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Mechanical Properties of SolidsStress and Strain

Reading time: ~20 min

Aman carries a heavy bag of vegetables home by a thin plastic string: it cuts painfully into his fingers and soon snaps. The same bag on a broad cloth strap is comfortable. The weight is the same; what changed is how much area shares it.

When forces act on a body that stays in equilibrium, the body is deformed, a little or a lot, depending on what it is made of and how large the force is. Often the change is too small to see, but it is always there. As soon as the body is deformed, forces appear inside it that try to undo the change: a restoring force, equal in size and opposite in direction to the applied force. The restoring force per unit area is called stress. If a force F acts at right angles to a cross-section of area A,

\text{stress} = \dfrac{F}{A} \qquad (8.1)

The SI unit of stress is N m⁻², also called the pascal (Pa), and its dimensional formula is [M L⁻¹ T⁻²], the same as that of pressure.

Riya hangs the same 20 kg load first from a steel wire 2 mm across, then from one 1 mm across. Compared with the thicker wire, the stress in the thinner wire is

twice as large
four times as large
the same, because the load is the same
a quarter as large

The area of a circle is πd²/4, so halving the diameter divides the area by 4 and multiplies the stress by 4. For the 1 mm wire, F = 20 × 9.8 = 196 N and A = π × (0.5 × 10⁻³ m)² = 7.85 × 10⁻⁷ m², so the stress is 2.5 × 10⁸ N m⁻². That is the stress at which steel stops springing back (Table 8.1 in Section 8.5), so the thin wire is in real danger.

Three Ways to Change a Solid

There are three ways in which a solid may change its dimensions when forces act on it, shown in Fig. 8.1 of the book. Below, one block does all of them. The changes are drawn hugely exaggerated: a real steel block would change by less than the width of a hair.

Tensile and compressive stress. Pull a cylinder by two equal and opposite forces at right angles to its ends (Fig. 8.1(a)), and the restoring force per unit area is a tensile stress. Push the ends together instead, and it is a compressive stress. Both are called longitudinal stress. Either way the length changes, by ΔL, and the fractional change is the longitudinal strain:

\text{longitudinal strain} = \dfrac{\Delta L}{L} \qquad (8.2)

Shearing stress. Apply two equal and opposite forces parallel to the end faces (Fig. 8.1(b)), and one face slides sideways relative to the other, by Δx. The restoring force per unit area is now a tangential or shearing stress, and the shearing strain is the sideways shift divided by the length:

\text{shearing strain} = \dfrac{\Delta x}{L} = \tan\theta \qquad (8.3)

where θ is the angle through which the side has turned from the vertical. Press a thick book down with your palm and push it sideways, and its pages do the same thing (Fig. 8.1(c)).

Hydraulic stress. A solid ball lowered deep into water is pressed equally from every side (Fig. 8.1(d)). The force of the water acts at right angles to every part of the surface, and the body is under hydraulic compression: its volume shrinks but its shape does not change. The restoring force per unit area is the hydraulic stress, equal in size to the pressure of the water. The strain is the volume strain:

\text{volume strain} = \dfrac{\Delta V}{V} \qquad (8.5)

Back on dry land, the ball returns to its old size. Every strain is a change in a dimension divided by the original dimension, so strain has no unit and no dimensions.

Why tan θ ≈ θ

Real shear angles are tiny. Here are θ itself (in radians) and tan θ, for angles up to 40°.

The green line is θ in radians and the blue curve is tan θ. For small angles the two are almost the same. At θ = 10°, θ = rad and tan θ = .

They differ by only about 1 %, and real shear angles in solids are far smaller than 10°. So we may write

\text{shearing strain} = \tan\theta \approx \theta \qquad (8.4)

One Wire, Two Pulls

A wire hangs from the ceiling and carries a weight F at its lower end. The ceiling pulls the top of the wire up with a force F, and the weight pulls the bottom down with a force F. What is the tensile stress at a cross-section of area A in the middle of the wire?

2F/A, because the wire is pulled at both ends
F/A
zero, because the two pulls cancel
F/(2A), because each end carries half the load

Imagine cutting the wire at that section. The part below is in equilibrium under the weight F pulling down and the part above pulling up, so the tension at the section is F, not 2F, and the tensile stress is F/A. One more point to remember: stress is not a vector. Unlike a force, it cannot be given one direction. What does have a direction is the force on the part of the body on a chosen side of a section.

Speed Round: Stress and Strain

Stress = F/A in N m⁻²; tensile and compressive stress change the length, shearing stress the shape, hydraulic stress the volume; every strain is a ratio with no unit.

Get five right to move on.

Sina