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Material Properties and Processing

7% of exam

Mechanical, thermal, and electrical properties, stress-strain behavior, ferrous and nonferrous metals and engineered materials, phase diagrams and heat treating, corrosion, and failure mechanisms.

6 concepts

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

\sigma

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is the applied load divided by the original cross-sectional area

A_0

—original, not current—and engineering strain

\varepsilon

is the change in gauge length divided by the original gauge length

L_0

. 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).

\sigma = \dfrac{F}{A_0}, \qquad \varepsilon = \dfrac{\Delta L}{L_0}

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

\sigma = E\varepsilon

is Hooke’s law.

E

is a stiffness, not a strength: steel (

E \approx 200\,\text{GPa}

) is about three times stiffer than aluminum (

E \approx 70\,\text{GPa}

) 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.

\sigma = E\,\varepsilon \qquad \Longrightarrow \qquad E = \dfrac{\sigma}{\varepsilon}\ \text{(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

S_y

is defined by the 0.2% offset method: draw a line parallel to the initial elastic slope but shifted to

\varepsilon = 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

S_u

, 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 = (L_f - L_0)/L_0 \times 100\,

or percent reduction of area. Modulus of resilience

U_r

is the elastic strain energy per unit volume—the triangular area under the elastic line,

U_r = S_y^2/(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.

U_r = \dfrac{S_y^{\,2}}{2E}, \qquad \%EL = \dfrac{L_f - L_0}{L_0}\times 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.

S_u\,[\text{psi}] \approx 500\,\text{BHN}, \qquad S_u\,[\text{MPa}] \approx 3.5\,\text{BHN}

True stress versus engineering stress

True stress

\sigma_T = F/A

uses the instantaneous area

A

, and true strain

\varepsilon_T = \ln(L/L_0) = \ln(1+\varepsilon)

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)

\sigma_T = \sigma(1+\varepsilon)

and

\varepsilon_T = \ln(1+\varepsilon)

. 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.

\sigma_T = \dfrac{F}{A} = \sigma(1+\varepsilon), \qquad \varepsilon_T = \ln(1+\varepsilon)

Exam strategy

Start every tensile problem by writing

A_0

and

L_0

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

S_y

, the peak is

S_u

, 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+\varepsilon)

.

Key equations

Engineering stress

\sigma = \dfrac{F}{A_0}

Load $F$ over ORIGINAL area $A_0$ (m² → Pa, in² → psi). The default for design data and offset yield.

Engineering strain

\varepsilon = \dfrac{\Delta L}{L_0}

Change in gauge length over original gauge length $L_0$

Hooke’s law (elastic)

\sigma = E\,\varepsilon

Valid below the proportional limit. $E$ = Young’s modulus (Pa or psi), the slope of the elastic line.

Modulus of resilience

U_r = \dfrac{S_y^{\,2}}{2E}

Percent elongation (ductility)

\%EL = \dfrac{L_f - L_0}{L_0}\times 100

True stress

\sigma_T = \dfrac{F}{A} = \sigma\,(1+\varepsilon)

True strain

\varepsilon_T = \ln\!\left(\dfrac{L}{L_0}\right) = \ln(1+\varepsilon)

Hardness–strength (plain-carbon steel)

S_u\,[\text{psi}] \approx 500\,\text{BHN}

Rough estimate of ultimate strength from Brinell number; $S_u\,[\text{MPa}]\approx 3.5\,\text{BHN}$

Worked examples

Reading modulus, yield, and ultimate from a tensile test

Problem. A round steel bar of diameter $12.5\,\text{mm}$ and gauge length $50.0\,\text{mm}$ is pulled in tension. At a load of $24.0\,\text{kN}$ (still elastic) the gauge length has stretched $0.0489\,\text{mm}$ . The bar first yields at $42.0\,\text{kN}$ and reaches a peak load of $58.0\,\text{kN}$ . Find the elastic modulus, the yield strength, and the ultimate tensile strength.

Solution. Original area:

A_0 = \tfrac{\pi}{4}(0.0125)^2 = 1.227\times10^{-4}\,\text{m}^2

E = \dfrac{\sigma}{\varepsilon} = \dfrac{195.6\times10^{6}}{9.78\times10^{-4}} = 200\ \text{GPa}

Resilience, ductility, and a true-stress correction

Problem. For the same bar ( $S_y = 342\,\text{MPa}$ , $E = 200\,\text{GPa}$ ), the broken halves reassemble to a gauge length of $61.0\,\text{mm}$

Common pitfalls

•Dividing by the necked (current) area to get ENGINEERING stress. Engineering stress always uses the original area $A_0$ ; only true stress uses the instantaneous area.
•Confusing yield with ultimate strength. $S_y$ governs the onset of permanent set (design limit); $S_u$ 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 $S_y$ and $S_u$ 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 $\varepsilon = 0.002$ , not the proportional limit.
•Using $\varepsilon_T = \varepsilon$ at large strain. True strain is $\ln(1+\varepsilon)$ ; 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.