3.4 Mechanics of Materials I: Stress, Strain & Axial Loading

Stress & Strain

Stress is internal force intensity — force per unit area. The two basic types tested on the FE Civil exam:

• Normal stress σ = P/A — axial force P over cross-sectional area A (units: psi or Pa). Tension is positive, compression negative.
• Average shear stress τ = V/A — shear force V parallel to the area (units: psi or Pa).

Strain is the deformation response. Normal strain ε = δ/L is the change in length δ divided by original length L (dimensionless). Shear strain γ is the change in angle (radians) between originally perpendicular faces.

Hooke's law and Young's modulus

In the elastic (linear) range, stress is proportional to strain:

• σ = E·ε

where E is Young's modulus (modulus of elasticity), the slope of the initial straight line on the stress-strain curve. Typical values: structural steel E ≈ 29,000 ksi (200 GPa); aluminum ≈ 10,000 ksi (69 GPa); concrete ≈ 3,000–5,000 ksi. For shear, the analog is τ = G·γ, with G the shear modulus. The three elastic constants are related by G = E / [2(1 + ν)]. Knowing any two of E, G, and ν fixes the third — a relationship the exam uses to test whether you understand that these constants are not independent.

Strain is dimensionless (in/in or m/m), but it is often reported in microstrain (µε = 10⁻⁶ in/in); watch for that factor when a problem gives strain in µε. Likewise, stress is sometimes given in MPa or ksi — always confirm the unit system before plugging into Hooke's law, because mixing psi with GPa is a frequent and costly slip.

Axial Deformation & Poisson's Ratio

The elongation (or shortening) of a prismatic bar under axial load is:

• δ = P·L / (A·E)

where P is axial force, L is length, A is cross-sectional area, and E is the modulus of elasticity. For a bar with several segments of different P, A, or E, sum the segments: δ = Σ(Pᵢ·Lᵢ)/(Aᵢ·Eᵢ). A frequent trap is forgetting to convert units — keep force, length, and E in a consistent system (kips, inches, ksi gives δ in inches). The product AE is the axial rigidity (stiffness against stretching): a larger AE means a stiffer bar that deforms less under the same load.

Tapered bars or bars with distributed loading require integration, δ = ∫P(x)dx/(A(x)E), rather than the simple formula, so confirm that P, A, and E are constant over each segment before applying δ = PL/AE.

Poisson's ratio

When a bar stretches axially it contracts laterally. Poisson's ratio quantifies this:

• ν = − (lateral strain) / (axial strain)

For most metals ν ≈ 0.25–0.35 (steel ≈ 0.30); the theoretical maximum is 0.5 (incompressible). A bar stretched axially by strain ε develops lateral strain −νε in each transverse direction. This is why a stretched bar gets thinner and a compressed one bulges, and it underlies the change in volume of a loaded member. Poisson's ratio also enters multi-axial (biaxial and triaxial) stress states through the generalized Hooke's law, where the strain in one direction depends on the stresses in all three directions, each weighted by ν.

Thermal strain and stress

A temperature change produces thermal strain ε = α·ΔT, where α is the coefficient of thermal expansion (≈ 6.5×10⁻⁶ /°F for steel). Free thermal deformation is:

• δ_thermal = α·ΔT·L

If a member is fully restrained so it cannot expand, the prevented strain produces thermal stress σ = E·α·ΔT — independent of length. This is why bridges need expansion joints.

If a member is only partially restrained, superpose the two effects: the free thermal growth α·ΔT·L minus any deformation the restraint allows equals the net deformation, and the residual prevented strain times E gives the stress. Note that thermal stress in a fully restrained member is independent of cross-sectional area and length — it depends only on E, α, and ΔT — which surprises candidates who expect a longer bar to be more stressed.

Stress-Strain Diagram & Factor of Safety

The stress-strain diagram from a tension test maps a material's behavior. Key landmarks for ductile steel:

The slope of the linear region is the modulus of elasticity E. Ductile materials (mild steel) show a long plastic plateau and large elongation before fracture; brittle materials (cast iron, concrete) fracture with little warning near the ultimate stress.

Factor of safety

Design keeps working stress below a fraction of the failure stress:

• Factor of Safety (FS) = (failure stress) / (allowable stress) = σ_failure / σ_allow

FS is based on yield for ductile materials and ultimate for brittle ones.

Worked Example — Axial Deflection

A steel rod (E = 29,000 ksi) is 60 in long with area A = 0.5 in² and carries P = 10 kips tension. Find the elongation.

• δ = PL/(AE) = (10·60) / (0.5·29,000) = 600 / 14,500 = 0.0414 in

Check the stress: σ = P/A = 10/0.5 = 20 ksi, well below steel's ~50 ksi yield, so the elastic formula validly applies.