Machine Design: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 worm with 4 start(s) turns at 1450 rpm and drives a worm wheel with 40 teeth. Find the speed n_w of the wheel.

z = 4N_w = 40n = 1450 rpm

n_w = rpm

2.Gears & gear trainsHard

In a compound gear train, gear A (50 teeth) on the input shaft turning at 1440 rpm drives gear B (72 teeth). Gear C (32 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 = 1440 rpmA = 50B = 72C = 32D = 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 = 400 N·md = 50 mmL = 1 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 20 kW at 3000 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 20 kWN = 3000 rpmτ_allow = 75 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 140 MPa, σᵧ = 120 MPa and τₓᵧ = 30 MPa (tension positive). The element is rotated 20° counter-clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = 140 MPaσ_y = 120 MPaτ_xy = 30 MPaθ = 20°

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A solid shaft of diameter 40 mm carries a bending moment of 1500 N·m together with a torque of 500 N·m. At the top surface, where the bending stress is tensile and largest, find the largest principal stress σ₁.

d = 40 mmM = 1500 N·mT = 500 N·m

σ₁ = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 25 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 40 mm, length 300 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.

segment 1: steel, Ø25 mm, 500 mm, E = 200 GPasegment 2: brass, Ø40 mm, 300 mm, E = 100 GPaδ = 0.4 mm

P = kN

8.Axial stress & strainHard

A steel segment (diameter 40 mm, length 300 mm, E = 200 GPa) and a segment made of brass (diameter 30 mm, length 500 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 40 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø40 mm, 300 mm, E = 200 GPasegment 2: brass, Ø30 mm, 500 mm, E = 100 GPaP = 40 kN

σ₂ = MPa

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 = 2.5 mE = 70 GPa (aluminium)d = 40 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of aluminium is pinned at both ends. It must carry 115 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 115 kNn = 2L = 2 mE = 70 GPa (aluminium)

d = mm

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 25 mm is followed 3 cycles later by a peak of 3.8 mm. Find the damping ratio ζ.

x₀ = 25 mmx₃ = 3.8 mmn = 3

ζ =

12.Mechanical vibrationsHard

A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 80 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency f_n of vertical vibration in Hz.

m = 10 kgk₁ = 40 kN/mk₂ = 80 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: n_w = 145 rpm

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 = 1450·4/40 = 145 rpm

Answer

n_w = 145 rpm

2.Gears & gear trains

Answer: n_out = 444.4 rpm

First stage

n_B = 1440·50/72 = 1000 rpm = n_C

Second stage

n_out = 1000·32/72 = 444.4 rpm

Answer

n_out = 444.4 rpm

total ratio = 3.24 : 1

3.Torsion of shafts

Answer: φ = 0.4669 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·1000/(80000·613592) = 0.008149 rad

φ = 0.008149·180/π = 0.4669°

Answer

φ = 0.4669°

4.Torsion of shafts

Answer: d = 16.29 mm

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s

T = P/ω = 20000/314.2 = 63.66 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·63662/(π·75)) = 16.29 mm

Answer

d = 16.29 mm

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

5.Mohr’s circle & stress transformation

Answer: σ_y′ = 103.1 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′ = 130 − 10·0.766 − 30·0.6428 = 103.1 MPa

Check

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

σ_x′ = 156.9 MPa, σ_x′ + σ_y′ = 260 = σ_x + σ_y

Answer

σ_y′ = 103.1 MPa

6.Mohr’s circle & stress transformation

Answer: σ₁ = 245.2 MPa

Stresses at the surface

σ = 32M/(πd³) = 32·1500000/(π·40³) = 238.7 MPa

τ = 16T/(πd³) = 16·500000/(π·40³) = 39.79 MPa

Mohr’s circle

C = σ/2 = 119.4 MPa

R = √(119.4² + 39.79²) = 125.8 MPa

σ₁ = C + R = 245.2 MPa

σ₂ = C − R = −6.457 MPa

Answer

σ₁ = 245.2 MPa

7.Axial stress & strain

Answer: P = 53.47 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 = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·490.9) = 5.093 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(100000·1257) = 2.387 × 10⁻⁶ mm/N

Load

P = 0.4/7.48 × 10⁻⁶ = 53474 N = 53.47 kN

Answer

P = 53.47 kN

8.Axial stress & strain

Answer: σ₂ = 8.171 MPa

Stiffnesses

k₁ = 200000·1257/300 = 837758 N/mm

k₂ = 100000·706.9/500 = 141372 N/mm

Share of segment 2

F₂ = 40000·141372/979130 = 5775 N

Stress

σ₂ = 5775/706.9 = 8.171 MPa (compression)

Answer

σ₂ = 8.171 MPa

9.Column buckling

Answer: P_cr = 28.35 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 round section has the same I about every axis.

I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴

Effective length

KL = 0.7·2500 = 1750 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·125664/1750² = 28349 N = 28.35 kN

Answer

P_cr = 28.35 kN

10.Column buckling

Answer: d = 72.17 mm

Required critical load

P_cr = n·P = 2·115000 = 230000 N

KL = 1·2000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·230000·2000²/(π³·70000))^(1/4) = 72.17 mm

Answer

d = 72.17 mm

Check: the slenderness KL/r is 110.8, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.09945

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.

Logarithmic decrement

δ = (1/3)·ln(25/3.8) = 0.628

Damping ratio

ζ = δ/√(4π² + δ²) = 0.628/√(39.48 + 0.3943) = 0.09945

Answer

ζ = 0.09945

12.Mechanical vibrations

Answer: f_n = 17.43 Hz

Equivalent stiffness

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

k_eq = 40 + 80 = 120 kN/m

Natural frequency

ω_n = √(120000/10) = 109.5 rad/s

f_n = 109.5/(2π) = 17.43 Hz

Answer

f_n = 17.43 Hz

ExamProblems · examproblems.com/courses/machine-design