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User Manual / Examples / Micromechanics

Micromechanics.
Solver-visible results.

Translate constituent, fiber-form, and published lamina data into solver-ready engineering properties. Every case identifies its inputs, calculated readout, provenance, and primary technical sources.

Traceable by design. Published values are transcribed from linked primary sources. CDS plots are calculated or reconstructed from the stated inputs and are clearly labeled.

01

Example class

Micromechanics

Translate constituent, fiber-form, and published lamina data into solver-ready engineering properties.

M01Published material data

T300/5208 orthotropic property baseline

Load a published carbon/epoxy lamina dataset and inspect the stiffness contrast that drives every downstream laminate calculation.

Sources and theory (1)Hu et al. (2015), T300/5208 property table
M01 T300/5208 orthotropic property baselineIllustrated example · not a live run
  1. Materials
  2. Micro
  3. Properties

1 Inputs

01
E1 181 GPa
02
E2 10.3 GPa
03
G12 7.17 GPa
04
G23 3.78 GPa

Keep these conditions unchanged when comparing with the answer.

2 Output visualization

GPaE1 = 181 GPa181E1E2 = 10.3 GPa10.3E2G12 = 7.17 GPa7.17G12G23 = 3.78 GPa3.78G23
3 What the result shows

The longitudinal modulus is 17.6× E2, making orientation the dominant design variable.

Result basisPublished material data. Values and trends reproduce this page’s example; this is not a fresh solver result or an additional validation claim.
Enlarge output visualization
GPaE1 = 181 GPa181E1E2 = 10.3 GPa10.3E2G12 = 7.17 GPa7.17G12G23 = 3.78 GPa3.78G23

The longitudinal modulus is 17.6× E2, making orientation the dominant design variable.

M02Published inputs · CDS comparison

T300/5208 strength-allowable map

Route tension, compression, and in-plane shear allowables into the failure-criterion block without hiding material-axis asymmetry.

Sources and theory (1)Hu et al. (2015), T300/5208 property table
M02 T300/5208 strength-allowable mapIllustrated example · not a live run
  1. Materials
  2. Micro
  3. Properties

1 Inputs

01
Xt/Xc 1500 MPa
02
Yt 40 MPa
03
Yc 246 MPa
04
S12 68 MPa

Keep these conditions unchanged when comparing with the answer.

2 Output visualization

strength allowable · MPaFiber 1Xc 1500 MPaXt 1500 MPaXc 1500Xt 1500Transverse 2Yc 246 MPaYt 40 MPaYc 246Yt 40Shear 12−S 68 MPa+S 68 MPa−S 68+S 68compressiontension
3 What the result shows

Transverse tensile strength is the controlling allowable at only 2.7% of Xt.

Result basisPublished inputs · CDS comparison. Values and trends reproduce this page’s example; this is not a fresh solver result or an additional validation claim.
Enlarge output visualization
strength allowable · MPaFiber 1Xc 1500 MPaXt 1500 MPaXc 1500Xt 1500Transverse 2Yc 246 MPaYt 40 MPaYc 246Yt 40Shear 12−S 68 MPa+S 68 MPa−S 68+S 68compressiontension

Transverse tensile strength is the controlling allowable at only 2.7% of Xt.

M03Reported performance thresholds

Aligned discontinuous TuFF translation

Represent a highly aligned short-fiber material as a micromechanics case instead of treating discontinuous reinforcement as automatically low performance.

Sources and theory (2)TuFF aligned-fiber composite patentHeider et al. (2019), TuFF microstructure
M03 Aligned discontinuous TuFF translationIllustrated example · not a live run
  1. Materials
  2. Micro
  3. Properties

1 Inputs

01
IM7/PEI
02
3 mm fiber
03
≥50% fiber volume
04
controlled alignment

Keep these conditions unchanged when comparing with the answer.

2 Output visualization

3 mm aligned fiberslongitudinal modulus≥138 GPalongitudinal strength≥2.07 GPareported patent thresholds · IM7 / PEI
3 What the result shows

Reported thresholds exceed 138 GPa modulus and 2.07 GPa longitudinal strength.

Result basisReported performance thresholds. Values and trends reproduce this page’s example; this is not a fresh solver result or an additional validation claim.
Enlarge output visualization
3 mm aligned fiberslongitudinal modulus≥138 GPalongitudinal strength≥2.07 GPareported patent thresholds · IM7 / PEI

Reported thresholds exceed 138 GPa modulus and 2.07 GPa longitudinal strength.

M04CDS parametric reproduction

Fiber-length efficiency sweep

Use the Cox shear-lag relation to see how load-transfer efficiency changes when an aligned discontinuous architecture moves from sub-millimeter to multi-millimeter fibers.

Sources and theory (2)Cox (1952), shear-lag formulationYu et al. (2014), 3 mm HiPerDiF alignment data
M04 Fiber-length efficiency sweepIllustrated example · not a live run
  1. Materials
  2. Micro
  3. Properties

1 Inputs

01
0.5–6 mm length sweep
02
β = 1.6 mm⁻¹
03
aligned-fiber case

Keep these conditions unchanged when comparing with the answer.

2 Output visualization

1 mm: efficiency 0.1700.173 mm: efficiency 0.5900.596 mm: efficiency 0.7920.79fiber length · mmlength efficiency ηL
3 What the result shows

The CDS parametric curve increases from ηL 0.05 at 0.5 mm to 0.79 at 6 mm.

Result basisCDS parametric reproduction. Values and trends reproduce this page’s example; this is not a fresh solver result or an additional validation claim.
Enlarge output visualization
1 mm: efficiency 0.1700.173 mm: efficiency 0.5900.596 mm: efficiency 0.7920.79fiber length · mmlength efficiency ηL

The CDS parametric curve increases from ηL 0.05 at 0.5 mm to 0.79 at 6 mm.

M05CDS r15 micromechanics sweep

Fiber-volume stiffness sweep

Run the continuous-fiber property route while increasing fiber volume fraction and retaining the same carbon-fiber and epoxy records.

Sources and theory (1)CDS elastic micromechanics implementation
M05 Fiber-volume stiffness sweepIllustrated example · not a live run
  1. Materials
  2. Micro
  3. Properties

1 Inputs

01
Ef1 230 GPa
02
Em 3.5 GPa
03
Vf 0.35–0.65
04
longitudinal ROM route

Keep these conditions unchanged when comparing with the answer.

2 Output visualization

E1 · GPa82.835%105.445%128.155%150.765%fiber volume fraction
3 What the result shows

The solver increases E1 from 82.8 to 150.7 GPa across the selected fiber-volume range.

Result basisCDS r15 micromechanics sweep. Values and trends reproduce this page’s example; this is not a fresh solver result or an additional validation claim.
Enlarge output visualization
E1 · GPa82.835%105.445%128.155%150.765%fiber volume fraction

The solver increases E1 from 82.8 to 150.7 GPa across the selected fiber-volume range.

M06CDS r15 constituent-compliance solve

Transverse modulus sensitivity

Compare the weaker transverse load path as reinforcement content rises, using the solver's constituent compliance route.

Sources and theory (1)CDS homogenization models
M06 Transverse modulus sensitivityIllustrated example · not a live run
  1. Materials
  2. Micro
  3. Properties

1 Inputs

01
Ef2 15 GPa
02
Em 3.5 GPa
03
Vf 0.35–0.65

Keep these conditions unchanged when comparing with the answer.

2 Output visualization

E2 · GPa4.6235%5.1745%5.8755%6.8265%fiber volume fraction
3 What the result shows

E2 increases from 4.62 to 6.82 GPa, far more slowly than E1 because matrix compliance remains dominant.

Result basisCDS r15 constituent-compliance solve. Values and trends reproduce this page’s example; this is not a fresh solver result or an additional validation claim.
Enlarge output visualization
E2 · GPa4.6235%5.1745%5.8755%6.8265%fiber volume fraction

E2 increases from 4.62 to 6.82 GPa, far more slowly than E1 because matrix compliance remains dominant.

M07CDS r15 architecture modifier

Void-content stiffness knockdown

Activate the void correction and quantify how manufacturing porosity reduces the resolved longitudinal ply modulus.

Sources and theory (1)CDS void and defect corrections
M07 Void-content stiffness knockdownIllustrated example · not a live run
  1. Materials
  2. Micro
  3. Properties

1 Inputs

01
Baseline E1 128.1 GPa
02
void fraction 0–5%
03
CDS void modifier

Keep these conditions unchanged when comparing with the answer.

2 Output visualization

E1 · GPa128.10%125.71%121.13%116.55%void volume fraction
3 What the result shows

The example correction reduces E1 to 116.5 GPa at 5% void content.

Result basisCDS r15 architecture modifier. Values and trends reproduce this page’s example; this is not a fresh solver result or an additional validation claim.
Enlarge output visualization
E1 · GPa128.10%125.71%121.13%116.55%void volume fraction

The example correction reduces E1 to 116.5 GPa at 5% void content.

M08CDS r15 orientation-tensor route

Fiber-alignment efficiency

Exercise the orientation correction with a controlled misalignment sweep for an otherwise identical aligned-fiber material.

Sources and theory (1)CDS orientation and architecture theory
M08 Fiber-alignment efficiencyIllustrated example · not a live run
  1. Materials
  2. Micro
  3. Properties

1 Inputs

01
Mean misalignment 0–15°
02
fourth-order longitudinal orientation factor

Keep these conditions unchanged when comparing with the answer.

2 Output visualization

normalized η1.000°0.985°0.9410°0.8715°mean misalignment
3 What the result shows

The normalized longitudinal efficiency falls from 1.00 to 0.87 at 15° misalignment.

Result basisCDS r15 orientation-tensor route. Values and trends reproduce this page’s example; this is not a fresh solver result or an additional validation claim.
Enlarge output visualization
normalized η1.000°0.985°0.9410°0.8715°mean misalignment

The normalized longitudinal efficiency falls from 1.00 to 0.87 at 15° misalignment.

M09CDS r15 textile architecture route

Woven crimp correction

Route a balanced woven reinforcement through the crimp modifier to separate ideal tow stiffness from the manufactured fabric response.

Sources and theory (1)CDS woven and textile models
M09 Woven crimp correctionIllustrated example · not a live run
  1. Materials
  2. Micro
  3. Properties

1 Inputs

01
Balanced 0/90 fabric
02
crimp angle 0–15°
03
equal tow fractions

Keep these conditions unchanged when comparing with the answer.

2 Output visualization

normalized E1.000°0.985°0.9410°0.8715°crimp angle
3 What the result shows

Increasing crimp progressively lowers the normalized in-plane stiffness to 0.87.

Result basisCDS r15 textile architecture route. Values and trends reproduce this page’s example; this is not a fresh solver result or an additional validation claim.
Enlarge output visualization
normalized E1.000°0.985°0.9410°0.8715°crimp angle

Increasing crimp progressively lowers the normalized in-plane stiffness to 0.87.

M10CDS r15 thermophysical property solve

Directional thermal conductivity

Resolve the longitudinal transport property set passed from micromechanics into the transient process solver.

Sources and theory (1)CDS thermal-property assembly
M10 Directional thermal conductivityIllustrated example · not a live run
  1. Materials
  2. Micro
  3. Properties

1 Inputs

01
kf1 10 W/m·K
02
km 0.2 W/m·K
03
Vf 0.40–0.70

Keep these conditions unchanged when comparing with the answer.

2 Output visualization

W/m·K4.1240%5.1050%6.0860%7.0670%fiber volume fraction
3 What the result shows

The longitudinal conductivity rises from 4.12 to 7.06 W/m·K as fiber content increases.

Result basisCDS r15 thermophysical property solve. Values and trends reproduce this page’s example; this is not a fresh solver result or an additional validation claim.
Enlarge output visualization
W/m·K4.1240%5.1050%6.0860%7.0670%fiber volume fraction

The longitudinal conductivity rises from 4.12 to 7.06 W/m·K as fiber content increases.