# Basics I — Welcome to LS-DYNA Examples

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A static tensile test is simulated using shell elements. One end of the specimen is constrained, while concentrated nodal loads are applied at the other end. Uniform stresses develop in the narrowed center section.

## Introduction

```Problem #1: Tensile Test

Objectives
* Learn how to activate LS-DYNA implicit mode.
* Learn how to select linear or nonlinear analysis.

Problem Description
A static tensile test is simulated using shell elements.
One end of the specimen is constrained, while concentrated
nodal loads are applied at the other end. Uniform stresses
develop in the narrowed center section.

Input Filename: tensile1.k

Procedure
Copy the input file to your local directory. Using an editor,
view the input file and answer the following questions:

1. Which shell element formulation is used?
2. How many steps are used to apply the load?

Run the simulation, and post-process using LS-POST. Record
below the total applied load and total tip displacement
(HINT: the ascii database NODFOR includes data for each
of the tip nodes).

3. Applied load Tip displacement

Modify the input deck to increase the applied load 100x,
and rerun the simulation. Record the new data:

4. Applied load Tip displacement

5. Is the simulation linear or nonlinear ?
(Does the tip displacement scale linearly with applied load?)

6. What types of nonlinearity exist in this problem?
( ) material ( ) geometric ( ) contact

Modify the *CONTROL_IMPLICIT_SOLUTION keyword to perform a
linear simulation, and record the results below for the large
applied load:

7. Applied load Tip displacement

8. What is the reaction force at the fixed end of the beam?

9. Why does the reaction force not equal the applied load?

Modify the input deck to use the linear element formulation
type 21, and repeat the simulation.

10. Does the reaction force now match the applied load?
( ) yes ( ) no```

## Keywords

```*BOUNDARY_SPC_NODE
*CONTROL_IMPLICIT_AUTO
*CONTROL_IMPLICIT_GENERAL
*CONTROL_IMPLICIT_SOLUTION
*CONTROL_IMPLICIT_SOLVER
*CONTROL_TERMINATION
*DATABASE_BINARY_D3PLOT
*DATABASE_EXTENT_BINARY
*DATABASE_NODAL_FORCE_GROUP
*DATABASE_NODFOR
*DATABASE_NODOUT
*DEFINE_CURVE
*ELEMENT_SHELL
*END
*KEYWORD
*load_node_point
*MAT_elastic
*NODE
*PART
*SECTION_SHELL
*SET_NODE_LIST
*TITLE```

## Reduced Input

```  *KEYWORD
*TITLE
implicit tensile test
\$
\$ test coupon: 200 mm long, 20 mm width, 2.67 mm thickness, 50.8 mm gauge length
\$
\$ units; mm, s, ton, N
\$
\$ A. Tabiei
\$
\$---+----1----+----2----+----3----+----4----+----5----+----6----+----7----+----8
\$
*CONTROL_TERMINATION
1.0000
\$
\$---+----1----+----2----+----3----+----4----+----5----+----6----+----7----+----8
\$
*CONTROL_IMPLICIT_GENERAL
\$   imflag       dt0      iefs   nstepsb       igs
1       1.0         0         0         0
\$
*CONTROL_IMPLICIT_SOLUTION
\$  nlsolvr    ilimit    maxref     dctol     ectol     rctol     lstol
0         0         0       0.0       0.0         0         0
\$    dnorm   divflag   inistif   nlprint
0         0         0         0
\$
*CONTROL_IMPLICIT_SOLVER
\$   lsolvr   prntflg    negeig
0         0         0
\$
*CONTROL_IMPLICIT_AUTO
\$    iauto    iteopt    itewin     dtmin     dtmax
0         0         0       0.0       0.0
\$
\$---+----1----+----2----+----3----+----4----+----5----+----6----+----7----+----8
\$
*DATABASE_EXTENT_BINARY
\$    neiph     neips    maxint    strflg    sigflg    epsflg    rltflg    engflg
1         1
\$   cmpflg    ieverp    beamip

\$---+----1----+----2----+----3----+----4----+----5----+----6----+----7----+----8
\$
*DATABASE_BINARY_D3PLOT
0.010
*DATABASE_NODFOR
0.010
*DATABASE_NODOUT
0.010
*DATABASE_NODAL_FORCE_GROUP
2
*SET_NODE_LIST
2
30,31,32,33,38,39
\$
\$---+----1----+----2----+----3----+----4----+----5----+----6----+----7----+----8
\$
*MAT_elastic
\$      MID        RO         E        PR
1 0.780E-08 0.207E+06 0.280E+00
\$
*SECTION_SHELL
1         0
2.670E-00 2.670E-00 2.670E-00 2.670E-00
\$
*PART
shell tensile strip
1         1         1
\$
\$---+----1----+----2----+----3----+----4----+----5----+----6----+----7----+----8
\$
*DEFINE_CURVE
1
0.00000000000000E+00 0.0000000000000E+00
1.00000000000000E+00 2.0000000000000E+02
\$
*load_node_point
\$     Node       DOF      LCID        SF
204         1         1         1
205         1         1         1
206         1         1         1
207         1         1         1
212         1         1       0.5
213         1         1       0.5
\$
*BOUNDARY_SPC_NODE
30         0         1         1         1         1         1         1
31         0         1         1         1         1         1         1
32         0         1         1         1         1         1         1
33         0         1         1         1         1         1         1
38         0         1         1         1         1         1         1
39         0         1         1         1         1         1         1
\$
\$---+----1----+----2----+----3----+----4----+----5----+----6----+----7----+----8
\$
*END```