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

An 80 mm pulley at 720 rpm drives a 300 mm pulley by a belt with 3 % slip. Find the speed n₂ of the driven pulley.

n₁ = 720 rpmd₁ = 80 mmd₂ = 300 mms = 3 %

n₂ = rpm

2.Gears & gear trainsHard

An open belt connects an 80 mm driving pulley at 1440 rpm to a 450 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 1000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 80 mmn₁ = 1440 rpmd₂ = 450 mmC = 500 mmμ = 0.25T₁ = 1000 N

P = kW

3.Torsion of shaftsMedium

A stepped solid brass shaft is fixed at A. Segment AB has diameter 50 mm and length 2 m; segment BC has diameter 25 mm and length 2 m. A torque of 100 N·m acts at the free end C. Find the angle of twist φ_C of end C, in degrees.

T = 100 N·md₁ = 50 mm, L₁ = 2 md₂ = 25 mm, L₂ = 2 mG = 39 GPa

φ_C = °

4.Torsion of shaftsHard

A solid brass shaft transmits 20 kW at 500 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 1° per m of length. Find the smallest diameter d that satisfies both limits.

P = 20 kWN = 500 rpmτ_allow = 60 MPaθ_allow = 1°/mG = 39 GPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 0 MPa, σᵧ = 20 MPa and τₓᵧ = −50 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 0 MPaσ_y = 20 MPaτ_xy = −50 MPa

τ_max = MPa

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 70 MPa, σᵧ = −30 MPa and τₓᵧ = −30 MPa (σ_z = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = 70 MPaσ_y = −30 MPaτ_xy = −30 MPaσ_Y = 460 MPa

n =

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 aluminium (diameter 30 mm, length 300 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.

segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 300 mm, E = 70 GPaδ = 0.3 mm

P = kN

8.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 70 mm, E = 70 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø30 mm, E = 200 GPa, σ_allow = 160 MPatube: Ø100/Ø70 mm, E = 70 GPa, σ_allow = 100 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 = 2.5 mE = 70 GPa (aluminium)b × h = 25 × 40 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of steel is fixed at both ends. It must carry 150 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 150 kNn = 2L = 3 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

An 80 kg mass on a spring with a stiffness of 12 kN/m has a viscous damper giving a damping ratio of 0.6. Find the damped natural frequency f_d of its free vibration in Hz.

m = 80 kgk = 12 kN/mζ = 0.6

f_d = Hz

12.Mechanical vibrationsHard

A 50 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 10 kN/m and k₂ = 80 kN/m. The mass stays level. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 50 kgk₁ = 10 kN/mk₂ = 80 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: n₂ = 186.2 rpm

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.

Speed without slip

720·80/300 = 192 rpm

With slip

n₂ = 192·0.97 = 186.2 rpm

Answer

n₂ = 186.2 rpm

2.Gears & gear trains

Answer: P = 2.708 kW

Angle of wrap

θ = 2.384 rad

Slack-side tension

e^μθ = e^0.5959 = 1.815, T₂ = 1000/1.815 = 551.1 N

Power

P = (1000 − 551.1)·6.032 = 2.708 kW

Answer

P = 2.708 kW

3.Torsion of shafts

Answer: φ_C = 8.141 °

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 moments

J₁ = π·50⁴/32 = 613592 mm⁴

J₂ = π·25⁴/32 = 38350 mm⁴

Twist of each segment

T = 100 N·m = 100000 N·mm

G = 39 GPa = 39000 MPa

φ_AB = 100000·2000/(39000·613592) = 0.008358 rad

φ_BC = 100000·2000/(39000·38350) = 0.1337 rad

Total twist

φ_C = 0.008358 + 0.1337 = 0.1421 rad = 8.141°

Answer

φ_C = 8.141°

4.Torsion of shafts

Answer: d = 48.9 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 20000/52.36 = 382 N·m

Diameter for strength

T = 382 N·m = 381972 N·mm

d_τ = ∛(16·381972/(π·60)) = 31.89 mm

Diameter for stiffness

θ = 1·π/180/1000 = 1.745 × 10⁻⁵ rad/mm

G = 39 GPa = 39000 MPa

d_φ = (32·381972/(π·39000·1.745 × 10⁻⁵))^1/4 = 48.9 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(31.89, 48.9) = 48.9 mm

Answer

d = 48.9 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: τ_max = 50.99 MPa

Radius of Mohr’s circle

τ_max = R = √(((0 − 20)/2)² + −50²) = 50.99 MPa

Answer

τ_max = 50.99 MPa

6.Mohr’s circle & stress transformation

Answer: n = 4.468

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 = 20 MPa, R = 58.31 MPa

σ₁ = 78.31 MPa, σ₂ = −38.31 MPa, σ₃ = 0

Von Mises

σ_vM = √(78.31² − 78.31·(−38.31) + (−38.31)²) = 103 MPa

n = 460/103 = 4.468

Answer

n = 4.468

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

7.Axial stress & strain

Answer: P = 32.9 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) = 300/(70000·706.9) = 6.063 × 10⁻⁶ mm/N

Load

P = 0.3/9.119 × 10⁻⁶ = 32899 N = 32.9 kN

Answer

P = 32.9 kN

8.Axial stress & strain

Answer: P_max = 337.4 kN

Areas

A_r = π·30²/4 = 706.9 mm²

A_t = π(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 70·10³·4006 = 4.218 × 10⁸ N

Governing strain

ε = min(0.0008, 0.001429) = 0.0008 (steel)

Largest load

P_max = 0.0008·4.218 × 10⁸ = 337407 N = 337.4 kN

Answer

P_max = 337.4 kN

9.Column buckling

Answer: P_cr = 23.03 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 = 0.5·2500 = 1250 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·52083/1250² = 23029 N = 23.03 kN

Answer

P_cr = 23.03 kN

10.Column buckling

Answer: d = 51.37 mm

Required critical load

P_cr = n·P = 2·150000 = 300000 N

KL = 0.5·3000 = 1500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·300000·1500²/(π³·200000))^(1/4) = 51.37 mm

Answer

d = 51.37 mm

Check: the slenderness KL/r is 116.8, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: f_d = 1.559 Hz

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.

Undamped natural frequency

ω_n = √(12000/80) = 12.25 rad/s

f_n = 12.25/(2π) = 1.949 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 1.949·√(1 − 0.6²) = 1.559 Hz

Answer

f_d = 1.559 Hz

12.Mechanical vibrations

Answer: k_eq = 90 kN/m

Equivalent stiffness

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

k_eq = 10 + 80 = 90 kN/m

Answer

k_eq = 90 kN/m

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