When you apply a force along the length of a structural member, internal forces develop throughout the material to resist that load. Stress and strain are the two quantities that let you quantify what's happening inside.

Stress ( $\sigma$ ) is the internal force per unit area that develops inside a material in response to an applied load. In axially loaded members, this stress acts perpendicular to the cross-section (meaning it pushes or pulls straight through the cut face of the member).

Strain ( $\epsilon$ ) is the deformation per unit length. It tells you how much the member stretches or shortens relative to its original size.

An axially loaded member is any structural element where the load acts along the longitudinal axis. This produces either:

Tension (pulling force → member elongates)
Compression (pushing force → member shortens)

The cross-sectional area of the member directly affects how stress distributes internally. Whether the cross-section is circular, rectangular, or irregular, the area is what you divide the force by to get stress.

Types of Stress and Strain

Tensile stress and strain develop when a pulling force elongates the member. Think of stretching a rubber band: the material gets longer and thinner.
Compressive stress and strain develop when a pushing force shortens the member, like compressing a spring.
Shear stress and strain can appear even in "axially loaded" members if the force isn't perfectly aligned with the longitudinal axis. This commonly shows up at bolted or pinned connections.
Thermal stresses and strains arise from temperature changes that cause expansion or contraction. These aren't from applied loads at all, but they still produce real deformation. Expansion joints in bridges exist specifically to accommodate this.

Calculating Stress and Strain

Fundamental Concepts, 9.7: Elasticity, Stress, Strain, and Fracture - Physics LibreTexts

Stress Calculation

The formula for normal stress in an axially loaded member is straightforward:

$\sigma = \frac{F}{A}$

where $F$ is the internal axial force and $A$ is the cross-sectional area.

Common units include pascals (Pa), megapascals (MPa), and pounds per square inch (psi). For reference, 1 MPa = 1 N/mm².

Two things to watch for:

The stress distribution is uniform across the cross-section only if the force is applied through the centroid and the cross-section is constant. This is the assumption you'll use for most problems in this course.
Stress concentrations occur at sudden geometric changes like holes, notches, or sharp corners. At these locations, the local stress can be several times higher than the average $F/A$ value.

Strain Calculation

Normal strain is calculated as:

$\epsilon = \frac{\Delta L}{L}$

where $\Delta L$ is the change in length and $L$ is the original (unloaded) length.

Strain is dimensionless since it's length divided by length. It's often expressed as a decimal (e.g., 0.002), a percentage (0.2%), or in microstrain (2000 $\mu$ strain).
A positive $\Delta L$ means elongation (tension), and a negative $\Delta L$ means contraction (compression).
$L$ is always the original length before any load is applied. This is your reference.

Stress-Strain Relationship for Elastic Materials

Hooke's Law

For most engineering materials loaded below a certain threshold, stress is directly proportional to strain. This is Hooke's Law:

$\sigma = E \cdot \epsilon$

The constant $E$ is the modulus of elasticity (also called Young's modulus). It's a material property that quantifies stiffness. Some typical values:

Steel: $E \approx 200 \text{ GPa}$ (29,000 ksi)
Aluminum: $E \approx 70 \text{ GPa}$ (10,000 ksi)
Concrete (compression): $E \approx 25\text{–}30 \text{ GPa}$

A higher modulus means the material is stiffer and deforms less under the same stress.

The elastic limit is the maximum stress a material can handle and still return to its original shape when unloaded. Beyond this point, permanent (plastic) deformation begins, and the material has yielded.

Stress-Strain Curve

Within the elastic region, the stress-strain curve is a straight line. The slope of that line equals $E$ . This is why a steeper curve means a stiffer material.

Key features you can read from a full stress-strain curve:

Proportional limit: where the curve stops being linear
Yield strength: where significant plastic deformation begins
Ultimate strength: the maximum stress the material can sustain
Fracture point: where the material breaks

The area under the stress-strain curve up to fracture represents the material's toughness, which is the total energy it can absorb before breaking.

Deformation of Axially Loaded Members

Deformation Calculation

By combining the stress and strain formulas with Hooke's Law, you get the axial deformation formula:

$\Delta L = \frac{FL}{AE}$

This is derived in three steps:

Start with stress: $\sigma = \frac{F}{A}$
Apply Hooke's Law: $\epsilon = \frac{\sigma}{E} = \frac{F}{AE}$
Solve for deformation: $\Delta L = \epsilon \cdot L = \frac{FL}{AE}$

This is one of the most-used equations in this unit. Make sure you can identify $F$ , $L$ , $A$ , and $E$ for each segment of a member.

For members with multiple segments (different forces, areas, or materials along the length), calculate $\Delta L$ for each segment separately and sum them:

$\Delta L_{total} = \sum \frac{F_i L_i}{A_i E_i}$

Keep your sign convention consistent: positive for tension/elongation, negative for compression/shortening.

Thermal Effects on Deformation

Temperature changes cause materials to expand or contract independently of any applied load. The thermal deformation is:

$\Delta L_{thermal} = \alpha \cdot L \cdot \Delta T$

where $\alpha$ is the coefficient of thermal expansion (a material property) and $\Delta T$ is the temperature change.

Some typical $\alpha$ values:

Steel: $\alpha \approx 12 \times 10^{-6} \text{ /°C}$
Aluminum: $\alpha \approx 23 \times 10^{-6} \text{ /°C}$

Aluminum expands nearly twice as much as steel for the same temperature change. This matters in structures exposed to large temperature swings (bridges, pipelines, aircraft components). If the member is free to expand, thermal deformation produces no stress. But if the member is constrained (fixed at both ends, for example), the prevented expansion generates internal thermal stress, which you'd calculate using the deformation compatibility approach.