STR2 STR3
; program control IP KLC 0 KOUT ILC 0 IEL1 IEU1 IEL2 IEU2 0 KPROB KGRAF KAR
; end of input data EN EN
| $\mathtt{KLC}$ | The key of load cases. | |
|---|---|---|
| $=1$ | process all load cases | |
| $=3$ | process selected load cases | |
| $\mathtt{KOUT}$ | The key of output. | |
| $=0$ | displacements | |
| $=1$ | displacements and stresses on all elements | |
| $=2$ | displacements and stresses on selected elements | |
| $=-1$ | strains on all elements | |
| $=-2$ | strains on selected elements | |
| $\mathtt{ILC}$ | The load case index, $0\le\mathtt{ILC}\le\mathtt{NLC}.$ The number of load cases $\mathtt{NLC}$ is determined automatically depending on $\mathtt{KPROB}.$ | |
| $\mathtt{IEL1},\mathtt{IEU1}$ $\mathtt{IEL2},\mathtt{IEU2}$ | Element selection for output. For $\mathtt{KOUT}=\pm2$ only the elements $\mathtt{IE}$ in range $\mathtt{IEL1}\le\mathtt{IE}\le\mathtt{IEU1}$ and $\mathtt{IEL2}\le\mathtt{IE}\le\mathtt{IEU2}$ are processed. | |
| $\mathtt{KPROB}$ | The key of problem type. | |
| $=0$ | elastostatic | |
| $=1$ | dynamic | |
| $=2$ | nonlinear | |
| $\mathtt{KGRAF}$ | The key of output for graphics. | |
| $=0$ | output only to the protocol (values in integration points) | |
| $=1$ | output only to the text file name.STR (values extrapolated to nodes) |
|
| $=2$ | output to both the protocol and the file name.STR |
|
| $=3$ | output only to the binary file name.STB in single precision (values extrapolated to nodes) |
|
| $\mathtt{KAR}$ | The integer quotient of arithmetic series used to select load cases, $\mathtt{KAR}\ge0.$ For $\mathtt{KLC}=3$ the value of $\mathtt{KAR}$ specifies the index of the last processed load case. | |
The program processes load cases according to the entered parameters in the following way:
| $\mathtt{KLC}=1,$ $\mathtt{KPROB}=0,$ $\mathtt{KAR}=0$ or $1$ | all (regardless of $\mathtt{ILC}$) |
| $\mathtt{KLC}=1,$ $\mathtt{KPROB}=0,$ $\mathtt{KAR}>1$ | first, then every $\mathtt{KAR}$-th |
| $\mathtt{KLC}=1,$ $\mathtt{KPROB}>0,$ $\mathtt{KAR}=0$ or $1$ | first to $\mathtt{ILC}$-th |
| $\mathtt{KLC}=1,$ $\mathtt{KPROB}>0,$ $\mathtt{KAR}>1$ | first, then every $\mathtt{KAR}$-th, until $\mathtt{ILC}$-th |
| $\mathtt{KLC}=3,$ $\mathtt{KAR}\le\mathtt{ILC}$ | only $\mathtt{ILC}$-th |
| $\mathtt{KLC}=3,$ $\mathtt{KAR}>\mathtt{ILC}$ | $\mathtt{ILC}$-th to $\mathtt{KAR}$-th |