name.i5

Program

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