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 belt runs on an 80 mm driving pulley at 1450 rpm. The tension is 400 N on the tight side and 200 N on the slack side. Find the power P transmitted, in kW.

d₁ = 80 mmn₁ = 1450 rpmT₁ = 400 NT₂ = 200 N

P = kW

2.Gears & gear trainsHard

In a compound gear train, gear A (32 teeth) on the input shaft turns at 960 rpm and drives gear B (80 teeth). Gear C (40 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 256 rpm. How many teeth N_D must gear D have?

n_in = 960 rpmA = 32B = 80C = 40n_out = 256 rpm

N_D =

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 15 mm is twisted by a torque of 30 N·m. The two ends of a gauge length of 200 mm rotate 2.66° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 30 N·md = 15 mmL = 200 mmφ = 2.66°

G = GPa

4.Torsion of shaftsHard

A solid brass shaft transmits 75 kW at 1500 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 0.5° per m of length. Find the smallest diameter d that satisfies both limits.

P = 75 kWN = 1500 rpmτ_allow = 60 MPaθ_allow = 0.5°/mG = 39 GPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 140 MPa, σᵧ = −60 MPa and τₓᵧ = 50 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 140 MPaσ_y = −60 MPaτ_xy = 50 MPa

τ_max = MPa

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 10 MPa, σᵧ = −80 MPa and τₓᵧ = 10 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 20° counter-clockwise from the x-axis.

σ_x = 10 MPaσ_y = −80 MPaτ_xy = 10 MPaθ = 20°

σ_θ = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 50 mm, length 400 mm, E = 120 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, Ø40 mm, 600 mm, E = 200 GPasegment 2: copper, Ø50 mm, 400 mm, E = 120 GPaδ = 0.2 mm

P = kN

8.Axial stress & strainHard

A steel segment (diameter 30 mm, length 600 mm, E = 200 GPa) and a segment made of copper (diameter 30 mm, length 200 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 40 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø30 mm, 600 mm, E = 200 GPasegment 2: copper, Ø30 mm, 200 mm, E = 120 GPaP = 40 kN

σ₂ = MPa

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 = 4 mE = 200 GPa (steel)b × h = 25 × 40 mm

P_cr = kN

10.Column bucklingHard

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

L = 6 mE = 200 GPa (steel)d = 30 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 20 kg mass is supported by a spring with a stiffness of 50 kN/m and a viscous damper with a damping coefficient of 160 N·s/m. Find the damping ratio ζ.

m = 20 kgk = 50 kN/mc = 160 N·s/m

ζ =

12.Mechanical vibrationsHard

A 25 kg mass hangs from spring k₃ = 10 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 25 kN/m and k₂ = 30 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 = 25 kN/mk₂ = 30 kN/mk₃ = 10 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: P = 1.215 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.

Belt speed

v = 6.074 m/s

Power

P = (400 − 200)·6.074 = 1215 W = 1.215 kW

Answer

P = 1.215 kW

2.Gears & gear trains

Answer: N_D = 60

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.

Intermediate shaft speed

n_B = n_C = 960·32/80 = 384 rpm

Teeth on gear D

N_D = C·n_C/n_out = 40·384/256 = 60

Answer

N_D = 60

3.Torsion of shafts

Answer: G = 26 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·15⁴/32 = 4970 mm⁴

φ = 2.66·π/180 = 0.04643 rad

Shear modulus

T = 30 N·m = 30000 N·mm

G = TL/(Jφ) = 30000·200/(4970·0.04643) = 26003 MPa

G = 26 GPa

Answer

G = 26 GPa

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

4.Torsion of shafts

Answer: d = 61.48 mm

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 75000/157.1 = 477.5 N·m

Diameter for strength

T = 477.5 N·m = 477465 N·mm

d_τ = ∛(16·477465/(π·60)) = 34.35 mm

Diameter for stiffness

θ = 0.5·π/180/1000 = 8.727 × 10⁻⁶ rad/mm

G = 39 GPa = 39000 MPa

d_φ = (32·477465/(π·39000·8.727 × 10⁻⁶))^1/4 = 61.48 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(34.35, 61.48) = 61.48 mm

Answer

d = 61.48 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: τ_max = 111.8 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.

Radius of Mohr’s circle

τ_max = R = √(((140 − −60)/2)² + 50²) = 111.8 MPa

Answer

τ_max = 111.8 MPa

6.Mohr’s circle & stress transformation

Answer: σ_θ = 5.9 MPa

Transformation equation

σ_θ = −35 + 45·0.766 + 10·0.6428 = 5.9 MPa

Answer

σ_θ = 5.9 MPa

7.Axial stress & strain

Answer: P = 48.96 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 = 1257 mm², A₂ = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·1257) = 2.387 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 400/(120000·1963) = 1.698 × 10⁻⁶ mm/N

Load

P = 0.2/4.085 × 10⁻⁶ = 48960 N = 48.96 kN

Answer

P = 48.96 kN

8.Axial stress & strain

Answer: σ₂ = 36.38 MPa

Stiffnesses

k₁ = 200000·706.9/600 = 235619 N/mm

k₂ = 120000·706.9/200 = 424115 N/mm

Share of segment 2

F₂ = 40000·424115/659734 = 25714 N

Stress

σ₂ = 25714/706.9 = 36.38 MPa (compression)

Answer

σ₂ = 36.38 MPa

9.Column buckling

Answer: P_cr = 13.11 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 = 40·25³/12 = 52083 mm⁴

Effective length

KL = 0.7·4000 = 2800 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·52083/2800² = 13113 N = 13.11 kN

Answer

P_cr = 13.11 kN

10.Column buckling

Answer: σ_cr = 6.294 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²

r = √(I/A) = √(39761/706.9) = 7.5 mm

KL = 0.7·6000 = 4200 mm

KL/r = 4200/7.5 = 560

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/560² = 6.294 MPa

Answer

σ_cr = 6.294 MPa

11.Mechanical vibrations

Answer: ζ = 0.08

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·√(50000·20) = 2000 N·s/m

Damping ratio

ζ = c/c_c = 160/2000 = 0.08

Answer

ζ = 0.08

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

12.Mechanical vibrations

Answer: f_n = 2.928 Hz

Equivalent stiffness

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

k₁₂ = 25 + 30 = 55 kN/m

k_eq = 55·10/(55 + 10) = 8.462 kN/m

Natural frequency

ω_n = √(8462/25) = 18.4 rad/s

f_n = 18.4/(2π) = 2.928 Hz

Answer

f_n = 2.928 Hz

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