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....................2012

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Contents:-

Subject page

Types of structures analysis 4

The advantages of methods 5Approximate method 6

Frame diagrams 7

Portal method 9

Assumptions 10

Frame analysis 11

Diagram &Results 13

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Approximate analysis of

statically indeterminate

structures

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Structural AnalysisThere are two types of structural analysis:-

Approximate Analysis methods. Exact Analysis methods.

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Structural AnalysisThe advantages of Approximate Analysis methods:-

Easy & fast to get the results.

Calculate the approximate size of structural members.

Then, we able to roughly estimate the cost of project.

And go back to client and architect to decide, weather to goahead with this design or use alternative solution.

The advantages of exact analysis methods:-

Giving the precise member size.

The design engineer will gain confidence that his works soaccurate, because its based on latest theories of analysis.

The disadvantages:-

Consuming time.

This method of analysis is costly.5university Of kufa

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Structural AnalysisApproximate Methods:-

Applications

Analysis building frame for vertical loads.

Assumptions:- The location of point of Inflection(PIs) on the

girders in 1/10 from the each girder ends.

The axial forces in the girders are assumed tobe zero.

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Structural AnalysisFrame

Diagram 10k 15k3k/fA B

A B

30f24f3f 3f

10k 15kPoints of Inflection

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Structural AnalysisSimple beam between

Points of inflection10k 15k3k/f

10f7f 7f47.5k 49.5k47.5k 49.5k

3f 3f

156k 162k

56.5k 58.5k

+-

-

304k

156k 162kMoment diagram

End Cantilever

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Structural AnalysisAnalysis of building frame for lateral loads:-

To treat the affect of lateral loads, wind & seismic, by use the following:-

Bracing, like K or X bracing.

Shear walls.

Moment resisting wind connection.

1.portal method 2. Cantilever Method

characteristics of Methods

No change in members sizes from top floor to bottom floor, however, forthe same loadings and spans the change in sizes will be due to big windmoments in lower floors.

Meanwhile, the columns sizes will be have big difference from top tobottom. So this fact has to be considered, otherwise large error in analysiswill occurred.

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Portal Method:-

At least three assumptions must be made. In the portal

method, the frame has to be divided into independentportals with three assumptions as follow:-

1. The columns bend in such manner that there is pointof inflection at mid depth.

2. The girders bend in such manner that there is point ofinflection at mid span.

3. The horizontal shear on each level are distributedbetween the columns, the shear divides among

columns in a ratio of one part to exterior columns andtwo parts to interior columns.

In our instance, there are 27 redundants

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Portal MethodFrame Analysis:-

1. Column Shear:-

The shear in each column in the various levelswere first obtained.

Total shear on 1st level=15 k

X+2x+2x+x=15 x=2.5k, 2x=5k

2. Column Moments:- the columns are assumedto have (PIs) at their mid depths, so theirmoment at top and bottom equal the column

shear times half column height. 11university Of kufa

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Portal Method3. Girder moments & Shears:- at any joint in the frame

the sum of the moments in the girders equal the sumof the moments in the columns. By bigenning at theupper left hand corner of the frame and working acrossfrom left to right , the girder moments were found inthis order DH, HL, LP, CG, GK, so on , and girder shears

may be found by divide the girder moment by half thegirder lengths.

4. Column Axial force:- equal girder shear force and add itup when it goes down.

And the axial force in interior columns equal to thedifference between adjacent girders shears whichequal zero, in this case.

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Portal MethodM=Moment

V=shearS=Axial Force

Column Depth=20

Girder Length=20

15k

30k

30k

A

B

C

D

E

F

G

H

I

J

K

L

M

N

O

PM=25

V=2.5 M=25V=2.5 M=25V=2.5V=2.5M=25

S=+2.5V=5.0

M=50

S=0V=5.0

M=50

S=0V=2.5

M=25

S=-2.5

M=100

V=10V=7.5M=75

S=+12.5

M=100

V=10V=15.0M=150

S=0

M=100

V=10V=15.0M=150

S=0V=7.5

M=75

S=-12.5

M=200V=20

V=12.5

M=125

S=+32.5

M=200V=20

V=25.0

M=250

S=0

M=200V=20

V=25.0

M=250

S=0V=12.5

M=125

S=-32.5

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