Material Behavior and Mechanical Testing

Introduction

The strength of materials depends not only on the applied load but also on how a material deforms and eventually fails. Mechanical testing provides the experimental basis for selecting materials, validating designs, establishing quality-control limits, and developing constitutive models for analysis. The principal tests considered in this chapter are tensile, compression, shear, and hardness tests.

A mechanical engineer must distinguish between load and stress, and between deformation and strain. Load is an external action, whereas stress describes the internal intensity of force. Deformation is a change in shape or size, whereas strain is deformation normalized by an original dimension. These distinctions allow test results from specimens of different sizes to be compared.

Learning Objectives

After studying this chapter, you should be able to:

• interpret tensile, compression, shear, and hardness tests;
• construct and explain engineering stress–strain relationships;
• distinguish elasticity, plasticity, yielding, and fracture;
• apply Hooke’s law and determine Young’s modulus, shear modulus, and Poisson’s ratio;
• explain ductility, brittleness, anisotropy, and temperature-dependent behavior;
• identify assumptions, limitations, and sources of error in mechanical testing;
• select suitable material data for engineering design.

Stress, Strain, and Material Response

Normal stress and strain

When an axial force acts perpendicular to a cross-sectional area, the average normal stress is

where σ is engineering normal stress, P is the applied axial force, and A₀ is the original cross-sectional area. The SI unit is pascal (Pa), although MPa is commonly used in engineering.

The corresponding engineering normal strain is

where ΔL is the change in gauge length and L₀ is the original gauge length. Strain is dimensionless and is often reported as mm/mm or as a percentage.

These expressions assume that the original dimensions are used. At large deformation, the current area and length change substantially, so true stress and true strain may be more appropriate:

Here A and L are the instantaneous area and gauge length. The true measures are particularly useful in metal forming and large plastic deformation. The engineering measures are normally used for standard tensile-test reporting.

Shear stress and shear strain

A force acting tangentially to an area produces shear stress:

where τ is average shear stress and V is the tangential force. For small elastic distortions, shear strain is

where Δx is the relative tangential displacement between two planes separated by distance L. Shear strain is measured in radians, although radians are dimensionless.

The Tensile Test

The tensile test is the most widely used method for determining the strength and deformation characteristics of metals, polymers, and many composite materials. A standardized specimen is gripped in a testing machine and pulled under controlled conditions while force and extension are recorded.

A typical engineering stress–strain curve for a ductile metal contains the following regions:

1. Linear elastic region: stress is approximately proportional to strain.
2. Elastic limit: the greatest stress for which the specimen returns essentially to its original dimensions after unloading.
3. Yielding: plastic deformation begins. Some materials show a distinct upper and lower yield point; others do not.
4. Strain hardening: additional plastic deformation requires increasing stress.
5. Uniform elongation and ultimate tensile strength: the maximum engineering stress is reached.
6. Necking: deformation becomes localized, reducing the load-carrying area.
7. Fracture: the specimen separates.

Figure: Idealized engineering stress–strain curve for a ductile material.

Elastic modulus and Hooke’s law

In the linear elastic region, uniaxial Hooke’s law is

where E is Young’s modulus, with units of Pa. The slope of the initial straight portion of a stress–strain curve is E. Hooke’s law applies only within the proportional elastic range. It should not be used after yielding or for materials whose response is strongly nonlinear.

The elastic strain energy density for a linearly elastic material is

where u is energy per unit volume, measured in J/m³. This relation assumes uniaxial linear elasticity.

Yield strength and offset method

For materials without a sharply defined yield point, the yield strength is commonly determined by the 0.2% offset method. A line parallel to the elastic portion is drawn from a strain of 0.002. Its intersection with the stress–strain curve defines the conventional yield strength. This is a reporting convention, not a fundamental discontinuity in the material.

Ultimate strength, ductility, and toughness

The ultimate tensile strength, σᵤ, is the maximum engineering stress. The fracture stress is the engineering stress at failure. Because necking reduces the actual area, the true stress at fracture may be much greater than the engineering fracture stress.

Percentage elongation is

where L_f is the final gauge length. Percentage reduction in area is

where A_f is the minimum final cross-sectional area. Both measures indicate ductility, but they are sensitive to specimen geometry and test procedure.

The toughness of a material is the energy absorbed per unit volume before fracture. In a tensile test it is represented by the total area under the stress–strain curve. Toughness differs from strength: a high-strength material may be brittle and absorb little energy before breaking.

Elasticity, Plasticity, and Poisson’s Ratio

Elastic deformation is recoverable upon unloading. Plastic deformation is permanent. Real materials may exhibit both responses simultaneously; unloading from a plastic state generally recovers only the elastic component.

When a specimen is stretched longitudinally, it usually contracts laterally. Poisson’s ratio is

where ν is Poisson’s ratio, εₜ is transverse strain, and εₗ is longitudinal strain. The negative sign makes ν positive for ordinary tensile loading because the two strains have opposite signs.

For an isotropic, homogeneous, linearly elastic material,

where G is the shear modulus. Also,

where K is the bulk modulus. These relationships require isotropy and linear elasticity. They do not generally apply to anisotropic composites or materials undergoing plastic deformation.

Compression Testing

In a compression test, a specimen is shortened by an axial force. Engineering compressive stress may be reported as

and compressive strain as

with a sign convention often taken as negative for compression. Many engineering reports use positive magnitudes for convenience.

Compression tests are particularly important for concrete, ceramics, foams, polymers, powders, and materials that are difficult to grip in tension. Ductile metals may not fracture in compression; instead they may expand laterally and undergo barreling. Friction between specimen ends and loading platens restrains lateral expansion and produces nonuniform deformation. Lubrication, specimen slenderness, parallel end faces, and alignment are therefore important.

For slender specimens, instability may occur before material crushing. Thus a compression test must distinguish material behavior from structural buckling. A short specimen is preferred when the objective is to measure compressive material properties.

Shear Testing

Shear properties are measured using direct shear, torsion, or specialized lap-shear tests. Under elastic conditions,

This is the shear form of Hooke’s law and is valid for linear elastic behavior.

A torsion test is useful for shafts and provides information about shear modulus, torsional strength, and failure mode. For a circular shaft in the elastic range,

where T is torque, r is radial distance from the center, and J is the polar second moment of area. For a solid circular shaft of radius R, J = πR⁴/2. The maximum elastic shear stress occurs at the outer surface, r = R. The formula assumes a circular prismatic shaft, small deformation, elastic behavior, and Saint-Venant torsion conditions.

Shear tests are also essential for adhesives, fasteners, welds, composite laminates, and bolted or bonded joints. Results can be affected by stress concentrations and by the fact that the nominal stress may differ substantially from the local stress near a joint edge.

Hardness Testing

Hardness is resistance to localized permanent deformation, usually indentation, scratching, or rebound. It is not a fundamental elastic constant, but it often correlates with strength and wear resistance.

Common methods include:

Rockwell testing reports hardness from indentation depth, whereas Brinell and Vickers calculations use indentation geometry. A hardness number is meaningful only with its scale, indenter, load, dwell time, and standard. For example, “HRC” and “HRB” are not interchangeable.

Hardness testing is fast and usually nondestructive apart from a small mark. However, surface finish, specimen thickness, curvature, decarburization, residual stress, and proximity of earlier indentations can affect results. Hardness should not be treated as a direct substitute for a tensile test unless a validated correlation exists for the specific material and condition.

Ductility, Brittleness, and Failure Behavior

A ductile material undergoes substantial plastic deformation before fracture. Ductility is valuable because visible deformation can warn of overload and because plastic redistribution can reduce the effect of local stress concentrations. Mild steel, annealed copper, and many aluminum alloys are commonly ductile under ordinary conditions.

A brittle material fractures with little plastic deformation. Brittle fracture can propagate rapidly, particularly from cracks or flaws. Ceramics, glass, gray cast iron, and some hardened steels may behave brittly.

Ductility is not an absolute label. It depends on temperature, strain rate, stress state, microstructure, specimen geometry, and environment. A material that is ductile in a slow tensile test may become brittle at low temperature or high loading rate. Notches increase triaxial stress and often reduce apparent ductility.

Fracture surfaces provide evidence of failure mode. Ductile fracture often shows necking and a fibrous cup-and-cone appearance; cleavage-like surfaces are associated with brittle fracture. Fractographic interpretation must consider loading history and material microstructure.

Anisotropy and Material Directionality

An isotropic material has the same properties in every direction. An anisotropic material has direction-dependent properties. Rolled metals may have different yield strengths and ductilities in the rolling and transverse directions. Fiber-reinforced composites are strongly anisotropic: stiffness and strength are generally greatest along the fibers.

For anisotropic materials, a single tensile test is insufficient. Specimens should be cut in relevant orientations, and the material coordinate system must be recorded. Composite design commonly uses longitudinal, transverse, and shear properties rather than one Young’s modulus and one tensile strength.

Anisotropy may result from crystal structure, solidification, rolling, forging, additive manufacturing, aligned fibers, or layered construction. Ignoring it can cause unsafe predictions, especially when loads are multiaxial or vary in direction.

Temperature- and Rate-Dependent Behavior

Temperature affects atomic mobility, phase stability, yield strength, ductility, creep, and fracture behavior. Many metals become stronger but less ductile as temperature decreases. Some body-centered-cubic steels exhibit a ductile-to-brittle transition over a temperature range. At elevated temperature, strength often decreases and time-dependent deformation becomes important.

Creep is progressive deformation under sustained stress, commonly significant at high homologous temperature. A simplified representation of creep includes primary, secondary, and tertiary stages. The secondary stage often has an approximately steady creep rate, which is important for design life.

Polymers are especially sensitive to temperature and strain rate because their molecular chains rearrange over time. At low temperature or high strain rate, a polymer may appear glassy and brittle; at higher temperature or lower rate, it may behave more ductilely or viscoelastically.

Consequently, test reports must state temperature, loading rate, specimen orientation, conditioning, and environment. Data obtained under one condition should not automatically be used under another.

Worked Example: Interpreting a Tensile Test

A steel specimen has an original diameter of 12 mm and gauge length of 50 mm. At a load of 45 kN, its gauge length is 50.15 mm. Determine the engineering stress and strain.

The original area is

Since 1 N/mm² = 1 MPa,

The extension is ΔL = 50.15 − 50.00 = 0.15 mm. Therefore,

or 0.30%. If this point lies in the linear elastic region, an approximate modulus based on this single point would be E = σ/ε ≈ 132.6 GPa. In practice, E should be determined from the slope of several reliable points in the initial linear region, not from one measurement.

Illustrative Case: Selecting a Material for a Press-Fit Shaft

Background

An engineering firm is replacing a steel shaft in a moderate-speed machine. The shaft experiences fluctuating torque and occasional overload. Production requires dimensional stability and economical inspection.

Problem

A candidate alloy has high hardness but limited tensile ductility. Another candidate has slightly lower hardness but higher elongation and toughness. The design team initially recommends the harder alloy on the assumption that hardness alone indicates superior performance.

Analysis

The shaft is subjected primarily to torsional shear stress, but stress concentrations occur at a shoulder. A hardness number can suggest wear resistance and may correlate with tensile strength within a controlled alloy family, but it does not directly establish torsional fatigue strength, fracture toughness, or notch sensitivity. Tensile testing, torsion or shear data, and fatigue information are required. The service temperature and heat-treatment condition must also match the test data.

Approach and outcome

The team specifies tensile tests in the longitudinal direction, torsional testing, hardness mapping across the heat-affected and machined regions, and metallographic verification. The tougher alloy is selected because it provides adequate strength while offering greater tolerance to overload and surface defects. Hardness remains a production-control check rather than the sole design criterion.

Learning points

• A single mechanical property rarely defines suitability.
• Test conditions must represent service conditions.
• Strength, toughness, ductility, wear resistance, and fatigue resistance are distinct properties.
• Standards and traceable calibration improve confidence in test results.

Testing Practice, Standards, and Limitations

Mechanical testing should follow an appropriate standard, such as a recognized ASTM, ISO, or national standard. The standard specifies specimen geometry, preparation, loading rate, calibration, environmental conditions, and reporting. Laboratories should document specimen identification, material condition, orientation, dimensions, test machine, extensometer, uncertainty, and failure location.

Important sources of error include machine compliance, grip slippage, misalignment, inaccurate dimensions, extensometer calibration, surface damage, residual stress, temperature variation, and incorrect strain-rate control. Repeat tests are useful, but repetition cannot correct a systematic error.

Engineering decisions should use statistically representative data rather than a single favorable result. Safety-critical applications may require traceability, certification, witness testing, and conservative design values. Ethical practice requires reporting anomalous failures rather than selectively discarding inconvenient data.

Digital image correlation, automated extensometry, instrumented indentation, and high-speed data acquisition increasingly supplement conventional tests. Machine learning may assist in identifying patterns in large test datasets, but it cannot replace physical validation, transparent uncertainty assessment, or engineering judgment. Models trained on one alloy, geometry, or temperature range should not be assumed valid outside that domain.

Chapter Summary

Mechanical testing links material structure and behavior to engineering design. Engineering stress is force divided by original area, while engineering strain is extension divided by original gauge length. Tensile tests reveal elastic response, yielding, strain hardening, ultimate strength, necking, fracture, ductility, and toughness. Hooke’s law relates stress and strain only within the applicable linear elastic range.

Compression tests are important for materials and components that behave differently in compression, but friction and buckling must be controlled. Shear tests determine resistance to tangential loading; in linear elasticity, shear stress equals shear modulus times shear strain. Hardness tests provide rapid localized measurements but require careful specification of scale and test conditions.

Elasticity is recoverable deformation, whereas plasticity is permanent deformation. Young’s modulus, shear modulus, bulk modulus, and Poisson’s ratio are related for isotropic linear-elastic materials. Ductility, brittleness, anisotropy, temperature, strain rate, and environment strongly influence measured behavior. Reliable design requires test data that match the material condition and service conditions, together with standards, calibration, traceability, and honest reporting.

Key Terms

Stress: Internal force intensity, commonly force divided by area.

Strain: Deformation normalized by an original dimension.

Engineering stress: Load divided by original cross-sectional area.

True stress: Load divided by instantaneous cross-sectional area.

Elasticity: Ability to recover deformation after unloading.

Plasticity: Permanent deformation remaining after unloading.

Young’s modulus: Slope of the linear elastic normal stress–strain curve.

Shear modulus: Ratio of elastic shear stress to shear strain.

Poisson’s ratio: Negative ratio of transverse strain to longitudinal strain.

Yield strength: Stress marking the onset of conventional plastic deformation.

Ultimate tensile strength: Maximum engineering stress in a tensile test.

Ductility: Capacity for plastic deformation before fracture.

Brittleness: Tendency to fracture with little plastic deformation.

Toughness: Energy absorbed per unit volume before fracture.

Hardness: Resistance to localized permanent deformation.

Anisotropy: Directional dependence of material properties.

Creep: Time-dependent deformation under sustained load.

Necking: Localized reduction in area during tensile deformation.

Engineering stress–strain curve: Graph relating nominal stress to nominal strain during a test.

Review Questions

Short Answer Questions

1. Define engineering stress and engineering strain.
2. State Hooke’s law for uniaxial loading and identify its limitation.
3. What is the purpose of the 0.2% offset method?
4. Distinguish ductility from toughness.
5. Why must the hardness scale be reported with a hardness value?

Long Answer Questions

1. Describe the complete engineering stress–strain response of a ductile metal from loading to fracture.
2. Explain the differences between tensile, compression, shear, and hardness tests.
3. Derive the relationship between Young’s modulus, shear modulus, and Poisson’s ratio for an isotropic linear-elastic material, and state its applicability.
4. Discuss how temperature and strain rate influence material behavior.
5. Explain why anisotropy must be considered when testing rolled metals and fiber-reinforced composites.

Analytical / Application Questions

1. A 20 mm diameter specimen carries a tensile load of 80 kN. Calculate its engineering stress and explain what additional information is needed to calculate strain.
2. A material has E = 210 GPa and ν = 0.30. Calculate its shear modulus using the isotropic elastic relation, and state why the result may be inappropriate for an anisotropic composite.
3. A designer chooses a material solely because it has a higher Brinell hardness number. Evaluate this decision for a notched shaft subjected to fluctuating torsion and propose a more appropriate testing program.

Multiple Choice Questions with Answers

Question 1

Which expression defines engineering normal stress?

A. σ = P/A₀

B. σ = ΔL/L₀

C. σ = Gγ

D. σ = P/L₀

Correct Answer: A. σ = P/A₀

Question 2

Hooke’s law for uniaxial linear elasticity is

A. ε = Eσ

B. σ = Eε

C. σ = νE

D. E = σ + ε

Correct Answer: B. σ = Eε

Question 3

A specimen extends from 50.00 mm to 50.20 mm. Its engineering strain is

A. 0.0004

B. 0.0040

C. 0.0400

D. 4.00

Correct Answer: B. 0.0040

Question 4

Which property is represented by the slope of the initial linear tensile stress–strain curve?

A. Toughness

B. Yield strength

C. Young’s modulus

D. Hardness

Correct Answer: C. Young’s modulus

Question 5

The 0.2% offset method is commonly used to determine

A. fracture toughness

B. conventional yield strength

C. Poisson’s ratio

D. hardness number

Correct Answer: B. conventional yield strength

Question 6

For an isotropic linear-elastic material, the relation between E, G, and ν is

A. G = E/[2(1 + ν)]

B. G = E(1 + ν)

C. G = 2E/(1 + ν)

D. G = E/[3(1 − 2ν)]

Correct Answer: A. G = E/[2(1 + ν)]

Question 7

Which test is primarily based on resistance to localized indentation?

A. Tensile test

B. Compression test

C. Hardness test

D. Torsion test

Correct Answer: C. Hardness test

Question 8

What commonly occurs after the ultimate tensile strength of a ductile specimen is reached in an engineering stress–strain test?

A. Elastic unloading only

B. Necking

C. Complete recovery

D. Zero strain

Correct Answer: B. Necking

Question 9

Why can a single tensile test be inadequate for a fiber-reinforced composite?

A. Composites have no measurable stress

B. The material may be anisotropic

C. The test cannot measure force

D. Poisson’s ratio is always zero

Correct Answer: B. The material may be anisotropic

Question 10

A 10 mm diameter specimen carries 31.4 kN. Its engineering stress is approximately

A. 40 MPa

B. 100 MPa

C. 400 MPa

D. 1,000 MPa

Correct Answer: C. 400 MPa