Physics · Ch 6 — Mechanical Properties of Solids

Young's modulus (Y)

6.5.1

Young's modulus (Y)

Young's modulus, named after the British physicist Thomas Young (1773-1829), is the modulus of elasticity associated with a change in length of an object such as a metal wire, rod, or beam under an applied deforming force — for this reason it is also called the elasticity of length. It applies only to solids, since only solids maintain a definite length in the first place.

Consider a metal wire of length $L$ and radius $r$ suspended vertically from a rigid support, with a load of mass $M$ hung from its free end. The weight $Mg$ acts as the deforming force, applied along the length of the wire. In equilibrium, the longitudinal (tensile) stress in the wire is

\text{Longitudinal stress} = \dfrac{F}{A} = \dfrac{Mg}{\pi r^2}

since the wire's circular cross-section has area $A = \pi r^2$ . This stress produces an elongation in the wire: if the wire's new length is $L + l$ , then $l$ is the extension, and the longitudinal strain is

\text{Longitudinal strain} = \dfrac{l}{L}

Young's modulus $Y$ is then defined as the ratio of longitudinal stress to longitudinal strain:

$Y = \dfrac{\text{Longitudinal stress}}{\text{Longitudinal strain}} = \dfrac{Mg/\pi r^2}{l/L} = \dfrac{MgL}{\pi r^2 l}$ …

Table 6.1Young's modulus of some familiar materials, from softest (lead) to stiffest (steel)

Material | Young's modulus Y ×10^10 Pa (N/m2)

Lead | 1.5

Glass (crown) | 6.0

Aluminium | 7.0

Silver | 7.6

Gold | 8.1

Brass | 9.0 …

Misc Ex.6.1Ratio of Young's modulus of brass and copper wires

Worked out. Worked example comparing a brass wire (length 4.5 m, cross-sectional area 3×10^-5 m²) and a copper wire (length 5.0 m, cross-sectional area 4×10^-5 m²), both stretched by the same load F and producing the same elongation l in both. The method writes Young's modulus for each wire as Y = FL/(Al), substitutes the given L and A for brass and copper separately (with the common F and l left as symbols since they cancel in the ratio), and divides the two expressions to eliminate F and l, obtaining the ratio Y_brass : Y_copper = 1.2 : 1 — illustrating how a ratio problem lets two of the four quant …

Misc Ex.6.2Young's modulus of a loaded wire from measured elongation

Worked out. Worked example finding the Young's modulus of the material of a wire of length 20 m and cross-sectional area 1.25×10^-4 m², subjected to a load of 2.5 kg (using 1 kgwt = 9.8 N) which produces an elongation of 1×10^-4 m. The method computes the applied force F = mg = 2.5×9.8 N, then substitutes L, A, F and the elongation l directly into Y = FL/(Al), arriving at Y = 3.92×10^10 N/m² — a direct numerical application of the Young's modulus formula derived earlier in this section, using a suspended-wire-with-hanging-load setup …

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