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 1.1 kW motor at 1440 rpm drives a worm with 3 start(s). The worm wheel has 72 teeth and the drive has an efficiency of 50%. Find the output torque T_w on the wheel shaft.

z = 3N_w = 72n = 1440 rpmP = 1.1 kWη = 50%

T_w = N·m

2.Gears & gear trainsHard

In a compound gear train, gear A (25 teeth) on the input shaft turning at 3000 rpm drives gear B (72 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D (72 teeth) on the output shaft. Find the output speed n_out.

n_in = 3000 rpmA = 25B = 72C = 12D = 72

n_out = rpm

3.Torsion of shaftsMedium

A solid steel 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 = 500 N·md = 30 mmL = 3 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 15 kW at 500 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 15 kWN = 500 rpmτ_allow = 60 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = −20 MPa, σᵧ = 90 MPa and τₓᵧ = 40 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = −20 MPaσ_y = 90 MPaτ_xy = 40 MPaσ_z = 0

τ_abs = MPa

6.Mohr’s circle & stress transformationHard

A solid shaft of diameter 30 mm carries a bending moment of 750 N·m together with a torque of 400 N·m. At the top surface, where the bending stress is tensile and largest, find the largest principal stress σ₁.

d = 30 mmM = 750 N·mT = 400 N·m

σ₁ = MPa

7.Axial stress & strainMedium

Find the axial strain ε in the steel bar, in microstrain (µε = 10⁻⁶).

P = 80 kNd = 12 mmL = 2 mE = 200 GPa

ε = µε

8.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 25 kNsegment 1: Ø30 mm, 1 m, brass (E = 100 GPa)segment 2: Ø12 mm, 0.5 m, brass (E = 100 GPa)

δ = mm

9.Column bucklingMedium

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

L = 2 mE = 70 GPa (aluminium)b × h = 25 × 60 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 60 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 60 kNn = 3L = 3 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

A 100 kg mass is supported by a spring with a stiffness of 10 kN/m and a viscous damper with a damping coefficient of 300 N·s/m. Find the damping ratio ζ.

m = 100 kgk = 10 kN/mc = 300 N·s/m

ζ =

12.Mechanical vibrationsHard

A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 100 kN/m and k₂ = 25 kN/m. The mass stays level. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 100 kgk₁ = 100 kN/mk₂ = 25 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: T_w = 87.54 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.

Wheel speed

n_w = n·z/N_w = 1440·3/72 = 60 rpm

Output torque

ω_w = 2π·60/60 = 6.283 rad/s

T_w = η·P/ω_w = 0.5·1100/6.283 = 87.54 N·m

Answer

T_w = 87.54 N·m

2.Gears & gear trains

Answer: n_out = 173.6 rpm

First stage

n_B = 3000·25/72 = 1042 rpm = n_C

Second stage

n_out = 1042·12/72 = 173.6 rpm

Answer

n_out = 173.6 rpm

total ratio = 17.28 : 1

3.Torsion of shafts

Answer: φ = 13.51 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 500·1000·3000/(80000·79522) = 0.2358 rad

φ = 0.2358·180/π = 13.51°

Answer

φ = 13.51°

4.Torsion of shafts

Answer: d = 28.97 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 15000/52.36 = 286.5 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·286479/(π·60)) = 28.97 mm

Answer

d = 28.97 mm

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

5.Mohr’s circle & stress transformation

Answer: τ_abs = 68.01 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. In three dimensions the absolute maximum shear stress is (σ_max − σ_min)/2 over all three principal stresses; in plane stress the third one is zero.

In-plane principal stresses

C = 35 MPa, R = √((−55)² + 40²) = 68.01 MPa

σ₁ = 103 MPa, σ₂ = −33.01 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(136, 103, 33.01)/2 = 68.01 MPa

Answer

τ_abs = 68.01 MPa

Here σ₁ and σ₂ have opposite signs, so the in-plane maximum R is also the absolute maximum.

6.Mohr’s circle & stress transformation

Answer: σ₁ = 301.8 MPa

Stresses at the surface

σ = 32M/(πd³) = 32·750000/(π·30³) = 282.9 MPa

τ = 16T/(πd³) = 16·400000/(π·30³) = 75.45 MPa

Mohr’s circle

C = σ/2 = 141.5 MPa

R = √(141.5² + 75.45²) = 160.3 MPa

σ₁ = C + R = 301.8 MPa

σ₂ = C − R = −18.86 MPa

Answer

σ₁ = 301.8 MPa

7.Axial stress & strain

Answer: ε = 3537 µε

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·12²/4 = 113.1 mm²

σ = P/A = 80·1000/113.1 = 707.4 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 707.4/200000 = 0.003537 = 3537 µε

Answer

ε = 3537 µε

8.Axial stress & strain

Answer: δ = 1.459 mm

Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.

Same force in both segments

The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.

A₁ = π·30²/4 = 706.9 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 35.37 MPa

σ₂ = 221 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 35.37·1000/100000 = 0.3537 mm

δ₂ = σ₂L₂/E₂ = 221·500/100000 = 1.105 mm

δ = δ₁ + δ₂ = 1.459 mm

Answer

δ = 1.459 mm

9.Column buckling

Answer: P_cr = 27.54 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 = 60·25³/12 = 78125 mm⁴

Effective length

KL = 0.7·2000 = 1400 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·78125/1400² = 27538 N = 27.54 kN

Answer

P_cr = 27.54 kN

10.Column buckling

Answer: d = 45.22 mm

Required critical load

P_cr = n·P = 3·60000 = 180000 N

KL = 0.5·3000 = 1500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·180000·1500²/(π³·200000))^(1/4) = 45.22 mm

Answer

d = 45.22 mm

Check: the slenderness KL/r is 132.7, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.15

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·√(10000·100) = 2000 N·s/m

Damping ratio

ζ = c/c_c = 300/2000 = 0.15

Answer

ζ = 0.15

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

12.Mechanical vibrations

Answer: k_eq = 125 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 = 100 + 25 = 125 kN/m

Answer

k_eq = 125 kN/m

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