A brief note on Shear Modulus, Bulk Modulus and Poisson's Ratio . Read it before a test. Try each question at the end before you open its answer.
The big ideas Shear : a force along a face tilts the sides through an angle θ ≈ Δ x h \theta \approx \dfrac{\Delta x}{h} θ ≈ h Δ x . G = F / A θ G = \dfrac{F/A}{\theta} G = θ F / A , about Y 3 \dfrac{Y}{3} 3 Y for metals. Bulk : a pressure on every side shrinks the volume. B = − p Δ V / V B = -\dfrac{p}{\Delta V/V} B = − Δ V / V p ; compressibility = 1 B = \dfrac{1}{B} = B 1 . Liquids and gases have a bulk modulus but no shear modulus. A gas at a steady temperature has B = p B = p B = p . Poisson's ratio : σ = Δ d / d Δ L / L \sigma = \dfrac{\Delta d/d}{\Delta L/L} σ = Δ L / L Δ d / d , from 0 0 0 to 0.5 0.5 0.5 ; a stretched rod changes volume by ε ( 1 − 2 σ ) \varepsilon(1 - 2\sigma) ε ( 1 − 2 σ ) . Links : Y = 2 G ( 1 + σ ) Y = 2G(1 + \sigma) Y = 2 G ( 1 + σ ) and Y = 3 B ( 1 − 2 σ ) Y = 3B(1 - 2\sigma) Y = 3 B ( 1 − 2 σ ) . The facts at a glance Material Y Y Y , G G G , B B B (GPa) Steel 200 200 200 , 80 80 80 , 160 160 160 Copper 110 110 110 , 42 42 42 , 140 140 140 Aluminium 70 70 70 , 25 25 25 , 75 75 75 Water none, none, 2.2 2.2 2.2
A worked example A steel ball 5.0 cm 5.0\ \text{cm} 5.0 cm in radius is lowered to where the extra pressure is 6.0 × 10 7 Pa 6.0 \times 10^{7}\ \text{Pa} 6.0 × 1 0 7 Pa . ( B = 1.6 × 10 11 Pa B = 1.6 \times 10^{11}\ \text{Pa} B = 1.6 × 1 0 11 Pa .)
Fraction. Δ V V = 6.0 × 10 7 Pa 1.6 × 10 11 Pa = 3.75 × 10 − 4 \dfrac{\Delta V}{V} = \dfrac{6.0 \times 10^{7}\ \text{Pa}}{1.6 \times 10^{11}\ \text{Pa}} = 3.75 \times 10^{-4} V Δ V = 1.6 × 1 0 11 Pa 6.0 × 1 0 7 Pa = 3.75 × 1 0 − 4 . Volume. V = 4 3 π ( 0.050 m ) 3 = 5.24 × 10 − 4 m 3 V = \tfrac43\pi(0.050\ \text{m})^3 = 5.24 \times 10^{-4}\ \text{m}^3 V = 3 4 π ( 0.050 m ) 3 = 5.24 × 1 0 − 4 m 3 , so Δ V = 1.96 × 10 − 7 m 3 \Delta V = 1.96 \times 10^{-7}\ \text{m}^3 Δ V = 1.96 × 1 0 − 7 m 3 , about 0.20 cm 3 0.20\ \text{cm}^3 0.20 cm 3 . Radius. It shrinks by a third of the fraction: 3.75 × 10 − 4 3 × 5.0 cm = 6.25 μ m \dfrac{3.75 \times 10^{-4}}{3} \times 5.0\ \text{cm} = 6.25\ \mu\text{m} 3 3.75 × 1 0 − 4 × 5.0 cm = 6.25 μ m . Practice with new numbers
A program to try The program works out the steel ball of the worked example.
Read it, then run it.
Make the ball of glass, B = 3.7 × 10 10 Pa B = 3.7 \times 10^{10}\ \text{Pa} B = 3.7 × 1 0 10 Pa . How many times more does it shrink?
Mistakes to avoid The wrong area in shear. Use the area of the face the force slides along. The angle as a distance. Multiply θ \theta θ by the height to get the shift. A percentage used as a fraction. Divide by 100 100 100 . The radius shrinking by the whole volume fraction. It shrinks by a third of it. 1 − σ 1 - \sigma 1 − σ for 1 − 2 σ 1 - 2\sigma 1 − 2 σ . There are two sideways directions. Check yourself A sideways force of 800 N 800\ \text{N} 800 N acts on a face of 0.020 m 2 0.020\ \text{m}^2 0.020 m 2 . What is the shear stress? Stuck? Show a hint Show the answerWhy does a liquid have no shear modulus? Stuck? Show a hint Show the answerWhat is the bulk modulus of a gas at 2.0 × 10 5 Pa 2.0 \times 10^{5}\ \text{Pa} 2.0 × 1 0 5 Pa , squeezed slowly? Stuck? Show a hint Show the answerA rod's length grows by 0.20 % 0.20\% 0.20% and its width shrinks by 0.07 % 0.07\% 0.07% . What is its Poisson's ratio? Stuck? Show a hint Show the answerA material has σ = 0.25 \sigma = 0.25 σ = 0.25 and G = 40 GPa G = 40\ \text{GPa} G = 40 GPa . What is its Y Y Y ? Stuck? Show a hint Show the answer