Ansys Mechanical APDL lecture 4 by Haydar Alsalami from IRAQ - Hilla, studied in JNTUH - INDIA .

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Ansys Mechanical APDL lecture 4 by Haydar Alsalami from IRAQ - Hilla, studied in JNTUH - INDIA . haydar2200@gmail.com - mobile : 00919059070644 - 009647801518389

Transcript of Ansys Mechanical APDL lecture 4 by Haydar Alsalami from IRAQ - Hilla, studied in JNTUH - INDIA .

1 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

Combination of Link and Beam:

2 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

Preprocessor:

Choosing structural to

update all mechanical

and physical properties

such as Loads, stresses ..

etc.

3 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

First choose LINK: 2D

if available for easy

solution, if not choose

3D

4 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

Second choose BEAM:

2D if available for easy

solution, if not choose 2

node 188.

5 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

Add Real Constants: Enter cross sectional

Area for LINK as in the problem figure.

For BEAM, if we use version 11 and below,

we can Add constants and enter Area of

BEAM, but here in version 14 , there are NO

constants for BEAM so we can Add constants

in Sections.

6 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

e: means exponential

2e11: means 200000000000

PRXY : start from 0 and less than 1,

here take it 0

If we have DENSITY we have to

enter it in the material properties.

This must be 2e11 in this example

7 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

Sections used with BEAMS and Lines.

We can’t use Sections with SOLIDS and

Areas.

Set ID=1 then choose Sub-type as in the problem, then

fill in the dimensions below.

B = 2.5e-3 , we know that after divide Area by h in the

figure.

H = 2.54e-3 directly from figure.

Nb and Nh mean number of

divisions for section, they give

smoothness and flexibility for

section, we assumed about 30,

they are same Segments in 3D

max.(it’s important).

8 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

New method for drawing by

using COMMAND LINE:

K : keypoint

NPT : number of point

X,Y,Z : coordinates

Then press ( Enter )

9 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

It will be like this.

10 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

New method for drawing by

using COMMAND LINE:

L : Line

1 : keypoint 1

2 : keypoint 2

This means connection

between 1 and 2

Then press (ENTER)

11 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

It will be like this.

12 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

From Mesh Attribute:

Choose Picked Lines,

then pick BEAM Lines

as in the fig. below, then

Apply.

13 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

Since we use version 14 , we have to choose

Element type as 2 BEAM188 as in the fig.

and left Real constant set number as 1 , this

means that we want BEAM with number 2 for

Lines that already picked , and LINK with

number 1 to the remaining Lines.

Press OK

ANSYS now understand

that picked lines as a

BEAM, and other lines as

a LINK.

Note: in version 11 and below we can choose from window

above BEAM as a number 2 in the Element type and number

2 in the Real constant for a picked lines, then Apply ….. then

pick remaining lines and choose LINK as a number 1 in the

Element type and number 1 in the Real constant.

14 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

Apply Meshing:

For BEAM let us say ( 10 )

For LINK enter ( 1 ) compulsory

Then Mesh: Pick All Lines.

15 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

All DOF

All DOF

Press UX, UY, UZ, ROTX,

ROTY.

And left only ROTZ , because

we have rotation about Z.

16 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

Apply force ( -3000 )

17 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

18 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

19 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

After choosing (By Sequence )

From left table, then choose

(SMISC) from right table, then in

the rectangle below enter number

2 as in the figure, it will be

(SMISC, 2) this code relate to

find (Shear Force at ith node)

20 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

SMISC, 2 Shear force at ith node

SMISC, 8 Shear force at jth node

SMISC, 6 Bending moment at ith node

SMISC, 12 Bending moment at jth node

NMISC, 1 Maximum stress at ith node

NMISC, 2 Minimum stress at ith node

NMISC, 3 Maximum stress at jth node

NMISC, 4 Minimum stress at jth node

21 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

It should be like this

22 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

Choose any code to Plot

element as in the figure.

23 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

Choose any code to

show in the List, or

choose many codes in

one time to show in the

List.

Attention: these results for

Elements not for Nodes.

24 | Haydar Alsalami \ JNTUH \ India mobile: 00919059070644 – Iraq mobile: 009647801518389 E-Mail: haydar2200@yahoo.com

In this window we select different

codes one for node I and another for

node J to see the difference Plot

between them a in the figure.

In this fig. we chose SMISC2 and

SMISC8 to see the difference

between the shear force at ith node

and shear force at jth node. Note: there must be a relationship between chosen codes for

I and j for example:

SMISC2 & SMISC8 or SMISC6 & SMISC12

NMISC1 & NMISC2 or NMISC3 & NMISC4