive). The element is rotated 20° clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = 100 MPaσ_y = −60 MPaτ_xy = 30 MPaθ = -20°

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A solid shaft of diameter 40 mm carries an axial compressive force of 90 kN together with a torque of 750 N·m. At the surface of the shaft, find the maximum in-plane shear stress τ_max.

d = 40 mmF = −90 kNT = 750 N·m

τ_max = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 25 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 30 mm, length 500 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.

segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: brass, Ø30 mm, 500 mm, E = 100 GPaδ = 0.5 mm

P = kN

8.Axial stress & strainHard

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 70 mm, E = 70 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø40 mm, E = 200 GPa, σ_allow = 150 MPatube: Ø100/Ø70 mm, E = 70 GPa, σ_allow = 70 MPa

P_max = kN

9.Column bucklingMedium

A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 1.5 mE = 200 GPa (steel)d = 30 mm

P_cr = kN

10.Column bucklingHard

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

L = 1.5 mE = 70 GPa (aluminium)d = 40 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 12 mm is followed 3 cycles later by a peak of 4.7 mm. Find the damping ratio ζ.

x₀ = 12 mmx₃ = 4.7 mmn = 3

ζ =

12.Mechanical vibrationsHard

A 20 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 15 kN/m and k₂ = 30 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency f_n of vertical vibration in Hz.

m = 20 kgk₁ = 15 kN/mk₂ = 30 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: T₂ = 3.519 N·m

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.

Motor torque

ω₁ = 2π·2900/60 = 303.7 rad/s

T₁ = P/ω₁ = 750/303.7 = 2.47 N·m

Output torque

T₂ = 2.47·30/20·0.95 = 3.519 N·m

Answer

T₂ = 3.519 N·m

2.Gears & gear trains

Answer: N_D = 36

Intermediate shaft speed

n_B = n_C = 2880·36/72 = 1440 rpm

Teeth on gear D

N_D = C·n_C/n_out = 18·1440/720 = 36

Answer

N_D = 36

3.Torsion of shafts

Answer: G = 80.17 GPa

Polar moment and angle

J = π·20⁴/32 = 15708 mm⁴

φ = 0.91·π/180 = 0.01588 rad

Shear modulus

T = 50 N·m = 50000 N·mm

G = TL/(Jφ) = 50000·400/(15708·0.01588) = 80166 MPa

G = 80.17 GPa

Answer

G = 80.17 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

4.Torsion of shafts

Answer: φ_C = 0.6681 °

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. When both ends are fixed, statics alone gives only T_A + T_B = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).

Equilibrium and compatibility

T_A + T_B = 200

T_A·a = T_B·b

T_A = 200·2/2.8 = 142.9 N·m

T_B = 200·0.8/2.8 = 57.14 N·m

Rotation of C

Use segment AC: it is fixed at A and carries T_A.

J = π·40⁴/32 = 251327 mm⁴

G = 39 GPa = 39000 MPa

φ_C = T_A·a/(GJ) = 142857·800/(39000·251327) = 0.01166 rad = 0.6681°

Answer

φ_C = 0.6681°

5.Mohr’s circle & stress transformation

Answer: σ_y′ = −22 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′ = 20 − 80·0.766 − 30·(−0.6428) = −22 MPa

Check

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

σ_x′ = 62 MPa, σ_x′ + σ_y′ = 40 = σ_x + σ_y

Answer

σ_y′ = −22 MPa

6.Mohr’s circle & stress transformation

Answer: τ_max = 69.6 MPa

Stresses at the surface

σ = F/A = −90000/(π·40²/4) = −90000/1257 = −71.62 MPa

τ = 16T/(πd³) = 16·750000/(π·40³) = 59.68 MPa

Mohr’s circle

C = σ/2 = −35.81 MPa

R = √((−35.81)² + 59.68²) = 69.6 MPa

Answer

τ_max = 69.6 MPa

7.Axial stress & strain

Answer: P = 49.36 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 = 490.9 mm², A₂ = 706.9 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·490.9) = 3.056 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 500/(100000·706.9) = 7.074 × 10⁻⁶ mm/N

Load

P = 0.5/1.013 × 10⁻⁵ = 49362 N = 49.36 kN

Answer

P = 49.36 kN

8.Axial stress & strain

Answer: P_max = 398.8 kN

Areas

A_r = π·40²/4 = 1257 mm²

A_t = π(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 70·10³·4006 = 5.317 × 10⁸ N

Governing strain

ε = min(0.00075, 0.001) = 0.00075 (steel)

Largest load

P_max = 0.00075·5.317 × 10⁸ = 398786 N = 398.8 kN

Answer

P_max = 398.8 kN

9.Column buckling

Answer: P_cr = 8.721 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 2·1500 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·39761/3000² = 8721 N = 8.721 kN

Answer

P_cr = 8.721 kN

10.Column buckling

Answer: σ_cr = 30.71 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 = 1·1500 = 1500 mm

KL/r = 1500/10 = 150

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/150² = 30.71 MPa

Answer

σ_cr = 30.71 MPa

11.Mechanical vibrations

Answer: ζ = 0.04967

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 = (1/3)·ln(12/4.7) = 0.3124

Damping ratio

ζ = δ/√(4π² + δ²) = 0.3124/√(39.48 + 0.09762) = 0.04967

Answer

ζ = 0.04967

12.Mechanical vibrations

Answer: f_n = 7.549 Hz

Equivalent stiffness

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

k_eq = 15 + 30 = 45 kN/m

Natural frequency

ω_n = √(45000/20) = 47.43 rad/s

f_n = 47.43/(2π) = 7.549 Hz

Answer

f_n = 7.549 Hz

ExamProblems · examproblems.com/courses/machine-design