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 6/13 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