Block 4 — Select and constrain
NSEL,S,LOC,X,0
D,ALL,ALL,0
ALLSEL,ALL
NSEL,S,LOC,X,0 replaces the current selection with the nodes located in x=0.
D,ALL,ALL,0 restricts all of its available degrees of freedom. Finally,
ALLSEL,ALL retrieves all entities.
Selection is status
Many commands act only on the selected one. Forget ALLSEL can make stages
ignoring much of the model without the problematic line appearing incorrect.
Block 5 — Distribute the total force
NSEL,S,LOC,X,1.0
*GET,n_tip,NODE,0,COUNT
F,ALL,FY,-1000/n_tip
ALLSEL,ALL
The end nodes are selected first. *GET count how many there are and save the result
en n_tip. The strength of each node is -1000/n_tip, so the sum remains
equal to − 1000N even if you change the mesh.
This idea is important: the input value represents a total force, not a force that must be repeated in full at each node.
Block 6 — Exit PREP7 and Solve
FINISH
/SOLU
ANTYPE,STATIC
SOLVE
FINISHFINISHexits the current processor and returns to BEGIN./SOLUgoes into Solution.ANTYPE,STATICdeclares a static analysis.SOLVErequests the solution of the current load step.- The second
FINISHcloses the solution stage.
That SOLVE finish only proves that the solver got an answer. We still have to
check if it correctly represents the physical problem.
Block 7 — Enter POST1
/POST1
SET,LAST
PLNSOL,U,SUM
PLNSOL,S,EQV
FINISH
/OUTPUT/POST1opens the general postprocessor.SET,LASTread the latest set of results available.PLNSOL,U,SUMrepresents the magnitude of the displacement.PLNSOL,S,EQVrepresents the equivalent stress./OUTPUTwith no arguments returns live text output to the Output Window.
Each plot replaces the previous one in the viewport. At the end you will see the equivalent stress.
You can rerun PLNSOL,U,SUM manually to restore the displacement contour.
PLNSOL,U,SUM). It connects the model to its results; it does not demonstrate convergence or analytical validation.
PLNSOL,S,EQV). A second type of visual inspection; it does not replace the linearity challenge tolerances.