problem 1 005

Software Verification PROGRAM NAME: SAP2000 REVISION NO.: 0

EXAMPLE 1-005 - 1

EXAMPLE 1-005 FRAME - DISPLACEMENT LOADING

EXAMPLE DESCRIPTION Using a portal frame, this example verifies SAP2000 for settlement and rotation of normal supports, skewed supports (settlement only) and spring supports. Note that for spring supports, the grounded end of the spring is displaced or rotated.

Six different models are created. The models are identical, except for the loading and the support condition at joint 4 as shown in the figure below. The results for various support reactions in each model are compared with independent hand calculations.

Important Note: Only bending deformations are considered in the analysis. Shear and axial deformations are ignored. In SAP2000 this is achieved by setting the property modification factor for area to 100,000 and setting the property modification factor for shear area to 0.

GEOMETRY AND PROPERTIES

144"

144"

Z

X

2

1

Y

2

3

3

41

Material PropertiesE = 29,000 k/in2

Section Propertiesb = 12 ind = 12 inA = 144 in2I = 1,728 in4

Support condition at joint 4 varies for each model


EXAMPLE 1-005 - 2

SUPPORT CONDITION AT JOINT 4 AND LOADING

U z=

-0.5"

Model A

U z=

-0.2"

Model B

0.01 rad

Model C

Model D Model E Model F

0.01 rad

10 k

U3 = -1"1

3

Joint local axes for skewed supports

cos = 0.8sin = 0.6

Model Support Condition at Joint 4 Loading

A Roller -0.5" Z displacement at joint 4

B Z direction translational spring, k = 10 kip/in -0.2" Z displacement at joint 4

C Roller 0.01 radian rotation about joint 1

D Rotational spring about Y axis, k = 80,000 kip-in/rad 0.01 radian rotation about joint 4

E Skewed roller 10 kip Z direction force at midpoint of frame element 2

F Skewed roller -1" joint local 3 direction displacement at joint 4


EXAMPLE 1-005 - 3

TECHNICAL FEATURES OF SAP2000 TESTED Settlement of support in frame structures Rotation of support in frame structures Settlement of support with linear (translational) spring Rotation of support with rotational spring Skewed supports Skewed support settlement

RESULTS COMPARISON Independent results are hand calculated using the unit load method described on page 244 in Cook and Young 1985.

Model Output

Parameter SAP2000 Independent Percent

Difference

Fz (jt. 1) kip 6.293 6.293 0% A. Support Settlement My (jt. 1) kip-in -906.250 -906.250 0%

Fz (jt. 1) kip 1.115 1.115 0% B. Spring Support Settlement My (jt. 1) kip-in -160.492 -160.492 0%

Fz (jt. 1) kip -18.125 -18.125 0% C. Support Rotation My (jt. 1) kip-in 2,610.000 2,610.000 0%

My (jt. 1 ) kip -473.469 -473.469 0% D. Spring Support Rotation Ry (jt. 4) rad. 0.00408 0.00408 0%

Fz (jt. 1) kip 5.811 5.811 0% E. Skewed Support F3 (jt. 4) kip 5.236 5.236 0%

Fz (jt. 1) kip 27.215 27.215 0% F. Skewed Support Settlement My (jt. 1) kip-in -3,918.919 -3,918.919 0%


EXAMPLE 1-005 - 4

COMPUTER FILES: Example 1-005a, Example 1-005b, Example 1-005c, Example 1-005d, Example 1-005e, Example 1-005f

CONCLUSION The SAP2000 results show an exact match with the independent results.


EXAMPLE 1-005 - 5

HAND CALCULATION


EXAMPLE 1-005 - 6


EXAMPLE 1-005 - 7


EXAMPLE 1-005 - 8


EXAMPLE 1-005 - 9


EXAMPLE 1-005 - 10


EXAMPLE 1-005 - 11

EXAMPLE 1-005Frame - Displacement LoadingExample DescriptionGeometry and PropertiesSupport Condition at Joint 4 and LoadingTechnical Features of SAP2000 TestedResults ComparisonComputer Files: Example 1-005a, Example 1-005b, Example 1-005c, Example 1-005d, Example 1-005e, Example 1-005fConclusionHand Calculation

problem 1 005

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