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04 — Tutorials

Worked FEA examples.

By Mircea-Vlad Radulescu — image placeholders below mark where step screenshots go; see Tutorials/StrutExample_M3d.docx for the original.

Goal

A simple landing gear strut example where a 3-member assembly (members O–A, O–B, O–C) resists vertical and drag force similar to those on a landing gear. Build a strut model similar to Section 2.3, Example 1 in Peery's Aircraft Structures, with single-point restraints and a single-point force at the tip of the strut. Find the loads in the two-force members O–A, O–B, O–C and compare the results with the hand-calculation equivalent.

🖼️ Screenshots — model overview Wireframe and shaded views of the finished strut model, same as the docx's own cover images. suggested: strut-00-overview.png

References

Peery, J. David. Aircraft Structures. Dover Publications (2011), originally McGraw-Hill Book Company Inc., 1950, New York.

Part 1

1 Nodes

Create a Node Set to represent the strut nodes O, A, B, C. Using the Node Create tool, enter locations directly in the command line (coordinate system is global/basic):

O 0, 0, 0 A 0, 30, 40 B 20, 30, 40 C 30, 0, 40

Then add 10 nodes in between each pair (Node-in-Between tool — pick the first and second node sets, Enter No > 10) for a finer visual mesh along each member. Deselect All before continuing.

🖼️ Screenshots needed Node Create — coordinates entered in the command line The four strut nodes (O, A, B, C) placed in the model Node-in-Between — picking the first/second node sets, count set to 10 suggested: strut-01a-nodes-coords.png, strut-01b-nodes-placed.png, strut-01c-nodes-between.png

2 Elements

Select the Beam Element tool, then the Element Create tool, to connect tip node O to each of A, B, and C in turn — three elements total.

🖼️ Screenshot needed Beam Element Create — the three O–A / O–B / O–C members picked suggested: strut-02-elements.png

3 Element Properties

Create a steel material: Create ▸ FEM Tools ▸ Material ▸ Isotropic. Then a round-tube beam section referencing it: Create ▸ FEM Tools ▸ Property ▸ Beam Tube.

Do (outer diameter) 1 in t (wall thickness) 0.1 in Di (inner diameter) 0.9 in A (area) 0.5969 in² J (torsion constant) 0.0039794 in⁴ I (moment of inertia) 0.2700984284 in⁴

🖼️ Screenshots needed Material ▸ Isotropic — steel property entry Property ▸ Beam Tube — section dims + Material ID reference suggested: strut-03-material.png, strut-03-property.png

4 Apply Element Properties

Tools ▸ Element Modify ▸ Property ID — select all three elements, Right Click ▸ Done, key in the Beam Tube property's ID.

🖼️ Screenshot needed Element Modify ▸ Property ID, with all three elements selected suggested: strut-04-apply-properties.png

5 Check for Coincident Nodes

Tools ▸ Checks ▸ Merge Coincident Nodes — box-select all nodes, Done, accept the default tolerance, then delete any duplicates it finds. Good practice before solving.

🖼️ Screenshot needed Merge Coincident Nodes — tolerance prompt / duplicate-node list suggested: strut-05-coincident.png

6 Load Set

LoadsBC ▸ Load Sets — add a Set ID and title, Create, then OK. The new set becomes the active one automatically.

🖼️ Screenshot needed Load Sets window — creating the active set suggested: strut-06-load-set.png

7 Boundary Condition Set

LoadsBC ▸ BC Sets — same pattern as Load Set.

🖼️ Screenshot needed BC Sets window — creating the active set suggested: strut-07-bc-set.png

8 Create Restraints

LoadsBC ▸ Create Restraint — pick nodes A, B, and C. Translation Restraint Vector 1 1 1 (fixed), Rotation Restraint Vector 0 0 0 (free) — a true pin at each, matching the two-force-member assumption.

🖼️ Screenshots needed Create Restraint — translation/rotation vector entry The three pinned supports (shown against a black background for clarity) suggested: strut-08-restraints.png

9 Create Forces

LoadsBC ▸ Create Force — pick node O. Force vector in the command line:

1000 0 4000 lbf (Fx, Fy, Fz)

If needed, refine the value afterward via Edit Object Values ▸ pick the force vector ▸ Entity Editor.

🖼️ Screenshots needed Create Force — vector entry at node O Entity Editor — force magnitude/direction fields suggested: strut-09-forces.png

10 Create Solution

Solution ▸ Create Solution — Solution Type Linear Static, Tolerance default. Create, then OK.

🖼️ Screenshot needed Create Solution window suggested: strut-10-solution.png

11 Create Solution Set

Solution ▸ Create Solution Step — select the Load Set and BC Set (no Temperature Set), add a Step Title. Create, then Activate, then OK.

🖼️ Screenshot needed Create Solution Step window suggested: strut-11-solution-step.png

12 Solve

Solution ▸ Solve. The run should take a second — console output ends with:

STARTING ASSY STARTING ITERATIVE SOLUTION ITER: 0 Err: 1.0000000000000000 ITER: 100 Err: 0.0281846423725408 ITER: 200 Err: 0.0016478307835453 ITER: 300 Err: 0.0000118095067473 FINAL ERROR ITER: 361 Err: 0.0000000004759823 FINISHED SOLUTION

Solved successfully, with a negligible final iteration error.

13 Post-Processing — Deformation

Deformed Results ▸ Select Displacements ▸ OK, then the Deformed Shape button to view it. If it's hard to see, use Change the Deformation to rescale (e.g. 10). Contour plot (Select Contour Results) gives per-node displacement components:

Min TX = -0.235437 Min TY = -0.235444 Max TZ = 0.228919

Deformation at point O is hard to estimate by hand (it depends on the interaction of all three deforming/rotating members) — not attempted here.

🖼️ Screenshots needed Deformed shape overlay (rescaled) TX / TY / TZ contour plots suggested: strut-13-deformation.png

14 Animation

With a deformed shape selected, click Animate to breathe the shape between undeformed and deformed.

15 Element Forces

Select Contour Results ▸ END B FX for axial load, then pick each member's end element individually:

O–A: -4,998.33 lbf O–B: 5,383.02 lbf O–C: -4,998.85 lbf

Close to the hand-calculation estimate — see Appendix 1 below.

🖼️ Screenshot needed END B FX contour, with each member's axial force picked suggested: strut-15-element-forces.png

16 END

This completes Part 1 of the tutorial. See the appendices below for the hand-calculation check.

Part 2 — eccentric load

Assuming the force is eccentric, 6 inches laterally off point O to simulate a wheel axle. Find the forces at A, B, and C assuming members O–B and O–C are two-force members (SPC 123) while O–A resists bending and torsion, and point A is a universal joint that can carry torsion but no bending (SPC 12356).

17 Nodes

Add one more node, 6 inches off O, to represent the eccentric load application point:

0, -6, 0

18–20 Elements, Element Properties, Apply Element Properties

Same pattern as Part 1, reusing the same material and Beam Tube property.

21 Check for Coincident Nodes

Same as Part 1 — good practice to repeat after adding the new node.

22–23 Load Set, Boundary Condition Set

Reuse the same Load Set and BC Set from Part 1 — just press OK without creating new ones.

24 Create Restraints

If modifying the Part 1 model directly, filter-select and delete the existing restraint at point A first, then recreate it:

A — Translation 1 1 1, Rotation 0 1 1 (universal joint — carries torsion, not bending) B — Translation 1 1 1, Rotation 0 0 0 (unchanged pin) C — Translation 1 1 1, Rotation 0 0 0 (unchanged pin)

25 Create Forces

Remove the force at O (if reusing the Part 1 model), then apply the same vector at the new eccentric node:

1000 0 4000 lbf

26–29 Create Solution, Create Solution Set, Solve, Post-Processing

Same steps as Part 1.

ITER: 0 Err: 1.0000000000000000 ITER: 100 Err: 0.0017885452628859 ITER: 200 Err: 0.0005264253932269 ITER: 300 Err: 0.0000069196846159 FINAL ERROR ITER: 352 Err: 0.0000000004318397 FINISHED SOLUTION

30 Deformation

Max TX = 0, Min TX = -0.359233 Max TY = 0.346809, Min TY = -0.289001 Max TZ = 0.824919, Min TZ = -0.230741

Larger than Part 1 — the eccentric load imparts bending and torsion that Part 1's pure axial load didn't.

31 Animation

Note the torsional deformation present here that Part 1 didn't have.

32 Element Forces

O–A: -5,634.04 lbf O–B: 6,844.48 lbf O–C: -5,974.20 lbf

Loads increase compared to Part 1, as expected. Note the FEM torsion output doesn't match the hand calc's torsion estimate — see Appendix 2.

33 END

This completes Part 2. See the appendices below for the hand-calculation check.

Appendix 1 — Element Force Check (Part 1)

Member forces estimated by hand, assuming each member takes only axial load along its own axis (i.e. as if the elements were rods rather than beams — close enough to approximately corroborate the model).

Tube section (Peery Example 1, Ch. 2.3)
Property	Value
Do (outer diameter)	1 in
t (wall thickness)	0.1 in
Di (inner diameter)	0.9 in
A (area)	0.5969 in²
J (torsion constant)	0.0039794 in⁴
I (moment of inertia)	0.2700984284 in⁴

Member geometry & direction cosines
Point	X	Y	Z	Length	x̂	ŷ	ẑ
A	0	30	40	50	0.000	0.600	0.800
B	20	30	40	53.85	0.371	0.557	0.743
C	30	0	40	50	0.600	0.000	0.800

Member axial force — hand calc vs. FEM (END B FX)
Member	Hand calc	FEM	% diff
O–A	-5,000.00 lbf	-4,998.33 lbf	-0.03%
O–B	5,385.16 lbf	5,383.02 lbf	-0.04%
O–C	-5,000.00 lbf	-4,998.85 lbf	-0.02%

🖼️ Screenshot needed Book snapshot — Peery Example 1, p.23 (hand-calc reference figure) suggested: strut-appendix1-book.png

Appendix 2 — Element Force Check (Part 2)

Hand calc assumes O–B and O–C take only axial load along their own axes, while O–A also carries torsion. The torsion load is resolved first, then the force-sum equations are set up with one unknown each. Note the FEM torsion output doesn't match the hand-calc torsion estimate (T vs. END B BX) — flagged in the original tutorial as a known discrepancy, not a modeling error to chase down.

Forces & moments — hand calc vs. FEM
	Hand calc	FEM	% diff
FA	-5,701 lbf	-5,634.04 lbf	-1.17%
FB	6,806.85 lbf	6,844.49 lbf	0.55%
FC	-6,000 lbf	-5,974.20 lbf	-0.43%
Ts	2,880 in·lbf	— (not separable in FEM output)	—
Tv	3,840 in·lbf	— (not separable in FEM output)	—
T	4,800 in·lbf	617.59 in·lbf	-87.13%

🖼️ Screenshot needed Book snapshot — Peery Example 2, p.25 (hand-calc reference figure) suggested: strut-appendix2-book.png

Need the full derivation? The complete matrix setup and intermediate steps are in Tutorials/StrutExample_M3d.docx in the nastran-visualizer-vite repo.

By Mircea-Vlad Radulescu — image placeholders below mark where step screenshots go; see Tutorials/Beam_w_WebExample_M3d.docx for the original.

Goal

Create a simple beam model (a truss-like structure) with single-point restraints and a single-point force at the tip. First create nodes and elements, then assign material properties and orientations. Forces and boundary conditions are set, a solution is created and solved, then results are post-processed for deformation and element forces — corroborated against a hand calculation.

🖼️ Screenshots — model overview Wireframe and shaded views of the finished beam model suggested: beam-00-overview.png

1 Nodes

Create a Node Set to represent the beam:

A 0, 0 B 0, 3 C 4, 3

Add 6 nodes in between (Node-in-Between tool) for a finer visual mesh along each segment. Deselect All before continuing.

🖼️ Screenshots needed Node Create — coordinates entered in the command line Node-in-Between — count set to 6 suggested: beam-01a-nodes-coords.png, beam-01b-nodes-between.png

2 Elements

Beam Element tool, then Element Create — connect A–B and B–C.

🖼️ Screenshot needed Beam Element Create — the two A–B / B–C members picked suggested: beam-02-elements.png

3 Element Properties

Create a steel material: Create ▸ FEM Tools ▸ Material ▸ Isotropic. Then an I-beam section referencing it: Create ▸ FEM Tools ▸ Property ▸ Beam I2.

🖼️ Screenshots needed Material ▸ Isotropic — steel property entry Property ▸ Beam I2 — section dims + Material ID reference suggested: beam-03-material.png, beam-03-property.png

4 Apply Element Properties

Tools ▸ Element Modify ▸ Property ID — select both elements, key in the Beam I2 property's ID.

🖼️ Screenshot needed Element Modify ▸ Property ID, both elements selected suggested: beam-04-apply-properties.png

5 Modify Element Vectors

Tools ▸ Element Modify ▸ Beam Up Vectors — box-select both elements, Done, Key In the up vector:

0 1 (up vector, X1 X2)

An I-beam's bending stiffness depends on which way it's facing, unlike the Strut example's round (axisymmetric) tube. Section properties can be refined afterward via Edit Properties Values ▸ pick an element ▸ Entity Editor.

🖼️ Screenshots needed Beam Up Vectors — Key In 0 1 Entity Editor — section property refinement suggested: beam-05-up-vectors.png

6 Check for Coincident Nodes

Tools ▸ Checks ▸ Merge Coincident Nodes — box-select all nodes, accept the default tolerance, delete any duplicates found.

🖼️ Screenshot needed Merge Coincident Nodes — tolerance prompt / duplicate list suggested: beam-06-coincident.png

7–8 Load Set, Boundary Condition Set

LoadsBC ▸ Load Sets, then LoadsBC ▸ BC Sets — add a Set ID/title, Create, OK for each. Both become active automatically.

🖼️ Screenshots needed Load Sets window BC Sets window suggested: beam-07-load-set.png, beam-08-bc-set.png

9 Create Restraints

LoadsBC ▸ Create Restraint — pick node A only. Translation Restraint Vector 1 1 1, Rotation Restraint Vector 1 1 1 — fully fixed, a true cantilever root.

🖼️ Screenshot needed Create Restraint — node A fully fixed suggested: beam-09-restraint.png

10 Create Forces

LoadsBC ▸ Create Force — pick node C (the free tip). Force vector in the command line:

0 -10000 (Fx, Fy)

Refine via Edit Object Values ▸ Entity Editor to a larger negative value, FY = -100000, for a clearer result.

🖼️ Screenshots needed Create Force — vector entry at node C Entity Editor — FY refined to -100000 suggested: beam-10-forces.png

11 Create Solution

Solution ▸ Create Solution — Linear Static, default tolerance. Create, then OK.

🖼️ Screenshot needed Create Solution window suggested: beam-11-solution.png

12 Create Solution Set

Solution ▸ Create Solution Step — select the Load Set and BC Set, add a Step Title. Create, then Activate, then OK.

🖼️ Screenshot needed Create Solution Step window suggested: beam-12-solution-step.png

13 Solve

Solution ▸ Solve. Console output:

SOLUTION: Lin Stat NO OF EQUATIONS:- 78 GLOBAL LOADS RESULTANT:- FX: 0.0000000000000000 FY: -100000.0000000000000000 FZ: 0.0000000000000000 MX: 0.0000000000000000 MY: 0.0000000000000000 MZ: -400000.0000000000000000 STARTING ASSY STARTING ITERATIVE SOLUTION ITER: 0 Err: 1.0000000000000000 FINAL ERROR ITER: 43 Err: 0.0000000000000141 FINISHED SOLUTION

Solved successfully, with a negligible final iteration error.

14 Post-Processing — Deformation

Deformed Results ▸ Select Displacements ▸ OK, then Deformed Shape to view it. Rescale (e.g. 100) if hard to see. Contour plot gives per-node displacement:

Max TX = 0.00124857 Min TY = -0.00492352

Close to the hand-calculation estimate — see Appendix 2 below.

🖼️ Screenshots needed Deformed shape overlay (rescaled) TX / TY contour plots suggested: beam-14-deformation.png

15 Animation

With a deformed shape selected, click Animate.

16 Element Forces

Select Contour Results ▸ END B FX for axial load:

Top element (B–C): EF1 = 128,915 Diagonal element (A–B): EF2 = -161,504

Both close to the hand-calculation estimate — see Appendix 1 below.

🖼️ Screenshot needed END B FX contour, with each member's axial force picked suggested: beam-16-element-forces.png

17 END

This completes the tutorial. See the appendices below for the hand-calculation check.

Appendix 1 — Element Force Check

Member forces estimated by hand, assuming the truss structure members take only axial load along their own axes (as if rods rather than beams — close enough to approximately corroborate the model).

Geometry & applied load
Quantity	Value
P (applied load)	-100,000
Height	3
Length	4
Angle Θ	0.644 rad (36.87°)

Summing forces about point C (ΣFx = F1 + F2·cos Θ = 0, ΣFy = P + F2·sin Θ = 0):

Member forces
Member	Force	Sense
E1 (= -F1, top/B–C)	133,333	Tension
E2 (= -F2, diagonal/A–B)	-166,667	Compression

🖼️ Screenshot needed Load/geometry diagram — two-force-member idealization (P, Θ, F1, F2) suggested: beam-appendix1-diagram.png

Appendix 2 — Element Deformation Check

The beam elements are estimated as axial springs (k = A·E/L) to get a first-order deformation estimate.

Section & material
Quantity	Value
dim1 (height)	0.4
dim2 / dim3 (flange width)	0.2 / 0.2
dim4 (web thickness)	0.01
dim5 / dim6 (flange thickness)	0.005 / 0.005
Area, A	0.002 m²
Modulus, E	70,000,000,000 Pa
L1 (B–C) / L2 (A–B)	4 / 5
k1 = A·E/L1	35,000,000
k2 = A·E/L2	28,000,000

Estimated deformation
Quantity	Value
δ1 = EF1/k1	0.0038
δ2 = EF2/k2	-0.0060
δx = δ1 + δ2·cos Θ (point C)	-0.0010
δy = δ2·sin Θ (point C)	-0.0036

Need the full derivation? The complete setup and intermediate steps are in Tutorials/Beam_w_WebExample_M3d.docx in the nastran-visualizer-vite repo.

MVR
Analytics

Engineering background,
practical mindset.

What I offer.

Web-based NASTRAN Workflow

Start a conversation.

Worked FEA examples.

MVRAnalytics

Engineering background,practical mindset.

What I offer.

Web-based NASTRAN Workflow

Start a conversation.

Worked FEA examples.

MVR
Analytics

Engineering background,
practical mindset.