PhysEduc › Class 11 Physics › Chapter 8: Mechanical Properties of Solids

Stress and strain

NCERT Class 11 Physics, Chapter 8: Mechanical Properties of Solids, §8.1–8.2. A short visual lesson: 3 narrated 3D videos, 1 interactive simulation and 2 quiz questions.

Elastic and plastic

Narrated video. Transcript:

In the chapter on rotation we treated bodies as perfectly rigid: no force could change their shape. Real bodies are not like that. Every solid can be stretched, squeezed or bent a little.

Watch a thick steel bar resting on two bricks. Stand a heavy load on its middle and it bends, very slightly, by an amount far too small to see without magnifying it.

Take the load off and the bar springs back to its original shape. This property, of regaining the original size and shape when the deforming force is removed, is called elasticity.

Now press your thumb into a lump of wet clay, or a piece of putty. The dent stays. Clay and putty have almost no tendency to recover their shape. Such substances are called plastic, and this behaviour is plasticity.

Why does a solid spring back? Zoom in to the atoms. Each atom is held to its neighbours by interatomic forces, and they sit in stable equilibrium positions.

Picture the bonds as tiny springs joining balls in a lattice. Pull the solid, and the springs between the atoms stretch. Push it, and they are squeezed.

Either way, the displaced atoms feel restoring forces that pull them back. Remove the outside force, and every atom returns to its old place. That is elasticity at the atomic scale.

Engineers need to know exactly how much a material deforms, and when it stops springing back. Bridges, flyovers, buildings, cranes and even bones depend on it. That is what this chapter is about.

Stress and longitudinal strain

Narrated video. Transcript:

Pull a cylinder at both ends with equal and opposite forces, F. It is stretched, and inside it a restoring force appears that is equal in size and opposite to the applied force.

The restoring force per unit area is called stress. If the force F acts on a cross-sectional area A, the stress is F divided by A.

Stress has the unit newton per square metre, which is also called the pascal. Its dimensions are M L⁻¹ T⁻², the same as pressure.

Why per unit area? Hang the same load on a thin wire and on a thick rod. The thin wire feels a much bigger stress, and it is the stress that decides how much the material deforms.

Watch the cylinder grow longer by ΔL, starting from length L. The change in length divided by the original length, ΔL over L, is called the longitudinal strain.

Strain is a ratio of two lengths, so it has no unit and no dimensions. A strain of 0.001 means the cylinder grew by one part in a thousand, whatever its size.

Pulling outwards gives a tensile stress, and the body gets longer. Pushing inwards gives a compressive stress, and it gets shorter. Both act normal to the cross-section, so together they are called longitudinal stress.

Let’s try numbers. A steel rod of radius 10 millimetres is pulled by 100 kilonewtons. The area is π r squared, 3.14 × 10⁻⁴ square metres, so the stress is about 3.18 × 10⁸ newtons per square metre.

The same rod, one metre long, stretches by about 1.6 millimetres. The strain is 1.6 × 10⁻³, only about 0.16 percent. Solids deform very little, even under large forces.

Shearing and hydraulic stress

Narrated video. Transcript:

Now apply the two equal and opposite forces differently: parallel to the faces, not across them. Hold the bottom of a block fixed and push its top face sideways.

The top face slides over by a small distance Δx, while the bottom stays put. The block leans over into a slanted shape. This is a shearing deformation.

The tangential force per unit area of the face, F over A, is called the shearing stress, or tangential stress.

The shearing strain is the sideways shift divided by the height, Δx over L. Look at the angle θ by which the vertical edge tilts: tan θ equals Δx over L.

In real solids θ is tiny, so tan θ is almost equal to θ in radians. The shearing strain is simply the angle of tilt, θ.

You can feel shear with a thick book. Press its cover with your palm and slide your hand. The pages slide over one another and the edge of the book slants.

The third kind: drop a solid ball deep into the sea. The water pushes on it from every side, at right angles to every bit of its surface. The ball is squeezed equally in all directions.

Its shape does not change, only its volume. The stress here is the pressure of the water, called hydraulic stress. The volume strain is ΔV over V, the change in volume over the original volume.

So there are three kinds of stress and strain: longitudinal, shearing and volume. Each is force per area divided into a ratio with no units. In the coming lessons we will connect stress and strain.

Try it: three ways to deform

Interactive simulation. Things to try:

  • Start with Tension: make the force negative to switch from tensile to compressive stress.
  • Halve the radius. The area becomes a quarter, so the stress is four times bigger.
  • Shear: the top face slides while the base stays fixed. The strain is the tilt angle θ.
  • Bulk: lower the sample into the ocean. The water squeezes it equally from all sides.
  • Real strains are tiny, so the drawing exaggerates them. The power of ten is shown at the bottom.

Which kind of stress?

A thick book lies flat on a table. You press your palm on its top cover and push sideways, while friction holds the bottom cover. The pages slide over each other and the book’s edge slants. What kind of stress acts on the book?

  1. Tensile stress
  2. Compressive stress
  3. Shearing stress
  4. Hydraulic stress
Answer

C. Your push is parallel to the top face, with an equal and opposite force from the table on the bottom face. Tangential forces cause a shearing stress, and the shearing strain is the tilt angle θ of the edge. Tensile and compressive stresses act normal to the face; hydraulic stress pushes in from all sides.

Stress and strain in a wire

A wire 2.0 m long with a cross-sectional area of 1.0 mm² supports a load of 10 kg and stretches by 0.98 mm. Take g = 9.8 m/s². What are the stress and the strain?

  1. Stress 9.8 × 10⁷ N/m², strain 4.9 × 10⁻⁴
  2. Stress 9.8 N/m², strain 4.9 × 10⁻⁴
  3. Stress 9.8 × 10⁷ N/m², strain 0.49
  4. Stress 1.0 × 10⁷ N/m², strain 4.9 × 10⁻⁴
Answer

A. The force is the weight: F = mg = 10 × 9.8 = 98 N. The area is 1.0 mm² = 1.0 × 10⁻⁶ m², so stress = 98 ÷ 10⁻⁶ = 9.8 × 10⁷ N/m². Strain = ΔL/L = 0.98 × 10⁻³ m ÷ 2.0 m = 4.9 × 10⁻⁴, with no unit. Forgetting to convert mm² or mm to metres, or using the mass instead of the weight, gives the other answers.

Summary

  • Stress: stress = F / A (N/m² = Pa)
  • Longitudinal strain: ΔL / L
  • Shearing strain: Δx / L = tan θ ≈ θ
  • Volume strain: ΔV / V
  • Elasticity: a body regains its size and shape when the force is removed. Plastic bodies like putty and clay stay deformed.
  • Stress is the restoring force per unit area. It has the unit N/m² (pascal) and dimensions [ML⁻¹T⁻²].
  • Strain is a ratio, so it has no unit and no dimensions.
  • Forces normal to the face give tensile or compressive (longitudinal) stress; forces along the face give shearing stress; pressure from all sides gives hydraulic stress.