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 0.75 kW motor at 2900 rpm drives a worm with 4 start(s). The worm wheel has 30 teeth and the drive has an efficiency of 60%. Find the output torque T_w on the wheel shaft.

z = 4N_w = 30n = 2900 rpmP = 0.75 kWη = 60%

T_w = N·m

2.Gears & gear trainsHard

In a compound gear train, gear A (16 teeth) on the input shaft turning at 2900 rpm drives gear B (32 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D (20 teeth) on the output shaft. Find the output speed n_out.

n_in = 2900 rpmA = 16B = 32C = 16D = 20

n_out = rpm

3.Torsion of shaftsMedium

A solid steel shaft 2.5 m long carries a torque of 200 N·m. The angle of twist between its ends must not exceed 0.5°. Find the smallest diameter d that meets this limit.

T = 200 N·mL = 2.5 mG = 80 GPaφ_allow = 0.5°

d = mm

4.Torsion of shaftsHard

A solid shaft must transmit 20 kW at 750 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 20 kWN = 750 rpmτ_allow = 75 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 50 MPa, σᵧ = 0 MPa and τₓᵧ = 10 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 50 MPaσ_y = 0 MPaτ_xy = 10 MPa

τ_max = MPa

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −40 MPa, σᵧ = −80 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 50° counter-clockwise from the x-axis.

σ_x = −40 MPaσ_y = −80 MPaτ_xy = 30 MPaθ = 50°

σ_θ = MPa

7.Axial stress & strainMedium

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 100 mm, inner diameter 70 mm, E = 100 GPa). Both are 300 mm long and are compressed together between rigid plates by 250 kN. Find the shortening δ.

P = 250 kNrod: Ø20 mm, E = 200 GPatube: Ø100/Ø70 mm, E = 100 GPaL = 300 mm

δ = mm

8.Axial stress & strainHard

A steel segment (diameter 30 mm, length 400 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 300 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø30 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 300 mm, E = 70 GPaP = 20 kN

σ₂ = MPa

9.Column bucklingMedium

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

L = 6 mE = 70 GPa (aluminium)b × h = 40 × 80 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of steel is pinned 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 40 kg mass is supported by a spring with a stiffness of 8 kN/m and a viscous damper with a damping coefficient of 570 N·s/m. Find the damping ratio ζ.

m = 40 kgk = 8 kN/mc = 570 N·s/m

ζ =

12.Mechanical vibrationsHard

A 5 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 20 kN/m and the lower one k₂ = 80 kN/m. Ignore damping and the masses of the springs. Find the natural frequency f_n of vertical vibration in Hz.

m = 5 kgk₁ = 20 kN/mk₂ = 80 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: T_w = 11.11 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 = 2900·4/30 = 386.7 rpm

Output torque

ω_w = 2π·386.7/60 = 40.49 rad/s

T_w = η·P/ω_w = 0.6·750/40.49 = 11.11 N·m

Answer

T_w = 11.11 N·m

2.Gears & gear trains

Answer: n_out = 1160 rpm

First stage

n_B = 2900·16/32 = 1450 rpm = n_C

Second stage

n_out = 1450·16/20 = 1160 rpm

Answer

n_out = 1160 rpm

total ratio = 2.5 : 1

3.Torsion of shafts

Answer: d = 51.97 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 = 200 N·m = 200000 N·mm

L = 2.5 m = 2500 mm

G = 80 GPa = 80000 MPa

φ = 0.5·π/180 = 0.008727 rad

J = TL/(Gφ) = 200000·2500/(80000·0.008727) = 716197 mm⁴

Diameter

d = (32J/π)^1/4 = 51.97 mm

Answer

d = 51.97 mm

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

4.Torsion of shafts

Answer: d = 25.86 mm

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 20000/78.54 = 254.6 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·254648/(π·75)) = 25.86 mm

Answer

d = 25.86 mm

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

5.Mohr’s circle & stress transformation

Answer: τ_max = 26.93 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 = √(((50 − 0)/2)² + 10²) = 26.93 MPa

Answer

τ_max = 26.93 MPa

6.Mohr’s circle & stress transformation

Answer: σ_θ = −33.93 MPa

Transformation equation

σ_θ = −60 + 20·−0.1736 + 30·0.9848 = −33.93 MPa

Answer

σ_θ = −33.93 MPa

7.Axial stress & strain

Answer: δ = 0.1619 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(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 100·10³·4006 = 4.634 × 10⁸ N

Shortening

δ = 250000·300/4.634 × 10⁸ = 0.1619 mm

Answer

δ = 0.1619 mm

8.Axial stress & strain

Answer: σ₂ = 9.003 MPa

Stiffnesses

k₁ = 200000·706.9/400 = 353429 N/mm

k₂ = 70000·706.9/300 = 164934 N/mm

Share of segment 2

F₂ = 20000·164934/518363 = 6364 N

Stress

σ₂ = 6364/706.9 = 9.003 MPa (compression)

Answer

σ₂ = 9.003 MPa

9.Column buckling

Answer: P_cr = 2.047 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 = 80·40³/12 = 426667 mm⁴

Effective length

KL = 2·6000 = 12000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·426667/12000² = 2047 N = 2.047 kN

Answer

P_cr = 2.047 kN

10.Column buckling

Answer: σ_cr = 7.896 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 = 1·2500 = 2500 mm

KL/r = 2500/5 = 500

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/500² = 7.896 MPa

Answer

σ_cr = 7.896 MPa

11.Mechanical vibrations

Answer: ζ = 0.5038

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·√(8000·40) = 1131 N·s/m

Damping ratio

ζ = c/c_c = 570/1131 = 0.5038

Answer

ζ = 0.5038

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

12.Mechanical vibrations

Answer: f_n = 9.003 Hz

Equivalent stiffness

Springs in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k₁ + 1/k₂.

1/k_eq = 1/20 + 1/80 → k_eq = 20·80/(20 + 80) = 16 kN/m

Natural frequency

ω_n = √(16000/5) = 56.57 rad/s

f_n = 56.57/(2π) = 9.003 Hz

Answer

f_n = 9.003 Hz

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