# EX. UK-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\UK\UK-24 Analysis of a Concrete Block Using Solid Elements.STD when you install the program.

### Example Problem No. 24

```    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
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
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
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 要素荷重の設定 - ソリッド .

```    UNIT KNS MMS
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
28 FY -1000.0
2 TO 4 22 TO 24 42 TO 44 FX 100.0
SUPPORT DISPLACEMENT
9 FX .0011
3 6 9 12 FACE 4 PRE GY -500.0
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`

## 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
SELF Y -1.0
28 FY -1000.0
2 TO 4 22 TO 24  42 TO 44 FX 100.0
SUPPORT DISPLACEMENT
9 FX .0011
UNIT POUND FEET
3 6 9 12 FACE 4 PRE GY -500.0
UNIT KNS MMS
1 1.0 2 1.0 3 1.0 4 1.0
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
``````

```                                                                  PAGE NO.    1
****************************************************
*                                                  *
*           Version  22.08.00.***                  *
*           Proprietary Program of                 *
*           Bentley Systems, Inc.                  *
*           Date=    OCT 19, 2021                  *
*           Time=    17:54: 3                      *
*                                                  *
*  Licensed to: Bentley Systems Inc                *
****************************************************
1. STAAD SPACE EXAMPLE PROBLEM USING SOLID ELEMENTS
INPUT FILE: UK-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
37. SELF Y -1.0
39. 28 FY -1000.0
42. 2 TO 4 22 TO 24  42 TO 44 FX 100.0
44. SUPPORT DISPLACEMENT
45. 9 FX .0011
46. UNIT POUND FEET
49. 3 6 9 12 FACE 4 PRE GY -500.0
50. UNIT KNS MMS
53. 1 1.0 2 1.0 3 1.0 4 1.0
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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 19,2021   TIME= 17:54: 3 ****
EXAMPLE PROBLEM USING SOLID ELEMENTS                     -- PAGE NO.   18
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