Physics · Class 11

Store energy in a wire and beam

The force rises from zero as a wire extends or a beam bends, so stored work is half final force times deflection. Beam geometry enters through its cross-section second moment: greater depth sharply reduces bending.

Open the interactive experiment →

Store energy in a wire and beam

மீள்தன்மை கம்பியில் சேமிக்கப்பட்ட ஆற்றல் ½FΔL. சட்டத்தின் தடிமன் அதிகரித்தால் வளைவு குறையும்.

Key notes, relationships and model limits

Key relationship

Uwire=½FΔL=F²L/(2AY); beam δ=FL³/(48YIsection); Isection=bd³/12; Ubeam=½Fδ.

Remember

Common trap

The material modulus alone does not specify the stiffness of a wire or beam.

Model limits

Compare a uniform tensile wire and a separate simply supported rectangular beam loaded at its centre. Linear elasticity and small beam deflection; shear deformation and material yield are excluded.

What you will learn

Compute elastic energy and compare axial stiffness with beam bending geometry.

Try the experiment

Try both scenarios, change one control, and predict the result before checking the graph and readouts.

Think of it this way

Change one input while keeping the others fixed, as you would in a controlled experiment.

Worked example

The wire at 100 N extends 0.001 m and stores 0.05 J. A 2 m simply supported beam 50 mm wide and 100 mm deep with Y=200 GPa bends 0.00002 m and stores 0.001 J under the same centre load.

Why this matters

The model connects store energy in a wire and beam with an observable reading. Check the equation and assumptions before applying it.

Before you start

SI units · algebra · graph reading

Check your understanding

If load doubles on the same elastic wire, how does stored energy change?

It doubles
It becomes four times as large
It stays constant

Read the explanation

Extension doubles and U=½FΔL therefore multiplies by four.

Course connection: Class XI Chapter 8. Consult the roadmap for CBSE, NEET, JEE Main and Advanced scope.