displacement method of analysis slope deflection...

Displacement Method of AnalysisSlope Deflection Method

Chapter 11

Displacement Method of Analysis

• Two main methods of analyzing indeterminate structure

– Force method • The method of consistent deformations & the equation of three

moment

• The primary unknowns are forces or moments

– Displacement method

• The slope-deflection method & the moment distribution method

• The primary unknown is displacements (rotation & deflection)

• It is particularly useful for the analysis of highly statically indeterminate structures

• Easily programmed on a computer & used to analyze a wide range of indeterminate structures


• Degree of Freedom

– The number of possible joint rotations & independent joint translations in a structure is called the degree of freedom of the structure

– In three dimensions each node on a frame can have at most three linear displacements & three rotational displacements.

– In two dimension each node can have at most two linear displacements & one rotational displacement.

DOF = n J– R

- n number of possible joint’s movements

- J number of joints

- R number of restrained movements


• Degree of Freedom

DOF = n J– R- n number of possible joint’s movements

- n = 2 for two dimensional truss structures- n = 3 for three dimensional truss structures- n= 3 for two dimensional frame structures- n= 6 for three dimensional frame structures

- J number of joints- R number of restrained movements

• Neglecting Axial deformationDOF = n J– R – m

- m number of members

Slope-Deflection Equations

•Derivation


• Angular Displacement at A, A

'

'

1 1 20 0

2 3

4

2 3

1 1 2

2&

0 02 3 2 3

AB B

AB A

AA

BA AB

A

B

A

A

B

M L M LM L L

EI EI

M L M LM L L L

E

EI EIM M

L L

I EI


• Angular Displacement at B, B

4

2

BA B

AB B

EIM

L

EIM

L

Slope-Deflection Equations• Relative Linear Displacement

1 2 10.

2 3 2 3

M L M LL L

EI EI

' 0.BM

2

6AB BA

EIM M M

L

Slope-Deflection Equations• Fixed-End Moment

1 12 0.

2 2

PL ML L

EI EI

0.yF

8

PLM

Fixed-End moment

FEM


( )

( )

AB AB

BA BA

M FEM

M FEM

Slope-Deflection EquationsFixed-End Moment


• Slope-deflection equations

– The resultant moment (adding all equations together)

• Lets represent the member stiffness as k = I/L &

• The span ration due to displacement as = /L

• Referring to one end of the span as near end (N) & the other end as the far end (F).

– Rewrite the equations

2 2 3 ( )AB A B AB

IM E FEM

L L

2 2 3 ( )BA B A BA

IM E FEM

L L

2 2 3 ( )

2 2 3 ( )

N N F N

F F N F

M EK FEM

M EK FEM


• Slope-deflection equations for pin supported End Span

– If the far end is a pin or a roller support

– Multiply the first equation by 2 and subtracting the second equation from it

2 2 3 ( ) '

0 2 2 3 0

N N F N

F N

M EK FEM

EK

3 ( ) 'N N NM EK FEM

Fixed End Moment Table

Fixed End Moment Table- Continue

Summary

2 2 3 ( )

2 2 3 ( )

N N F N

F F N F

M EK FEM

M EK FEM

3 ( ) 'N N NM EK FEM


– MN, MF = the internal moment in the near & far end of the span. • Considered positive when acting in a clockwise direction

– E, k = modulus of elasticity of material & span stiffness k = I/L

– N, F = near & far end slope of the span at the supports in radians.

• Considered positive when acting in a clockwise direction

– = span ration due to a linear displacement / L.

• If the right end of a member sinks with respect the the left end the sign is positive

Slope-Deflection EquationsSteps to analyzing beams using this method

– Find the fixed end moments of each span (both ends left & right)

– Apply the slope deflection equation on each span & identify the unknowns

– Write down the joint equilibrium equations

– Solve the equilibrium equations to get the unknown rotation & deflections

– Determine the end moments and then treat each span as simply supported beam subjected to given load & end moments so you can workout the reactions & draw the bending moment & shear force diagram

Draw the bending moment & shear force diagram.

Fixed End Moment

2

4 42 . 2 .

8

2 66 . 6 .

12

AB BA

BC CB

FEM t m FEM t m

FEM t m FEM t m

Slope Deflection Equations

0

0

0

0

2 2 0 2 0.5 24

2 2 0 2 24

2 42 2 0 6 6

6 3

2 22 2 0 6 6

6 3

A

A

C

C

AB A B AB B

BA A B BA B

BC B C BC B

CB B C CB B

IM E M EI

IM E M EI

IM E M EI

IM E M EI

4m 6m

4t2t/m

A B C2II

Example 1

Joint Equilibrium Equations

1.70.5 2 1.15 .

1.72 3.7 .

3.7 .

2 1.76 7.1 .

3

AB

BA

BC

CB

M EI t mEI

M EI t mEI

M t m

M EI t mEI

0

4 1.72 6 0.

3

BA BC

B B B

M M

EI EIEI

Joint B

Substituting in slope deflection equations

Computing The Reactions

RA

1.15

4t

RB1

3.7

1

1.15 4 2 3.71.36

4

4 1.36 2.64

A

B

R t

R t

2

2 6 3 3.7 7.15.43

6

2 6 5.43 6.57

B

C

R t

R t

RB2

3.7

2t/m

RC

7.1

6.57

5.43

2.64

1.36

--

++

7.1

3.7

1.15

3.79

1.57+

+

---

• Determine the internal moments in the beam at the supports.

Fixed End Moment2

2

2

2

2

60 4 226.67 .

6

60 2 453.33 .

6

30 6135 .

8

AB

BA

BC

FEM kN m

FEM kN m

FEM kN m


0

0

12 2 0 26.67 26.67

6 3

22 2 0 53.33 53.33

6 3

13 0 135 135

6 2

A

A

AB A B AB B

BA A B BA B

BC B BC B

IM E M EI

IM E M EI

IM E M EI

60kN

6m 6m

30kN/m

A B CII

4m

Example 2

0

2 1 753.33 135 0 81.67

3 2 6

70

BA BC

B B B

B

M M

EI EI EI

EI


1(70) 26.67 3.33 .

3

2(70) 53.33 100 .

3

1(70) 135 100 .

2

AB

BA

BC

M kN m

M kN m

M kN m

Joint B


Example 3

• Example 2: Determine the internal moments in the beam at the supports

– Support A; downward movement of 0.3cm & clockwise rotation of 0.001 rad. Support B; downward movement of 1.2cm. Support C downward movement of 0.6cm. EI = 5000 t.m2

6m 6m

A B C

II

Fixed End Moment

0.AB BA BC CBFEM FEM FEM FEM

Displacements:

0.001

1.2 0.30.0015

600

0.6 1.20.001

600

A

AB

BC


2 50002 0.001 3 0.0015

6

2 50002 0.001 3 0.0015

6

AB B

BA B

M

M

3 5000

0.0016

BC BM


0

7 0.004 0

0.00057

BA BC

B

B

M M

rad

Joint B


2 50002 0.001 0.00057 3 0.0015 3.22 .

6

2 50002 0.00057 0.001 3 0.0015 3.93 .

6

3 50000.00057 0.001 3.93 .

6

AB

BA

BC

M t m

M t m

M t m

Example 1b

0.0AB BAFEM FEM

0.0A C

Equilibrium Equations

Example 2b

Example 3b

Example 4

• Analysis of Frames Without Sway– The side movement of the end of a column in a frame is called SWAY.

– Determine the moment at each joint of the frame. EI is constant

25 24 880 .

96BCFEM kN m

Fixed End Moment0.AB BAFEM FEM

25 24 880 .

96CBFEM kN m

0.CD DCFEM FEM

A= 0. 2 0. 0. 0

12AB B

IM E

0.1667AB BM EI


A= 0. 2 2 0 0. 0

12BA B

IM E

0.333BA BM EI

Example 5

0.333 0.5 0.25 80 0.C C BEI EI EI

2 2 0 808

BC B C

IM E

0.5 0.25 80BC B CM EI EI

2 2 0 808

CB C B

IM E

0.5 0.25 80CB C BM EI EI

D= 0. 2 2 0 0. 0

12CD C

IM E

0.333CD CM EI

D= 0. 2 0 0. 0

12DC C

IM E

0.1667DC CM EI


0BA BCM M

Joint B

0.333 0.5 0.25 80 0.B B CEI EI EI 0.833 0.25 80B CEI EI

0CB CDM M

Joint C

0.833 0.25 80C BEI EI

22.9 .

45.7 .

45.7 .

45.7 .

45.7 .

22.9 .

AB

BA

BC

CB

CD

DC

M kN m

M kN m

M kN m

M kN m

M kN m

M kN m


Two equation & two unknown 137.1

B CEI

45.7 45.7

22.9

82.3

22.9

Example 6

2I

2I

II

I

A

B

E

C

F

D

– Draw the bending moment diagram

30kN20kN48kN/m

2

2

20 1 28.89 .

3BAFEM kN m

Fixed End Moment2

2

20 2 14.44 .

3ABFEM kN m

248 464 .

12BCFEM kN m

64 .CBFEM kN m

0BE EB CF FCFEM FEM FEM FEM

30 2 60 .CDM kN m

A= 0.Slope Deflection Equations

2 2 0 4.443

AB A B

IM E

24.44

3AB BM EI

4m3m

4m 2m3m

1m

Example 6b

A= 0. 2 2 0 8.89

3BA A B

IM E

48.89

3BA BM EI

2

2 2 0 644

BC B C

IM E

2 64BC B CM EI EI

2

2 2 0 644

CB B C

IM E

2 64CB B CM EI EI

2 2 0 03

BE B E

IM E

4

3BE BM EI

E= 0.

2 2 0 03

EB E B

IM E

2

3EB BM EI

E= 0.

2

2 2 0 04

CF C F

IM E

2CF CM EI

F= 0.

2

2 2 0 04

FC F C

IM E

FC CM EI

F= 0.


0BA BC BEM M M

4.67 55.11B CEI EI

Joint B

4 48.89 2 64 0

3 3B B C BEI EI EI EI

0CB CF CDM M M

4 4B CEI EI

Joint C

2 64 2 60 0B C CEI EI EI

Two equation & two unknown

4.18CEI 12.70BEI


212.7 4.44 4.03 .

3ABM kN m

412.70 8.89 25.83

3BAM kNm

2 12.7 4.18 64 42.77 .BCM kN m

12.7 2 ( 4.18) 64 68.34 .CBM kN m

4.18

8.35

60

68.35

42.77

25.83

16.94

4.03

8.47

412.70 16.94 .

3BEM kN m

212.70 8.47 .

3EBM kN m

2 ( 4.18) 8.35 .CFM EI kN m

4.18 .FCM kN m

Slope Deflection (Frame with Sway) • Analysis of Frames with Sway

Fixed End Moment

2 0. 3 012 12

AB B

IM E


1 1

6 24BEI EI

As there is no span loading in any

of the member FEM for all the

members is zero

2 2 0. 3 012 12

BA B

IM E

1 1

3 24BEI EI

– Draw the bending moment diagram. EI constant

Example 7

0. 4 22 2 3 0

15 15 15 15BC B C B C

IM E EI EI

2 4

2 2 0 015 15 15

CB C B B C

IM E EI EI

2 12 2 0 3 0.

18 18 9 54CD C C

IM E EI EI

1 12 0 3 0

18 18 9 54DC C C

IM E EI EI


0BA BCM M Joint B MB = 0

1 1 4 20.

3 24 15 15B B CEI EI EI EI

14.4 3.2 0B CEI EI EI 1

0CB CDM M

Joint C MC = 0

Three unknown & just two equations so we need

another equilibrium equation. Let take Fx = 0.

2 4 2 10.

15 15 9 54B C CEI EI EI EI

7.2 26.4 0B CEI EI EI 2

40 0.A DH H

12

BA ABA

M MH

18

CD DCD

M MH

40 18 1.5 0.BA AB CD DCM M M M

1 1 1 1720 1.5

3 24 6 24B BEI EI EI EI

2 1 1 10.

9 54 9 54C CEI EI EI EI

0.162 0.75 0.333 720B CEI EI EI 3

Now solve the three equation

1 14.4 3.2 0

1 7.2 26.4 0

0.162 0.75 0.333 720

B

C

EI

EI

EI

136.2CEI

438.9BEI

6756.6EI


208 .ABM kN m

135 .BAM kN m

135 .BCM kN m

95 .CBM kN m

95 .CDM kN m

110 .DCM kN m

-

-+

-

+

+

110

135

135

208

95

95

220 15375 .

12BCFEM kN m

Fixed End Moment

0.AB BAFEM FEM

220 15375 .

12CBFEM kN m

0.CD DCFEM FEM

2 0. 3 012 12

AB B

IM E


1 1

6 24BEI EI


Example 7

2 2 0. 3 012 12

BA B

IM E

1 1

3 24BEI EI

0.2 2 3 375

15 15BC B C

IM E

4 2375

15 15B CEI EI

2 2 0 37515

CB C B

IM E

2 4375

15 15B CEI EI

2 2 0 3 0.18 18

CD C

IM E

2 0 3 018 18

DC C

IM E


0BA BCM M Joint B MB = 0

1 1 4 2375 0.

3 24 15 15B B CEI EI EI EI

14.4 3.2 9000B CEI EI EI 1

2 1

9 54CEI EI

1 1

9 54CEI EI

Three unknown & just two equations so we need

another equilibrium equation. Let take Fx = 0.

0CB CDM M Joint C MC = 0

2 4 2 1375 0.

15 15 9 54B C CEI EI EI EI

7.2 26.4 20250B CEI EI EI 2

40 0.A DH H

12

BA ABA

M MH

18

CD DCD

M MH

40 18 1.5 0.BA AB CD DCM M M M

1 1 1 1720 1.5


2 1 1 10.


0.162 0.75 0.333 720B CEI EI EI 3

1 14.4 3.2 9000

0 7.2 23.2 29250

0 1.58 0.185 2178

B

C

EI

EI

EI

Now solve the three equation

1 14.4 3.2 9000

1 7.2 26.4 20250

0.162 0.75 0.333 720

B

C

EI

EI

EI

1 14.4 3.2 9000

0. 7.2 23.2 29250

0. 0. 5.29 4257

B

C

EI

EI

EI

R1-R2

0.162R1-R3

0.22R2-R3

4257804.7

5.29CEI

1469.6BEI

9587.2EI Substituting in slope deflection equations

1 11467.2 9587.2 155 .

6 24ABM kN m

1 11469.6 9587.2 90 .

3 24BAM kN m

4 21469.6 ( 804.7) 375 90 .

15 15BCM kN m

2 41469.6 ( 804.7) 375 356 .

15 15CBM kN m

2 1( 804.7) 9587.2 356 .

9 54CDM kN m

1 1( 804.7) 9587.2 267 .

9 54DCM kN m

267

356

90

90

155

356

--

-

Example 8


Fixed End Moment

As there is no span loading in any of the

member FEM for all the members is zero

12 0. 3 05 5

AB B

IM E


1

2 6

5 25BEI EI

12 2 0 3 05 5

BA B

IM E

1

4 6

5 25BEI EI

Example 9

02 2 3 0

7 7BE B E

IM E

4 2

7 7B EEI EI

02 2 3 0

7 7EB E B

IM E

2 4

7 7B EEI EI

12 0. 3 05 5

FE E

IM E

1

2 6

5 25EEI EI

12 2 0 3 05 5

EF E

IM E

1

4 6

5 25EEI EI

22 2 3 05 5

BC B C

IM E

2

4 2 6

5 5 25B CEI EI EI

22 2 3 05 5

CB C B

IM E

2

2 4 6

5 5 25B CEI EI EI

02 2 3 0

7 7CD C D

IM E

4 2

7 7C DEI EI

02 2 3 0

7 7DC D C

IM E

2 4

7 7C DEI EI

22 2 3 05 5

DE D E

IM E

2

4 2 6

5 5 25D EEI EI EI

22 2 3 05 5

ED E D

IM E

2

2 4 6

5 5 25D EEI EI EI


0BA BC BEM M M

Joint B MB = 0

1 2

4 6 4 2 6 4 20

5 25 5 5 25 7 7B B C B EEI EI EI EI EI EI EI

1 2380 70 50 42 42 0B C EEI EI EI EI EI 1

0EB ED EFM M M

Joint E ME = 0

2 1

2 4 2 4 6 4 60

7 7 5 5 25 5 25B E D E EEI EI EI EI EI EI EI

1 250 70 380 42 42 0B D EEI EI EI EI EI 2

0CB CDM M

Joint C MC= 0

2

2 4 6 4 20

5 5 25 7 7B C C DEI EI EI EI EI

270 240 50 42 0B C DEI EI EI EI 3

0DC DEM M

Joint D MD = 0

2

2 4 4 2 60

7 7 5 5 25C D D EEI EI EI EI EI

250 240 70 42 0C D EEI EI EI EI 4

40 0B EH H

Top story FX = 0

5

CB BCB

M MH

5

DE EDE

M MH

HB HE

40 80 0A FH H

Bottom story FX = 0

5

BA ABA

M MH

5

EF FEF

M MH

HA HF

2 2

2 4 6 4 2 6200

5 5 25 5 5 25B C B CEI EI EI EI EI EI

26 6 6 6 4.8 1000B C D EEI EI EI EI EI

2 2

4 2 6 2 4 60

5 5 25 5 5 25D E D EEI EI EI EI EI EI

5

1 1

4 6 2 6600


1 1

4 6 2 60

5 25 5 25E EEI EI EI EI

130 30 24 15000B EEI EI EI 6

6 unknown and 6 equation

1

2

380 70 0 50 42 42 0

50 0 70 380 42 42 0

70 240 50 0 0 42 0

0 50 240 70 0 42 0

6 6 6 6 0 4.8 1000

30 0 0 30 24 0 15000

B

C

D

E

EI

EI

EI

EI

EI

EI

171.79BEI

79.80CEI

79.80DEI

171.79EEI

1 1054.46EI

2 837.29EI


184.4 .ABM kN m

115.6 .BAM kN m

147.2 .BEM kN m

147.2 .EBM kN m

184.4 .FEM kN m

115.6 .EFM kN m

31.6 .BCM kN m

68.4 .CBM kN m

68.4 .CDM kN m

68.4 .DCM kN m

68.4 .DEM kN m

31.6 .EDM kN m

-

-

-

-

++

- ++

++

-

147.2

31.6

68.4

31.6

184.4

115.6

147.2

115.6

184.4

68.4

68.4

68.4

Before we start let us discus the relative

displacement () of each span

Degree of freedom

DOF = 3x4-6-3 = 3

That means we got three unknown &

we need three equations


Example 10

AB

The relative displacement () for span AB

is equal AB – 0. = AB (clockwise)

The relative displacement () for span BC is

equal 0. – BC = –BC (counterclockwise)

The relative displacement () for span CD

is equal 0. – (–CD)= CD (clockwise)

BC

CD

A

B

D

C

60o

take AB =

BC = AB sin30 = 0.5

CD = AB cos30 = 0.866

Let us build a relationship between AB, BC & CD

So in the slope deflection equations we will use;

as the relative displacement of span AB.

–0.5 as the relative displacement of span BC.

0.866as the relative displacement of span CD.

AB

BC

CD60o

60o

B =30o

22 42.667 .

12BCFEM t m

Fixed End Moment

0.AB BAFEM FEM

2.667.CBFEM m

0.CD DCFEM FEM

2 0. 3 03 3

AB B

IM E


2 2

3 3BEI EI

2 2 0 3 03 3

BA B

IM E

4 2

3 3BEI EI

0.52 2 3 2.667

4 4BC B C

IM E

1 32.667

2 16B CEI EI EI

0.52 2 3 2.667

4 4CB B C

IM E

1 32.667

2 16B CEI EI EI

0.8662 2 0 3 0

6 6CD C

IM E

2 0.866

3 6CEI EI

0.8662 0 3 0

6 6DC C

IM E

1 0.866

3 6CEI EI


0BA BCM M

Joint B MB = 0

4 2 1 32.667 0

3 3 2 16B B CEI EI EI EI EI

112 24 23 128B CEI EI EI 1

0CB CDM M

Joint C MC = 0

1 3 2 0.8662.667 0

2 16 3 6B C CEI EI EI EI EI

24 80 2.072 128B CEI EI EI 2

34.5 3.1 38.375 144B CEI EI EI 3

Third Equilibrium Equations (Method 1)

21 22 12.92 11.92 96 0AB BA CD DCM M M M

6.92m

6.0m

8.0

3.0 2.0 2.0

6.0

3.0

8 t

11 12.923 6

8 2 0.0

DC CDAB BAAB DC

M MM MM M

0A DH H

Third equation FX = 0

From the free body diagram for column CD

6

CD DCD

M MH

Free body diagram for column AB

D

C

HD

CDM

DCM

A

B

HA

ABM

BAM

1.5m

2.6

m

VA

2.6 1.5 0A BA AB AH M M V

Free body diagram for Beam BC

VB

C

CBMB

BCM 2t/m

4 2 4 2 0B BC CBV M M

1.54

2.6 2.6 4

BC CBBA AB

A

M MM MH

1.54 0.

2.6 2.6 4 6

BC CBBA AB CD DCM MM M M M

4

4

BC CB

A B

M MV V

Third Equilibrium Equations (Method 2)

24 9 10.4 144BA AB BC CB CD DCM M M M M M

4 2 2 2 124 9

3 3 3 3 2B B B CEI EI EI EI EI EI

2 0.866 1 0.86610.4 144


3 1 32.667 2.667

16 2 16B CEI EI EI EI

34.5 3.1 38.375 144B CEI EI EI 3

Solving the three equation

112 24 23 128

24 80 2.072 128

34.5 3.1 38.375 144

B

C

EI

EI

EI

3.11BEI

2.71CEI

6.77EI


112 24 23 128

24 80 2.072 128

34.5 3.1 38.375 144

B

C

EI

EI

EI

3.11BEI

2.71CEI

6.77EI


2.44 .ABM t m

0.36 .BAM t m

0.36 .BCM t m

2.78 .CBM t m

2.78 .CDM t m

1.88 .DCM t m A

B

D

C

_

_

+

+

2.78

1.88

2.78

0.360.36

2.442.8

Before we start let us discus the relative

displacement () of each span


Example

The relative displacement () for span AB

is equal AB – 0. = AB (clockwise)

The relative displacement () for span BC is equal

(– 2) – 1 = – (1+ 2) = – BC (counterclockwise)

The relative displacement () for span CD

is equal 0. – (–CD)= CD (clockwise)

BC = 1+ 2

take AB =

BC = 2(AB cos) = 2 5/8.6 = 1.163

CD = AB =

Let us build a relationship between AB, BC & CD

So in the slope deflection equations we will use;

as the relative displacement of span AB.

–1.163 as the relative displacement of span BC.

as the relative displacement of span CD.

A

B

D

C

CD

AB

1

2

AB

BC

CD

B

1

2

1 = 2 & CD = AB because of

the symmetry in the geometry

24 716.33 .

12BCFEM kN m

Fixed End Moment

0.AB BAFEM FEM

16.33 .CBFEM kN m

0.CD DCFEM FEM

2 0. 3 08.6 8.6

AB B

IM E


2 6

8.6 73.96BEI EI

2 2 0 3 08.6 8.6

BA B

IM E

4 6

8.6 73.96BEI EI

1.1632 2 3 16.33

7 7BC B C

IM E

4 2 6.97816.33

7 7 49B CEI EI EI

1.1632 2 3 16.33

7 7CB C B

IM E

2 4 6.97816.33

7 7 49B CEI EI EI

2 2 0 3 08.6 8.6

CD C

IM E

4 6

8.6 73.96CEI EI

2 0 3 08.6 8.6

DC C

IM E

2 6

8.6 73.96CEI EI


0BA BCM M

Joint B MB = 0

4 6 4 2 6.97816.33 0

8.6 73.96 7 7 49B B CEI EI EI EI EI

103.65 28.57 6.13 1633B CEI EI EI 1

0CB CDM M

Joint C MC = 0

2 4 6.978 4 616.33 0

7 7 49 8.6 73.96B C CEI EI EI EI EI

28.57 103.65 6.13 1633B CEI EI EI 2

6 0A DH H

Third equation FX = 0

From the free body diagram for column CD

Free body diagram for column AB

A

B

HA

ABM

BAM

5m

7m

VA

7 5 0A BA AB AH M M V

Free body diagram for Beam BC

7 4 7 3.5 0B BC CBV M M

514

7 7 7

BC CBBA AB

A

M MM MH

14

7

BC CB

A B

M MV V

VB

C

CBMB

BCM 4kN/m

VC

C

HD

CDM

DCM

5m

7m

D

VD

7 5 0D CD DC DH M M V

7 4 7 3.5 0C BC CBV M M

14

7

BC CB

D C

M MV V

6 49 7 5 7 5 0BA AB BC CB CD DC BC CBM M M M M M M M

3.688 3.688 5.12 294B CEI EI EI 3


103.65 28.57 6.13 1633

28.57 103.65 6.13 1633

3.688 3.688 5.12 294

B

C

EI

EI

EI

18.897BEI

24.603CEI

61.532EI

514

7 7 7

CD DC BC CB

D

M M M MH

6 0A DH H

7 7 10 294BA AB CD DC BC CBM M M M M M

2 6 4 6 2 67 7

8.6 73.96 8.6 73.96 8.6 73.96B B CEI EI EI EI EI EI

4 6 4 2 6.978

10 16.338.6 73.96 7 7 49

C B CEI EI EI EI EI

2 4 6.97816.33 294

7 7 49B CEI EI EI


0.6 .ABM kN m

3.8 .BAM kN m

3.8 .BCM kN m

16.43 .CBM kN m

16.43 .CDM kN m

10.71 .DCM kN m

_

+

10.71

3.8

16.43

16.43

0.6

3.8

14.4

A

B

D

C

displacement method of analysis slope deflection...

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