sap2000 handout-ces6116.pdf

8/16/2019 SAP2000 Handout-CES6116.pdf

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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 PerezSwapnil Chogle


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SAP2000 Tutorial CES 6116 Finite Element Analysis CEE-UCF

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Problem # 1 (Example 5.7, pg 219 of Textbook)

Determine the joint displacements, member end forces, and support reactions for the beamshown 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) AnalysisPhase III) Post-processing

Phase I) Pre-processing

• Setting up the model Geometryo Choose units: KN-mo 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 elemento

Select All > Edit > Change Label > Re-label Selected items > Ok.

• Define Materialo Choose units: N-mm


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o Concrete > Click Modify>Type of Material>Isotropic>Analysis Property data >Ok.

• Define Frame Sectiono Choose units: KN-mmo 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.

• Assign Joint Restrainso Select joints to which you want to assign restrains.o J1 > Fast Restrains > Fixedo

J2 > U2, R1, R3o

J3, J5 > U2, U3, R1, R3o

J4 > Fast Restrains > Free

• Assign Loadso 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 loadvalue > 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 onDiagram > Ok

o Display > Show Element forces / stresses > Frames > Member Force Diagram forFrames > Load 1 Load Case > Moment 3-3 > Scaling > Auto > Show values onDiagram > OkTo 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)


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Problem # 1 (KN-m)

Joint displacement


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PROBLEM # 1 (INPUT DATA)(KN-m UNITS)

S T A T I C L O A D C A S E S

STATIC CASE SELF WT

CASE TYPE FACTOR

LOAD1 DEAD 1.0000

J O I N T D A T A

JOINT GLOBAL-X GLOBAL-Y GLOBAL-Z RESTRAINTS ANGLE-A ANGLE-B ANGLE-C

J1 0.00000 0.50000 0.00000 1 1 1 1 1 1 0.000 0.000 0.000J2 6.00000 0.50000 0.00000 0 1 0 1 0 1 0.000 0.000 0.000

J3 10.00000 0.50000 0.00000 0 1 1 1 0 1 0.000 0.000 0.000J4 15.00000 0.50000 0.00000 0 0 0 0 0 0 0.000 0.000 0.000J5 20.00000 0.50000 0.00000 0 1 1 1 0 1 0.000 0.000 0.000

F R A M E E L E M E N T D A T A

FRAME JNT-1 JNT-2 SECTION ANGLE RELEASES SEGMENTS R1 R2 FACTOR LENGTH

M1 J1 J2 CONCA 0.000 000000 4 0.000 0.000 1.000 6.000M2 J2 J3 CONCB 0.000 000000 4 0.000 0.000 1.000 4.000

M3 J3 J4 CONCB 0.000 000000 4 0.000 0.000 1.000 5.000M4 J4 J5 CONCB 0.000 000000 4 0.000 0.000 1.000 5.000

M A T E R I A L P R O P E R T Y D A T A

MAT MODULUS OF POISSON'S THERMAL WEIGHT PER MASS PER LABEL ELASTICITY RATIO COEFF UNIT VOL UNIT VOL

CONC 28000000.0 0.200 0.000 0.000 0.000

F R A M E S E C T I O N P R O P E R T Y D A T A

SECTION MAT SECTION DEPTH FLANGE FLANGE WEB FLANGE

FLANGE LABEL LABEL TYPE WIDTH THICK THICK WIDTH THICK

TOP TOP BOTTOM BOTTOMCONCA CONC 0.685 0.325 0.000 0.000 0.000 0.000

CONCB CONC 0.600 0.325 0.000 0.000 0.000 0.000


SECTION AREA TORSIONAL MOMENTS OF INERTIA SHEAR AREAS

LABEL INERTIA I33 I22 A2 A3

CONCA 0.223 5.505E-03 8.705E-03 1.960E-03 0.186 0.186

CONCB 0.195 4.540E-03 5.850E-03 1.716E-03 0.163 0.163


SECTION SECTION MODULII PLASTIC MODULII RADII OF GYRATION LABEL S33 S22 Z33 Z22 R33 R22

CONCA 2.542E-02 1.206E-02 3.812E-02 1.809E-02 0.198 9.382E-02CONCB 1.950E-02 1.056E-02 2.925E-02 1.584E-02 0.173 9.382E-02


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SECTION TOTAL TOTAL LABEL WEIGHT MASS

CONCA 0.000 0.000CONCB 0.000 0.000

S H E L L S E C T I O N P R O P E R T Y D A T A

SECTION MAT SHELL MEMBRANE BENDING MATERIAL LABEL LABEL TYPE THICK THICK ANGLE

SSEC1 CONC 4 1.000E-03 1.000E-03 0.000

S H E L L S E C T I O N P R O P E R T Y D A T A

SECTION TOTAL TOTAL LABEL WEIGHT MASS

SSEC1 0.000 0.000

J O I N T F O R C E S Load Case LOAD1

JOINT GLOBAL-X GLOBAL-Y GLOBAL-Z GLOBAL-XX GLOBAL-YY GLOBAL-ZZ

J3 0.000 0.000 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 VALUE-B

M1 FORCE GLOBAL-Z 0.0000 -30.0000 0.2500 -22.5000M1 FORCE GLOBAL-Z 0.2500 -22.5000 0.7500 -7.5000

M1 FORCE GLOBAL-Z 0.7500 -7.5000 1.0000 0.0000

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 FORCE GLOBAL-Z 0.0000 -200.0000

M3 FORCE GLOBAL-Z 1.0000 -150.0000


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PROBLEM #1 (RESULTS)

(KN-mm Units)

J O I N T D I S P L A C E M E N T S

JOINT LOAD U1 U2 U3 R1 R2 R3

J1 LOAD1 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

J2 LOAD1 0.0000 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

J O I N T R E A C T I O N S

JOINT LOAD F1 F2 F3 M1 M2 M3

J1 LOAD1 0.0000 0.0000 146.4303 0.0000 -281767.375 0.0000

J2 LOAD1 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

J3 LOAD1 0.0000 0.0000 243.3161 0.0000 0.0000 0.0000

J5 LOAD1 0.0000 0.0000 50.2536 0.0000 0.0000 0.0000

F R A M E E L E M E N T F O R C E S

FRAME LOAD LOC P V2 V3 T M2 M3

M1 LOAD1

0.00 0.00 -146.43 0.00 0.00 0.00 -281767.381500.00 0.00 -107.06 0.00 0.00 0.00 -93059.413000.00 0.00 -78.93 0.00 0.00 0.00 45023.58

4500.00 0.00 -62.06 0.00 0.00 0.00 149356.56

6000.00 0.00 -56.43 0.00 0.00 0.00 236814.56

M2 LOAD10.00 0.00 -56.43 0.00 0.00 0.00 236814.56

1000.00 0.00 143.57 0.00 0.00 0.00 93244.882000.00 0.00 143.57 0.00 0.00 0.00 -50324.80

3000.00 0.00 143.57 0.00 0.00 0.00 -193894.474000.00 0.00 143.57 0.00 0.00 0.00 -337464.16

M3 LOAD1

0.00 0.00 -99.75 0.00 0.00 0.00 -247464.14

1250.00 0.00 -99.75 0.00 0.00 0.00 -122781.132500.00 0.00 -99.75 0.00 0.00 0.00 1901.89

3750.00 0.00 -99.75 0.00 0.00 0.00 126584.91

5000.00 0.00 50.25 0.00 0.00 0.00 251267.92

M4 LOAD1

0.00 0.00 50.25 0.00 0.00 0.00 251267.921250.00 0.00 50.25 0.00 0.00 0.00 188450.94

2500.00 0.00 50.25 0.00 0.00 0.00 125633.963750.00 0.00 50.25 0.00 0.00 0.00 62816.98

5000.00 0.00 50.25 0.00 0.00 0.00 0.00


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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 planeframe shown in fig 7.10 (a), due to the combined effect of the loading shown andsettlement 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 – thetranslations 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 numbers4 through 9, as shown in fig 7.10 (b)

Procedure to model a structure in SAP2000 Nonlinear

Phase I) Pre-processingPhase II) Analysis

Phase III) Post-processing

Phase I) Pre-processing• Setting up the model Geometryo

Choose units: kip-fto 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 elemento Select All > Edit > Change Label > Re-label Selected items > Ok.

• Define Materialo Choose units: kip-ino Concrete > Click Modify>Type of Material>Isotropic>Analysis Property data >Ok.

• Define Frame Sectiono Choose units: kip-in


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o Define > Frame Section > Import I section > c:\sap200n\sections.pro > chooseW12x40 > 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 Restrainso Select joints to which you want to assign restrains.o J1, J3 > Fast Restrains > Fixedo J2 > Fast Restrains > Free

• Assign Loadso

Choose units: kip-fto

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 > Inputload 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 forFrames > Load 1 Load Case > Shear 2-2 > Scaling > Auto > Show values onDiagram > Ok

o Display > Show Element forces / stresses > Frames > Member Force Diagram forFrames > Load 1 Load Case > Moment 3-3 > Scaling > Auto > Show values onDiagram > OkTo 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)


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Problem # 2 (kip-in)

Joint Displacement


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Problem # 3 kip-in)

Joint displacement


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PROBLEM # 2 (RESULTS)

(KIP-in UNITS)



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 0.0000



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



M1 LOAD10.00 -104.92 -18.40 0.00 0.00 0.00 -1211.08

134.03 -24.36 21.71 0.00 0.00 0.00 1254.97

268.05 -24.36 21.71 0.00 0.00 0.00 -1655.38

M2 LOAD10.00 -30.29 -12.12 0.00 0.00 0.00 -155.38

60.00 -30.29 -4.62 0.00 0.00 0.00 347.08

120.00 -30.29 2.88 0.00 0.00 0.00 399.55180.00 -30.29 10.38 0.00 0.00 0.00 2.01

240.00 -30.29 17.88 0.00 0.00 0.00 -845.52


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PROBLEM # 3 (RESULTS)(KIP-in UNITS)



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



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



M1 LOAD1

0.00 -98.77 -20.79 0.00 0.00 0.00 -1419.40134.16 -18.27 19.46 0.00 0.00 0.00 1370.31268.33 -18.27 19.46 0.00 0.00 0.00 -1239.99

M2 LOAD1

0.00 -25.57 -7.64 0.00 0.00 0.00 260.0160.00 -25.57 -1.440E-01 0.00 0.00 0.00 493.66

120.00 -25.57 7.36 0.00 0.00 0.00 277.30

180.00 -25.57 14.86 0.00 0.00 0.00 -389.06240.00 -25.57 22.36 0.00 0.00 0.00 -1505.42

sap2000 handout-ces6116.pdf

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