Elasticity and Mechanical Properties of Solids

Work Done in Stretching a Wire

Conditions: Within elastic limit, linear Hooke's law regime

• Let a wire of length L and cross-sectional area A be stretched by an infinitesimal amount dx under a stretching force F.
• The work done for this small displacement is dW = F dx.
• Substituting $F = (Y A x) / L from Hooke's law$, integrate dW from x = 0 to $x = \DeltaL$.
• $W = \int_{0}^{\delta L} \frac{YA x}{L} dx = \frac{YA}{L} \left[\frac{x^2}{2}\right]_{0}^{\delta L} = \frac{1}{2} \frac{YA (\delta L)^2}{L}$.

Result: $W = (1/2) F \DeltaL = (1/2) \times Stress \times Strain \times Volume$

Relation between Elastic Constants (Y, K, and η)

Conditions: Isotropic, homogeneous elastic material

• Express Young's modulus in terms of Bulk modulus K and rigidity modulus η using strain components under triaxial loading.
• The fundamental relations connecting elastic constants are $Y = 9K\eta / (3K + \eta)$ and $Y = 2\eta(1 + \sigma)$.
• Eliminate Poisson's ratio (σ) to establish the direct thermodynamic identity among Y, K, and η.

Result: $\sigma = \frac{3K - 2\eta}{6K + 2\eta}$ and \frac{9}{Y} = \frac{1}{K} + \frac{3}{\eta}

Stress-Strain Curve for a Metallic Wire

Draw: A graph with tensile strain on the x-axis and tensile stress on the y-axis, showing proportionality limit, elastic limit, yield point, ultimate tensile strength point, and fracture point.

Interpret: Region up to yield point is elastic; beyond yield point, plastic deformation occurs, ending in necking and fracture.

Exam use: Identifying ultimate tensile strength and distinguishing ductile from brittle materials.

Searle's Apparatus Experiment

Draw: Schematic setup showing two parallel wires (experimental and reference) loaded with weights, equipped with a spirit level and micrometer screw.

Interpret: Eliminates thermal expansion errors by using a reference wire while measuring minute elongations ΔL corresponding to applied load F.

Exam use: Determining Young's modulus of a metallic wire in laboratory settings.

Hysteresis Loop for Cyclic Loading

Draw: Loading and unloading stress-strain curves forming a closed loop during cyclic elastic-plastic deformation.

Interpret: The area enclosed by the loop represents mechanical energy dissipated as heat per unit volume per cycle.

Exam use: Understanding damping capacity and energy loss in vibrating mechanical structures.

A steel wire of length 2 m and cross-sectional area 1 mm² is stretched by 2 mm. If $Y = 2 \times 10^{11} N/m^{2}$, calculate the tension developed in the wire.

• Identify given parameters: L = 2 m, $A = 1 mm^{2} = 1 \times 10^{-6} m^{2}$, Δ$L = 2 mm = 2 \times 10^{-3} m$, $Y = 2 \times 10^{11} N/m^{2}$.
• Apply Young's modulus formula: $F = (Y A \DeltaL) / L$.
• Substitute values: $F = (2 \times 10^{11} \times 1 \times 10^{-6} \times 2 \times 10^{-3}) / 2$.
• Calculate result: F = 200 N.

Answer: 200 N

Find the elastic potential energy stored in a steel wire of length 4 m and cross-sectional area 0.5 cm² when stretched by 1 mm. Given $Y = 2 \times 10^{11} N/m^{2}$.

• Convert units: L = 4 m, $A = 0.5 cm^{2} = 0.5 \times 10^{-4} m^{2}$, Δ$L = 1 \times 10^{-3} m$.
• Use energy formula: $U = (1/2) \times (Y A / L) \times (\DeltaL)^{2}$.
• Substitute values: $U = 0.5 \times [(2 \times 10^{11} \times 0.5 \times 10^{-4}) / 4] \times (1 \times 10^{-3})^{2}$.
• Calculate result: $U = 0.5 \times (2.5 \times 10^{8}) \times (1 \times 10^{-6}) = 0.125 J$.

Answer: 0.125 J

For a given isotropic material, the Young's modulus is 2.4 times its rigidity modulus. Calculate the Poisson's ratio of the material.

• Use the relation between Young's modulus and rigidity modulus: $Y = 2\eta(1 + \sigma)$.
• Given condition: $Y = 2.4 \eta$.
• Substitute the condition into the relation: 2.4 $\eta = 2\eta(1 + \sigma)$.
• Cancel η and solve for σ: $1.2 = 1 + \sigma$, which yields $\sigma = 0.2$.

Answer: 0.2