FE Mechanics of Materials: Stress and Strain Review

Success on the FE exam hinges on a rock-solid understanding of fundamental principles, and few are as critical as stress and strain. These concepts form the bedrock of mechanical design, failure analysis, and material selection. Mastering them is not just about memorizing formulas; it's about developing the intuition to predict how real-world components will behave under load, a skill directly tested across multiple exam problems.

Foundational Definitions: Stress and Strain

Stress is defined as the intensity of internal force acting on a specific plane within a material. It is force per unit area. The two primary types you must distinguish are normal stress and shear stress. Normal stress ($\sigma$) acts perpendicular to the cross-sectional area and is caused by forces that tend to elongate or shorten a member. For a prismatic bar with axial force $P$ and cross-sectional area $A$, the average normal stress is calculated as $\sigma =P/A$. Shear stress ($\tau$) acts parallel to the cross-sectional area and is caused by forces that tend to slide one part of the material past another. The average shear stress for a simple connection is $\tau =V/A$, where $V$ is the shear force.

Strain is the measure of deformation. Normal strain ($ϵ$) is the change in length per original length, expressed as $ϵ=\delta /L$, where $\delta$ is the total elongation or contraction. It is a dimensionless quantity. Shear strain ($\gamma$) measures the change in angle (in radians) between two lines that were originally perpendicular within the material. Understanding that stress is the cause (load) and strain is the effect (deformation) is your first critical step.

Material Behavior and the Stress-Strain Diagram

The relationship between stress and strain for a material is experimentally determined by a tensile test and plotted on a stress-strain diagram. For ductile materials like mild steel, this curve reveals key properties. The initial linear region is governed by Hooke's Law: $\sigma =Eϵ$. Here, $E$ is the elastic modulus (or Young's modulus), the slope of the linear portion, representing the material's stiffness. The point where the curve first deviates from linearity is the proportional limit. Slightly beyond this is the yield strength, often denoted ${\sigma }_Y$, a critical design property marking the onset of significant plastic (permanent) deformation.

After yielding, the material strain-hardens until reaching the ultimate tensile strength, the maximum stress on the curve. Finally, necking occurs, leading to fracture. For brittle materials like cast iron, the diagram shows a much smaller or non-existent plastic region, with fracture occurring shortly after the proportional limit. The FE exam frequently tests your ability to identify these points on a given diagram.

Key Material Properties and Thermal Effects

Beyond the elastic modulus and yield strength, you must know Poisson's ratio ($\nu$). When a material is stretched longitudinally, it contracts laterally. Poisson's ratio is the negative ratio of lateral strain to axial strain: $\nu =-{ϵ}_{lateral}/{ϵ}_{axial}$. For most metals, $\nu \approx 0.33$. This property is essential for understanding multi-axial deformation.

Materials also deform due to temperature changes. Thermal strain is given by ${ϵ}_T=\alpha \Delta T$, where $\alpha$ is the coefficient of thermal expansion. If a member is free to expand, this strain does not produce stress. However, if expansion or contraction is fully or partially restrained, significant thermal stress develops. The total strain in a restrained member is the sum of the mechanical strain ($\sigma /E$) and the thermal strain, often leading to the equation $\sigma =E({ϵ}_{total}-\alpha \Delta T)$.

Statically Indeterminate Axial Loading

A statically indeterminate axial member or system has more unknown support reactions or internal forces than available equilibrium equations. Solving these problems is a high-yield FE topic and requires a three-step compatibility-geometry approach.

1. Equilibrium: Write the relevant static equilibrium equation(s).
2. Geometry of Deformation: Based on the physical constraints, establish a compatibility equation relating the deformations of the parts. (e.g., The total elongation of bar A plus the contraction of bar B must equal zero because the connecting plate is rigid).
3. Force-Deformation (Constitutive): Relate the deformations in the compatibility equation back to forces using $\delta =PL/AE$ (and ${\delta }_T=\alpha \Delta TL$ if thermal effects are present).

You then solve the system of equations from steps 1, 2, and 3 simultaneously. The core challenge is correctly formulating the compatibility condition; a misstep here is the most common error.

Common Pitfalls

1. Confusing Stress Types: Using an area parallel to the force to calculate normal stress, or using a cross-sectional area for single-shear in a bolt. Remember: For normal stress from an axial force, use the area perpendicular to the force. For shear stress in a bolt, use the cross-sectional area of the bolt shank, not the head.
2. Misapplying Hooke's Law: Applying $\sigma =Eϵ$ beyond the proportional limit or for materials that have already yielded. Hooke's Law is only valid in the linear elastic region of the stress-strain curve.
3. Incorrect Compatibility Equations: In statically indeterminate problems, assuming all members elongate equally when they don't. Carefully sketch the deformed geometry based on the physical constraints (e.g., a rigid bar will rotate about a pin, causing different displacements at different connection points).
4. Ignoring Sign Conventions: Treating thermal stress algebraically incorrectly. Remember: A temperature increase ($+\Delta T$) with full restraint induces compressive stress (negative $\sigma$) if the material expands positively with heat. Consistently define tension and elongation as positive.

Summary

• Stress ($\sigma ,\tau$) is force per unit area (cause), while strain ($ϵ,\gamma$) is deformation per original dimension (effect).
• The stress-strain diagram defines material properties: the slope of the linear region is the elastic modulus ($E$), the point of plastic onset is the yield strength (${\sigma }_Y$), and the negative ratio of lateral to axial strain is Poisson's ratio ($\nu$).
• Thermal strain is ${ϵ}_T=\alpha \Delta T$; restraint against this deformation induces thermal stress.
• Solve statically indeterminate axial loading problems by combining 1) Equilibrium, 2) Compatibility of deformations, and 3) Force-Deformation ($\delta =PL/AE$) relationships.
• On the FE exam, always check that you are using the correct area for stress calculations and that your assumptions about material behavior (linear elastic) are valid for the problem context.