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Elastic Properties of Solids

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Elastic Properties of Solids - Lesson Note

Elastic Properties of Solids

Lesson Note for Edwin Ogie Library

Objectives

  • Define elasticity and explain its importance in materials.
  • State and explain Hooke’s Law.
  • Derive and apply the formula F = K e for elastic materials.
  • Calculate the elastic (or stiffness) constant K of a material.
  • Solve various problems involving force, extension, and elastic constants.

Elasticity

Elasticity is the property of a material that enables it to return to its original shape and size once an external force (such as stretching or compressing) is removed.

Elastic Limit: The maximum deformation a material can undergo without permanent change.

Hooke’s Law

Hooke’s Law states that, within the elastic limit, the extension e of a material is directly proportional to the applied force F. This relationship is expressed as:

F = K e

Where:

  • F = Applied force (N)
  • e = Extension (m)
  • K = Elastic (or stiffness) constant (N/m)

A higher K value indicates a stiffer material that is less prone to deformation.

Worked Examples

Example 1: Elastic Constant of an Elastic Cord

Problem: An elastic cord is stretched by a load of 8.0 N causing an extension of 250 cm. Calculate the elastic constant K.

e = 250 cm = 2.5 m
F = 8.0 N

Using F = K e:
K = F / e = 8.0 N / 2.5 m = 3.2 N/m

Example 2: Extension Change in a Spring

Problem: A spring extends to 1.86 cm when a force of 0.2 N (from a 20 g mass) is applied. The original length of the spring is 1.0 cm. Determine the new extension when the force increases to 0.3 N (additional 10 g).

Initial extension:
e1 = 1.86 cm - 1.00 cm = 0.86 cm = 0.0086 m
F1 = 0.2 N

K = F1 / e1 = 0.2 N / 0.0086 m ≈ 23.26 N/m

New extension:
e2 = F2 / K = 0.3 N / 23.26 N/m ≈ 0.0129 m

Example 3: Force Constant of a Metal Wire

Problem: A metal wire extends by 4 mm when a force of 12 N is applied. Find the force constant K.

e = 4 mm = 0.004 m
F = 12 N

K = F / e = 12 N / 0.004 m = 3000 N/m

Example 4: Extension of a Rubber Band

Problem: A rubber band stretches by 0.03 m under a force of 15 N. Determine the extension when the force is increased to 25 N.

K = 15 N / 0.03 m = 500 N/m
e2 = 25 N / 500 N/m = 0.05 m

Example 5: Force Applied to a Steel Spring

Problem: A steel spring with a stiffness constant K = 400 N/m extends by 6 cm. Find the force applied.

e = 6 cm = 0.06 m
F = K e = 400 N/m × 0.06 m = 24 N

Example 6: Extension of a Copper Wire with Reduced Diameter

Problem: A copper wire extends 2 cm under a force of 10 N. A second wire, made of the same material but with half the diameter, is subjected to the same force. Find the extension of the second wire.

Reduction in diameter → Area decreases by factor of 4
e2 = 4 × e1 = 4 × 2 cm = 8 cm

Conclusion

Elasticity allows materials to return to their original shape after deformation.

Hooke’s Law (F = K e) shows that within the elastic limit, the extension of a material is directly proportional to the applied force.

The elastic constant K quantifies the stiffness of a material; higher values correspond to stiffer materials.

Practice Assignment

  1. A spring with a stiffness constant of 150 N/m is stretched by 5 cm. Calculate the force required.
  2. A wire extends by 3 mm when a 5 N force is applied. Determine its elastic constant.
  3. If a force of 20 N stretches a rubber band by 0.08 m, what is the extension when the force is reduced to 10 N?
  4. A 50 g mass causes a spring to extend by 2 cm. What will be the extension when a 100 g mass is used?
  5. A metal rod extends by 1.5 cm under a force of 30 N. Determine its stiffness constant.

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