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 5.5 kW motor running at 1440 rpm drives a pinion with 12 teeth, which meshes with a gear with 18 teeth. The gear pair has an efficiency of 96%. Find the output torque T₂ on the gear shaft.

P = 5.5 kWn₁ = 1440 rpmN₁ = 12N₂ = 18η = 96%

T₂ = N·m

2.Gears & gear trainsHard

In a planetary gear set the sun gear has 30 teeth and each planet gear has 15 teeth. The planet carrier is held fixed and the sun is driven at 600 rpm. Find the speed n_r of the ring gear (magnitude).

N_s = 30N_p = 15n_s = 600 rpm

n_r = rpm

3.Torsion of shaftsMedium

A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 40 mm and length 2.5 m; segment BC has diameter 30 mm and length 1.5 m. A torque of 250 N·m acts at the free end C. Find the angle of twist φ_C of end C, in degrees.

T = 250 N·md₁ = 40 mm, L₁ = 2.5 md₂ = 30 mm, L₂ = 1.5 mG = 26 GPa

φ_C = °

4.Torsion of shaftsHard

A hollow shaft must transmit 10 kW at 1000 rpm. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter d_o.

P = 10 kWN = 1000 rpmd_i/d_o = 0.7τ_allow = 75 MPa

d_o = mm

5.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = −40 MPa, σᵧ = −50 MPa and τₓᵧ = −20 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = −40 MPaσ_y = −50 MPaτ_xy = −20 MPaσ_z = 0

τ_abs = MPa

6.Mohr’s circle & stress transformationHard

A solid shaft of diameter 30 mm carries an axial compressive force of 100 kN together with a torque of 100 N·m. At the surface of the shaft, find the largest principal stress σ₁.

d = 30 mmF = −100 kNT = 100 N·m

σ₁ = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 20 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 60 mm, length 400 mm, E = 100 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, Ø20 mm, 500 mm, E = 200 GPasegment 2: brass, Ø60 mm, 400 mm, E = 100 GPaδ = 0.3 mm

P = kN

8.Axial stress & strainHard

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

P = 80 kNsegment 1: Ø35 mm, 2 m, steel (E = 200 GPa)segment 2: Ø20 mm, 0.5 m, steel (E = 200 GPa)

δ = mm

9.Column bucklingMedium

A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 2 mE = 200 GPa (steel)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 = 1 mE = 70 GPa (aluminium)d = 20 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 2 kg mass on a spring with a stiffness of 25 kN/m has a viscous damper giving a damping ratio of 0.5. Find the damped natural frequency f_d of its free vibration in Hz.

m = 2 kgk = 25 kN/mζ = 0.5

f_d = Hz

12.Mechanical vibrationsHard

A 20 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 20 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 = 20 kgk₁ = 20 kN/mk₂ = 25 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: T₂ = 52.52 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 = 5500/150.8 = 36.47 N·m

Output torque

T₂ = 36.47·18/12·0.96 = 52.52 N·m

Answer

T₂ = 52.52 N·m

2.Gears & gear trains

Answer: n_r = 300 rpm

Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (n_s − n_c)/(n_r − n_c) = −N_r/N_s: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.

Ring gear teeth

N_r = N_s + 2·N_p = 30 + 2·15 = 60

Willis equation with the carrier fixed

With n_c = 0 the sun drives the ring through the planets, which act as idlers.

n_r = −n_s·N_s/N_r = −600·30/60 = −300 rpm

|n_r| = 300 rpm

Answer

n_r = 300 rpm

The ring turns opposite to the sun.

3.Torsion of shafts

Answer: φ_C = 15.87 °

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·40⁴/32 = 251327 mm⁴

J₂ = π·30⁴/32 = 79522 mm⁴

Twist of each segment

T = 250 N·m = 250000 N·mm

G = 26 GPa = 26000 MPa

φ_AB = 250000·2500/(26000·251327) = 0.09565 rad

φ_BC = 250000·1500/(26000·79522) = 0.1814 rad

Total twist

φ_C = 0.09565 + 0.1814 = 0.277 rad = 15.87°

Answer

φ_C = 15.87°

4.Torsion of shafts

Answer: d_o = 20.43 mm

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 10000/104.7 = 95.49 N·m

Outer diameter from the allowable stress

T = 95.49 N·m = 95493 N·mm

1 − k⁴ = 1 − 0.7⁴ = 0.7599

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·95493/(π·75·0.7599)) = 20.43 mm

d_i = 0.7·20.43 = 14.3 mm

Answer

d_o = 20.43 mm

A solid shaft would need d = 18.65 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: τ_abs = 32.81 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 = −45 MPa, R = √(5² + (−20)²) = 20.62 MPa

σ₁ = −24.38 MPa, σ₂ = −65.62 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(41.23, 24.38, 65.62)/2 = 32.81 MPa

Answer

τ_abs = 32.81 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

6.Mohr’s circle & stress transformation

Answer: σ₁ = 2.472 MPa

Stresses at the surface

σ = F/A = −100000/(π·30²/4) = −100000/706.9 = −141.5 MPa

τ = 16T/(πd³) = 16·100000/(π·30³) = 18.86 MPa

Mohr’s circle

C = σ/2 = −70.74 MPa

R = √((−70.74)² + 18.86²) = 73.21 MPa

σ₁ = C + R = 2.472 MPa

σ₂ = C − R = −143.9 MPa

Answer

σ₁ = 2.472 MPa

7.Axial stress & strain

Answer: P = 32.01 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 = 314.2 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·314.2) = 7.958 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 400/(100000·2827) = 1.415 × 10⁻⁶ mm/N

Load

P = 0.3/9.372 × 10⁻⁶ = 32009 N = 32.01 kN

Answer

P = 32.01 kN

8.Axial stress & strain

Answer: δ = 1.468 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·35²/4 = 962.1 mm², A₂ = π·20²/4 = 314.2 mm²

σ₁ = 83.15 MPa

σ₂ = 254.6 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 83.15·2000/200000 = 0.8315 mm

δ₂ = σ₂L₂/E₂ = 254.6·500/200000 = 0.6366 mm

δ = δ₁ + δ₂ = 1.468 mm

Answer

δ = 1.468 mm

9.Column buckling

Answer: P_cr = 32.9 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 = 1·2000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·66667/2000² = 32899 N = 32.9 kN

Answer

P_cr = 32.9 kN

10.Column buckling

Answer: σ_cr = 69.09 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 = 0.5·1000 = 500 mm

KL/r = 500/5 = 100

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/100² = 69.09 MPa

Answer

σ_cr = 69.09 MPa

11.Mechanical vibrations

Answer: f_d = 15.41 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 = √(25000/2) = 111.8 rad/s

f_n = 111.8/(2π) = 17.79 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 17.79·√(1 − 0.5²) = 15.41 Hz

Answer

f_d = 15.41 Hz

12.Mechanical vibrations

Answer: k_eq = 45 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 = 20 + 25 = 45 kN/m

Answer

k_eq = 45 kN/m

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