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 1.5 kW motor running at 1440 rpm drives a pinion with 20 teeth, which meshes with a gear with 32 teeth. The gear pair has an efficiency of 96%. Find the torque T₁ on the motor shaft.

P = 1.5 kWn₁ = 1440 rpmN₁ = 20N₂ = 32η = 96%

T₁ = 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 (45 teeth). Gear C (36 teeth) is fixed to the same shaft as B and drives gear D (60 teeth) on the output shaft. Find the output speed n_out.

n_in = 3000 rpmA = 25B = 45C = 36D = 60

n_out = rpm

3.Torsion of shaftsMedium

A solid brass shaft 2.5 m long carries a torque of 400 N·m. The angle of twist between its ends must not exceed 3°. Find the smallest diameter d that meets this limit.

T = 400 N·mL = 2.5 mG = 39 GPaφ_allow = 3°

d = mm

4.Torsion of shaftsHard

A hollow shaft transmits 100 kW at 1450 rpm. Find the maximum shear stress τ_max in the shaft.

P = 100 kWN = 1450 rpmd_o = 40 mmd_i = 28 mm

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 50 MPa, σᵧ = 20 MPa and τₓᵧ = 50 MPa (tension positive). The element is rotated 45° clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = 50 MPaσ_y = 20 MPaτ_xy = 50 MPaθ = -45°

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A solid shaft of diameter 30 mm carries a bending moment of 300 N·m together with a torque of 1000 N·m. At the top surface, where the bending stress is tensile and largest, find the maximum in-plane shear stress τ_max.

d = 30 mmM = 300 N·mT = 1000 N·m

τ_max = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 30 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 40 mm, length 500 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, 500 mm, E = 200 GPasegment 2: bronze, Ø40 mm, 500 mm, E = 110 GPaδ = 0.2 mm

P = kN

8.Axial stress & strainHard

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of bronze (outer diameter 120 mm, inner diameter 50 mm, E = 110 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 = 200 MPatube: Ø120/Ø50 mm, E = 110 GPa, σ_allow = 50 MPa

P_max = kN

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 = 6 mE = 70 GPa (aluminium)b × h = 20 × 100 mm

P_cr = kN

10.Column bucklingHard

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

L = 4 mE = 70 GPa (aluminium)d = 25 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 15 kg mass is supported by a spring with a stiffness of 80 kN/m and a viscous damper with a damping coefficient of 880 N·s/m. Find the damping ratio ζ.

m = 15 kgk = 80 kN/mc = 880 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 = 12 kN/m and k₂ = 50 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 = 12 kN/mk₂ = 50 kN/mk₃ = 10 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: T₁ = 9.947 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.

Motor torque

ω₁ = 2π·1440/60 = 150.8 rad/s

T₁ = P/ω₁ = 1500/150.8 = 9.947 N·m

Answer

T₁ = 9.947 N·m

2.Gears & gear trains

Answer: n_out = 1000 rpm

First stage

n_B = 3000·25/45 = 1667 rpm = n_C

Second stage

n_out = 1667·36/60 = 1000 rpm

Answer

n_out = 1000 rpm

total ratio = 3 : 1

3.Torsion of shafts

Answer: d = 47.26 mm

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.

Required polar moment

T = 400 N·m = 400000 N·mm

L = 2.5 m = 2500 mm

G = 39 GPa = 39000 MPa

φ = 3·π/180 = 0.05236 rad

J = TL/(Gφ) = 400000·2500/(39000·0.05236) = 489708 mm⁴

Diameter

d = (32J/π)^1/4 = 47.26 mm

Answer

d = 47.26 mm

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

4.Torsion of shafts

Answer: τ_max = 68.97 MPa

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 100000/151.8 = 658.6 N·m

Polar second moment of area

J = π(40⁴ − 28⁴)/32 = 190984 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 658572·20/190984 = 68.97 MPa

Answer

τ_max = 68.97 MPa

5.Mohr’s circle & stress transformation

Answer: σ_y′ = 85 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.

Transformation equation for the y′ face

σ_y′ = 35 − 15·0 − 50·(−1) = 85 MPa

Check

The sum of the normal stresses does not change with rotation.

σ_x′ = −15 MPa, σ_x′ + σ_y′ = 70 = σ_x + σ_y

Answer

σ_y′ = 85 MPa

6.Mohr’s circle & stress transformation

Answer: τ_max = 196.9 MPa

Stresses at the surface

σ = 32M/(πd³) = 32·300000/(π·30³) = 113.2 MPa

τ = 16T/(πd³) = 16·1000000/(π·30³) = 188.6 MPa

Mohr’s circle

C = σ/2 = 56.59 MPa

R = √(56.59² + 188.6²) = 196.9 MPa

Answer

τ_max = 196.9 MPa

7.Axial stress & strain

Answer: P = 27.96 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 = 706.9 mm², A₂ = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·706.9) = 3.537 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 500/(110000·1257) = 3.617 × 10⁻⁶ mm/N

Load

P = 0.2/7.154 × 10⁻⁶ = 27957 N = 27.96 kN

Answer

P = 27.96 kN

8.Axial stress & strain

Answer: P_max = 581.6 kN

Areas

A_r = π·40²/4 = 1257 mm²

A_t = π(120² − 50²)/4 = 9346 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 110·10³·9346 = 1.279 × 10⁹ N

Governing strain

ε = min(0.001, 0.0004545) = 0.0004545 (bronze)

Largest load

P_max = 0.0004545·1.279 × 10⁹ = 581552 N = 581.6 kN

Answer

P_max = 581.6 kN

9.Column buckling

Answer: P_cr = 2.611 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 = 100·20³/12 = 66667 mm⁴

Effective length

KL = 0.7·6000 = 4200 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·66667/4200² = 2611 N = 2.611 kN

Answer

P_cr = 2.611 kN

10.Column buckling

Answer: σ_cr = 6.747 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 = 0.5·4000 = 2000 mm

KL/r = 2000/6.25 = 320

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/320² = 6.747 MPa

Answer

σ_cr = 6.747 MPa

11.Mechanical vibrations

Answer: ζ = 0.4017

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·√(80000·15) = 2191 N·s/m

Damping ratio

ζ = c/c_c = 880/2191 = 0.4017

Answer

ζ = 0.4017

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

12.Mechanical vibrations

Answer: f_n = 2.954 Hz

Equivalent stiffness

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

k₁₂ = 12 + 50 = 62 kN/m

k_eq = 62·10/(62 + 10) = 8.611 kN/m

Natural frequency

ω_n = √(8611/25) = 18.56 rad/s

f_n = 18.56/(2π) = 2.954 Hz

Answer

f_n = 2.954 Hz

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