Elastic Deformation, Stress and Strain

Why a solid can deform

A bridge girder looks rigid, yet it bends slightly as traffic passes. A steel wire stretches under a hanging mass, and a rubber band changes length visibly. The rigid-body model used earlier is an approximation: it neglects changes in distances between parts of the body. Real solids can change length, shape or volume when forces act. Whether that change is harmless, recoverable or a route to failure depends on the material and the loading.

Elasticity is the tendency to recover the original size and shape after the deforming load is removed. Gently stretch a spring, then release it: within a suitable loading range it returns. Plastic deformation leaves a permanent change. A lump of putty pressed into a new shape has little tendency to recover its previous shape and is close to an ideal plastic. A metal can behave elastically under a small load and plastically under a sufficiently large one. “Elastic” and “plastic” therefore describe response under specified conditions, not two labels that can always be assigned permanently to an object.

At the microscopic level, deformation changes the separations and relative arrangement of neighbouring particles. Internal interactions oppose the imposed change and can restore the original arrangement after a small deformation. Large deformations can rearrange the structure irreversibly. We will describe the observable response using force, dimensions and material constants rather than attempting an atomic calculation.

Stress measures the force transmitted through a section

Imagine cutting a hanging wire across an interior section and isolating the lower part. The upper part pulls it upward. If the wire’s own weight is negligible and the suspended load is F, equilibrium of this lower part requires an upward internal force F. At the ceiling there is another force F, opposite to the load, but the two end forces are not added to make the tension 2F. They act on different boundaries. The force transmitted across any one section is F.

Stress describes internal force per unit area. For a uniform rod in static equilibrium under a uniformly distributed axial load, its normal stress magnitude can be found from the applied force divided by cross-sectional area. Concentrated contact, nonuniform shape or bending can make stress vary across a section; the simple F/A then represents an average, not every local value. A thinner wire under the same force has greater stress because less area shares the load.

Stress is not an ordinary vector like force: describing an internal load requires specifying the surface across which it acts as well as the force direction. In this chapter we use particular normal and shear stress components, often just their magnitudes. Equal units do not make pressure and every type of stress interchangeable; uniform pressure is one special loading condition.

Longitudinal loading: stretching or compression

An axial pull acts normally to the cross-section and causes tensile stress. An axial push causes compressive stress. Both are longitudinal loading because the principal length change is along the load direction. Let L be the unloaded length and ΔL = final length − initial length. The fractional change ΔL/L is longitudinal strain. It compares the deformation with the original size: a 1 mm extension means much more for a 10 mm specimen than for a 10 m wire.

Shear changes relative positions of faces

Put a thick book flat on a table, hold its bottom and push its top sideways. The pages slide relative to one another and the upright side becomes inclined. A solid block behaves similarly under tangential loading. The force acts parallel to a face, unlike the normal force in a stretching experiment. Shear stress magnitude is the tangential force divided by the area of the loaded face.

The right triangle formed by the original vertical side, its lateral displacement and the sloping side gives tan θ = Δx/L. For a small angle, the tangent and angle in radians nearly coincide: at 10°, θ is about 0.1745 rad and tan θ about 0.1763, a difference near 1%. Do not put the number 10 into a small-angle formula simply because an angle was stated in degrees. A fixed base can provide both force and moment reactions; an isolated pair of opposite tangential forces alone can form a couple. The complete loading/support arrangement must maintain the stated equilibrium.

Hydraulic compression changes volume

Now surround a solid sphere by fluid and raise the pressure equally on all sides. Each small force is normal to the local surface, so there is no preferred direction of loading. An isotropic sphere becomes smaller without changing its spherical shape. This is hydraulic compression. The hydraulic stress magnitude equals the applied pressure increase Δp. All-sided loading must be distinguished from a force acting on only two ends of a rod.

Units and choosing the correct deformation

Every strain here is a ratio of like dimensions, so its dimensional formula is 1, conventionally written M⁰L⁰T⁰. A strain of 0.002 means 0.2%, not 0.002%. A modulus will connect a particular stress to its corresponding strain. Before calculating, identify whether the question concerns axial extension, tangential distortion or all-sided compression; then identify the loaded area and the original dimension in the strain.

Quick Quiz