Materials Science · Additive manufacturing and multiscale characterisation

Defects, texture and anisotropy

Useful background: Processing routes and microstructural evolution; Measuring and checking material claims. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Explain why additive direction and defects affect measured response.
Distinguish crystallographic texture from grain shape and oriented defects.
Resolve a declared uniaxial stress onto an inclined plane without calling it a local flaw-tip stress.
Interpret matched directional measurements and state what additional evidence a proposed mechanism needs.

Layerwise manufacture can create directionality

Texture, elongated grains, layer interfaces and defect shapes can create anisotropic behaviour. Lack-of-fusion defects and gas pores have different geometries and origins. A single total pore fraction does not distinguish them.

Imagine two printed metal coupons with the same measured void volume fraction. One has mostly rounded voids; the other contains thin, extended unbonded regions. Pulling across an extended region and pulling along it are different loading situations. To explain a directional test result, first identify which structure is directional and how the load meets it. The word printed does not supply a unique strength or preferred loading direction.

Here z is the recorded build direction, normal to the deposited layer planes, and x is one recorded in-plane direction. Scan directions may rotate between layers, so x need not be every layer’s scan direction. Anisotropy means a response depends on direction relative to the material structure under specified conditions. It is different from spatial heterogeneity: a response may vary from place to place even when a small region has no preferred direction. Both can occur together.

Crystal orientation, grain shape and a layer boundary are different observations

A grain is a region with a common crystal orientation, apart from its internal imperfections. Crystallographic texture describes the distribution of lattice orientations across many grains relative to the specimen axes. A preferred distribution means some orientations occur more often than in a random orientation distribution. Texture is not surface roughness, and a long grain outline does not by itself identify its lattice directions.

The same column-shaped grain outlines can enclose differently oriented lattices. Conversely, roughly equiaxed grains can have a preferred lattice orientation. Crystal stiffness and the stress required to activate a slip system can depend on crystallographic direction. A preferred orientation distribution can therefore make the aggregate’s elastic or plastic response directional even when there are very few pores. Grain shape and the arrangement of phases can create additional directional barriers or failure paths; they are separate variables.

During fusion-based additive processing, crystals can grow from existing solid into a newly solidifying region, inheriting its lattice orientation. Selection among growing orientations and changes in heat-flow direction can produce preferred orientations and elongated structures. Side-branching can also change a cell’s growth direction within one grain; a change in the drawn growth path need not be a new grain. The outcome depends on the alloy, location and processing history.

A deposited layer line is a process coordinate, not proof of an unbonded interface. Solidification and growth can cross previous melt boundaries. Identify any unbonded region, segregated phase, orientation change or other actual feature before treating a visible track or layer boundary as a weak plane. Direction and size can change observed behaviour provides preparation on directional properties; Layerwise processing and local thermal history develops the preceding process account.

Follow the defect’s origin, shape and location

In laser powder bed fusion, incomplete melting or inadequate overlap and filling between adjacent regions can leave lack-of-fusion defects: irregular unbonded regions, sometimes with incompletely melted particles and thin extensions. Such regions can follow the arrangement of tracks or layers. The relevant mechanism is failure to consolidate the intended region; a single nominal energy-per-volume number does not uniquely specify that outcome.

Gas already associated with the feedstock or gas entrapped in the liquid can leave comparatively rounded pores when trapped during solidification. A different route is an unstable deep vapour depression, called a keyhole: collapse or pinch-off can trap a bubble that remains as a pore. Increasing input energy is therefore not a universal cure for porosity. Shape and location are useful clues, but rounded does not uniquely mean feedstock gas, and lack-of-fusion regions are not all identical flat ellipses. A causal assignment needs process and structural evidence together.

For geometric bookkeeping only, compare a spherical void of radius r = 0.100 mm with an oblate ellipsoid whose semiaxes are a = b = 0.200 mm and c = 0.0250 mm. A semiaxis is half a full diameter. Their volumes are V_s = (4π/3)r³ and V_e = (4π/3)abc. Both are 0.00418879 mm³ because abc = r³. The flat void is twice as wide and one quarter as thick as the sphere when full diameters are compared.

If each hypothetical 100 mm³ specimen contains 100 nonoverlapping voids of the stated type, both have void volume 0.418879 mm³ and porosity p = V_void/V_envelope ≈ 0.00418879, or 0.419%. That equality gives no equality of largest flaw dimension, curvature, distance from a free surface or orientation. The central circular area of the sphere is πr² ≈ 0.0314 mm²; the broad central face section of the ellipsoid is πab ≈ 0.1257 mm², four times larger. This is a geometrical area comparison, not a fourfold strength or crack-driving-force law.

Original schematic comparison. The upper pair shows identical four elongated grain outlines, with different projected lattice-axis orientations on the left and identical projections on the right. These are two-dimensional illustrative lattice projections, not measured orientation distributions or texture intensities. Grain outline alone therefore cannot identify crystallographic texture. Below, a central sphere section and the central a–c edge section of an oblate ellipsoid are drawn at the same length scale: sphere radius 0.100 millimetre; ellipsoid semiaxes a and b 0.200 millimetre and c 0.0250 millimetre. The circular radius is half the ellipsoid broad semiaxis, and four times its short semiaxis. Both ideal void volumes are 0.00418879 cubic millimetre. The ellipsoid broad a–b central section, which is not drawn, has four times the spherical central-section area. Neither equal void volume nor this area ratio supplies a strength prediction.

Resolve the load before predicting a failure mode

Use a deliberately limited model: a uniform uniaxial tensile stress σ acts along a load direction in otherwise intact material. Choose an imaginary plane with unit normal n, and let θ be the angle between n and the load direction, from 0° to 90°. The normal is perpendicular to the plane. At θ = 0° the plane is perpendicular to the load; at θ = 90° the plane is parallel to it.

Take an inclined patch of area A_s. Its area projected onto a plane perpendicular to the load is A_s cos θ. Force balance in the uniform uniaxial field gives a transmitted force magnitude σ A_s cos θ along the load direction. Dividing by A_s gives traction magnitude σ cos θ. Resolving that force a second time gives the normal component σ_n = σ cos² θ and the shear magnitude τ = σ sin θ cos θ. Stress has units of force per area: 1 MPa = 1 N/mm².

For σ = 120 MPa, a plane normal to the load has σ_n = 120 MPa and τ = 0. At θ = 60°, σ_n = 30.0 MPa and τ ≈ 52.0 MPa. Less resolved opening does not mean no resolved loading; shear is largest at 45° in this uniaxial model. At 90° both components on that particular plane vanish, but other planes, other flaws and other stress components can still matter.

These are resolved components of the nominal surrounding stress, not the actual traction on a traction-free pore surface and not the amplified stress at a sharp flaw edge. For an existing flat flaw they help distinguish opening from sliding tendencies. Flaw size, tip shape, interactions, residual stress, local plasticity and resistance are still required to predict initiation or growth. Cracks and stress concentration develops that next step. Rotating a part may change which features are loaded unfavourably; it does not erase the features.

Original uniaxial projection model at σ = 120 MPa; θ is the angle between the load and the plane normal.
θ	Normal σ_n (MPa)	Shear magnitude τ (MPa)
0°	120	0
30°	90.0	52.0
45°	60.0	60.0
60°	30.0	52.0
90°	0	0

Original surrounding-stress projection model, not the actual traction on a traction-free pore face. Nominal uniaxial tension is 120 MPa and points vertically. In the first sketch the plane is horizontal, its normal points vertically, and theta is zero: resolved normal stress is 120 MPa and shear is zero. In the second sketch the plane normal is 60 degrees from the vertical load and perpendicular to its inclined plane: resolved normal stress is 30 MPa and shear magnitude about 52 MPa. Sketch arrow lengths are schematic. In the lower graph theta is the angle between load and plane normal, from zero to 90 degrees. The solid blue curve with circles is normal stress divided by nominal stress, cosine squared theta; the long-dashed orange curve with square markers is shear magnitude divided by nominal stress, sine theta times cosine theta. At zero degrees these are one and zero; at 45 degrees both are one half; at 90 degrees both are zero. The normalised shear has its maximum one half at 45 degrees. The graph resolves loading only; no flaw-edge stress or failure threshold is plotted.

Anisotropy belongs to a named response and a supported comparison

The following original teaching record uses separate, matched destructive tensile coupons: same alloy and heat-treated state, envelope geometry, gauge length, surface finish, temperature, strain-rate definition and endpoint. R has rounded voids; P has thin extended defects whose broad planes are approximately normal to z. Both have the same pore inventory above, p ≈ 0.419%. These are invented observations, not measured properties of a named alloy and not predictions of the projection model.

E is the small-strain axial elastic modulus measured over the same declared fit interval. Fracture elongation here is final permanent gauge-length increase divided by the same original gauge length; it is not a local strain at a pore. Three results are shown per direction so the actual observed spread remains visible. No population confidence interval or design allowable is supplied.

For P, the mean E is 190 GPa in both directions, but mean fracture elongation is 12% in x and 6% in z. Relative to the x mean, the z mean is lower by (12 − 6)/12 = 50%, a six-percentage-point difference. Thus elastic similarity in these observations does not establish similar failure deformation. R has equal 18% elongation means here, although no finite set of tests establishes full isotropy at every angle or for fatigue.

The P defect-plane orientation and the stronger nominal opening for z loading are consistent with the lower z elongation, but the table alone does not uniquely establish the cause. Compare actual defect dimensions/orientations/locations, fracture origins and paths, and crystal-orientation and phase observations. Different texture, surface damage or residual stress could also influence the result. Controlled comparisons can strengthen a mechanism account; one scalar density measurement cannot replace that evidence.

Invented matched-coupon record. Each bracket contains three separate results; the geometric inventory is common to both sets.
Set / load direction	E results (GPa)	Fracture elongation results (%)
R / x	[188, 190, 192]	[17, 18, 19]
R / z	[188, 190, 192]	[17, 18, 19]
P / x	[188, 190, 192]	[11, 12, 13]
P / z	[188, 190, 192]	[5, 6, 7]

Orientation belongs with the test

Record loading relative to build and structural directions, surface finish and post-processing. An apparently isotropic property in one measurement does not establish isotropy for every response or failure mode.

Record the full relationship: build axes, scan pattern when relevant, extraction location, loading direction, defect-plane normal, material and surface state, and the measured response. “Vertical” without a reference frame is incomplete. Rotating a specimen in the test fixture while retaining the same tensile axis relative to its structure does not create a new material loading direction.

A heat treatment can change phase distribution, dislocation structure, texture or residual stress without eliminating a pore or an oxide-lined unbonded region. A pressure-assisted treatment and a heat treatment without that pressure are not interchangeable. Reinspect the features and compare the final measured response rather than assuming every directional effect disappears. This lesson supplies no heat-treatment schedule or acceptable defect limit.

A result should therefore name what changed and what was observed: for example, similar measured elastic moduli but different fracture elongations for two loading directions in the stated finished condition. Observation from atomic to component scales develops how complementary measurements sample different features; Quality evidence and reproducible processing handles the wider production claim.

Worked examples

Work through the case

Two specimens share porosity fraction but contain spherical pores versus flat lack-of-fusion defects. Must tensile response match?

Compare defect shape and orientation.
Flat defects can act differently under opening load.
Assess the relevant local stress and crack process.

No. Geometry and orientation of defects matter.

Equal porosity hides a different broad section

Use the sphere/ellipsoid inventory above. Check the porosity equality and the broad-section area ratio. Does either number determine tensile strength?

The common single-void volume is (4π/3)(0.100)³ = 0.00418879 mm³. The ellipsoid product is 0.200 × 0.200 × 0.0250 = 0.00100 mm³, giving the same volume.
One hundred voids occupy 0.418879 mm³ in the 100 mm³ envelope: p = 0.00418879, or 0.418879%.
The broad-section ratio is (πab)/(πr²) = (0.200²)/(0.100²) = 4. Orientation, edge shape and interactions were not specified by p.

The porosity is equal and the stated broad section is four times larger for the oblate void. Neither is a tensile-strength prediction.

Two projections, two different areas

A_s = 2.00 mm² is an imaginary inclined patch at θ = 60° in the 120 MPa uniaxial field. Find projected area, transmitted force and the normal/shear components.

The area perpendicular to the load is A_s cos 60° = 1.00 mm². The transmitted force magnitude is 120 N/mm² × 1.00 mm² = 120 N, directed along the load.
Its normal component is 120 cos 60° = 60.0 N; its shear magnitude is 120 sin 60° ≈ 103.9 N.
Divide these components by the actual patch area 2.00 mm²: σ_n = 30.0 MPa and τ ≈ 52.0 MPa. The traction magnitude is 60.0 MPa, distinct from its 30.0 MPa normal component.

Projected area 1.00 mm²; transmitted force 120 N; resolved normal/shear forces 60.0/103.9 N, giving 30.0/52.0 MPa. Dividing by the projected area again would use the wrong denominator.

Similar stiffness, different failure deformation

Use the P records with a 25.0 mm original gauge length. Calculate the mean permanent extensions at fracture for x and z. Does equal mean E imply these extensions must match?

The mean fracture elongations are (11 + 12 + 13)/3 = 12% and (5 + 6 + 7)/3 = 6%.
Mean permanent extensions are 25.0 × 0.12 = 3.00 mm and 25.0 × 0.06 = 1.50 mm. Their difference is 1.50 mm; z is 50% lower relative to x.
Both mean E values are 190 GPa. E measures the specified small-strain elastic slope; fracture elongation includes the later deformation and failure process.

Mean permanent extensions are 3.00 mm in x and 1.50 mm in z. Similar measured elastic slopes do not establish similar ductility or a unique failure mechanism.

Try the reasoning

Why identify build direction?
Show the answer and reasoning
It relates test loading to process-induced structure.
Directional effects can otherwise be hidden.
Can heat treatment remove every processing defect?
Show the answer and reasoning
No.
Some pores or geometric defects can persist.
Two sections have the same long grain outlines, but diffraction finds different distributions of lattice orientation. A report says their texture is identical because their grains look equally elongated. Is that justified?
Show the answer and reasoning
No. Grain outline and lattice orientation distribution are different observations.
Texture concerns the orientations of the crystals relative to specimen axes, not only the boundary shape. A directional modulus or slip response may differ despite matching outlines. Conversely, equiaxed grains need not have a random orientation distribution.
Keep a = b = 0.200 mm in the ellipsoid model but halve c to 0.0125 mm, keeping 100 voids and the 100 mm³ envelope. Find the new porosity and broad-section area. Does the smaller p establish a safer specimen?
Show the answer and reasoning
p ≈ 0.20944%; the broad section remains about 0.12566 mm². A safety ranking does not follow.
Volume is proportional to c, so total void volume and p halve. The section πab is unchanged. The thinner edge geometry, orientation, resistance and actual stress state still matter; lower void volume does not uniquely bound failure.
A flaw-normal direction is 45° from a nominal uniaxial tensile load of 80.0 MPa. Find the resolved normal and shear magnitudes. Does their equality prove equal opening and sliding crack-growth resistance?
Show the answer and reasoning
σ_n = 40.0 MPa and τ = 40.0 MPa; no equality of resistance is established.
Both projection factors are one half at 45°. The values describe nominal loading components, while crack-tip fields and mode-dependent resistance require additional geometry/material information. They are not actual tractions on a free crack face.
Compare 120 MPa at θ = 60° with 40.0 MPa at θ = 30°. Do the planes have the same nominal normal loading? Compare the shear magnitudes and interpret the result.
Show the answer and reasoning
Both have σ_n = 30.0 MPa. Their shear magnitudes are about 52.0 MPa and 17.3 MPa, respectively.
Normal factors are 1/4 and 3/4, so both products are 30 MPa. The shear factor is √3/4 for both angles, but the applied stresses differ by three. Equal nominal opening does not specify the whole loading state or equal failure behaviour.
A new P coupon tested at a direction halfway between x and z has not been measured. May its fracture elongation be assigned the arithmetic mean 9% from the two means above?
Show the answer and reasoning
No. The two directional means do not supply an angular fracture law.
The projection model is not a fracture-elongation model, and texture, defect populations and deformation can interact nonlinearly. A new-direction observation or a supported validated constitutive/failure model is needed; endpoint interpolation is an extra assumption.
After a treatment, the same specimen class has a more uniform crystal-orientation distribution, but tomography still finds the oriented thin defects. A report predicts equal fracture elongation in all directions because texture was reduced. What is missing?
Show the answer and reasoning
The remaining directional defects and other changed states must be evaluated; reducing texture alone does not establish isotropic failure.
A treatment can change one source of anisotropy while leaving another. Check flaw and fracture-path evidence, phase/surface/residual-stress state and matched response measurements. Neither the original density nor the improved texture substitutes for those observations.

What to carry forward

Describe defect morphology, not only amount.
Retain orientation and finished state.
Crystal orientation distribution, grain shape and defect-plane orientation are distinct sources of directional behaviour.
Resolve a stated surrounding stress onto the relevant plane before discussing opening or sliding; projection is not a local flaw-tip or failure model.
Equal porosity does not imply equal flaw geometry or equal performance.
Name the response, axes and finished condition; similar elastic moduli need not imply similar fracture deformation.
Use matched observations and mechanism-specific evidence rather than a universal build-direction ranking.

References

Teaching resources and research supporting this lesson’s explanations and examples.