ATLANTIC 2.0
EXAMPLES FOUND ON INSTALLATION FILE
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1) File EXMPL01.GN4 
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This is a simple portal frame. It shows also various types of loadings: nodal concentrated
loads, beam concentrated loads, beam distributed loads, thermal loads, settlements.


2) File EXMPL02.GN4 
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This is a simple truss beam, as you can find in many industrial buildings.
It is usefull for showing how to get a truss structure (beam release).


3) File EXMPL03.GN4 
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This example shows how to study a beam on elastic foundation.


4) File EXMPL04.GN4 
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Here you can see how to set the local orientation system for a beam not parallel
to any coordinate plane, using the tecnique of "3-axis paralle to P1-P2".


5) File EXMPL05.GN4
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Just to show you a quite large model ( 866 nodes and 1600 beams ). It should be a model
of an industrial building. 
However with the Unregistered version of the program, you can manage structures up
to 10 nodes. So you can just see the model in ATANTIC-Pre. You can modify it and save, but
you are not allowed to create input for ATLANTIC-Solver and examine results in ATLANTIC-Post.
With the Registered version you can do all that, obviously ! 


Here is one more example related to AISC-ASD verifications.

6) File EXMPL06.GN4 - Flexural I beam
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Let's consider a simply supported beam, 6 meters ( 19.69 ft ) length.
The profile is European IPE 300.
The steel is Fe43C EN-10025 ( Fy = 27.5 KN/cm^2 -> 39.8 ksi )


Loads:
Load 1: Dead load        q =  5 KN/m   ( 342.6 lb/ft )
Load 2: Live Load        q = 10 KN/m   ( 685.2 lb/ft )

Laterally unbraced length (Lb):     3.0 m  ( 9.85 ft )



qTOT =  5 +  10  =     15   KN/m    ( 1027.8 lb/ft   )
M = 15 x 6^2 / 8 =     67.5 KN*m    (   49.8 kips*ft )     


Width-Thickness ratio: 

bf = 150   mm (  5.91  in )
tf =  10.7 mm (  0.421 in )                             

b/t = 150 * 0.5 / 10.7 = 7.0

65 / Fy^1/2 = 65 / 39.8^1/2 = 10.3    -> COMPACT


		    
Lc = min( 76 bf / Fy^(1/2)  ; 20000 / ( (d/Af) Fy ) ) );  (cfr. F1.1 )
Lc = 1.80 m  ( 5.91 ft )

Lb > Lc


Allowable bending stress in tension ( cfr. [F1-5] ):

Fbt = 0.60 Fy = 0.60 x 27.50 = 16.5 KN/cm^2  ( 23.93 ksi )

Allowable bending stress in compression ( value from [F1-6] or [F1-7] ed [F1-8] ):


Fbc = 14.76 KN/cm^2 ( 21.40 ksi )  (  ( form [F1-8]).

Cb has been put equal to 1.

So we have:

compression stress:  fb =  67.5 / 557 = 12.12 KN/cm^2 < 14.76

Stress rate:         12.12 / 14.76 = 0.82


And hereafter you will find the program printout (file EXMPL06.PRV)
showing the verifications:



 Beam:   1 - (nodes:    1/   2) -  (--------  ) - Profile:          IPE300
 --------------------------------------------------------------------

 length:       600.000 cm
 Mat.: Fe430  -  Fy:      27.46 KN/cm^2 -  Fu:      42.17 KN/cm^2

 Shape features:
 Area:                      53.80 cm^2 - Net area:                  45.73 cm^2
 Shear area dir. 2:         19.78 cm^2 - Shear area dir. 3:         32.10 cm^2
 J2:                      8356.00 cm^4 - J3:                       604.00 cm^4
 W2:                       557.10 cm^3 - W3:                        80.50 cm^3
 i2 (radius of gyrat.):     12.46 cm   - i3 (radius of gyrat.):      3.35 cm

 COMPACT Section

 l dir. 2:                 600.00 cm   - l dir. 3:                  600.00 cm
 K dir. 2:                   1.00      - K dir. 3:                    1.00 
 Kl/r dir. 2:                48.2      - Kl/r dir. 3:                179.1

 Lateral buckling parameters:
 h (profile heigth):        30.00 cm   - b (flange width):           15.00 cm
 s  (flange thick.):         1.07 cm   - Lb:                        300.00 cm
                                         Lc:                        180.64 cm

 Displacement parameters  (if applicable):
 L.L.      displacements:         1/ 400 of length
 L.L.+D.L. displacements:         1/ 300 of length


 AISC 90 CHECK RESULTS

 3) Bending dir. 2  (Chapter F)
  fb2cI =      12.12 KN/cm^2  - x =   300.00 cm  - Comb:   2
  Fb2cI =      14.75 KN/cm^2  -  cfr.: [ F1-8 ]  - Cb:   1.00
  M3 =        -67.50 KN*m
  fb2tI =      12.12 KN/cm^2  - x =   300.00 cm  - Comb:   2
  Fb2tI =      16.48 KN/cm^2  -  cfr.: [ F1-5 ]  - Cb:   1.00
  M3 =        -67.50 KN*m
  fb2cII=       0.00 KN/cm^2  - x =     0.00 cm  - Comb:   0
  Fb2cII=      19.62 KN/cm^2  -  cfr.: [ F1-8 ]  - Cb:   1.00
  M3 =          0.00 KN*m
  fb2tII=       0.00 KN/cm^2  - x =     0.00 cm  - Comb:   0
  Fb2tII=      21.91 KN/cm^2  -  cfr.: [ F1-5 ]  - Cb:   1.00
  M3 =          0.00 KN*m

 5) Shear dir. 2  (Chapter F)
  fv I  =       2.27 KN/cm^2  - x =     0.00 cm  - Comb:   2
  Fv I  =      10.98 KN/cm^2 -   cfr.: [ F4-1 ] 
  T =          45.00 KN
  fv2 II =       0.00 KN/cm^2  - x =     0.00 cm  - Comb:   0
  Fv II =      14.61 KN/cm^2 -   cfr.: [ F4-1 ] 
  T =           0.00 KN

 7) Axial compression and bending (Chapter H - [ H1-2 ])
  I)   f/F  tot =   0.82  - Comb:   2
   f/F  N   =   0.00  -  f/F  M2 =   0.00  -  f/F  M3 =   0.82 
  N =           0.00 KN - M2 =       0.00 KN*m - M3  =     -67.50 KN*m

 8) Axial tension and bending (Chapter H - [ H1-3 ])
  I)   f/F  tot =   0.74  - Comb:   2
   f/F  N   =   0.00  -  f/F  M2 =   0.00  -  f/F  M3 =   0.74 
  N =           0.00 KN - M2 =       0.00 KN*m - M3  =     -67.50 KN*m

 9) Axial compression and bending (Chapter H - [ H1-1 ])
  I)   f/F  tot =   0.82  - Comb:   2
   f/F  N   =   0.00  -  f/F  M2 =   0.00  -  f/F  M3 =   0.82 
  Cm dir.2:     1.00  - Cm dir.3:    0.40
  N =           0.00 KN - M2 =       0.00 KN*m - M3  =     -67.50 KN*m

 10) Displacements check
  L.L. displacements (dir. 2):            0.98 cm ( l/   614)
  D.L.+L.L. displacements(dir. 2) :       1.47 cm ( l/   409)






