University of Central Florida
Department of Civil and Environmental
Engineering
CES 6116 Finite Element Analysis
“SAP2000 Tutorial”
Instructor:
Dr. F. Necati Catbas
Assisted by former students:
José J Perez
Swapnil Chogle
SAP2000 Tutorial
CES 6116 Finite Element Analysis
CEE-UCF
Problem # 1 (Example 5.7, pg 219 of Textbook)
Determine the joint displacements, member end forces, and support reactions for the beam
shown in fig 5.18 (a), using the matrix stiffness method.
Fig 5.18
Solution:
Analytical model: See fig 5.18 (b). The beam has four degree of freedom (numbered 1 to 4)
and four restrained coordinates (numbered 5 to 8)
Procedure to model a structure in SAP2000 Nonlinear
Phase I) Pre-processing
Phase II) Analysis
Phase III) Post-processing
Phase I) Pre-processing
• Setting up the model Geometry
o Choose units: KN-m
o File > New Model > Coordinate System Definition > Cartesian > Number of grid >
Grid Spacing > Ok.
o Draw > Edit Grid (to change the grid) > Ok.
o Select the “Snap” icon to snap to point, midpoint, etc.
o Draw > Draw frame element
o Select All > Edit > Change Label > Re-label Selected items > Ok.
•
Define Material
o Choose units: N-mm
2/13
SAP2000 Tutorial
CES 6116 Finite Element Analysis
CEE-UCF
o Concrete > Click Modify>Type of Material>Isotropic>Analysis Property data >Ok.
•
Define Frame Section
o Choose units: KN-mm
o Add Rectangular Section > Dimensions > Depth > Width > Section properties >
Moment of inertia about 3 > Ok > Ok > Ok.
o Repeat the above step if ‘Moment of Inertia is varying’ (as in problem 1).
o Select the member > Assign > frame > Section.
o View > Set Element.
•
o
o
o
o
o
•
Assign Joint Restrains
Select joints to which you want to assign restrains.
J1 > Fast Restrains > Fixed
J2 > U2, R1, R3
J3, J5 > U2, U3, R1, R3
J4 > Fast Restrains > Free
Assign Loads
o Select members (frame) to which you want to assign loads.
o Select M1 > Assign > Frame Static Load > Trapezoidal > Load Case Name > Load
1 > Force Type and Direction > Forces > Gravity > Trapezoidal Load > Input load
value > Options > Ok.
o Select J2 > Assign > Joint Static Load > Forces > Load Case Name > Load 1 >
Force Global Z > Input load value (-ve force in downward direction) > Ok.
o Repeat the above step for 150 KN concentrated force at Joint J4.
o Select J3 > Assign > Joint Static Load > Forces > Load Case Name > Load 1 >
Moment Global YY > Input load value (+ ve Moment Clockwise) > Ok.
Phase II) Analysis
•
Analyze > Run.
o Analysis Complete > Check for Error / Warning > Ok
Phase III) Post-Processing
•
Check deflected shape > Ok.
o Display > Show Deformed Shape
o Display > Show Element forces / stresses > Joints > Joint Reaction Forces > Load 1
Load Case > Reactions > Ok > 3D (to view the moments at supports)
o Display > Show Element forces / stresses > Frames > Member Force Diagram for
Frames > Load 1 Load Case > Shear 2-2 > Scaling > Auto > Show values on
Diagram > Ok
o Display > Show Element forces / stresses > Frames > Member Force Diagram for
Frames > Load 1 Load Case > Moment 3-3 > Scaling > Auto > Show values on
Diagram > Ok
To view the output table
o File > Print Output Table > Type of Analysis Result > Select Load Case > Print to
File > File Name > Ok (it will create output as a text file)
o Option > Windows > Four (all results)
3/13
SAP2000 Tutorial
CES 6116 Finite Element Analysis
Problem # 1 (KN-m)
Joint displacement
4/13
CEE-UCF
SAP2000 Tutorial
CES 6116 Finite Element Analysis
CEE-UCF
PROBLEM # 1 (INPUT DATA)
(KN-m UNITS)
STATIC LOAD CASES
STATIC
CASE
CASE
TYPE
SELF WT
FACTOR
LOAD1
DEAD
1.0000
JOINT DATA
JOINT
GLOBAL-X
GLOBAL-Y
J1
J2
J3
J4
J5
0.00000
6.00000
10.00000
15.00000
20.00000
0.50000
0.50000
0.50000
0.50000
0.50000
GLOBAL-Z
RESTRAINTS ANGLE-A ANGLE-B ANGLE-C
0.00000
0.00000
0.00000
0.00000
0.00000
111111
010101
011101
000000
011101
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
FRAME ELEMENT DATA
FRAME JNT-1 JNT-2 SECTION
M1
M2
M3
M4
J1
J2
J3
J4
J2
J3
J4
J5
ANGLE RELEASES SEGMENTS
CONCA
CONCB
CONCB
CONCB
0.000
0.000
0.000
0.000
000000
000000
000000
000000
4
4
4
4
R1
R2 FACTOR LENGTH
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.000
1.000
1.000
1.000
6.000
4.000
5.000
5.000
MATERIAL PROPERTY DATA
MAT
LABEL
MODULUS OF
ELASTICITY
CONC
28000000.0
POISSON'S
RATIO
0.200
THERMAL WEIGHT PER
COEFF
UNIT VOL
0.000
MASS PER
UNIT VOL
0.000
0.000
FRAME SECTION PROPERTY DATA
SECTION
FLANGE
LABEL
MAT
SECTION
LABEL
TYPE
CONCA
CONCB
CONC
CONC
DEPTH
FLANGE
FLANGE
0.685
0.600
WIDTH
TOP
0.325
0.325
THICK
TOP
0.000
0.000
WEB
THICK
0.000
0.000
FLANGE
WIDTH
BOTTOM
0.000
0.000
THICK
BOTTOM
0.000
0.000
FRAME SECTION PROPERTY DATA
SECTION
LABEL
AREA
TORSIONAL
INERTIA
CONCA
CONCB
0.223
0.195
5.505E-03
4.540E-03
MOMENTS OF INERTIA
I33
I22
8.705E-03
5.850E-03
1.960E-03
1.716E-03
SHEAR AREAS
A2
A3
0.186
0.163
0.186
0.163
FRAME SECTION PROPERTY DATA
SECTION
LABEL
CONCA
CONCB
SECTION MODULII
S33
S22
2.542E-02
1.950E-02
1.206E-02
1.056E-02
PLASTIC MODULII
Z33
Z22
3.812E-02
2.925E-02
5/13
1.809E-02
1.584E-02
RADII OF GYRATION
R33
R22
0.198
0.173
9.382E-02
9.382E-02
SAP2000 Tutorial
CES 6116 Finite Element Analysis
CEE-UCF
FRAME SECTION PROPERTY DATA
SECTION
LABEL
TOTAL
WEIGHT
TOTAL
MASS
CONCA
CONCB
0.000
0.000
0.000
0.000
SHELL SECTION PROPERTY DATA
SECTION
LABEL
MAT
LABEL
SSEC1
CONC
SHELL
TYPE
MEMBRANE
THICK
BENDING
THICK
4 1.000E-03
MATERIAL
ANGLE
1.000E-03
0.000
SHELL SECTION PROPERTY DATA
SECTION
TOTAL
LABEL WEIGHT
SSEC1
0.000
TOTAL
MASS
0.000
J O I N T F O R C E S Load Case LOAD1
JOINT GLOBAL-X GLOBAL-Y
J3
0.000
0.000
GLOBAL-Z GLOBAL-XX GLOBAL-YY GLOBAL-ZZ
0.000
0.000
90.000
0.000
F R A M E S P A N D I S T R I B U T E D L O A D S Load Case LOAD1
FRAME
TYPE
DIRECTION
DISTANCE-A
VALUE-A
DISTANCE-B
M1
M1
M1
FORCE
FORCE
FORCE
GLOBAL-Z
GLOBAL-Z
GLOBAL-Z
0.0000
0.2500
0.7500
-30.0000
-22.5000
-7.5000
0.2500
0.7500
1.0000
F R A M E S P A N P O I N T L O A D S Load Case LOAD1
FRAME
TYPE
DIRECTION
DISTANCE
VALUE
M2
M3
FORCE
FORCE
GLOBAL-Z
GLOBAL-Z
0.0000
1.0000
-200.0000
-150.0000
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VALUE-B
-22.5000
-7.5000
0.0000
SAP2000 Tutorial
CES 6116 Finite Element Analysis
CEE-UCF
PROBLEM #1 (RESULTS)
(KN-mm Units)
JOINT DISPLACEMENTS
JOINT LOAD
U1
U2
U3
R1
R2
R3
J1 LOAD1
0.0000
0.0000
0.0000
0.0000
J2 LOAD1
0.0000
0.0000
-4.7375
0.0000 -5.544E-04
0.0000
J3 LOAD1
0.0000
0.0000
0.0000
0.0000 6.745E-04
0.0000
J4 LOAD1
0.0000
0.0000
-9.8336
0.0000 6.164E-04
0.0000
J5 LOAD1
0.0000
0.0000
0.0000
0.0000 -3.219E-03
0.0000
0.0000
0.0000
JOINT REACTIONS
JOINT LOAD
F1
F2
F3
M1
J1 LOAD1
0.0000
0.0000 146.4303
J2 LOAD1
0.0000
0.0000
J3 LOAD1
0.0000
0.0000 243.3161
0.0000
0.0000
0.0000
J5 LOAD1
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
M2
M3
0.0000 -281767.375
0.0000
50.2536
0.0000
0.0000
0.0000
FRAME ELEMENT FORCES
FRAME LOAD
LOC
P
V2
V3
T
M2
M3
M1 LOAD1
0.00
1500.00
3000.00
4500.00
6000.00
0.00
0.00
0.00
0.00
0.00
-146.43
-107.06
-78.93
-62.06
-56.43
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
-281767.38
-93059.41
45023.58
149356.56
236814.56
M2 LOAD1
0.00
1000.00
2000.00
3000.00
4000.00
0.00
0.00
0.00
0.00
0.00
-56.43
143.57
143.57
143.57
143.57
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
236814.56
93244.88
-50324.80
-193894.47
-337464.16
M3 LOAD1
0.00
1250.00
2500.00
3750.00
5000.00
0.00
0.00
0.00
0.00
0.00
-99.75
-99.75
-99.75
-99.75
50.25
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
-247464.14
-122781.13
1901.89
126584.91
251267.92
M4 LOAD1
0.00
1250.00
2500.00
3750.00
5000.00
0.00
0.00
0.00
0.00
0.00
50.25
50.25
50.25
50.25
50.25
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
251267.92
188450.94
125633.96
62816.98
0.00
7/13
SAP2000 Tutorial
CES 6116 Finite Element Analysis
CEE-UCF
Problem # 2 & # 3 (Example 6.6, pg 300 & Example 7.4, pg 380 of Textbook)
Determine the joint displacements, member end forces, and support reactions for the plane
frame shown in fig 7.10 (a), due to the combined effect of the loading shown and
settlement of 1 in of the left support. Use the matrix stiffness method.
Fig 7.10
Solution:
Analytical model: See fig 7.10 (b). The frame has three degrees of freedom – the
translations in the X and Y directions, and the rotation, of joint 2 – which are numbered 1,
2, and 3, respectively. The six restrained coordinates of the frame are identified by numbers
4 through 9, as shown in fig 7.10 (b)
Procedure to model a structure in SAP2000 Nonlinear
Phase I) Pre-processing
Phase II) Analysis
Phase III) Post-processing
Phase I) Pre-processing
• Setting up the model Geometry
o Choose units: kip-ft
o File > New Model > Coordinate System Definition > Cartesian > Number of grid >
Grid Spacing > Ok.
o Draw > Edit Grid (to change the grid) > Ok.
o Select the “Snap” icon to snap to point, midpoint, etc.
o Draw > Draw frame element
o Select All > Edit > Change Label > Re-label Selected items > Ok.
•
Define Material
o Choose units: kip-in
o Concrete > Click Modify>Type of Material>Isotropic>Analysis Property data >Ok.
•
Define Frame Section
o Choose units: kip-in
8/13
SAP2000 Tutorial
CES 6116 Finite Element Analysis
CEE-UCF
o Define > Frame Section > Import I section > c:\sap200n\sections.pro > choose
W12x40 > Check section property > Moment of inertia about 3 > Check material,
Steel >> Ok.
o Repeat the above step if ‘Moment of Inertia is varying’ (as in problem 1).
o Select the member > Assign > frame > Section.
o View > Set Element.
•
Assign Joint Restrains
o Select joints to which you want to assign restrains.
o J1, J3 > Fast Restrains > Fixed
o J2 > Fast Restrains > Free
•
Assign Loads
o Choose units: kip-ft
o Select members (frame) to which you want to assign loads.
o Select M1 > Assign > Frame Static Load > Point and uniform > Load Case Name >
Load 1 > Load Type and Direction > Forces > Gravity > Point Load > Distance 0.5
> Input load value > Options > Ok.
o Select M2 > Assign > Frame Static Load > Point and uniform > Load Case Name >
Load 1 > Load Type and Direction > Forces > Gravity > Uniform Load > Input
load value > Ok.
o Select J2 > Assign > Joint Static Load > Forces > Load Case Name > Load 1 >
Moment Global YY > Input load value (+ ve Moment Clockwise) > Ok.
To settlement of 1 in at joint J1 (only for problem # 7.4)
o Choose units: kip-in
o Select J1 > Assign joint spring > Ok
o Assign > Joint Static Loads > Displacement > Ground Displacement > Input > Ok
Phase II) Analysis
•
Analyze > Run.
o Analysis Complete > Check for Error / Warning > Ok
Phase III) Post-Processing
•
Check deflected shape > Ok.
o Display > Show Deformed Shape
o Display > Show Element forces / stresses > Joints > Joint Reaction Forces > Load 1
Load Case > Reactions > Ok > 3D (to view the moments at supports)
o Display > Show Element forces / stresses > Frames > Member Force Diagram for
Frames > Load 1 Load Case > Shear 2-2 > Scaling > Auto > Show values on
Diagram > Ok
o Display > Show Element forces / stresses > Frames > Member Force Diagram for
Frames > Load 1 Load Case > Moment 3-3 > Scaling > Auto > Show values on
Diagram > Ok
To view the output table
o File > Print Output Table > Type of Analysis Result > Select Load Case > Print to
File > File Name > Ok (it will create output as a text file)
o Option > Windows > Four (all results)
9/13
SAP2000 Tutorial
CES 6116 Finite Element Analysis
Problem # 2 (kip-in)
Joint Displacement
10/13
CEE-UCF
SAP2000 Tutorial
CES 6116 Finite Element Analysis
Problem # 3 kip-in)
Joint displacement
11/13
CEE-UCF
SAP2000 Tutorial
CES 6116 Finite Element Analysis
CEE-UCF
PROBLEM # 2 (RESULTS)
(KIP-in UNITS)
JOINT DISPLACEMENTS
JOINT LOAD
U1
U2
U3
R1
R2
R3
J1 LOAD1
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
J2 LOAD1
0.0212
0.0000
-0.0671
0.0000 2.658E-03
0.0000
J3 LOAD1
0.0000
0.0000
0.0000
0.0000
0.0000
F2
F3
0.0000
JOINT REACTIONS
JOINT LOAD
F1
M1
M2
M3
J1 LOAD1
30.2943
0.0000 102.1244
0.0000
-1211.0766
0.0000
J3 LOAD1
-30.2943
0.0000
17.8756
0.0000
845.5204
0.0000
V2
V3
T
-104.92
-24.36
-24.36
-18.40
21.71
21.71
0.00
0.00
0.00
0.00
0.00
0.00
0.00 -1211.08
0.00 1254.97
0.00 -1655.38
-30.29
-30.29
-30.29
-30.29
-30.29
-12.12
-4.62
2.88
10.38
17.88
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
FRAME ELEMENT FORCES
FRAME
M1
LOAD LOC
M2
M3
LOAD1
0.00
134.03
268.05
M2
P
LOAD1
0.00
60.00
120.00
180.00
240.00
12/13
-155.38
347.08
399.55
2.01
-845.52
SAP2000 Tutorial
CES 6116 Finite Element Analysis
CEE-UCF
PROBLEM # 3 (RESULTS)
(KIP-in UNITS)
JOINT DISPLACEMENTS
JOINT LOAD
U1
U2
U3
R1
R2
R3
J1 LOAD1
0.0000
0.0000
-1.0000
0.0000
0.0000
0.0000
J2 LOAD1
0.0179
0.0000
-1.0603
0.0000 -6.061E-04
0.0000
J3 LOAD1
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
JOINT REACTIONS
JOINT LOAD
F1
F2
F3
M1
M2
M3
J1 LOAD1
25.5745
0.0000
97.6440
0.0000 -1419.4026
0.0000
J3 LOAD1
-25.5745
0.0000
22.3560
0.0000 1505.4177
0.0000
P
V2
V3
T
M2
M3
0.00
134.16
268.33
-98.77
-18.27
-18.27
-20.79
19.46
19.46
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
-1419.40
1370.31
-1239.99
0.00
60.00
120.00
180.00
240.00
-25.57
-7.64
-25.57 -1.440E-01
-25.57
7.36
-25.57
14.86
-25.57
22.36
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
260.01
493.66
277.30
-389.06
-1505.42
FRAME ELEMENT FORCES
FRAME
LOAD
M1
LOAD1
M2
LOC
LOAD1
13/13