Machine Design: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

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

a = 56.25 mmm = 1.5 mmi = 1.5

N₂ =

2.Gears & gear trainsHard

A 4 kW motor at 2900 rpm drives gear A (12 teeth), which meshes with gear B (36 teeth). Gear C (16 teeth) on the same shaft as B drives gear D (80 teeth) on the output shaft. Each gear mesh has an efficiency of 96%. Find the output torque T_out.

P = 4 kWn_in = 2900 rpmA = 12B = 36C = 16D = 80η_mesh = 96%

T_out = N·m

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 = 250 N·md = 40 mmL = 1 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 5 kW at 500 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.

P = 5 kWN = 500 rpmτ_allow = 80 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 110 MPa, σᵧ = 20 MPa and τₓᵧ = 10 MPa (tension positive). Find the angle θ_p from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σ_x = 110 MPaσ_y = 20 MPaτ_xy = 10 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = −10 MPa, σᵧ = 10 MPa and τₓᵧ = −50 MPa (σ_z = 0). The material has a yield strength of 250 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σ_x = −10 MPaσ_y = 10 MPaτ_xy = −50 MPaσ_Y = 250 MPa

n =

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 bronze (diameter 40 mm, length 400 mm, E = 110 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, Ø30 mm, 300 mm, E = 200 GPasegment 2: bronze, Ø40 mm, 400 mm, E = 110 GPaδ = 0.2 mm

P = kN

8.Axial stress & strainHard

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 60 mm, E = 70 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø40 mm, E = 200 GPa, σ_allow = 250 MPatube: Ø100/Ø60 mm, E = 70 GPa, σ_allow = 50 MPa

P_max = kN

9.Column bucklingMedium

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

L = 1.5 mE = 70 GPa (aluminium)b × h = 25 × 100 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 = 25 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed 4 cycles later by a peak of 4 mm. Find the damping ratio ζ.

x₀ = 50 mmx₄ = 4 mmn = 4

ζ =

12.Mechanical vibrationsHard

A 20 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 100 kN/m and k₂ = 10 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 = 100 kN/mk₂ = 10 kN/m

f_n = Hz

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 = 2·56.25/1.5 = 75

Split by the ratio

N₁ = 75/(1 + 1.5) = 30

N₂ = 1.5·30 = 45

Answer

N₂ = 45

2.Gears & gear trains

Answer: T_out = 182.1 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.

Output speed

n_out = 2900·(12/36)·(16/80) = 193.3 rpm

ω_out = 2π·193.3/60 = 20.25 rad/s

Output power and torque

P_out = 4000·0.96² = 3686 W

T_out = P_out/ω_out = 3686/20.25 = 182.1 N·m

Answer

T_out = 182.1 N·m

3.Torsion of shafts

Answer: φ = 1.461 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 250·1000·1000/(39000·251327) = 0.02551 rad

φ = 0.02551·180/π = 1.461°

Answer

φ = 1.461°

4.Torsion of shafts

Answer: d = 18.25 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 5000/52.36 = 95.49 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·95493/(π·80)) = 18.25 mm

Answer

d = 18.25 mm

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

5.Mohr’s circle & stress transformation

Answer: θ_p = 6.264 °

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θ_p = atan2(2·10, 110 − 20) = 12.53°

Angle on the element

θ_p = 6.264°

Answer

θ_p = 6.264°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: n = 2.451

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 = 0 MPa, R = 50.99 MPa

σ₁ = 50.99 MPa, σ₂ = −50.99 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 102 MPa

n = 250/102 = 2.451

Answer

n = 2.451

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

7.Axial stress & strain

Answer: P = 39.87 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) = 400/(110000·1257) = 2.894 × 10⁻⁶ mm/N

Load

P = 0.2/5.016 × 10⁻⁶ = 39874 N = 39.87 kN

Answer

P = 39.87 kN

8.Axial stress & strain

Answer: P_max = 430.8 kN

Areas

A_r = π·40²/4 = 1257 mm²

A_t = π(100² − 60²)/4 = 5027 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 70·10³·5027 = 6.032 × 10⁸ N

Governing strain

ε = min(0.00125, 0.0007143) = 0.0007143 (aluminium)

Largest load

P_max = 0.0007143·6.032 × 10⁸ = 430847 N = 430.8 kN

Answer

P_max = 430.8 kN

9.Column buckling

Answer: P_cr = 159.9 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 = 100·25³/12 = 130208 mm⁴

Effective length

KL = 0.5·1500 = 750 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·130208/750² = 159924 N = 159.9 kN

Answer

P_cr = 159.9 kN

10.Column buckling

Answer: σ_cr = 4.819 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·25⁴/64 = 19175 mm⁴

Radius of gyration and slenderness

A = 490.9 mm²

r = √(I/A) = √(19175/490.9) = 6.25 mm

KL = 2·2000 = 4000 mm

KL/r = 4000/6.25 = 640

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/640² = 4.819 MPa

Answer

σ_cr = 4.819 MPa

11.Mechanical vibrations

Answer: ζ = 0.09999

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 = (1/4)·ln(50/4) = 0.6314

Damping ratio

ζ = δ/√(4π² + δ²) = 0.6314/√(39.48 + 0.3987) = 0.09999

Answer

ζ = 0.09999

12.Mechanical vibrations

Answer: f_n = 11.8 Hz

Equivalent stiffness

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

k_eq = 100 + 10 = 110 kN/m

Natural frequency

ω_n = √(110000/20) = 74.16 rad/s

f_n = 74.16/(2π) = 11.8 Hz

Answer

f_n = 11.8 Hz

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