Solve and demonstrate equilibrium

MAPDL may terminate SOLVE without us having understood the model. In M06 we will convert the solution into a testable statement: the applied force and its moment They must be exactly balanced by the fixed support.

Objectives

By completing M06 you will be able to:

• Explicitly declare a linear static analysis.
• Distinguish load step, substep and result set.
• Explain what states it connects SOLVE.
• Programmatically check that a solution is available.
• Read nodal reactions using *GET.
• Reconstruct a moment from translational forces.
• Use SPOINT and FSUM with a conscious reference point.
• Accept or reject the balance through quantitative tolerances.

Prerequisites and downloads

• Have completed M05: Beam must have audited loads and constraints.
• Recognize pattern selection → check → application → restore.
• Remember the global convention: X longitudinal, Y vertical, and Z transverse.

How to use this lesson

Recommendation: Calculate RFY and RMZ on paper before SOLVE. Balance is the first physical validation after an apparently correct solution.

Session map

Prediction — Solve first on paper

The total force is FY=-1000 N and acts one meter from the fixed support. Before open MAPDL, the free body diagram requires:

ΣFY = 0  →  RFY + (−1000) = 0  →  RFY = +1000 N

ΣMZ = 0  →  RMZ + 1·(−1000) = 0  →  RMZ = +1000 N·m

These figures are a prediction, not a finite element result. The objective of macro will be to try to refute them by reading the real solution.

Mental model — Four states, not one button

Do not treat solving as merely flipping a switch. The process contains four different states:

1. Prepared database: mesh, properties, constraints and loads.
2. Solution definition: type of analysis, increments and results requested.
3. Solution: MAPDL assembles and solves the system of equations.
4. Active set: a specific state of the results file is loaded into memory.

Physics — What does “linear static” mean?

The adjective contains three decisions:

• Structural: we seek displacements and forces that satisfy equilibrium and compatibility.
• Static: we do not include inertial terms or a physical time history.
• Linear: stiffness, geometry, and constitutive behavior do not change with the response.

The idealized system can be expressed as:

[K]{u} = {F}

Therefore, if we double the load without changing the model, displacements, reactions, and stresses must double. The challenge will use that proportionality as a second check.

APDL — State the hypothesis

/SOLU
ANTYPE,STATIC
NLGEOM,OFF
KBC,1
NSUBST,1
OUTRES,ALL,LAST

• ANTYPE,STATIC selects a static analysis.
• NLGEOM,OFF makes linear geometry explicit.
• KBC,1 applies the load stepwise in the single substep.
• NSUBST,1 defines a single substep for this linear case.
• OUTRES,ALL,LAST saves the requested final state.

Some options match the defaults. Writing them here is not padding; it turns the lesson assumptions into reviewable text.

Load step, substep and set

Load step

The load state to solve. It may represent a physical phase or a different condition.

Substep

Numerical increment within a load step. In nonlinear analysis it will have a decisive role.

Result set

Snapshot stored in the results file and recoverable later.

They are not synonyms. A load step can contain many substeps and not all of them have to have been written as sets.

Before SOLVE — Restore the model

ALLSEL,ALL
SOLVE
FINISH

The active selection is part of the MAPDL state. If a tip selection carries over from load application into the solution, the solved model may not be the one you thought you had prepared. That is why ALLSEL,ALL is an explicit precondition.

Check that a result exists

Enter postprocessing, count the result sets, and only then attempt to activate the last one:

/POST1
*GET,n_sets,ACTIVE,0,SET,NSET
solution_available=0

*IF,n_sets,GE,1,THEN
  SET,LAST
  solution_available=1
*ENDIF

This sequence prevents the rest of the audit from operating on nonexistent results when SOLVE has failed to generate a result set.

SET,LAST — Activating is not solving

SET,LAST does not recalculate the structure. It reads the latest available set and makes its values active in the results database. If multiple states exist, choosing the set deliberately is as important as choosing the nodes.

*GET,result_time,ACTIVE,0,SET,TIME

Saving the time value also identifies which state was audited, even in a static analysis where that “time” works as a marker for the load step.

Reactions — Read the support response

Reactions are queried at constrained nodes. We do not need fixed IDs because the physical region is stored in the fixed_nodes component.

CMSEL,S,fixed_nodes
node_id=0

*DO,j,1,n_fixed
  node_id=NDNEXT(node_id)
  *GET,rfx_node,NODE,node_id,RF,FX
  *GET,rfy_node,NODE,node_id,RF,FY
  *GET,rfz_node,NODE,node_id,RF,FZ

  rfx=rfx+rfx_node
  rfy=rfy+rfy_node
  rfz=rfz+rfz_node
*ENDDO

RF returns reactions in the nodal coordinate system. In the working example, all nodal systems remain aligned with the global Cartesian system.

The hidden moment in translational forces

SOLID185 has UX, UY and UZ, but not ROTX, ROTY nor ROTZ. This does not prevent transmitting moment. The reactions distributed over the support face form a force couple.

Each nodal reaction contributes:

{M} = {r} × {RF}

Mx = ry·RFz − rz·RFy
My = rz·RFx − rx·RFz
Mz = rx·RFy − ry·RFx

Adding these contributions gives the reaction moment at the fixed support without introducing a rotational degree of freedom that the element does not possess.

Choose the sum point

We will use the center of the fixed-support face:

x_ref=0
y_ref=beam_h/2
z_ref=beam_b/2

A moment is always referred to a point. Changing that point can change its components, although the force system remains physically equivalent. Choosing the support center eliminates artificial eccentricities associated with the block's geometric origin.

FSUM and SPOINT — An independent check

SPOINT,0,x_ref,y_ref,z_ref
CMSEL,S,fixed_nodes
FSUM
*GET,fsum_fy,FSUM,0,ITEM,FY
*GET,fsum_mz,FSUM,0,ITEM,MZ

FSUM sums the nodal contributions of elements connected to the selected set. SPOINT defines the global point about which moments are calculated. Do not assume that these internal contributions use the same sign convention as RF.

Force audit

force_error=ABS(rfy+tip_force)/ABS(tip_force)
force_parasitic=(ABS(rfx)+ABS(rfz))/ABS(tip_force)

The “plus” sign is not arbitrary: the load and its reaction must be opposite. The second indicator detects unintended resultants in X or Z that would remain hidden if we examined only RFY.

Moment audit

external_mz=(beam_l-x_ref)*tip_force
target_mz=-external_mz
moment_error=ABS(rmz+external_mz)/ABS(external_mz)
moment_parasitic=(ABS(rmx)+ABS(rmy))/ABS(external_mz)

Checking force without checking moment leaves the free-body diagram incomplete. A load applied at the wrong position can retain exactly the same resultant and produce a different moment.

Acceptance contract

The base case is only approved when it simultaneously meets:

• 126 nodes, 40 elements, 6 fixed nodes and 6 loaded nodes.
• Exactly one set of results available.
• Force error less than 0.5 %.
• Moment error less than 0.5 %.
• Parasitic resultant force and moment less than 0.5 %.
• passes=1.

The threshold is much larger than the expected residual in this linear problem. Its pedagogical purpose is to clearly separate a passing case from a modeling error.

CSV — Reproducible evidence

The macro generates m06_equilibrium_audit.csv:

case,n_nodes,n_elements,n_fixed,n_tip,n_sets,result_time,target_fy,
rfx,rfy,rfz,rmx,rmy,rmz,target_mz,force_error,moment_error,
fsum_fy,fsum_mz,passes

The file preserves inputs, topology, active state, results, and decision. A screenshot of the deformed shape does not provide that traceability.

Guided practice

1. Write the predictions RFY and RMZ without running MAPDL.
2. Run 06_start.mac in blocks and review the selection before SOLVE.
3. Check the .out file to confirm that the solution completed.
4. Activate the last set and list the reactions of the fixed component.
5. Reconstruct the moment about the center of the support.
6. Compare your CSV with 06_expected_results.csv.
7. Run the entire solution only after justifying each sign.

Bug hunting — Five false balances

Open 06_bug_hunt.mac and diagnose each case:

1. Spread selection: An attempt is made to solve without recovering the entire model.
2. Missing set: reactions are queried without activating a result set.
3. Wrong sign: reaction and charge are compared as if they had the same meaning.
4. global FSUM: a sum close to zero is interpreted as absence of reaction.
5. Non-existent moment: wanted RF,MZ in nodes without rotation and the force torque is ignored.

Don't fix the code first. For each error write: symptom, physical cause, state of MAPDL involved and minimal test that would confirm the diagnosis.

Challenge — Two states, one linear law

Modify the case to solve two load steps:

• Load step 1: FY=-500 N, TIME=1.
• Load step 2: FY=-1000 N, TIME=2.

You must keep both sets and demonstrate:

RFY₂ / RFY₁ = 2
RMZ₂ / RMZ₁ = 2

Each state must close force and moment with an error less than 0.5 %. The challenge is completed using a script, without selecting entities from the interface.

Self-assessment

1. What is the difference between a load step and a result set?
2. Why is it executed ALLSEL,ALL before SOLVE?
3. What shows that n_sets>0 and what doesn't it prove?
4. Why do RFY and tip_force have opposite signs?
5. How does a face of nodes with only translational DOFs transmit a moment?
6. Why should the sum point be declared?
7. What error can be hidden if you only check forces?

Evidence of learning

• m06_equilibrium_audit.csv with passes=1.
• Extract from .out confirming a finished solution.
• Manual calculation of RFY and RMZ.
• Explanation of the sign convention relating RF and FSUM.
• CSV of the challenge with two states and ratios equal to two.
• Reasoned diagnosis of the five defects.

Exit checklist

• ☐ I declare the type of analysis and the geometric hypothesis.
• ☐ I restore all selections before solving.
• ☐ I check that a set exists before consulting results.
• ☐ I consciously activate the state I want to audit.
• ☐ I add reactions exclusively in the fixed component.
• ☐ I check force and moment with respect to the same point.
• ☐ I am not looking for non-existent rotational reactions in SOLID185.
• ☐ My CSV preserves inputs, results, errors and decision.

Technical traceability

The lesson uses Structural Analysis Guide and Basic Analysis Guide for static solution flow. The syntax of ANTYPE, NLGEOM, KBC, NSUBST, OUTRES, SOLVE, SET, SPOINT, FSUM, NDNEXT and *GET is contrasted with Command Reference 2024 R1.