A brief note on Stress, Strain and Hooke's Law . Read it before a test. Try each question at the end before you open its answer.
The big ideas Elastic and plastic : an elastic body returns to its shape when the force goes; a plastic one does not. Stress = F A = \dfrac{F}{A} = A F , the restoring force per unit area, in pascal. Strain = Δ L L = \dfrac{\Delta L}{L} = L Δ L , with no unit. Hooke's law : within the proportional limit, stress is in proportion to strain. The stress-strain curve : proportional limit, elastic limit, yield point, ultimate strength, breaking point. Past the elastic limit some stretch stays. Young's modulus : Y = F L A Δ L Y = \dfrac{FL}{A\,\Delta L} Y = A Δ L F L , a property of the material. Ductile materials flow a long way before breaking; brittle ones do not. The facts at a glance Quantity Formula Stress F A \dfrac{F}{A} A F Strain Δ L L \dfrac{\Delta L}{L} L Δ L Young's modulus F L A Δ L \dfrac{FL}{A\,\Delta L} A Δ L F L Stretch of a wire F L A Y \dfrac{FL}{AY} A Y F L A wire as a spring k = Y A L k = \dfrac{YA}{L} k = L Y A Breaking load breaking stress × A \times A × A
A worked example A brass wire 2.8 m 2.8\ \text{m} 2.8 m long and 0.90 mm 0.90\ \text{mm} 0.90 mm across holds a 45 N 45\ \text{N} 45 N load. ( Y = 9.0 × 10 10 Pa Y = 9.0 \times 10^{10}\ \text{Pa} Y = 9.0 × 1 0 10 Pa .)
Area. A = π ( 0.45 × 10 − 3 m ) 2 = 6.36 × 10 − 7 m 2 A = \pi(0.45 \times 10^{-3}\ \text{m})^2 = 6.36 \times 10^{-7}\ \text{m}^2 A = π ( 0.45 × 1 0 − 3 m ) 2 = 6.36 × 1 0 − 7 m 2 . Stress. 45 N 6.36 × 10 − 7 m 2 = 7.07 × 10 7 Pa \dfrac{45\ \text{N}}{6.36 \times 10^{-7}\ \text{m}^2} = 7.07 \times 10^{7}\ \text{Pa} 6.36 × 1 0 − 7 m 2 45 N = 7.07 × 1 0 7 Pa . Strain. 7.07 × 10 7 Pa 9.0 × 10 10 Pa = 7.86 × 10 − 4 \dfrac{7.07 \times 10^{7}\ \text{Pa}}{9.0 \times 10^{10}\ \text{Pa}} = 7.86 \times 10^{-4} 9.0 × 1 0 10 Pa 7.07 × 1 0 7 Pa = 7.86 × 1 0 − 4 . Stretch. 7.86 × 10 − 4 × 2.8 m = 2.20 × 10 − 3 m 7.86 \times 10^{-4} \times 2.8\ \text{m} = 2.20 \times 10^{-3}\ \text{m} 7.86 × 1 0 − 4 × 2.8 m = 2.20 × 1 0 − 3 m , that is 2.20 mm 2.20\ \text{mm} 2.20 mm . Practice with new numbers
A program to try The program works out the brass wire of the worked example.
Read it, then run it.
Make the wire of steel, Y = 2.0 × 10 11 Pa Y = 2.0 \times 10^{11}\ \text{Pa} Y = 2.0 × 1 0 11 Pa . How much less does it stretch?
Mistakes to avoid The diameter used as the radius. Halve it first; the area goes as r 2 r^2 r 2 . Millimetres left in a formula in metres. Change the stretch and the diameter to metres. Force for stress. A wire breaks at a stress, so a thicker wire holds more. Hooke's law past the proportional limit. It holds only on the straight part. Y Y Y depending on the size of the wire. It belongs to the material. Check yourself A load gives a wire a stress of 5.0 × 10 7 Pa 5.0 \times 10^{7}\ \text{Pa} 5.0 × 1 0 7 Pa and a strain of 2.5 × 10 − 4 2.5 \times 10^{-4} 2.5 × 1 0 − 4 . What is Y Y Y ? Stuck? Show a hint Show the answerTwo wires of the same metal have the same length; one has three times the diameter. The same load hangs on each. Compare their stretches. Stuck? Show a hint Show the answerWhich point on the stress-strain curve marks the greatest stress a wire can bear? Stuck? Show a hint Show the answerIs glass ductile or brittle? Stuck? Show a hint Show the answerA wire stretches 0.8 mm 0.8\ \text{mm} 0.8 mm under a load. The same metal, twice as long and with the same diameter, carries the same load. How far does it stretch? Stuck? Show a hint Show the answer