PhysicsMechanical properties of solidsClass 11

Elasticity: stress, strain and Young’s modulus: interactive simulation

Pull on a wire and it stretches. For small loads the stretch is proportional to the load and vanishes when you let go; pull harder and the wire yields, flows and finally snaps.

Useful for: JEE Main, JEE Advanced, NEET, CBSE Boards, A-level Physics · 3 modes in one lab

What you can explore

  1. 1

    Stress–strain curve

    Proportional limit, elastic limit, yield, ultimate strength and fracture for ductile, brittle and rubber-like materials.
  2. 2

    Young's modulus

    Extension of steel, iron, copper and aluminium wires under a load, plus self-weight elongation.
  3. 3

    Wires in series & parallel

    Two wires end to end or side by side as springs with k = YA/L.

Try this

Walk a sample along its stress–strain curve, release it to see the permanent set, find how much real wires stretch under a load, and connect two wires end to end or side by side.

  1. 1

    Stretch the ductile metal slowly past 0.14 % strain, then release.

    What you should see

    Up to the elastic limit it returns to zero; beyond it the unloading line is parallel to the elastic slope and leaves a permanent set.

  2. 2

    Switch to glass.

    What you should see

    It breaks with no plastic region at all — brittle.

  3. 3

    In Young's modulus, hang 10 kg on a 2 m steel wire 1 mm thick.

    What you should see

    Stress ≈ 125 MPa and ΔL ≈ 1.25 mm. Copper (Y ≈ 120 GPa) stretches about 2.08 mm.

  4. 4

    Halve the diameter.

    What you should see

    The area falls to a quarter, so the extension becomes four times larger.

  5. 5

    In Wires in series & parallel, hang 20 kg on steel and copper wires.

    What you should see

    End to end: 3.33 mm total. Side by side: only 0.78 mm, with steel carrying most of the load.

Formulas behind it

Stress and strain

σ=FA,ε=ΔLL\sigma = \frac{F}{A},\qquad \varepsilon = \frac{\Delta L}{L}

Young’s modulus

Y=σε=FLA ΔLY = \frac{\sigma}{\varepsilon} = \frac{FL}{A\,\Delta L}

Elongation under its own weight

ΔL=ρgL22Y\Delta L = \frac{\rho g L^2}{2Y}

Elastic energy stored

U=12FΔL=12 stress×strain×volumeU = \tfrac{1}{2}F\Delta L = \tfrac{1}{2}\,\text{stress}\times\text{strain}\times\text{volume}

Wire as a spring

k=YALk = \frac{YA}{L}

Common mistakes

  • Using the diameter instead of the radius in A = πr².
  • Forgetting to convert mm² to m² — a factor of 10⁶.
  • Calling a rubber band “more elastic” than steel. In physics steel is more elastic: it has a far larger Young’s modulus.
  • Using the full length L for the self-weight elongation without the factor ½.
Practise 118 questionsCheck your answer instantly, with a worked solution.Stuck on a elasticity: stress, strain and young’s modulus question?Snap it and get a whiteboard video solution.

FAQ

Why are girders made with an I-shaped cross-section?

Bending stresses are largest at the top and bottom faces. Putting most of the material there gives high stiffness for little weight.

What is the difference between the proportional limit and the elastic limit?

Up to the proportional limit stress is directly proportional to strain. Between it and the elastic limit the material still returns to its original length, but not along a straight line.

What is Poisson’s ratio?

The ratio of lateral strain to longitudinal strain when a wire is stretched. For most metals it is between 0.25 and 0.35.

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