Assume a thin cylindrical wall, uniform internal pressure and closed ends carrying the pressure load. The radial stress is small compared with the membrane stresses when $t\ll r$.

For an end cap, the outward pressure force is $P\pi r^2$. The wall's axial tensile force is ${\sigma }_z(2\pi rt)$. Equilibrium therefore givesFor a longitudinally split half-cylinder of length $L$, pressure has resultant $P(2rL)$ across the projected rectangular area. Two cut wall sections each carry ${\sigma }_{\theta }tL$, soThus hoop stress is twice the closed-end axial stress in this model. An open-ended tube without an external end load does not acquire the same pressure-induced axial membrane stress.

The danger is identifying an etched height interval with the original planar phase intersection. Preferential ferrite dissolution exposes inclined cementite faces and can produce overhangs. A conventional AFM tip cannot follow a reentrant underside; its finite size also broadens ridge tops and may not enter the intervening troughs. A thresholded height width can consequently differ from both the true plate thickness and its original surface intersection.

A uniform inclination at an ideal flat cut does not itself invalidate a line-fraction estimate: both phases acquire the same geometrical width factor, which cancels in their ratio. The problem is differential relief and probe accessibility after etching, together with uncertain local angles and limited sampling. Using only the apparent ridge occupancy can overestimate cementite. An independently polished section, calibrated probe geometry, several directions and colonies, or another phase-sensitive method is needed to test the inference.

The lamellar interfaces now make $60^{\circ }$ with the surface normal, or $30^{\circ }$ with the surface itself. Draw inclined alternating plates rather than vertical stripes, with preferentially etched ferrite below the more resistant cementite.

The upper panel shows an inclined microstructure intersecting the original plane. Differential etching can expose a shallow inclined cementite face on one side of a ridge and an undercut or overhang on the other. The lower panel shows its accessible height profile, with additional rounding from a finite tip. It is therefore possible to obtain an asymmetric ridge even from an initially planar section through parallel lamellae.

For an ideal unetched planar cut, a plate of normal thickness $d$ has apparent surface widthThe same factor applies to the ferrite spacing. Inclination alone broadens all intersections together; it does not by itself change the true phase-width ratio.

Pearlite is a microstructural arrangement of two phases, ferrite and cementite, rather than a single phase. Ferrite is an iron-rich solid solution; cementite is the carbide $\mathrm{F}{\mathrm{e}}_3\mathrm{C}$. A pearlite colony contains alternating lamellae of these constituents formed by transformation of austenite.

A thermodynamic phase has a specified crystal structure and composition range with locally homogeneous properties on the appropriate scale. The word pearlite instead describes how distinct phases are arranged. Its lamellar spacing and orientation can change while the identities of its two phases remain the same.

SEM with energy-dispersive X-ray spectroscopy can identify gold versus silver through their characteristic X-ray energies. Acquire spectra and maps from representative regions: two compositionally distinct regions would support the two-phase interpretation of B. With standards or validated matrix corrections it can estimate local compositions and, with representative sampling, an overall composition. Surface topography and the interaction volume limit spatial interpretation.

X-ray fluorescence provides a complementary more bulk-averaged elemental measurement. Gold and silver produce different characteristic lines; calibrated intensities can yield their mass fractions, with corrections for absorption, excitation and specimen geometry. It does not by itself determine whether an element occupies one or two crystallographic phases.

Both methods supply elemental information missing from a lattice-parameter-only identification. A purely morphological micrograph could reveal phase separation but would not alone give a reliable numerical chemical composition.

Sample B should show two FCC peak families, one from a copper-rich phase and one from a silver-rich phase. Their lattice parameters are set by the compositions of the coexisting phases, not by one lattice parameter at the overall alloy composition.

For corresponding Miller indices the silver-rich phase has a larger lattice parameter and hence a larger interplanar spacing, so its reflection occurs at smaller $2\theta$. Some reflections may overlap or be weak. Relative intensities depend on phase fractions, atomic scattering factors, absorption and texture; a raw peak-height ratio is not directly a mass fraction.

At room temperature the mutual solubilities in the supplied Cu–Ag diagram are small, so the two families would be near the elemental copper and silver positions. Slow cooling is important: a metastable supersaturated single phase would be a different interpretation. Quantitative phase or composition analysis would require calibrated diffraction and an appropriate refinement, not merely counting the visible peaks.

Sample A is the Cu–Au alloy. At the indicated copper-rich composition, the supplied Cu–Au diagram permits one copper-rich solid-solution phase. The Cu–Ag diagram instead gives two equilibrium solid phases at room temperature. A single consistent set of FCC peaks therefore identifies A under the stated equilibrium assumptions; similar pure-metal lattice parameters alone would not distinguish gold from silver.

Use Vegard's law as an approximate atomic-fraction interpolation:With the higher-angle estimate from part(a),This is about $5.1$ atomic percent, not weight percent. Using rounded molar masses $M_{\mathrm{Au}}\simeq 197$ and $M_{\mathrm{Cu}}\simeq 63.55\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$,Using the first peak alone instead gives roughly $12.6wt\%$. The justified conclusion is therefore about $13$–$14wt\%$ gold, with uncertainty from angle rounding, calibration and the approximate linear lattice-composition relation.

For a cubic phase, $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$ and Bragg's law givesThe first six FCC-allowed squared-index sums are $3,4,8,11,12,16$, from indices all odd or all even. The labels1–6 are thereforeRemember that the plot reports $2\theta$, not $\theta$. The first peak givesThe last peak provides an independent estimate,Thus $a\simeq 0.3638\mathrm{nm}$ is appropriate to the supplied precision. The two estimates differ by about $0.08\%$; high-angle data are less sensitive to a fixed angular error, so the $400$ estimate is useful for the composition calculation. The printed angles do not justify many more significant figures or an exact fit of all six positions.

The voltage is a mechanically mediated effect: nickel magnetostriction strains the bonded central layer, and the direct piezoelectric effect converts stress into electrical displacement. An estimate of $|d_{31}|$ requires a strain-transfer and electrical-boundary assumption.

Let $x$ run along the applied field and $z$ through the layers. Assume saturated, approximately volume-preserving isotropic nickel magnetostriction, perfect bonding, mechanically dominant nickel faces and a high-impedance voltage measurement. Relative to the random-domain state,Treat the thin central layer as in plane stress, with these imposed in-plane strains, negligible converse-effect backreaction and isotropic elastic constants. ThensoFor a transversely polarized layer,The open-circuit increment has $\Delta D_3\simeq 0$, and $|E_3|=|V|/t=800\mathrm{V}{\mathrm{m}}^{-1}$. HenceThe observed voltage magnitude does not specify the sign without electrode polarity and poling direction. A one-dimensional approximation retaining only axial stress $E\Lambda$ instead gives $4.78\mathrm{pC}{\mathrm{N}}^{-1}$. That is a different mechanical model. Nickel stiffness and thickness, interface compliance, central-layer tensor coefficients and electrical loading are not specified, so the experiment does not determine a unique exact intrinsic coefficient without these assumptions.

The 020 reflection is extinguished by the diamond basis, even though it satisfies the face-centered lattice parity rule. Write the diamond structure as an FCC lattice with identical atoms at basis positions $(0,0,0)$ and $(1/4,1/4,1/4)$.

The structure factor separates into lattice and basis terms:At $(h,k,l)=(0,2,0)$, all indices are even, so the first bracket is $4$. The second bracket is $1+e^{i\pi }=0$. The two identical sublattices scatter with opposite phases and cancel:For all-even reflections of diamond, the additional allowed condition is $h+k+l$ divisible by four. This is a systematic absence, not an absence of the corresponding family of geometrical planes.

The metallic conventional cell has four tin atoms, whereas the conventional diamond cubic structure cell has eight. Therefore the volume per atom isThe fractional change for metallic to nonmetallic tin isThe atom count is unchanged, so this is also the unconstrained bulk volume change for complete transformation under the model. If it were macroscopically isotropic, the corresponding free linear expansion would be $(1.2627{)}^{1/3}-1\simeq 8.1\%$.

Such a large transformation strain creates stresses where transformed and untransformed regions constrain one another. Cracking, loss of mechanical integrity and disintegration can follow: the phenomenon is tin pest. The calculation establishes the volume mismatch, not the transformation rate; temperature, nucleation and impurities affect when a real specimen changes.