Mechanical, thermal, and electrical properties, stress-strain behavior, ferrous and nonferrous metals and engineered materials, phase diagrams and heat treating, corrosion, and failure mechanisms.
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A. Properties
The Stress–Strain Diagram and Mechanical Properties
Read the engineering stress–strain curve to extract modulus, yield, ultimate strength, ductility, resilience, and toughness—and know when true stress takes over.
Almost everything you will ever calculate about a metal—whether a column buckles, a bolt yields, a shaft survives a million cycles—traces back to a single curve produced by pulling a polished bar apart in a tensile machine and plotting load against stretch. The FE Mechanical exam leans on this curve constantly, and the points are lost not in exotic places but in the basics: confusing yield with ultimate strength, dividing force by the necked area when the definition says use the original area, or treating the area under the elastic line (resilience) as if it were the area under the whole curve (toughness). The FE Reference Handbook’s Materials Science/Structure of Matter chapter defines engineering stress, engineering strain, and the modulus; this concept turns those bare definitions into the fluent reading of a diagram that the exam rewards.
Engineering stress and engineering strain
The tensile test reduces a messy three-dimensional specimen to two scalars. Engineering stress σ is the applied load divided by the original cross-sectional area A0—original, not current—and engineering strain ε is the change in gauge length divided by the original gauge length L0. Because both denominators are fixed at their undeformed values, the engineering curve is something you can build directly from machine readings without measuring the shrinking diameter. Stress carries units of pressure (MPa or psi); strain is dimensionless (often quoted as in/in or as a percent).
σ=A0F,ε=L0ΔL
The elastic region and Young’s modulus
Below the proportional limit the curve is a straight line, and the material returns to its original length when unloaded. The slope of that line is the elastic modulus E (Young’s modulus), and the relationship σ=Eε is Hooke’s law. E is a stiffness, not a strength: steel (E≈200GPa) is about three times stiffer than aluminum (E≈70GPa) regardless of their yield strengths, and heat treatment barely moves E even as it doubles the yield. Read E from the slope, never from any single high-stress point on the curve.
σ=Eε⟹E=εσ(elastic region only)
Yield strength and the 0.2% offset
Yield marks the transition from recoverable (elastic) to permanent (plastic) deformation. Many engineering metals lack a sharp yield point, so yield strength Sy is defined by the 0.2% offset method: draw a line parallel to the initial elastic slope but shifted to ε=0.002, and take its intersection with the curve. That construction gives a repeatable number even for a smoothly rounded knee. Yield strength—not ultimate strength—is the property that governs whether a part returns to shape, so it anchors most design checks.
Ultimate strength, necking, and fracture
Past yield the material strain-hardens and the engineering stress rises to a peak: the ultimate tensile strength Su, the maximum stress the original area ever sees. At that peak a neck forms—deformation localizes—and from there the load required to keep stretching falls even though the local material is still hardening, so the engineering curve droops down to the fracture point. The droop is an artifact of dividing a falling load by the fixed original area; the true stress in the neck keeps climbing right up to fracture.
Ductility, resilience, and toughness
Three areas-and-distances on the curve carry their own names. Ductility is how far the material stretches permanently before it breaks, reported as percent elongation %EL=(Lf−L0)/L0×100 or percent reduction of area. Modulus of resilience Ur is the elastic strain energy per unit volume—the triangular area under the elastic line, Ur=Sy2/(2E)—and measures how much energy a spring can store and give back. Toughness is the total area under the entire curve up to fracture, the energy absorbed per unit volume before breaking; a tough material needs both decent strength and decent ductility, which is why a high-strength brittle ceramic is not tough.
Ur=2ESy2,%EL=L0Lf−L0×100
Hardness and its link to strength
Hardness is resistance to localized plastic indentation, measured by pressing a defined indenter under a known load and reading the dent size (Brinell, Rockwell, Vickers). It is fast, cheap, and nearly nondestructive, which is why it is used for quality control in place of a full tensile test. For plain-carbon steels the handbook gives an approximate proportionality between Brinell hardness number and tensile strength, useful for a quick estimate when no tensile data are at hand.
Su[psi]≈500BHN,Su[MPa]≈3.5BHN
True stress versus engineering stress
True stress σT=F/A uses the instantaneous area A, and true strain εT=ln(L/L0)=ln(1+ε) accumulates over the changing length. Because the area shrinks under tension, true stress always exceeds engineering stress once plastic flow starts, and (before necking, where volume is conserved) σT=σ(1+ε) and εT=ln(1+ε). The exam expects you to know which curve you are on: design tables and the offset yield use engineering values, while metal-forming and necking analysis use true values.
σT=AF=σ(1+ε),εT=ln(1+ε)
Exam strategy
Start every tensile problem by writing A0 and L0 and labeling which area you are dividing by—original for engineering, current for true. For modulus, use two points on the straight initial line, never a single point past yield. Memorize the small map of the curve: slope is E, the 0.2% offset gives Sy, the peak is Su, area-under-elastic-line is resilience, area-under-everything is toughness. If a problem quotes a strain greater than a few percent and asks for stress “in the material,” it almost certainly wants true stress; multiply engineering stress by (1+ε).
Key equations
Engineering stressσ=A0F
Load F over ORIGINAL area A0 (m² → Pa, in² → psi). The default for design data and offset yield.
Engineering strainε=L0ΔL
Change in gauge length over original gauge length L0. Dimensionless; multiply by 100 for percent.
Hooke’s law (elastic)σ=Eε
Valid below the proportional limit. E = Young’s modulus (Pa or psi), the slope of the elastic line.
Modulus of resilienceUr=2ESy2
Elastic strain energy per unit volume (J/m³). Area of the triangle under the elastic line up to yield.
Percent elongation (ductility)%EL=L0Lf−L0×100
Permanent strain after fracture, reassembled gauge length Lf. A measure of ductility.
True stressσT=AF=σ(1+ε)
Load over INSTANTANEOUS area A. The (1+ε) form assumes constant volume, valid before necking.
True strainεT=ln(L0L)=ln(1+ε)
Logarithmic strain; additive over increments. Used in metal-forming and necking analysis.
Rough estimate of ultimate strength from Brinell number; Su[MPa]≈3.5BHN. Steels only.
Worked examples
Reading modulus, yield, and ultimate from a tensile test
Problem. A round steel bar of diameter 12.5mm and gauge length 50.0mm is pulled in tension. At a load of 24.0kN (still elastic) the gauge length has stretched 0.0489mm. The bar first yields at 42.0kN and reaches a peak load of 58.0kN. Find the elastic modulus, the yield strength, and the ultimate tensile strength.
Solution. Original area: A0=4π(0.0125)2=1.227×10−4m2 (122.7mm2).
Elastic point: σ=24,000/1.227×10−4=195.6MPa and ε=0.0489/50.0=9.78×10−4.
E=σ/ε=195.6×106/9.78×10−4=2.00×1011Pa=200GPa.
Yield: Sy=42,000/1.227×10−4=342MPa.
Ultimate: Su=58,000/1.227×10−4=473MPa.
Sanity check: E=200GPa is textbook steel, and Sy<Su as required for a ductile metal. Answers: E=200GPa, Sy=342MPa, Su=473MPa.
E=εσ=9.78×10−4195.6×106=200GPa
Resilience, ductility, and a true-stress correction
Problem. For the same bar (Sy=342MPa, E=200GPa), the broken halves reassemble to a gauge length of 61.0mm. (a) Find the modulus of resilience and the percent elongation. (b) At the ultimate point the engineering stress is 473MPa at an engineering strain of 0.114. Estimate the true stress there.
Solution. (a) Resilience: Ur=Sy2/(2E)=(342×106)2/(2⋅2.00×1011)=2.92×105J/m3=292kJ/m3.
Ductility: %EL=(61.0−50.0)/50.0×100=22.0%.
(b) True stress (volume-conserved form): σT=σ(1+ε)=473(1+0.114)=527MPa.
Sanity check: Ur in the hundreds of kJ/m³ is typical for steel; true stress exceeds engineering stress as it must under tension. Answers: Ur=293kJ/m3, %EL=22.0%, σT≈527MPa.
σT=σ(1+ε)=473(1.114)=527MPa
Common pitfalls
•Dividing by the necked (current) area to get ENGINEERING stress. Engineering stress always uses the original area A0; only true stress uses the instantaneous area.
•Confusing yield with ultimate strength. Sy governs the onset of permanent set (design limit); Su is the peak the bar can carry before necking. They are different numbers.
•Reading E off a point well up the curve. Modulus is the slope of the straight elastic line only—pick two low-stress points.
•Calling the area under the elastic line ‘toughness.’ That triangle is RESILIENCE; toughness is the area under the ENTIRE curve to fracture.
•Assuming heat treatment changes stiffness. Quenching and tempering move Sy and Su a lot but barely change E—stiffness is set by atomic bonding.
•Forgetting the 0.2% offset for metals without a sharp yield. Yield strength is the offset-line intersection at ε=0.002, not the proportional limit.
•Using εT=ε at large strain. True strain is ln(1+ε); the difference is negligible below ~1% but real once plastic flow is large.
References
NCEES FE Reference Handbook — Materials Science/Structure of Matter (Mechanical Properties)
NCEES FE Reference Handbook — Mechanics of Materials (Uniaxial Loading and Deformation)
Callister & Rethwisch, Materials Science and Engineering: An Introduction — tensile properties chapter
C. Ferrous metals
Ferrous and Nonferrous Metals and Alloys
Distinguish carbon and alloy steels, cast irons, and stainless grades from the major nonferrous alloys (Al, Cu, Ti, Mg) by composition, properties, and typical use.
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E. Engineered materials
Engineered Materials: Polymers, Ceramics and Composites
Contrast thermoplastics and thermosets, ceramics and glasses, and apply the composite rule of mixtures for modulus, strength, and density along and across the fibers.
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F. Manufacturing processes
Manufacturing Processes: Casting, Forming, Machining and Additive
Survey casting, bulk and sheet forming, machining and tolerances, joining, and additive manufacturing—and how each process drives a part’s properties and cost.
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G. Phase diagrams, phase transformation, and heat treating
Phase Diagrams, Heat Treatment and Failure Mechanisms
Use binary phase diagrams and the lever rule, read the iron–carbon diagram, choose annealing/quenching/tempering, and recognize fatigue, creep, and fracture failures.
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I. Corrosion mechanisms and control
Corrosion Mechanisms, Control and Materials Selection
Recognize galvanic, uniform, and pitting corrosion, read the galvanic series, apply cathodic protection and coatings, and select a material for strength, weight, cost, and environment.
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