One number for stretching
The modulus of elasticity for stretching or compressing along a length has its own name. The ratio of tensile (or compressive) stress to longitudinal strain is Young's modulus, written \(Y\). For the steel wire of the last lesson it was \(2 \times 10^{11}\,\text{Pa}\). Every wire of that steel has the same \(Y\), whatever its length or thickness.
Young's modulus of a material is the ratio of the tensile or compressive stress to the longitudinal strain it produces, within the proportional limit.
Put the definitions of stress and strain into this ratio, and a formula for the extension of any wire follows. The derivation below takes three steps.
\[ Y = \dfrac{\text{stress}}{\text{strain}} \]
Why. the definition of Young's modulus, within the proportional limit.
\[ Y = \dfrac{F/A}{\Delta L/L} = \dfrac{FL}{A\,\Delta L} \]
Why. stress is the load over the area, and strain is the extension over the original length.
\[ \Delta L = \dfrac{FL}{AY} \]
Read it. solve for the extension. It grows with the load and the length, and falls with the area and the modulus.
How stiff are materials?
Young's modulus measures stiffness: how large a stress a material needs for a given strain. Some rough values:
- steel, \(2.0 \times 10^{11}\,\text{Pa}\)
- copper, \(1.1 \times 10^{11}\,\text{Pa}\)
- brass, \(1.0 \times 10^{11}\,\text{Pa}\)
- aluminium, \(0.7 \times 10^{11}\,\text{Pa}\)
- glass, about \(0.7 \times 10^{11}\,\text{Pa}\)
- soft rubber, about \(10^6\,\text{Pa}\)
Steel is about two hundred thousand times stiffer than soft rubber. A strain of \(10^{-3}\) needs \(2 \times 10^8\,\text{Pa}\) in steel, but only about \(10^3\,\text{Pa}\) in rubber.
The lift of the first lesson hangs from a steel cable \(30\,\text{m}\) long with a metal cross-section of \(2\,\text{cm}^2\). Eight people of \(60\,\text{kg}\) each step in. Take \(Y = 2 \times 10^{11}\,\text{Pa}\) and \(g = 10\,\text{m/s}^2\). By how much does the cable lengthen?
The extra tension is the weight of the people, \(F = 8 \times 60 \times 10 = 4800\,\text{N}\). \(L = 30\,\text{m}\) and \(A = 2 \times 10^{-4}\,\text{m}^2\).
\(\Delta L = FL/AY\).
\(\Delta L = 4800 \times 30/(2 \times 10^{-4} \times 2 \times 10^{11}) = 144\,000/(4 \times 10^7) = 3.6 \times 10^{-3}\,\text{m}\).
The cable lengthens by \(3.6\,\text{mm}\), and the car sinks by that much. The strain is only \(1.2 \times 10^{-4}\), far below the elastic limit.
An aluminium wire is \(2\,\text{m}\) long with a cross-section of \(2\,\text{mm}^2\). For aluminium \(Y = 0.7 \times 10^{11}\,\text{Pa}\). What force stretches it by \(0.5\,\text{mm}\), in newtons?
A steel wire is replaced by a steel wire of twice the diameter. What happens to Young's modulus?
Longer stretches more, thicker stretches less
For one material under one load, \(\Delta L = FL/AY\) says the extension depends on the wire through \(L/A\). The area of a round wire is \(A = \pi d^2/4\), so \(\Delta L \propto L/d^2\). Twice the length gives twice the extension, because there are twice as many bonds in line. Twice the diameter gives four times the area to share the load, so a quarter of the extension.
Try it in the sim above. With steel, \(100\,\text{N}\), \(2\,\text{m}\) and \(1\,\text{mm}\), the extension is \(1.27\,\text{mm}\). Set the length to \(1\,\text{m}\), and it halves to \(0.64\,\text{mm}\) while the strain does not change. Set the diameter to \(2\,\text{mm}\), and the extension falls to a quarter, \(0.32\,\text{mm}\). Switch to aluminium, and the whole line tilts up by \(2/0.7\), almost three times.
Wires P and Q are made of the same copper and carry the same load. P is \(1\,\text{m}\) long with a diameter of \(1\,\text{mm}\). Q is \(3\,\text{m}\) long with a diameter of \(3\,\text{mm}\). Find the ratio of their extensions, \(\Delta L_Q/\Delta L_P\).
The load and \(Y\) are the same, so \(\Delta L \propto L/d^2\).
\(\Delta L_Q/\Delta L_P = (L_Q/L_P)(d_P/d_Q)^2 = 3 \times (1/3)^2 = 1/3\).
Q is three times as long, which triples its extension, but nine times as thick in area, which divides it by nine.
Two wires of the same steel carry the same load. Wire B is twice as long as wire A and has twice its diameter. What is \(\Delta L_B/\Delta L_A\)?
Two wires are made from equal volumes of the same metal, so the longer one is thinner. Wire B is twice as long as wire A. Under the same load, what is \(\Delta L_B/\Delta L_A\)?
Two wires, end to end
Join a steel wire and a brass wire end to end, fix the top, and hang a load from the bottom. Cut the combination anywhere, and the part below the cut is held by the full load. So the same tension acts in both wires, even though they are different. Each stretches by its own \(FL/AY\), and the total extension is the sum.
In the end-to-end chip of Fig 3.1 the load is \(200\,\text{N}\). For steel \(AY = 2 \times 10^{-6} \times 2 \times 10^{11} = 4 \times 10^5\,\text{N}\), so it stretches \(200 \times 2/(4 \times 10^5) = 1.0\,\text{mm}\). Brass, with half the modulus, stretches \(2.0\,\text{mm}\). The total is \(3.0\,\text{mm}\). The softer wire takes the larger share of the stretch.
Two wires, side by side
Now hang both wires from the ceiling, side by side, and fix their lower ends to a light rigid bar. The load hangs from the bar at the point that keeps it horizontal. Then both wires stretch by the same amount. They also have the same length, so the strains are equal. The tensions are not equal: the tension in each wire is \(T = AY\,\Delta L/L\), larger in the stiffer wire.
The steel and brass wires of Fig 3.1 hang side by side and hold a light rigid bar level. A load of \(600\,\text{N}\) hangs from the bar. Find the tension in each wire, their common extension, and where on the bar the load must hang.
Equal lengths and equal extensions mean equal strains. Then \(T/AY\) is the same for both, and the areas are equal, so \(T_s/T_b = Y_s/Y_b = 2\).
\(T_s + T_b = 600\,\text{N}\) and \(T_s = 2T_b\).
\(T_b = 200\,\text{N}\) and \(T_s = 400\,\text{N}\). Each wire stretches \(\Delta L = T_b L/(AY_b) = 200 \times 2/(2 \times 10^5) = 2.0\,\text{mm}\).
For the bar not to turn, the torques about the load point must balance: \(400\,x = 200\,(d - x)\), so \(x = d/3\). The load hangs one third of the way from the steel wire to the brass wire.
A steel wire and an aluminium wire, each \(1\,\text{m}\) long with a cross-section of \(1\,\text{mm}^2\), are joined end to end and carry \(140\,\text{N}\). Take \(Y = 2 \times 10^{11}\,\text{Pa}\) for steel and \(0.7 \times 10^{11}\,\text{Pa}\) for aluminium. What is the total extension, in mm?
A copper wire and a steel wire of the same length and area hold a rigid bar level, side by side. Which wire carries the larger tension?
Thomas Young was a physician in London with an enormous range. He showed the interference of light and helped to read the Rosetta Stone. In his lectures on natural philosophy, printed in 1807, he described the stiffness of a material by one number. His "modulus of elasticity" was measured as the height of a column of the material that would compress it by its own length. Later writers turned it into stress over strain, and gave it his name.
A crane at a port lifts a container of \(30\,\text{t}\) on four falls of steel rope, each with \(5\,\text{cm}^2\) of metal and \(40\,\text{m}\) long. Each fall carries \(75\,000\,\text{N}\), a stress of \(1.5 \times 10^8\,\text{Pa}\). By \(FL/AY\) each rope stretches \(3\,\text{cm}\). So as the crane starts to wind, the container does not rise at once. The ropes must first stretch by those few centimetres.
- Thinking a thicker wire has a larger Young's modulus.\(Y\) belongs to the material. The wire's stiffness \(YA/L\) changes with its size; \(Y\) does not.
- Scaling the extension with \(1/d\) instead of \(1/d^2\).The load is shared by the area, \(\pi d^2/4\).
- Giving two wires end to end different tensions.End to end, every cut carries the whole load, so the tension is the same in both.
- Giving two wires side by side different extensions.A level rigid bar makes the extensions equal; the tensions differ instead.