EX. US-24 Analysis of a Concrete Block Using Solid Elements
This is an example of the analysis of a structure modeled using solid finite elements. This example also illustrates the method for applying an enforced displacement on the structure.
This problem is installed with the program by default to C:\Users\Public\Public Documents\STAAD.Pro CONNECT Edition\Samples\Sample Models\US\US-24 Analysis of a Concrete Block Using Solid Elements.STD when you install the program.
STAAD SPACE
*EXAMPLE PROBLEM USING SOLID ELEMENTS
Every STAAD input file has to begin with the word STAAD. The word SPACE signifies that the structure is a space frame and the geometry is defined through X, Y, and Z axes. The comment line which begins with an asterisk is an optional title to identify this project.
UNIT KNS MET
The units for the data that follows are specified above.
JOINT COORDINATES
1 0.0 0.0 2.0 4 0.0 3.0 2.0
5 1.0 0.0 2.0 8 1.0 3.0 2.0
9 2.0 0.0 2.0 12 2.0 3.0 2.0
21 0.0 0.0 1.0 24 0.0 3.0 1.0
25 1.0 0.0 1.0 28 1.0 3.0 1.0
29 2.0 0.0 1.0 32 2.0 3.0 1.0
41 0.0 0.0 0.0 44 0.0 3.0 0.0
45 1.0 0.0 0.0 48 1.0 3.0 0.0
49 2.0 0.0 0.0 52 2.0 3.0 0.0
The joint number followed by the X, Y, and Z coordinates are specified above. The coordinates of some of those nodes are generated utilizing the fact that they are equally spaced between the extremities.
ELEMENT INCIDENCES SOLID
1 1 5 6 2 21 25 26 22 TO 3
4 21 25 26 22 41 45 46 42 TO 6 1 1
7 5 9 10 6 25 29 30 26 TO 9 1 1
10 25 29 30 26 45 49 50 46 TO 12 1 1
The incidences of solid elements are defined above. The word SOLID is used to signify that these are 8-node solid elements as opposed to 3-noded or 4-noded plate elements. Each line contains the data for generating 3 elements. For example, element number 1 is first defined by all of its 8 nodes. Then, increments of 1 to the joint number and 1 to the element number (the defaults) are used for generating incidences for elements 2 and 3. Similarly, incidences of elements 4, 7 and 10 are defined while those of 5, 6, 8, 9, 11 and 12 are generated.
UNIT MMS
DEFINE MATERIAL START
ISOTROPIC STEEL
E 210
POISSON 0.25
DENSITY 7.5e-008
ALPHA 6e-006
DAMP 0.03
TYPE STEEL
STRENGTH FY 0.25 FU 0.4 RY 1.5 RT 1.2
END DEFINE MATERIAL
CONSTANTS
MATERIAL STEEL ALL
UNIT METERL
The DEFINE MATERIAL command is used to specify material properties and the CONSTANT is used to assign the material to all members.
PRINT ELEMENT INFO SOLID LIST 1 TO 5
This command will enable us to obtain, in a tabular form, the details of the incidences and material property values of elements 1 to 5.
SUPPORTS
1 5 21 25 29 41 45 49 PINNED
9 ENFORCED
The above lines contain the data for supports for the model. The ENFORCED support condition is used to declare a point at which an enforced displacement load is applied later (see load case 3).
LOAD 1
SELF Y -1.0
JOINT LOAD
28 FY -1000.0
The above data describe a static load case. It consists of selfweight loading and a joint load, both in the negative global Y direction.
LOAD 2
JOINT LOADS
2 TO 4 22 TO 24 42 TO 44 FX 100.0
Load case 2 consists of several joint loads acting in the positive global X direction.
LOAD 3
SUPPORT DISPLACEMENT
9 FX 0.0011
Load case 3 consists of an enforced displacement along the global X direction at node 9. The displacement in the other enforced support degrees of freedom will default to zero.
UNIT POUND FEET
LOAD 4
ELEMENT LOAD SOLIDS
3 6 9 12 FACE 4 PRE GY -500.0
In Load case 4, a pressure load of 500 pounds/sq.ft is applied on Face # 4 of solid elements 3, 6, 9 and 12. Face 4 is defined as shown in the following table :
|
Face Number |
Surface Joints | |||
|---|---|---|---|---|
| f1 | f2 | f3 | f4 | |
|
1 front |
Jt 1 |
Jt 4 |
Jt 3 |
Jt 2 |
|
2 bottom |
Jt 1 |
Jt 2 |
Jt 6 |
Jt 5 |
|
3 left |
Jt 1 |
Jt 5 |
Jt 8 |
Jt 4 |
|
4 top |
Jt 4 |
Jt 8 |
Jt 7 |
Jt 3 |
|
5 right |
Jt 2 |
Jt 3 |
Jt 7 |
Jt 6 |
|
6 back |
Jt 5 |
Jt 6 |
Jt 7 |
Jt 8 |
The above table, and other details of this type of loading can be found in TR.32.3.2 Element Load Specification - Solids .
UNIT KNS MMS
LOAD 5
REPEAT LOAD
1 1.0 2 1.0 3 1.0 4 1.0
Load case 5 illustrates the technique employed to instruct STAAD to create a load case which consists of data to be assembled from other load cases already specified earlier. We want the program to analyze the structure for loads from cases 1 through 4 acting simultaneously. In other words, the above instruction is the same as the following:
LOAD 5
SELF Y -1.0
JOINT LOAD
28 FY -1000.0
2 TO 4 22 TO 24 42 TO 44 FX 100.0
SUPPORT DISPLACEMENT
9 FX .0011
ELEMENT LOAD SOLIDS
3 6 9 12 FACE 4 PRE GY -500.0
LOAD COMB 10
1 1.0 2 1.0
Load case 10 is a combination load case, which combines the effects of cases 1 & 2. While the syntax of this might look very similar to that of the REPEAT LOAD case shown in case 5, there is a fundamental difference. In a REPEAT LOAD case, the program computes the displacements by multiplying the inverted stiffness matrix by the load vector built for the REPEAT LOAD case. But in solving load combination cases, the program merely calculates the end results (displacements, forces, reactions) by gathering up the corresponding values from the individual components of the combination case, factoring them, and then algebraically summing them up. This difference in approach is quite important in that non-linear problems such as PDELTA ANALYSIS, MEMBER TENSION, and MEMBER COMPRESSION situations, changes in support conditions etc. should be handled using REPEAT LOAD cases, not load combination cases.
PERFORM ANALYSIS PRINT STATICS CHECK
A static equilibrium report, consisting of total applied loading and total support reactions from each primary load case is requested along with the instructions to carry out a linear static analysis.
PRINT JOINT DISPLACEMENTS LIST 8 9
Global displacements at nodes 8 and 9 are obtained using the above command.
UNIT KNS METER
PRINT SUPPORT REACTIONS
Reactions at the supports are obtained using the above command.
UNIT NEWTON MMS
PRINT ELEMENT JOINT STRESS SOLID LIST 4 6
This command requests the program to provide the element stress results at the nodes of elements 4 and 6. The results will be printed for all the load cases. The word SOLID is used to signify that these are solid elements as opposed to plate or shell elements.
FINISH
The STAAD run is terminated.
Input File
STAAD SPACE EXAMPLE PROBLEM USING SOLID ELEMENTS
UNIT KNS MET
JOINT COORDINATES
1 0.0 0.0 2.0 4 0.0 3.0 2.0
5 1.0 0.0 2.0 8 1.0 3.0 2.0
9 2.0 0.0 2.0 12 2.0 3.0 2.0
21 0.0 0.0 1.0 24 0.0 3.0 1.0
25 1.0 0.0 1.0 28 1.0 3.0 1.0
29 2.0 0.0 1.0 32 2.0 3.0 1.0
41 0.0 0.0 0.0 44 0.0 3.0 0.0
45 1.0 0.0 0.0 48 1.0 3.0 0.0
49 2.0 0.0 0.0 52 2.0 3.0 0.0
ELEMENT INCIDENCES SOLID
1 1 5 6 2 21 25 26 22 TO 3
4 21 25 26 22 41 45 46 42 TO 6 1 1
7 5 9 10 6 25 29 30 26 TO 9 1 1
10 25 29 30 26 45 49 50 46 TO 12 1 1
UNIT MMS
DEFINE MATERIAL START
ISOTROPIC STEEL
E 210
POISSON 0.25
DENSITY 7.5e-008
ALPHA 6e-006
DAMP 0.03
TYPE STEEL
STRENGTH FY 0.25 FU 0.4 RY 1.5 RT 1.2
END DEFINE MATERIAL
CONSTANTS
MATERIAL STEEL ALL
UNIT METER
PRINT ELEMENT INFO SOLID LIST 1 TO 5
SUPPORTS
1 5 21 25 29 41 45 49 PINNED
9 ENFORCED BUT MX MY MZ
LOAD 1
SELF Y -1.0
JOINT LOAD
28 FY -1000.0
LOAD 2
JOINT LOADS
2 TO 4 22 TO 24 42 TO 44 FX 100.0
LOAD 3
SUPPORT DISPLACEMENT
9 FX .0011
UNIT POUND FEET
LOAD 4
ELEMENT LOAD SOLIDS
3 6 9 12 FACE 4 PRE GY -500.0
UNIT KNS MMS
LOAD 5
REPEAT LOAD
1 1.0 2 1.0 3 1.0 4 1.0
LOAD COMB 10
1 1.0 2 1.0
PERFORM ANALYSIS PRINT STAT CHECK
PRINT JOINT DISPLACEMENTS LIST 8 9
UNIT KNS METER
PRINT SUPPORT REACTIONS
UNIT NEWTON MMS
PRINT ELEMENT JOINT STRESS SOLID LIST 4 6
FINISH
STAAD Output File
PAGE NO. 1 **************************************************** * * * STAAD.Pro CONNECT Edition * * Version 22.12.00.*** * * Proprietary Program of * * Bentley Systems, Inc. * * Date= OCT 27, 2022 * * Time= 15: 9:43 * * * * Licensed to: Bentley Systems Inc * **************************************************** 1. STAAD SPACE EXAMPLE PROBLEM USING SOLID ELEMENTS INPUT FILE: US-24 Analysis of a Concrete Block Using Solid Elements.STD 2. UNIT KNS MET 3. JOINT COORDINATES 4. 1 0.0 0.0 2.0 4 0.0 3.0 2.0 5. 5 1.0 0.0 2.0 8 1.0 3.0 2.0 6. 9 2.0 0.0 2.0 12 2.0 3.0 2.0 7. 21 0.0 0.0 1.0 24 0.0 3.0 1.0 8. 25 1.0 0.0 1.0 28 1.0 3.0 1.0 9. 29 2.0 0.0 1.0 32 2.0 3.0 1.0 10. 41 0.0 0.0 0.0 44 0.0 3.0 0.0 11. 45 1.0 0.0 0.0 48 1.0 3.0 0.0 12. 49 2.0 0.0 0.0 52 2.0 3.0 0.0 13. ELEMENT INCIDENCES SOLID 14. 1 1 5 6 2 21 25 26 22 TO 3 15. 4 21 25 26 22 41 45 46 42 TO 6 1 1 16. 7 5 9 10 6 25 29 30 26 TO 9 1 1 17. 10 25 29 30 26 45 49 50 46 TO 12 1 1 18. UNIT MMS 19. DEFINE MATERIAL START 20. ISOTROPIC STEEL 21. E 210 22. POISSON 0.25 23. DENSITY 7.5E-008 24. ALPHA 6E-006 25. DAMP 0.03 26. TYPE STEEL 27. STRENGTH FY 0.25 FU 0.4 RY 1.5 RT 1.2 28. END DEFINE MATERIAL 29. CONSTANTS 30. MATERIAL STEEL ALL 31. UNIT METER 32. PRINT ELEMENT INFO SOLID LIST 1 TO 5 ELEMENT INFO SOLID LIST EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 2 ELEMENT NODE-1 NODE-2 NODE-3 NODE-4 NODE-5 NODE-6 NODE-7 NODE-8 1 1 5 6 2 21 25 26 22 2 2 6 7 3 22 26 27 23 3 3 7 8 4 23 27 28 24 4 21 25 26 22 41 45 46 42 5 22 26 27 23 42 46 47 43 EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 3 MATERIAL PROPERTIES. -------------------- ALL UNITS ARE - KNS METE ELEMENT YOUNG'S MODULUS MODULUS OF RIGIDITY DENSITY ALPHA 1 2.1000002E+08 0.0000000E+00 7.5000E+01 6.0000E-06 2 2.1000002E+08 0.0000000E+00 7.5000E+01 6.0000E-06 3 2.1000002E+08 0.0000000E+00 7.5000E+01 6.0000E-06 4 2.1000002E+08 0.0000000E+00 7.5000E+01 6.0000E-06 5 2.1000002E+08 0.0000000E+00 7.5000E+01 6.0000E-06 33. SUPPORTS 34. 1 5 21 25 29 41 45 49 PINNED 35. 9 ENFORCED BUT MX MY MZ 36. LOAD 1 37. SELF Y -1.0 38. JOINT LOAD 39. 28 FY -1000.0 40. LOAD 2 41. JOINT LOADS 42. 2 TO 4 22 TO 24 42 TO 44 FX 100.0 43. LOAD 3 44. SUPPORT DISPLACEMENT 45. 9 FX .0011 46. UNIT POUND FEET 47. LOAD 4 48. ELEMENT LOAD SOLIDS 49. 3 6 9 12 FACE 4 PRE GY -500.0 50. UNIT KNS MMS 51. LOAD 5 52. REPEAT LOAD 53. 1 1.0 2 1.0 3 1.0 4 1.0 54. LOAD COMB 10 55. 1 1.0 2 1.0 56. PERFORM ANALYSIS PRINT STAT CHECK P R O B L E M S T A T I S T I C S ----------------------------------- NUMBER OF JOINTS 36 NUMBER OF MEMBERS 0 NUMBER OF PLATES 0 NUMBER OF SOLIDS 12 NUMBER OF SURFACES 0 NUMBER OF SUPPORTS 9 EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 4 Using 64-bit analysis engine. SOLVER USED IS THE IN-CORE ADVANCED MATH SOLVER TOTAL PRIMARY LOAD CASES = 5, TOTAL DEGREES OF FREEDOM = 84 TOTAL LOAD COMBINATION CASES = 1 SO FAR. EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 5 STATIC LOAD/REACTION/EQUILIBRIUM SUMMARY FOR CASE NO. 1 CENTER OF FORCE BASED ON Y FORCES ONLY (MMS ). (FORCES IN NON-GLOBAL DIRECTIONS WILL INVALIDATE RESULTS) X = 0.999999993E+03 Y = 0.228947364E+04 Z = 0.999999993E+03 TOTAL APPLIED LOAD 1 ***TOTAL APPLIED LOAD ( KNS MMS ) SUMMARY (LOADING 1 ) SUMMATION FORCE-X = 0.00 SUMMATION FORCE-Y = -1900.00 SUMMATION FORCE-Z = 0.00 SUMMATION OF MOMENTS AROUND THE ORIGIN- MX= 1900000.15 MY= 0.00 MZ= -1900000.15 TOTAL REACTION LOAD 1 ***TOTAL REACTION LOAD( KNS MMS ) SUMMARY (LOADING 1 ) SUMMATION FORCE-X = 0.00 SUMMATION FORCE-Y = 1900.00 SUMMATION FORCE-Z = 0.00 SUMMATION OF MOMENTS AROUND THE ORIGIN- MX= -1900000.15 MY= -0.00 MZ= 1900000.15 MAXIMUM DISPLACEMENTS ( CM /RADIANS) (LOADING 1) MAXIMUMS AT NODE X = -1.21106E-04 23 Y = -1.15439E-03 28 Z = 1.21106E-04 7 RX= 0.00000E+00 0 RY= 0.00000E+00 0 RZ= 0.00000E+00 0 STATIC LOAD/REACTION/EQUILIBRIUM SUMMARY FOR CASE NO. 2 CENTER OF FORCE BASED ON X FORCES ONLY (MMS ). (FORCES IN NON-GLOBAL DIRECTIONS WILL INVALIDATE RESULTS) X = 0.000000000E+00 Y = 0.199999999E+04 Z = 0.999999993E+03 EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 6 TOTAL APPLIED LOAD 2 ***TOTAL APPLIED LOAD ( KNS MMS ) SUMMARY (LOADING 2 ) SUMMATION FORCE-X = 900.00 SUMMATION FORCE-Y = 0.00 SUMMATION FORCE-Z = 0.00 SUMMATION OF MOMENTS AROUND THE ORIGIN- MX= 0.00 MY= 900000.03 MZ= -1800000.06 TOTAL REACTION LOAD 2 ***TOTAL REACTION LOAD( KNS MMS ) SUMMARY (LOADING 2 ) SUMMATION FORCE-X = -900.00 SUMMATION FORCE-Y = -0.00 SUMMATION FORCE-Z = 0.00 SUMMATION OF MOMENTS AROUND THE ORIGIN- MX= 0.00 MY= -900000.03 MZ= 1800000.06 MAXIMUM DISPLACEMENTS ( CM /RADIANS) (LOADING 2) MAXIMUMS AT NODE X = 2.22892E-03 4 Y = 7.83934E-04 44 Z = 9.49033E-05 10 RX= 0.00000E+00 0 RY= 0.00000E+00 0 RZ= 0.00000E+00 0 STATIC LOAD/REACTION/EQUILIBRIUM SUMMARY FOR CASE NO. 3 TOTAL APPLIED LOAD 3 ***TOTAL APPLIED LOAD ( KNS MMS ) SUMMARY (LOADING 3 ) SUMMATION FORCE-X = 0.0000000E+00 SUMMATION FORCE-Y = 0.0000000E+00 SUMMATION FORCE-Z = 0.0000000E+00 SUMMATION OF MOMENTS AROUND THE ORIGIN- MX= 0.0000000E+00 MY= 0.0000000E+00 MZ= 0.0000000E+00 TOTAL REACTION LOAD 3 ***TOTAL REACTION LOAD( KNS MMS ) SUMMARY (LOADING 3 ) SUMMATION FORCE-X = 1.6182536E-11 SUMMATION FORCE-Y = -5.0570426E-12 SUMMATION FORCE-Z = 2.6296621E-11 SUMMATION OF MOMENTS AROUND THE ORIGIN- MX= 5.5900954E-08 MY= 3.9459497E-08 MZ= -3.2882914E-09 EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 7 MAXIMUM DISPLACEMENTS ( CM /RADIANS) (LOADING 3) MAXIMUMS AT NODE X = 1.10000E-01 9 Y = -1.21497E-02 6 Z = 1.61372E-02 24 RX= 0.00000E+00 0 RY= 0.00000E+00 0 RZ= 0.00000E+00 0 STATIC LOAD/REACTION/EQUILIBRIUM SUMMARY FOR CASE NO. 4 CENTER OF FORCE BASED ON Y FORCES ONLY (MMS ). (FORCES IN NON-GLOBAL DIRECTIONS WILL INVALIDATE RESULTS) X = 0.999999993E+03 Y = 0.299999998E+04 Z = 0.999999993E+03 TOTAL APPLIED LOAD 4 ***TOTAL APPLIED LOAD ( KNS MMS ) SUMMARY (LOADING 4 ) SUMMATION FORCE-X = 0.00 SUMMATION FORCE-Y = -95.76 SUMMATION FORCE-Z = 0.00 SUMMATION OF MOMENTS AROUND THE ORIGIN- MX= 95760.52 MY= 0.00 MZ= -95760.52 TOTAL REACTION LOAD 4 ***TOTAL REACTION LOAD( KNS MMS ) SUMMARY (LOADING 4 ) SUMMATION FORCE-X = 0.00 SUMMATION FORCE-Y = 95.76 SUMMATION FORCE-Z = 0.00 SUMMATION OF MOMENTS AROUND THE ORIGIN- MX= -95760.52 MY= 0.00 MZ= 95760.52 MAXIMUM DISPLACEMENTS ( CM /RADIANS) (LOADING 4) MAXIMUMS AT NODE X = 3.17652E-06 50 Y = -3.35288E-05 28 Z = -3.17652E-06 50 RX= 0.00000E+00 0 RY= 0.00000E+00 0 RZ= 0.00000E+00 0 STATIC LOAD/REACTION/EQUILIBRIUM SUMMARY FOR CASE NO. 5 EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 8 CENTER OF FORCE BASED ON X FORCES ONLY (MMS ). (FORCES IN NON-GLOBAL DIRECTIONS WILL INVALIDATE RESULTS) X = 0.000000000E+00 Y = 0.199999999E+04 Z = 0.999999993E+03 CENTER OF FORCE BASED ON Y FORCES ONLY (MMS ). (FORCES IN NON-GLOBAL DIRECTIONS WILL INVALIDATE RESULTS) X = 0.999999993E+03 Y = 0.232356609E+04 Z = 0.999999993E+03 TOTAL APPLIED LOAD 5 ***TOTAL APPLIED LOAD ( KNS MMS ) SUMMARY (LOADING 5 ) SUMMATION FORCE-X = 900.00 SUMMATION FORCE-Y = -1995.76 SUMMATION FORCE-Z = 0.00 SUMMATION OF MOMENTS AROUND THE ORIGIN- MX= 1995760.67 MY= 900000.03 MZ= -3795760.73 TOTAL REACTION LOAD 5 ***TOTAL REACTION LOAD( KNS MMS ) SUMMARY (LOADING 5 ) SUMMATION FORCE-X = -900.00 SUMMATION FORCE-Y = 1995.76 SUMMATION FORCE-Z = 0.00 SUMMATION OF MOMENTS AROUND THE ORIGIN- MX= -1995760.67 MY= -900000.03 MZ= 3795760.73 MAXIMUM DISPLACEMENTS ( CM /RADIANS) (LOADING 5) MAXIMUMS AT NODE X = 1.10000E-01 9 Y = -1.23568E-02 6 Z = 1.61372E-02 24 RX= 0.00000E+00 0 RY= 0.00000E+00 0 RZ= 0.00000E+00 0 ************ END OF DATA FROM INTERNAL STORAGE ************ 57. PRINT JOINT DISPLACEMENTS LIST 8 9 JOINT DISPLACE LIST 8 EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 9 JOINT DISPLACEMENT (CM RADIANS) STRUCTURE TYPE = SPACE ------------------ JOINT LOAD X-TRANS Y-TRANS Z-TRANS X-ROTAN Y-ROTAN Z-ROTAN 8 1 0.0000 -0.0002 -0.0001 0.0000 0.0000 0.0000 2 0.0020 0.0000 -0.0000 0.0000 0.0000 0.0000 3 0.0193 -0.0049 0.0089 0.0000 0.0000 0.0000 4 0.0000 -0.0000 0.0000 0.0000 0.0000 0.0000 5 0.0213 -0.0052 0.0088 0.0000 0.0000 0.0000 10 0.0020 -0.0002 -0.0001 0.0000 0.0000 0.0000 9 1 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 2 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 3 0.1100 0.0000 -0.0000 0.0000 0.0000 0.0000 4 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 5 0.1100 0.0000 -0.0000 0.0000 0.0000 0.0000 10 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 ************** END OF LATEST ANALYSIS RESULT ************** 58. UNIT KNS METER 59. PRINT SUPPORT REACTIONS SUPPORT REACTION EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 10 SUPPORT REACTIONS -UNIT KNS METE STRUCTURE TYPE = SPACE ----------------- JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z 1 1 27.47 128.97 -27.47 0.00 0.00 0.00 2 -72.24 -232.67 42.18 0.00 0.00 0.00 3 -2022.70 -302.04 -1192.39 0.00 0.00 0.00 4 1.52 6.63 -1.52 0.00 0.00 0.00 5 -2065.94 -399.11 -1179.21 0.00 0.00 0.00 10 -44.76 -103.70 14.70 0.00 0.00 0.00 5 1 0.00 236.52 -54.44 0.00 0.00 0.00 2 -62.32 11.42 -0.05 0.00 0.00 0.00 3 -16410.02 7434.80 -2287.95 0.00 0.00 0.00 4 0.00 11.97 -2.98 0.00 0.00 0.00 5 -16472.33 7694.71 -2345.41 0.00 0.00 0.00 10 -62.32 247.94 -54.49 0.00 0.00 0.00 21 1 54.44 236.52 -0.00 0.00 0.00 0.00 2 -159.92 -450.84 -0.00 0.00 0.00 0.00 3 -3341.67 -2923.60 -1877.00 0.00 0.00 0.00 4 2.98 11.97 -0.00 0.00 0.00 0.00 5 -3444.18 -3125.95 -1877.00 0.00 0.00 0.00 10 -105.49 -214.32 -0.00 0.00 0.00 0.00 25 1 -0.00 438.06 -0.00 0.00 0.00 0.00 2 -138.00 9.51 -0.00 0.00 0.00 0.00 3 -19197.98 5248.66 -10975.25 0.00 0.00 0.00 4 -0.00 21.34 -0.00 0.00 0.00 0.00 5 -19335.98 5717.57 -10975.25 0.00 0.00 0.00 10 -138.00 447.56 -0.00 0.00 0.00 0.00 29 1 -54.44 236.52 0.00 0.00 0.00 0.00 2 -170.27 431.34 0.00 0.00 0.00 0.00 3 3902.73 512.05 3842.64 0.00 0.00 0.00 4 -2.98 11.97 0.00 0.00 0.00 0.00 5 3675.05 1191.87 3842.64 0.00 0.00 0.00 10 -224.70 667.85 0.00 0.00 0.00 0.00 41 1 27.47 128.97 27.47 0.00 0.00 0.00 2 -72.24 -232.67 -42.18 0.00 0.00 0.00 3 -891.15 -2739.86 -1598.54 0.00 0.00 0.00 4 1.52 6.63 1.52 0.00 0.00 0.00 5 -934.39 -2836.93 -1611.72 0.00 0.00 0.00 10 -44.76 -103.70 -14.70 0.00 0.00 0.00 45 1 -0.00 236.52 54.44 0.00 0.00 0.00 2 -62.32 11.42 0.05 0.00 0.00 0.00 3 -430.44 -752.46 -237.57 0.00 0.00 0.00 4 -0.00 11.97 2.98 0.00 0.00 0.00 5 -492.75 -492.56 -180.10 0.00 0.00 0.00 10 -62.32 247.94 54.49 0.00 0.00 0.00 49 1 -27.47 128.97 27.47 0.00 0.00 0.00 2 -81.35 226.24 45.03 0.00 0.00 0.00 3 -778.26 2073.77 1192.39 0.00 0.00 0.00 4 -1.52 6.63 1.52 0.00 0.00 0.00 EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 11 SUPPORT REACTIONS -UNIT KNS METE STRUCTURE TYPE = SPACE ----------------- JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z 5 -888.61 2435.62 1266.41 0.00 0.00 0.00 10 -108.83 355.21 72.50 0.00 0.00 0.00 9 1 -27.47 128.97 -27.47 0.00 0.00 0.00 2 -81.35 226.24 -45.03 0.00 0.00 0.00 3 39169.49 -8551.31 13133.66 0.00 0.00 0.00 4 -1.52 6.63 -1.52 0.00 0.00 0.00 5 39059.14 -8189.46 13059.64 0.00 0.00 0.00 10 -108.83 355.21 -72.50 0.00 0.00 0.00 ************** END OF LATEST ANALYSIS RESULT ************** 60. UNIT NEWTON MMS 61. PRINT ELEMENT JOINT STRESS SOLID LIST 4 6 ELEMENT JOINT STRESS SOLID EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 12 ELEMENT STRESSES UNITS= NEWTMMS ------------------------------------------------------------------------------- NODE/ NORMAL STRESSES SHEAR STRESSES ELEMENT LOAD CENTER SXX SYY SZZ SXY SYZ SZX ------------------------------------------------------------------------------- 4 1 21 -0.144 -0.449 -0.155 -0.006 -0.011 0.000 4 1 25 -0.132 -0.368 -0.132 -0.011 -0.011 0.005 4 1 26 -0.009 -0.377 -0.009 -0.003 -0.003 0.005 4 1 22 -0.012 -0.449 -0.005 0.002 -0.018 0.009 4 1 41 -0.152 -0.484 -0.152 -0.015 -0.015 -0.005 4 1 45 -0.155 -0.449 -0.144 -0.011 -0.006 -0.000 4 1 46 -0.005 -0.449 -0.012 -0.018 0.002 0.009 4 1 42 0.007 -0.475 0.007 -0.023 -0.023 0.014 4 1 CENTER -0.075 -0.437 -0.075 -0.011 -0.011 0.005 S1= -0.070 S2= -0.080 S3= -0.438 SE= 0.363 DC= 0.707 -0.041 0.707 -0.707 -0.000 0.707 4 2 21 0.176 1.021 0.284 0.217 0.014 0.005 4 2 25 0.154 -0.006 0.022 0.251 0.014 -0.029 4 2 26 -0.028 0.053 -0.015 0.253 0.016 -0.002 4 2 22 -0.054 1.031 0.103 0.219 0.012 -0.036 4 2 41 0.189 1.034 0.321 0.258 0.038 0.029 4 2 45 0.162 -0.006 0.054 0.223 -0.010 -0.005 4 2 46 -0.225 -0.016 -0.051 0.221 -0.008 -0.026 4 2 42 -0.247 0.976 0.071 0.255 0.036 -0.060 4 2 CENTER 0.016 0.511 0.099 0.237 0.014 -0.015 S1= 0.606 S2= 0.101 S3= -0.082 SE= 0.617 DC= 0.372 0.928 0.014 -0.106 0.027 0.994 EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 13 ELEMENT STRESSES UNITS= NEWTMMS ------------------------------------------------------------------------------- NODE/ NORMAL STRESSES SHEAR STRESSES ELEMENT LOAD CENTER SXX SYY SZZ SXY SYZ SZX ------------------------------------------------------------------------------- 4 3 21 0.900 5.181 1.884 4.989 5.058 0.396 4 3 25 -0.893 -5.740 -1.294 6.615 1.429 -1.229 4 3 26 5.251 -3.282 3.647 5.654 0.468 3.274 4 3 22 5.379 5.974 1.830 4.029 6.019 1.649 4 3 41 2.148 9.507 2.550 0.107 5.891 1.229 4 3 45 2.276 4.348 1.292 -1.518 0.596 -0.396 4 3 46 -1.334 3.555 2.982 -0.558 -0.364 2.442 4 3 42 -3.127 7.049 -0.756 1.067 6.851 0.816 4 3 CENTER 1.325 3.324 1.517 2.548 3.244 1.023 S1= 7.030 S2= 0.411 S3= -1.275 SE= 7.604 DC= 0.425 0.744 0.516 0.809 -0.055 -0.586 4 4 21 -0.008 -0.024 -0.008 -0.001 -0.001 -0.000 4 4 25 -0.008 -0.022 -0.008 -0.001 -0.001 0.000 4 4 26 0.001 -0.022 0.001 -0.001 -0.001 0.000 4 4 22 0.001 -0.024 0.001 -0.001 -0.001 0.000 4 4 41 -0.008 -0.026 -0.008 -0.001 -0.001 -0.000 4 4 45 -0.008 -0.024 -0.008 -0.001 -0.001 -0.000 4 4 46 0.001 -0.024 0.001 -0.001 -0.001 0.000 4 4 42 0.001 -0.026 0.001 -0.001 -0.001 0.000 4 4 CENTER -0.004 -0.024 -0.004 -0.001 -0.001 0.000 S1= -0.003 S2= -0.004 S3= -0.024 SE= 0.021 DC= 0.705 -0.070 0.705 -0.707 0.000 0.707 4 5 21 0.925 5.729 2.005 5.199 5.061 0.402 4 5 25 -0.878 -6.136 -1.412 6.854 1.431 -1.254 4 5 26 5.215 -3.629 3.624 5.903 0.481 3.277 4 5 22 5.313 6.532 1.928 4.248 6.011 1.622 EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 14 ELEMENT STRESSES UNITS= NEWTMMS ------------------------------------------------------------------------------- NODE/ NORMAL STRESSES SHEAR STRESSES ELEMENT LOAD CENTER SXX SYY SZZ SXY SYZ SZX ------------------------------------------------------------------------------- 4 5 41 2.176 10.031 2.710 0.349 5.913 1.254 4 5 45 2.275 3.869 1.194 -1.307 0.579 -0.402 4 5 46 -1.564 3.066 2.920 -0.356 -0.371 2.425 4 5 42 -3.366 7.524 -0.677 1.299 6.864 0.770 4 5 CENTER 1.262 3.373 1.537 2.774 3.246 1.012 S1= 7.193 S2= 0.379 S3= -1.400 SE= 7.856 DC= 0.435 0.745 0.505 -0.764 0.008 0.645 4 10 21 0.032 0.572 0.129 0.211 0.004 0.005 4 10 25 0.022 -0.374 -0.110 0.240 0.004 -0.024 4 10 26 -0.038 -0.325 -0.024 0.250 0.013 0.003 4 10 22 -0.067 0.582 0.098 0.221 -0.006 -0.027 4 10 41 0.036 0.550 0.168 0.242 0.023 0.024 4 10 45 0.007 -0.455 -0.090 0.213 -0.016 -0.005 4 10 46 -0.230 -0.465 -0.063 0.203 -0.006 -0.017 4 10 42 -0.240 0.501 0.078 0.233 0.013 -0.046 4 10 CENTER -0.060 0.073 0.023 0.227 0.004 -0.011 S1= 0.243 S2= 0.024 S3= -0.230 SE= 0.410 DC= 0.600 0.800 -0.017 -0.024 0.039 0.999 6 1 23 0.329 0.394 0.413 -0.043 -0.127 -0.060 6 1 27 -0.071 -1.739 -0.071 -0.099 -0.099 -0.005 6 1 28 -0.676 -1.849 -0.676 -0.553 -0.553 -0.005 6 1 24 -0.166 0.394 0.140 -0.498 0.328 0.051 6 1 43 -0.097 -0.200 -0.097 -0.182 -0.182 -0.115 6 1 47 0.413 0.394 0.329 -0.127 -0.043 -0.060 6 1 48 0.140 0.394 -0.166 0.328 -0.498 0.051 EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 15 ELEMENT STRESSES UNITS= NEWTMMS ------------------------------------------------------------------------------- NODE/ NORMAL STRESSES SHEAR STRESSES ELEMENT LOAD CENTER SXX SYY SZZ SXY SYZ SZX ------------------------------------------------------------------------------- 6 1 44 -0.259 -0.089 -0.259 0.273 0.273 0.106 6 1 CENTER -0.049 -0.287 -0.049 -0.113 -0.113 -0.005 S1= 0.027 S2= -0.044 S3= -0.368 SE= 0.365 DC= 0.631 -0.451 0.631 -0.707 -0.000 0.707 6 2 23 -0.032 0.112 -0.001 0.030 -0.002 0.016 6 2 27 -0.001 -0.025 -0.046 0.073 -0.013 -0.027 6 2 28 -0.096 -0.003 -0.065 0.083 -0.003 -0.035 6 2 24 -0.085 0.177 0.109 0.040 -0.012 -0.078 6 2 43 -0.152 0.158 0.052 0.136 -0.023 -0.005 6 2 47 -0.140 -0.041 -0.013 0.092 0.008 -0.049 6 2 48 -0.496 -0.105 -0.119 0.082 0.019 -0.014 6 2 44 -0.464 0.136 0.076 0.125 -0.033 -0.057 6 2 CENTER -0.183 0.051 -0.001 0.083 -0.007 -0.031 S1= 0.081 S2= -0.001 S3= -0.213 SE= 0.263 DC= 0.314 0.928 -0.202 -0.060 0.232 0.971 6 3 23 -2.744 -0.535 -0.041 -0.327 -0.468 0.408 6 3 27 -3.140 -0.556 -1.018 0.642 0.296 -0.560 6 3 28 1.815 0.568 0.607 0.402 0.056 0.214 6 3 24 1.900 0.279 0.654 -0.567 -0.228 -0.755 6 3 43 0.636 -0.478 0.687 -0.031 -0.313 0.563 6 3 47 0.721 0.942 0.191 -0.999 0.141 -0.405 6 3 48 -0.136 0.128 -0.121 -0.759 -0.099 0.058 6 3 44 -0.531 -1.602 -0.555 0.210 -0.073 -0.910 6 3 CENTER -0.185 -0.157 0.050 -0.179 -0.086 -0.173 S1= 0.143 S2= -0.010 S3= -0.424 SE= 0.508 DC= -0.484 0.038 0.874 -0.507 0.802 -0.316 EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 16 ELEMENT STRESSES UNITS= NEWTMMS ------------------------------------------------------------------------------- NODE/ NORMAL STRESSES SHEAR STRESSES ELEMENT LOAD CENTER SXX SYY SZZ SXY SYZ SZX ------------------------------------------------------------------------------- 6 4 23 0.000 -0.024 0.000 -0.000 -0.000 -0.000 6 4 27 0.000 -0.024 0.000 -0.000 -0.000 -0.000 6 4 28 -0.000 -0.024 -0.000 -0.000 -0.000 -0.000 6 4 24 -0.000 -0.024 -0.000 -0.000 -0.000 0.000 6 4 43 0.000 -0.024 0.000 -0.000 -0.000 -0.000 6 4 47 0.000 -0.024 0.000 -0.000 -0.000 -0.000 6 4 48 -0.000 -0.024 -0.000 -0.000 -0.000 0.000 6 4 44 -0.000 -0.024 -0.000 -0.000 -0.000 0.000 6 4 CENTER 0.000 -0.024 0.000 -0.000 -0.000 -0.000 S1= 0.000 S2= -0.000 S3= -0.024 SE= 0.024 DC= -0.707 0.000 0.707 0.707 -0.002 0.707 6 5 23 -2.448 -0.052 0.370 -0.340 -0.596 0.364 6 5 27 -3.211 -2.343 -1.135 0.616 0.185 -0.592 6 5 28 1.043 -1.309 -0.134 -0.068 -0.500 0.174 6 5 24 1.649 0.826 0.902 -1.025 0.089 -0.782 6 5 43 0.387 -0.545 0.642 -0.077 -0.518 0.443 6 5 47 0.994 1.271 0.506 -1.034 0.106 -0.514 6 5 48 -0.492 0.393 -0.406 -0.349 -0.578 0.096 6 5 44 -1.255 -1.580 -0.739 0.608 0.167 -0.861 6 5 CENTER -0.417 -0.417 0.001 -0.209 -0.206 -0.209 S1= 0.117 S2= -0.208 S3= -0.741 SE= 0.750 DC= -0.265 -0.255 0.930 0.705 -0.710 0.006 6 10 23 0.296 0.507 0.412 -0.013 -0.128 -0.044 6 10 27 -0.071 -1.764 -0.117 -0.025 -0.112 -0.032 6 10 28 -0.773 -1.853 -0.741 -0.470 -0.556 -0.039 6 10 24 -0.251 0.572 0.249 -0.458 0.316 -0.027 EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 17 ELEMENT STRESSES UNITS= NEWTMMS ------------------------------------------------------------------------------- NODE/ NORMAL STRESSES SHEAR STRESSES ELEMENT LOAD CENTER SXX SYY SZZ SXY SYZ SZX ------------------------------------------------------------------------------- 6 10 43 -0.249 -0.043 -0.045 -0.046 -0.205 -0.121 6 10 47 0.272 0.354 0.315 -0.034 -0.035 -0.109 6 10 48 -0.356 0.289 -0.285 0.410 -0.479 0.037 6 10 44 -0.724 0.047 -0.184 0.398 0.239 0.050 6 10 CENTER -0.232 -0.236 -0.050 -0.030 -0.120 -0.036 S1= 0.011 S2= -0.212 S3= -0.316 SE= 0.289 DC= -0.080 -0.428 0.900 0.888 -0.441 -0.131 62. FINISH *********** END OF THE STAAD.Pro RUN *********** **** DATE= OCT 27,2022 TIME= 15: 9:43 **** EXAMPLE PROBLEM USING SOLID ELEMENTS -- PAGE NO. 18 ************************************************************ * For technical assistance on STAAD.Pro, please visit * * http://www.bentley.com/en/support/ * * * * Details about additional assistance from * * Bentley and Partners can be found at program menu * * Help->Technical Support * * * * Copyright (c) Bentley Systems, Inc. * * http://www.bentley.com * ************************************************************