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

A 5.5 kW motor running at 2900 rpm drives a pinion with 16 teeth and a module of 2.5 mm. Find the tangential force F_t on the pinion teeth.

P = 5.5 kWn = 2900 rpmN₁ = 16m = 2.5 mm

F_t = N

2.Gears & gear trainsHard

In a planetary gear set the sun gear has 30 teeth and each planet gear has 30 teeth. The planet carrier is held fixed and the sun is driven at 1200 rpm. Find the speed n_r of the ring gear (magnitude).

N_s = 30N_p = 30n_s = 1200 rpm

n_r = rpm

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 10 mm is twisted by a torque of 20 N·m. The two ends of a gauge length of 250 mm rotate 3.65° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 20 N·md = 10 mmL = 250 mmφ = 3.65°

G = GPa

4.Torsion of shaftsHard

A solid shaft must transmit 10 kW at 750 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 10 kWN = 750 rpmτ_allow = 60 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 50 MPa, σᵧ = 100 MPa and τₓᵧ = −10 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 50 MPaσ_y = 100 MPaτ_xy = −10 MPa

τ_max = MPa

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 70 MPa, σᵧ = 0 MPa and τₓᵧ = 50 MPa (σ_z = 0). The material has a yield strength of 350 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = 70 MPaσ_y = 0 MPaτ_xy = 50 MPaσ_Y = 350 MPa

n =

7.Axial stress & strainMedium

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 80 mm, inner diameter 60 mm, E = 110 GPa). Both are 400 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.

P = 150 kNrod: Ø20 mm, E = 200 GPatube: Ø80/Ø60 mm, E = 110 GPaL = 400 mm

δ = mm

8.Axial stress & strainHard

A steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 500 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 60 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø40 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 500 mm, E = 70 GPaP = 60 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 = 3 mE = 70 GPa (aluminium)b × h = 25 × 80 mm

P_cr = kN

10.Column bucklingHard

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

L = 2.5 mE = 200 GPa (steel)d = 20 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 12 kg mass is supported by a spring with a stiffness of 12 kN/m and a viscous damper with a damping coefficient of 40 N·s/m. Find the damping ratio ζ.

m = 12 kgk = 12 kN/mc = 40 N·s/m

ζ =

12.Mechanical vibrationsHard

A 25 kg mass hangs from spring k₃ = 100 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 10 kN/m and k₂ = 80 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency f_n of vertical vibration in Hz.

m = 25 kgk₁ = 10 kN/mk₂ = 80 kN/mk₃ = 100 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: F_t = 905.5 N

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.

Torque on the pinion

ω = 2π·2900/60 = 303.7 rad/s

T = P/ω = 5500/303.7 = 18.11 N·m

Pitch diameter

d = m·N = 2.5·16 = 40 mm

Tangential force

F_t = 2T/d = 2·18.11/0.04 = 905.5 N

Answer

F_t = 905.5 N

2.Gears & gear trains

Answer: n_r = 400 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 = 30 + 2·30 = 90

Willis equation with the carrier fixed

With n_c = 0 the sun drives the ring through the planets, which act as idlers.

n_r = −n_s·N_s/N_r = −1200·30/90 = −400 rpm

|n_r| = 400 rpm

Answer

n_r = 400 rpm

The ring turns opposite to the sun.

3.Torsion of shafts

Answer: G = 79.95 GPa

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 moment and angle

J = π·10⁴/32 = 981.7 mm⁴

φ = 3.65·π/180 = 0.0637 rad

Shear modulus

T = 20 N·m = 20000 N·mm

G = TL/(Jφ) = 20000·250/(981.7·0.0637) = 79947 MPa

G = 79.95 GPa

Answer

G = 79.95 GPa

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

4.Torsion of shafts

Answer: d = 22.11 mm

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 10000/78.54 = 127.3 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·127324/(π·60)) = 22.11 mm

Answer

d = 22.11 mm

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

5.Mohr’s circle & stress transformation

Answer: τ_max = 26.93 MPa

Radius of Mohr’s circle

τ_max = R = √(((50 − 100)/2)² + −10²) = 26.93 MPa

Answer

τ_max = 26.93 MPa

6.Mohr’s circle & stress transformation

Answer: n = 3.143

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 = 35 MPa, R = 61.03 MPa

σ₁ = 96.03 MPa, σ₂ = −26.03 MPa, σ₃ = 0

Von Mises

σ_vM = √(96.03² − 96.03·(−26.03) + (−26.03)²) = 111.4 MPa

n = 350/111.4 = 3.143

Answer

n = 3.143

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

7.Axial stress & strain

Answer: δ = 0.1969 mm

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·20²/4 = 314.2 mm²

A_t = π(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 110·10³·2199 = 3.047 × 10⁸ N

Shortening

δ = 150000·400/3.047 × 10⁸ = 0.1969 mm

Answer

δ = 0.1969 mm

8.Axial stress & strain

Answer: σ₂ = 16.22 MPa

Stiffnesses

k₁ = 200000·1257/600 = 418879 N/mm

k₂ = 70000·706.9/500 = 98960 N/mm

Share of segment 2

F₂ = 60000·98960/517839 = 11466 N

Stress

σ₂ = 11466/706.9 = 16.22 MPa (compression)

Answer

σ₂ = 16.22 MPa

9.Column buckling

Answer: P_cr = 7.996 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 = 80·25³/12 = 104167 mm⁴

Effective length

KL = 1·3000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·104167/3000² = 7996 N = 7.996 kN

Answer

P_cr = 7.996 kN

10.Column buckling

Answer: σ_cr = 31.58 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·20⁴/64 = 7854 mm⁴

Radius of gyration and slenderness

A = 314.2 mm²

r = √(I/A) = √(7854/314.2) = 5 mm

KL = 0.5·2500 = 1250 mm

KL/r = 1250/5 = 250

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/250² = 31.58 MPa

Answer

σ_cr = 31.58 MPa

11.Mechanical vibrations

Answer: ζ = 0.0527

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·√(12000·12) = 758.9 N·s/m

Damping ratio

ζ = c/c_c = 40/758.9 = 0.0527

Answer

ζ = 0.0527

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

12.Mechanical vibrations

Answer: f_n = 6.928 Hz

Equivalent stiffness

Only k₁ and k₂ are in parallel. That pair is in series with k₃, so combine them as reciprocals.

k₁₂ = 10 + 80 = 90 kN/m

k_eq = 90·100/(90 + 100) = 47.37 kN/m

Natural frequency

ω_n = √(47368/25) = 43.53 rad/s

f_n = 43.53/(2π) = 6.928 Hz

Answer

f_n = 6.928 Hz

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