Problem 4 001-programa

Software Verification SAP2000 2

PROGRAM NAME: REVISION NO.:

EXAMPLE 4-001 ASOLID – SOIL SUPPORTING UNIFORMLY LOADED CIRCULAR FOOTING PROBLEM DESCRIPTION In this example, the asolid element is used to study distribution of vertical displacement and vertical stress in the soil beneath a uniformly loaded circular footing. The results from the axisymmetric SAP2000 model are compared with hand-calculated results.

Only a small portion of the semi-infinite soil medium is modeled. The radius of the footing is 10 feet. A one radian segment of a circular column of soil that is 250 feet deep and has a 100-foot radius is modeled. The soil is assumed to have a modulus of elasticity of 2,000 ksf and a Poisson’s ratio of 0.25. The footing exerts a uniform 5 ksf load on the soil.

CLSym 10' 10'

90' Soil surface

250'

5 ksf soil surface loading from footing

EXAMPLE 4-001 - 1

Software Verification PROGRAM NAME: REVISION NO.:

SAP2000 2

GEOMETRY, PROPERTIES AND LOADING 10'

90' Soil Properties E = 2,000 k/ft2 ν = 0.25

Z 5 ksf X

Elements 5 feet wide by 1 foot deep Elements 1 foot wide by 1 foot deep

250'

Ux restraint along this edge, typical

Elements 5 feet wide by 5 feet deep

Elements 1 foot wide by 5 feet deep

Uz restraint along base, typical

EXAMPLE 4-001 - 2

Software Verification PROGRAM NAME: REVISION NO.:

SAP2000 2

TECHNICAL FEATURES OF SAP2000 TESTED  Analysis using asolid elements  Asolid surface pressure load  Incompatible bending modes for asolid objects RESULTS COMPARISON The independent results are calculated using data published in Poulos and Davis 1974. Results are presented separately for the models with and without the incompatible bending modes option.

With Incompatible Modes Output Parameter

Vertical Location

SAP2000

Independent

Percent Difference*

z = 0 ft

-0.0467

-0.0469

0%

Uz ft

z = -5 ft

-0.0374

-0.0376

-1%

z = -10 ft

-0.0284

-0.0286

-1%

At x = 0 ft

z = -40 ft

-0.0094

-0.0095

-1%

z = -80 ft

-0.0047

-0.0048

-2%

z = 0 ft

-4.993

-5.000

0%

Vertical stress ksf

z = -5 ft

-4.298

-4.553

-6%

z = -10 ft

-3.285

-3.232

-2%

At x = 0 ft

z = -40 ft

-0.433

-0.435

0%

z = -80 ft

-0.122

-0.115

+6%

EXAMPLE 4-001 - 3

Software Verification PROGRAM NAME: REVISION NO.:

SAP2000 2

Without Incompatible Modes Output Parameter

Vertical Location

SAP2000

Independent

Percent Difference*

z = 0 ft

-0.0464

-0.0469

-1%

Uz ft

z = -5 ft

-0.0371

-0.0376

-1%

z = -10 ft

-0.0282

-0.0286

-1%

At x = 0 ft

z = -40 ft

-0.0094

-0.0095

-1%

z = -80 ft

-0.0047

-0.0048

-2%

z = 0 ft

-5.260

-5.000

+5%

Vertical stress ksf

z = -5 ft

-4.265

-4.553

-6%

z = -10 ft

-3.127

-3.232

-3%

At x = 0 ft

z = -40 ft

-0.436

-0.435

0%

z = -80 ft

-0.122

-0.115

+6%

COMPUTER FILES: Example 4-001-incomp, Example 4-001-comp CONCLUSIONS The SAP2000 results show an acceptable comparison with the independent results both with and without the incompatible bending modes option.

The results for this example problem are not controlled by bending behavior.

Thus the incompatible bending modes option has little effect on the results.

In general, when bending is a significant contributor to the results of an example problem, models that do not use the incompatible bending modes option require a much more refined mesh to obtain results comparable to similar models that use the incompatible bending modes option.

In general, the incompatible bending modes option should always be used for asolid element models.

EXAMPLE 4-001 - 4

Software Verification PROGRAM NAME: REVISION NO.:

SAP2000 2

HAND CALCULATION

EXAMPLE 4-001 - 5

Software Verification PROGRAM NAME: REVISION NO.:

SAP2000 2

EXAMPLE 4-001 - 6