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Chapter 1

Introduction to Marc/Mentat

1.1 General Comments

A finite elements analysis involves different steps and program modules. A user normally defines in a graphical interphase, the so-called pre-processor, the computational model to be solved. The geometry can be either created in the pre-processor or imported from external computer-aided design programs (CAD), see Table 1.1.

Table 1.1 Some common CAD packages

Name Company Web page AutoCAD Autodesk http://www.autodesk.com/

CATIA Dassault Systèmes http://www.3ds.com/

Pro/ENGINEER Parametric Technology http://www.ptc.com (PTC Creo) Corporation

Solid Edge Siemens PLM http://www.plm.automation. siemens.com

SolidWorks Dassault Systèmes http://www.3ds.com/

Most of the commercial finite element pre-processors have specific import fil- ters to open third-party CAD files, e.g. a prt-file in the case of Pro/ENGINEER (Pro Creo). If there is no import filter for a specific CAD package available, the import of a geometry file is still possible via a neural file format such as ACIS, IGES, STL or STEP. It should be noted here that the generation of complex geometry is in general much easier in specialized CAD programs than in finite element pre-processors. Once the geometry is available in the pre-processors, the user must create and optimize the finite element mesh and assign geometrical and material properties to the elements. In a follow- ing step, the initial and boundary conditions must be defined and solution

2 1 Introduction to Marc/Mentat

parameters chosen. In a next step, a text file (ASCII) is generated which con- tains all the information that is required to solve the problem by the solver. The solver itself is normally not visible to the user and simply running in the background. Once the solution is obtained, the user can import the results in the so-called post-processor1 and analyze the results. Figure 1.1 shows the general steps which are involved in a finite element analysis.

Fig. 1.1 The Marc/Mentat package

1 The pre- and post-processor is in most of the finite element packages identical.

1.2 Graphical User Interface 3

1.2 Graphical User Interface

The Marc/Mentat graphical user interface has several noteworthy objects, to which we will refer by the following names:

cf. Fig. 1.2: 1©: Dropdown Menu 2©: Function Buttons 3©: Main Menu Tabs 4©: Tab Sections 5©: Model Navigator 6©: Graphic Interface 7©: Graphic Interface Navigation Menu 8©: Command Line Dialog

cf. Fig. 1.3: 9©: Function Dialog Windows 10©: Buttons in Dialog Windows 11©: Input Fields in Dialog Windows

When describing a certain operation in Marc/Mentat, the following notation will be used when referring to a specific object:

2©: Function Buttons 3©: Main Menu Tabs 4©: Tab Sections 8©: Command Line Dialog 9©: Go to: section\subsection 10©: Buttons in Dialog Windows 11©: Set <variable> = <value>

Typical buttons of the graphic interface navigation menu (see 7© in Fig. 1.2) are shown in Fig. 1.4.

4 1 Introduction to Marc/Mentat

Fig. 1.2 Overview of the Marc/Mentat user interface

1.2 Graphical User Interface 5

Fig. 1.3 Example of a function dialog window

Fig. 1.4 Graphic interface navigation menu (see Æ in Fig. 1.2)

6 1 Introduction to Marc/Mentat

Fig. 1.5 Schematic drawing of a mouse

When using Marc/Mentat you will have to perform different clicks with your mouse (see Fig. 1.5):

• Left click (LC), to pick items. • Right click (RC), to confirm selections. • Middle click (MC), by pressing the scroll/middle button, to undo selec-

tions.

1.3 Using Units 7

1.3 Using Units

1.3.1 SI Base Units

The International System of Units (SI)2 must be used in scientific publica- tions to express physical units. This system consists of the seven base quan- tities — length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity — and their respective base units are the meter, kilogram, second, ampere, kelvin, mole, and candela.3

1.3.2 Coherent SI derived Units

A coherent SI derived unit is defined uniquely as a product of powers of base units that include no numerical factor other than 1. Table 1.2 gives some examples of derived units and their expression in terms of base units.

Table 1.2 Example of coherent SI derived units

Quantity Coherent Derived Unit Name Symbol in terms of in terms of

other SI units SI base units

Celsius temperature degree Celsius ◦C K

Energy, work joule J N m m2 kg s−2

Force newton N m kg s−2

Plane angle radian rad 1 m/m

Power watt W J/s m2 kg s−3

Pressure, stress pascal Pa N/m2 m−1 kg s−2

1.3.3 Consistent Units

The application of a finite element code does normally not require that a specific system of units is selected. A finite element code keeps through an

2 The original name is known in French as: Système International d’Unités. 3 More information on units can be found in the brochures of the Bureau International des Poids et Mesures (BIPM): www.bipm.org/en/si.

8 1 Introduction to Marc/Mentat

analysis consistent units and requires only that a user assigns the absolute measure without specifying a specific unit. Thus, the units considered by the user during the pre-processing phase are maintained for the post-processing phase. The user must assure that the considered units are consistent, i.e. they fit each other. The following Table 1.3 shows an example of consistent units

Table 1.3 Example of consistent units

Property Unit

Length mm

Area mm2

Force N

Pressure MPa = N

mm2

Moment Nmm

Moment of inertia mm4

E-Modulus MPa = N

mm2

Density Ns2

mm4

Time s

Mass 103kg

Pay attention to the unit of the density. The following example shows the conversion of the density of steel:

%St = 7.8 kg

dm3 = 7.8× 103

m3 = 7.8× 10−6

mm3 . (1.1)

With

1 N = 1 m kg

s2 = 1× 103

mm kg

s2 und 1 kg = 1× 10−3

Ns2

mm (1.2)

follows the consistent density to:

%St = 7.8× 10−9 Ns2

mm4 . (1.3)

Since literature reports time by time also other units, the following Table 1.5 shows an example of consistent English units: Pay attention to the conversion of the density:

1.3 Using Units 9

Table 1.4 Example of consistent English units

Property Unit

Length in

Area in2

Force lbf

Pressure psi = lbf

in2

Moment lbf in

Moment of inertia in4

E-Modulus psi = lbf

in2

Density lbf sec2

in4

Time sec

%St = 0.282 lb

in3 = 0.282

in3 × 0.00259

lbf sec2

in = 0.73038× 10−3

lbf sec2

in4 .

(1.4)

1.3.4 Conversion of Important English Units to the Metric System

10 1 Introduction to Marc/Mentat

Table 1.5 Conversion of important U.S. customary units and British Imperial units (’English units’) to Metric units (m: meter; cm: centimeter; g: gram; N: newton; J: joule; W: watt)

Type English unit Conversion

Length inch 1 in = 0.025400 m

foot 1 ft = 0.304800 m

yard 1 yd = 0.914400 m

mile (statute) 1 mi = 1609.344 m

mile (nautical ) 1 nm = 1852.216 m

Area square inch 1 sq in = 1 in2 = 6.45160 cm2

square foot 1 sq ft = 1 ft2 = 0.092903040 m2

square yard 1 sq yd = 1 yd2 = 0.836127360 m2

square mile 1 sq mi = 1 mi2 = 2589988.110336 m2

acre 1 ac = 4046.856422400 m2

Volume cubic inch 1 cu in = 1 in3 = 0.000016387064 m3

cubic foot 1 cu ft = 1 ft3 = 0.028316846592 m3

cubic yard 1 cu yd = 1 yd3 = 0.764554857984 m3

Mass ounce 1 oz = 28.349523125 g

pound (mass) 1 lbm = 453.592370 g

short ton 1 sh to = 907184.74 g

long ton 1 lg to = 1016046.9088 g

Force pound-force 1 lbf = 1 lbF = 4.448221615260500 N

poundal 1 pdl = 0.138254954376 N

Stress pound-force per square inch 1 psi = 1 lbf in2

= 6894.75729316837 N m2

pound-force per square foot 1 lbf ft2

= 47.880258980336 N m2

Energy British thermal unit 1 Btu = 1055.056 J

calorie 1 cal = 4185.5 J

Power horsepower 1 hp = 745.699871582270 W

Chapter 2

Rods and Trusses

2.1 Definition of Rod Elements

The definition of rod elements is summarized in Table 2.1. The derivation in lectures normally starts with the introduction of an elemental coordinate system (x) which is aligned with the principal axis of the element. Based on the definition of this element, deformations (u1x, u2x) can only occur along the principal axis. Assuming linear interpolation functions for the displacements, a constant elemental stress and strain is obtained.

Table 2.1 Definition of rod elements

Simplified Definition (derivation in lecture)

Definitions Degrees of Freedom

Material: E u1x, u2x Geometry: L,A

General Definition (MSC Marc element type 9)

Definitions Degrees of Freedom

Material: E Node 1: u1X , u1Y , u1Z Geometry: A Node 2: u2X , u2Y , u2Z Node 1: X1, Y1, Z1 Node 2: X2, Y2, Z2

12 2 Rods and Trusses

The implementation of a rod element in a commercial finite element code is more general, i.e. based on the global coordinate system (X,Y, Z). Thus, the geometry is defined based on the global coordinates of each node (Xi, Yi, Zi) and the cross-sectional area A. In such a configuration, each node has three degrees of freedom, i.e. the three displacements expressed in the global coordi- nate system: uiX , uiY , uiZ . Nevertheless, the stress and strain are uniaxial in the truss member. In the case of the linear straight truss (MSC.Marc element type 9), the stiffness matrix is obtained based on an one-point integration rule whereas the mass matrix is obtained based on a two-point integration rule.

2.2 Basic Examples

2.2.1 1D Rod - Fixed Displacement

Problem description: Given is a rod of length L = 1.0 and constant axial tensile stiffness given by E = 20 and an area A = 0.5 as shown in Fig. 2.1. At the left-hand side there is a fixed support and at the right-hand side there is a prescribed displace- ment of u0 = 0.5. Discretize the problem with a single rod element (MSC Marc element type 9) and calculate the reaction force at the right-hand node.

Fig. 2.1 Schematic drawing of a single rod loaded by an end displacement u0

Marc solution:

Under File → Save As..., save file as ”bar disp”.

2.2 Basic Examples 13

Constructing the mesh

1. Under Geometry & Mesh :

Basic Manipulation select Ge- ometry and Mesh (see Fig. 2.3). 2. Under Mesh\Nodes, select Add. 3. 0,0,0 Enter 1,0,0 Enter. 4. Under Mesh\Elements, select Line (2) (see Fig. 2.2). 5. Press Add. 6. In 6©, select the two nodes with a (LC). 7. Press OK.

Fig. 2.2 Geometry and mesh dialog window

Fig. 2.3 Geometry and mesh

Setting the Geometric Properties

8. Under Geometric Properties : Geometric Properties select New (Struc-

tural). 9. Select 3D → Truss (see Fig. 2.4). 10. Set Properties\Area = 0.5. 11. Under Entities\Elements, press Add.

12. In 6© select the element with a (LC), then (RC). 13. Press OK.

Setting the Material Properties

14 2 Rods and Trusses

Fig. 2.4 Geometric properties

14. Under Material Properties : Material Properties select New→ Finite Stiffness → Standard.

15. Set Other Properties\Young’s Modulus = 20 (see Fig. 2.6) 16. Under Entities\Elements, press Add

17. In 6© select the element with a (LC), then (RC). 18. Press OK

Setting the Boundary Conditions

Fixed Support

19. Under Boundary Conditions : Boundary Conditions select New (Struc-

tural) → Fixed Displacement (see Fig. 2.7). 20. Under Properties tick Displacement X, Displacement Y and Displace-

ment Z. 21. Under Entities\Nodes, press Add.

22. Under 6© select the left most node (0,0,0) with a (LC), then (RC). 23. Press OK.

Displacement Boundary Condition

24. Under Boundary Conditions : Boundary Conditions select New (Struc-

tural) → Fixed Displacement. Set Name = disp1. 25. Under Properties tick Displacement X. Set Displacement X = 0.5 (see

Fig. 2.8). 26. Under Entities\Nodes, press Add.

2.2 Basic Examples 15

Fig. 2.5 Material properties

Fig. 2.6 Material properties

16 2 Rods and Trusses

Fig. 2.7 Boundary conditions dialog window

Fig. 2.8 Entering the displacement boundary condition

2.2 Basic Examples 17

Running the Job

29. Under Jobs : Jobs select New → Structural. 30. Press Check; See in 8© if there are any errors. 31. If there are none, press Run.

32. In ”Run Job”, press Advanced Job Submission (see Fig. 2.9). 33. Press Save Model. 34. Press Write Input File. Press OK. 35. Press Submit 1.

36. Wait until Status = Complete. 37. Press Open Post File (Model Plot Results Menu).

Viewing the model

38. Under Deformed Shape\Style, select Deformed and Original 39. Under Scalar Plot\Style, select Numerics (see Fig. 2.10). 40. Under Scalar Plot, press Scalar and select Displacement X. 41. Under Scalar Plot, press Scalar and select Reaction Force X. 42. Press OK.

Fig. 2.10 Job result dialog window

Result: The reaction force at the right-hand end of the rod is found to be 5.

2.2.2 1D Rod - Fixed Point Load

Problem description: Given is a rod of length L = 1.0 with a constant axial tensile stiffness given

18 2 Rods and Trusses

Fig. 2.9 Advanced job submission

by E = 20 and A = 0.5 as shown in Fig. 2.11. At the left-hand side there is a fixed support and the right-hand side is loaded by a single force F0 = 5. Use a single rod element to determine the elongation of the right-hand end.

Marc solution:

The steps for this example are the same as the one of the previous exam- ple, 2.2.1 except for steps 24-28 and the file name, which should be ‘bar force’

2.2 Basic Examples 19

Fig. 2.11 Schematic drawing of a single rod loaded by an end load F0

(see Sect. 2.2.1). The following steps have to replace these:

Setting Fixed Point Load

24. Under Boundary Conditions : Boundary Conditions select New (Struc-

tural) → Point Load. Set Name = force1. 25. Under Properties, tick Force X. Set at 5. 26. Under Entities\Nodes, press Add (see Fig. 2.12).

Fig. 2.12 Entering fixed point load value

Result: The resulting displacement on the right-hand end of the rod is found to be 0.5.

20 2 Rods and Trusses

2.2.3 1D Rod - Multiple Loadcases

Problem description: Given is a rod of length L = 1.0 with a constant axial tensile stiffness given by E = 20 and A = 0.5 as shown in Fig. 2.13. At the left-hand side there is a fixed support and the right-hand side is

I) elongated by a given displacement u0 = 0.5 and II) loaded by a single force F0 = 5.

Discretize the problem with a single rod element to determine:

a) the reaction force and b) the displacement at the right-hand end.

Fig. 2.13 Schematic drawing of a single rod with different load cases

Marc solution: For this example, involving loadcases, the steps are equal to those in ex-

ample 1, from steps 1-28 (see Sect. 2.2.1). Save as ‘bar twoloads’. Add the following steps:

Setting the Additional Boundary Condition

Force Boundary Condition

29. Under Boundary Conditions : Boundary Conditions select New (Struc-

tural) → Point Load. Set Name = force1. 30. Under Properties, tick Force X. Set at 5. 31. Under Entities\Nodes, press Add.

33. Press OK.

2.2 Basic Examples 21

Defining Loadcase 1 - Fixed Displacement

34. Under Loadcases : Loadcases select New → Static. 35. Set Name = fixed displacement (see Fig. 2.14). 36. Press Loads. Untick force1. Press OK. 37. Set Stepping Procedure\#Steps = 1. Press OK.

Fig. 2.14 Defining loadcase 1

Defining Loadcase 2 - Fixed Point Load

38. Under Loadcases : Loadcases select New → Static. 39. Set Name = fixed force. 40. Press Loads. Untick displ1. Press OK.

22 2 Rods and Trusses

41. Set Stepping Procedure\#Steps = 1 Press OK.

Running the two jobs

42. Under Jobs : Jobs select New → Structural. Set Name = force. (see Fig. 2.15)

43. Under Available, select fixed force. 44. Press Initial Loads. Untick displ1. Press OK. 45. Press Check; See in 8© if there are any errors. 46. If there are none, press Run.

47. In ‘Run Job’, press Advanced Job Submission. 48. Press Save Model. 49. Press Write Input File. Press OK. 50. Press Submit 1.

51. Wait until Status = Complete. Press OK. 52. Press OK.

53. Under Jobs : Jobs select New → Structural. Set Name = displacement. 54. Under Available, select fixed displacement. 55. Press Initial Loads. Untick force1. Press OK. 56. Press Check; See in (8) if there are any errors. 57. If there are none, press Run.

58. In ‘Run Job’, press Advanced Job Submission. 59. Press Save Model. 60. Press Write Input File. Press OK. 61. Press Submit 1.

62. Wait until Status = Complete. 63. Press Open Post File (Model Plot Results Menu).

Viewing the model

64. Under Deformed Shape\Style, select Deformed and Original. 65. Under Scalar Plot\Style, select Numerics.

66. Press Scalar and select Reaction Force X. Press OK. 67. Press OK.

Open Job fixed force

68. Under File→ Results→Open. Open bar twoloads force.t16 (see Fig. 2.16) 69. In the window ‘Model Plot Results’, under Scalar Plot, press Scalar

and select Displacement X. 70. Press OK.

Results: Loadcase a: The reaction force on the right-hand end of the rod, is found to

2.2 Basic Examples 23

Fig. 2.15 Running Job 1

be 5. Loadcase b: The resulting displacement on the right-hand end of the rod is found to be 0.5.

24 2 Rods and Trusses

Fig. 2.16 Opening fixed force result case