understand that the area under the force–extension graph represents the work done

Deformation of Solids – Stress, Strain, Modulus and Work Done

1. Fundamental Quantities

• Stress (σ) – internal resisting force per unit original cross‑sectional area.

    σ = F ⁄ A  [unit: pascal (Pa) = N m⁻²]

    Typical engineering units: kPa, MPa (10⁶ Pa), GPa (10⁹ Pa).

• Strain (ε) – relative change in length of a material.

    ε = ΔL ⁄ L₀  [unit: – (dimensionless)]

• Young’s Modulus (E) – constant of proportionality between stress and strain in the linear (elastic) region.

    E = σ ⁄ ε  [unit: pascal (Pa); usually expressed in GPa].

2. Key Points on the Stress–Strain Graph

3. Converting a Force–Extension Graph to a Stress–Strain Graph

For a uniform wire of original length L₀ and cross‑sectional area A:

\[

\sigma = \frac{F}{A}, \qquad \varepsilon = \frac{x}{L_{0}}

\]

Both axes are scaled by the same constant factors, so the shape of the curve is unchanged. The slope of the linear part of the σ‑ε graph is Young’s modulus E.

4. Work Done – Area Under the Curve

The mechanical work required to stretch a specimen from x = 0 to x = x_f is

\[

W = \int{0}^{xf} F(x)\,dx

\]

or, using stress and strain,

\[

W = \int{0}^{\varepsilonf} \sigma(\varepsilon)\,A\,L_{0}\,d\varepsilon .

\]

Thus the numerical value of the area under a force–extension (or stress–strain) curve equals the work done in joules (J). The interpretation of this area depends on the region of the curve:

• Elastic region (linear) – area = \(\tfrac12\) base × height = \(\tfrac12 kx_f^{2}\). This energy is stored as elastic potential energy and is fully recoverable on unloading.
• Plastic region (non‑linear or constant‑force) – area beyond the elastic limit represents energy dissipated as heat, internal friction, and permanent deformation. It is not recovered when the load is removed.
• Total area up to fracture – called the material’s toughness; it is the energy per unit volume that the material can absorb before breaking.

5. Example – Calculating Work from a Force–Extension Graph

Problem statement: A nylon thread (A = 1.5 × 10⁻⁶ m², L₀ = 0.30 m) is loaded as shown.

- Elastic region: straight line from (0, 0) to (0.005 m, 12 N).

- Plastic flow: force remains 12 N from 0.005 m to 0.015 m.

Solution:

1. Elastic work (triangular area):

    \(W_{\text{elastic}} = \tfrac12 \times 12\;\text{N} \times 0.005\;\text{m} = 0.030\;\text{J}\).

2. Plastic work (rectangular area):

    \(W_{\text{plastic}} = 12\;\text{N} \times (0.015-0.005)\;\text{m} = 0.120\;\text{J}\).

3. Total work: \(W_{\text{total}} = 0.030 + 0.120 = 0.150\;\text{J}\).
4. Young’s modulus from the elastic part:

    Spring constant \(k = F/x = 12/0.005 = 2400\;\text{N m}^{-1}\).

    \(E = \dfrac{kL_{0}}{A}= \dfrac{2400 \times 0.30}{1.5\times10^{-6}} = 4.8\times10^{8}\;\text{Pa} = 0.48\;\text{GPa}\).

6. Practical Determination of Young’s Modulus (Tensile Test)

1. Clamp a uniform wire/rod of known original length L₀ and cross‑section A.
2. Hang a series of calibrated masses; record the corresponding extensions x. Force = mg (use g = 9.81 m s⁻²).
3. Plot Force vs Extension. The slope of the straight‑line elastic portion gives the spring constant k.
4. Convert to a Stress vs Strain plot using σ = F/A and ε = x/L₀. The slope of the linear region is Young’s modulus E.
5. Repeat the whole procedure at least three times and take the average – this satisfies the AO3 requirement for experimental reliability.
6. To obtain the energy absorbed in plastic deformation, extend the specimen beyond the elastic limit, record the full force–extension curve, and calculate the area under the curve beyond the elastic limit (e.g. by dividing into simple geometric shapes or using numerical integration).

7. Summary Checklist

• Stress = F/A; Strain = ΔL/L₀ (both use original dimensions).
• Young’s modulus E = σ/ε – slope of the linear part of a stress–strain graph (units: Pa, usually GPa).
• Limit of proportionality ≠ elastic limit — the former marks the end of strict Hooke’s law, the latter the end of fully recoverable deformation.
• Yield point marks the start of permanent (plastic) deformation; ultimate tensile strength is the maximum stress before necking; fracture point ends the curve.
• Work done = area under the force–extension (or stress–strain) curve.

   – Elastic area → stored (recoverable) energy.

   – Plastic area → dissipated energy.

   – Total area up to fracture → material toughness.

• Experimental determination of E requires a clean, straight‑line elastic region and careful conversion from force–extension to stress–strain.

8. Practice Questions

1. A copper wire (A = 2.0 × 10⁻⁶ m², L₀ = 0.40 m) obeys Hooke’s law up to a force of 30 N, at which the extension is 0.006 m.

    (a) Calculate Young’s modulus (express your answer in GPa).

    (b) Determine the elastic potential energy stored at this load.

2. A polymer specimen shows a linear region up to 0.008 m (force = 8 N) followed by a constant‑force region of 8 N up to 0.020 m.

    Calculate the total work done and state the fraction of the energy that is stored elastically versus dissipated plastically.

3. Explain why the area under a stress–strain curve up to the fracture point represents the toughness of a material. Relate this to elastic and plastic work.
4. Data for a brass rod (A = 3.0 × 10⁻⁶ m², L₀ = 0.25 m):

   Force (N)	Extension (mm)
   0	0
   10	0.04
   20	0.08
   30	0.12
   40	0.16

    Plot the stress–strain graph, estimate Young’s modulus, and calculate the work done up to the 40 N load.

5. Describe a modification to the standard tensile‑test set‑up that would allow you to measure the energy absorbed during plastic deformation (i.e., the area under the curve beyond the yield point). Include any additional instrumentation required.

9. Key Terms