CHAPTER 2 ANALYSIS OF INDETERMINATE STRUCTURES: SLOPE ...

KIoT, Department of Civil Engineering

Lecture Note: Theory of Structures - II 1

CHAPTER 2

ANALYSIS OF INDETERMINATE STRUCTURES: SLOPE-DEFLECTION METHOD

1.1. INTRODUCTION

In the force (or compatibility) group of methods of structural analysis, such as the method of

consistent displacements and the method of least work, the unknowns are forces. By

comparison, in the slope-deflection method (which is one of the classical formulations of the

displacement group of methods) the unknowns are displacements. In this method, the

moments at the ends of a member are expressed in terms of the displacements of these ends.

The said member- end moments are made up of the following components:

The end moments due to external loads on the member with the member ends

assumed fixed, and

The end moments caused by the actual member-end displacements (rotations and

translations).

This method takes into account only the bending deformations of structures, and

consequently, is used to analyze indeterminate structures, made up of moment-resisting

members such as continuous beams and rigid-jointed frames.

In using the slope-deflection method, a slope-deflection equation is written for every member

of the given structure, expressing the end moments in terms of the member-end

displacements. Next, joint equilibrium equation is written for every joint capable of

undergoing rotation. The expressions on the right hand sides of the slope-deflection

equations are then substituted into the joint equilibrium equations. The resulting equations

are solved for the joint rotations. Finally, the values of the joint rotations are back-substituted

into the slope-deflection equations to yield the required values of the member-end moments.

Although the slope-deflection method is by itself an important method for the analysis of

indeterminate beams and frames, a good understanding of its basic principles provides a very



useful introduction to the matrix stiffness method of analysis, a method which forms the

bedrock of most computer software currently used for structural analysis.

The slope-deflection method uses algebraic procedure and it is therefore important to adopt a

sign convention for the forces and displacements.

SIGN CONVENTION

Moment is considered positive if it acts counterclockwise at the end of a member or

clockwise at a joint adjacent to a member. This is illustrated in Fig.1.1.

End rotation is positive if it is counterclockwise in direction.

A chord rotation is positive if it is counterclockwise in direction.

Fig.1.1 Sign Convention for end moments

Note

All end moments, end rotations and chord rotations shown in Fig.1.2(b) are positive by the

above sign convention.

1.2. DERIVATION OF THE SLOPE-DEFLECTION EQUATION

When load is applied to a continuous beam or a rigid-jointed frame, moments are induced at

the ends of the members. The slope-deflection equation expresses the relationship between

the moments at the end of a member and the displacements of the member ends as well as the

external loads applied to the member.

To develop this relationship, let us consider a typical member AB (of constant flexural

rigidity EI) of a rigid-jointed structure shown in Fig.1.2(a). Under the applied load let the end

moments developed and the deformed shape of the structure be as shown in Fig.1.2(b).

+ -



Deformed axis of member AB

(OR Elastic curve)

B

A B

L

A Chord

(a) (b)

EI = Const

Fig.1.2 Member-end displacements: (a) Typical member of a rigid-jointed frame

(b) Assumed displacement pattern of member AB

The notations in Fig.1.2(b) have the following meanings:

MAB = moment at end A of member AB;

MBA = moment at end B of member AB;

A, B = respective rotations of ends A and B of the member with reference to the un

deformed (horizontal) axis of the member;

= relative translation between the two ends of the member in the direction perpendicular

to the un-deformed axis of the member;

= rotation of the member's chord (i.e., the straight line connecting the member's ends

after deformation) due to the relative translation . The deformations are small and hence

=L

EXPRESSIONS FOR MEMBER MOMENTS

To derive expression for member moments in terms of the external load and the

deformations, we proceed by considering their effects on the moment one at a time. We shall

use moment-area method for the determination of the expressions for displacements.

A

MBA

MAB

L

B



1. End moments due to rotation A (B = = P = 0).

In Fig.1.3(a) represents a member AB for which there is no rotation at end B (fixed end),

no relative translations of the ends (the chord joining the two ends of the member after

deformation remains horizontal) and there is no external load (P = 0).

L

(a)

(b)

Fig.1.3: (a) Member AB with the applied displacement and induced end moments;

(b) M/EI diagram for the end moments

According to the second moment-area theorem, the tangential deviation of a point A on

the elastic curve from the tangent at another point B on the elastic curve, measured in the

direction perpendicular to the originally straight member, is equal to the moment of the area

of the M/EI diagram between A and B taken about A.

Referring to Fig.1.3(b) and using the above theorem, the distance between A and a tangent

drawn at B (this distance is zero since a tangent drawn at B is a horizontal line that coincides

with the un-deformed axis of the beam due to the fact that the slope at B is zero) is equal to

the area of the M/EI diagram between A and B taken about A. Thus:

03

2

232

LL

EI

MLL

EI

M BAAB

MBA

B

EI

MBA

MAB

EI

M AB

A A



From here, 2

ABAB

MM (1.1)

According to the first moment-area theorem, the angle in radians or the change in slope

between the tangents at two points A and B on the elastic curve of an originally straight

member is equal to the area of the M/EI diagram between points A and B. Hence,

EI

LM

EI

LM BAABA

22 [Note that ABA since 0B ]

Taking cognisance of eqn (1.1), the above expression reduces to:

,4

L

EIM A

AB

L

EIM A

BA

2 (1.2)

2. End moments due to rotation B (A = = P = 0)

Fig.1.4 represents a member that satisfies the above conditions.

Fig.1.4 Given member with applied displacement and induced end moments

Analogous to Case1 (Fig.1.3),

,4

L

EIM B

BA

L

EIM B

AB

2 (1.3)

L

B

MBA

B A

MAB



2b. Modified or adjusted end-moments for member with far end hinged

Consider the beam shown in Fig.1.4b which has a hinged far end. The relationship between

the applied moment MAB and the rotation A can be obtained by using the moment-area

method (as was used for the case of member with far end fixed: Fig.1.4 above).

(i)

(ii)

Fig.1.4b: (i) Member with far end hinged, (ii) M/EI diagram for the end moment

From Fig.1.4b(i), it is clear that

L

BAA

According to the second moment area theorem,

BA = moment of M/EI diagram between A and B taken about B.

BAEI

MLLL

EI

M

33

2

2

1 2

But EI

LM

L

BAA

3

AL

EIM

3 (1.3b)

EI=Const.

L

A MAB

MBA = 0

A B

BA

Tangent at A

M/EI



Eqn (1.3b) represents the modified or adjusted end moment at A due to rotation A when the

far end of the member is hinged.

3. End moments due to a relative joint displacement (A = B = P = 0)

A member with a relative joint displacement but no joint rotations is shown in Fig.1.5.

From the first moment-area theorem, the change in slope between A and B (referring to

Fig.1.5(a) this change in slope is zero since there is no angular rotation at either of the ends)

is equal to the area of the M/EI diagram between A and B.

(a)

(b)

Fig.1.5: (a) Member with joint displacement, (b) M/EI diagram for the end moments

Thus:

022

EI

LM

EI

LM ABBA

OR MBA = MAB (1.4)

AB

MAB

A

MBA

B

L

MBA/EI

MAB/EI



Applying the second moment-area theorem, the distance between B and A , measured

vertically from the tangent drawn at A is equal to the moment of the area of the M/EI

diagram, taken about B.

Hence

3

2

232)(

L

EI

LML

EI

LM ABBAAB

Combining the above expression with eqn (1.4), we obtain:

2

6

L

EIMM BAAB

(1.5)

4a End moments due to external loads acting on the member (A = B = = 0)

The fixed-end beam of Fig.1.6 represents this situation. Since the ends of the member are

fixed against rotation and translation, the member end moments are known as fixed-end

moments (FEM). They develop only as a result of the external loads. The fixed-end

moments may be obtained using the method of consistent displacements and moment-area

theorems, or indeed any convenient method of analysis of indeterminate structures. For ease

of reference, fixed-end moments for some common loading cases are usually given as

appendix in standard text books.

Fig.1.6 Fixed-end moments due to external loads

SLOPE-DEFLECTION EQUATION

To obtain the total end moment of a member AB, it is necessary to sum up the various end

moments due to external (FEMs) and those due to rotations and translations. This will be

achieved by combining eqns (1.2), (1.3), (1.5) and the fixed-end moments (FEMs).

FEMAB

w

A

P P

B

FEMBA



Hence:

ABBAAB FEMLL

EIM

)

32(

2 (1.6a)

BAABBA FEMLL

EIM

)

32(

2 (1.6b)

A close look at eqns (1.6) reveals that the two equations have the same form and that one can

be obtained from the other by swarping the subscripts A and B. Consequently, they can be

combined into one equation as follows:

nffnnf FEML

EIM 32

2 (1.7)

Eqn (1.7) is known as the slope-deflection equation. The subscript "n" refers to the near end

of the member where the moment Mnf acts while the subscript "f" refers to the far (or other)

end of the member.

,L

where is the relative translation of the supports.

MODIFIED SLOPE-DEFLECTION EQUATION

Where the far end of a member is hinged, it is sometimes convenient to obtain the end

moments using the modified slope deflection equation, which is as follows:

2

3 hrrhrrh

FEMFEM

L

EIM

(1.7a)

where the subscript "r" refers to the rigidly connected end of the member where the moment

Mrh is applied, and the subscript "h" refers to the hinged end of the member. Obviously,

moment at the far hinged end Mhr = 0 since a hinge cannot support moment.

0hrM



1.3. ANALYSIS OF CONTINUOUS BEAMS

In the analysis of continuous beams, slope-deflection equations are written for each span in

terms of the unknown displacements. The unknown displacements are the rotations of

members over supports (n and f in eqn (1.7)) and support translations (represented by in

eqn (1.7)) when a support undergoes translation. After writing slope-deflection equations for

the spans, joint equilibrium equation is written for each of the supports that is free to rotate.

A set of simultaneous equations result if more than one joint can rotate. However, if only one

joint can rotate only one equation with one unknown joint rotation results. The resulting

equation(s) is(are) solved for the displacement(s). The values or expressions for the

displacements are then back-substituted into the slope-deflection equations, which are then

solved to give member-end moments. The above procedure is illustrated in the following

example.

Example 1.1

Determine the member-end moments of the continuous beam ABC, fixed at ends A and C

and continuous over support B as shown in Fig.1.7a.

(b)

(a)

Fig.1.7 (a) Given beam and loading, (b) Free body diagram of joint B

SOLUTION

First, we evaluate the fixed-end moments (Table 1.2) as follows:

mkNFEM LABAB .24

12

418

12

22

LBC = 6m

3m

w = 18kN/m

P = 60kN

LAB = 4m

EI = Const

A

B

C MBA MBC

B

MBA



mkNFEMBA .24

mkNPL

FEM BCBC .45

8

660

8

mkNFEMCB .45

We now write the slope-deflection equation for each span using eqn.(1.7):

ABBA

AB

AB FEML

EIM 32

2

Notice that in the given beam, A = 0 (fixed support) and the chord rotation = 0 (since

there is no support translation, i.e., = 0).

244

2 B

AB

EIM

OR 245.0 BAB EIM (1)

BAAB

AB

BA FEML

EIM 32

2

Observe that while considering joint A, "a" was the near end ("n" in eqn 1.7) and "B" was the

far end ("f" in eqn.1.7) but when considering joint B, "B" became the near ("n") end and "A"

became the far ("f") end. Here again, = 0.

2424

2 BBA

EIM OR 24 BBA EIM (2)

Similarly, for span BC:

BCCB

BC

BC FEML

EIM 32

2

= 4526

2B

EI OR 4567.0 BBC EIM (3)

and

45322

BC

BC

CBL

EIM

= 456

2BEI

OR 4533.0 BCB EIM (4)



Next, we isolate B (Fig.1.7b) and write the equilibrium equation for the free-body as follows:

0 BCBA MM (5)

Substituting the values for MBA and MBC from eqns (2) and (3) into eqn (5) we obtain:

04567.024 BB EIEI OR EI

B

57.12

Substitution of the value of B into eqns (1) to (4) yields the end moments as follows:

mkNEI

EIM AB .7.172457.12

5.0

mkNEI

EIMBA .6.362457.12

mkNEI

EIMBC .6.364557.12

67.0

mkNEI

EIMCB .2.494557.12

33.0

STRUCTURES WITH OVERHANGS

In a case where a beam has an overhang, to analyse the beam, the overhang is first replaced

by the equivalent moment applied on the adjacent support. The procedure then becomes the

same as before except that the overhanging member has a known end moment. The following

example will illustrate the steps involved.

Example 1.2

Determine the support moments for the beam shown in Fig.1.8

Fig.1.8 Given beam and loading

18kN/m

A B

3m

4m 6m 1.5m

C D

60kN 12kN

EI = Const



SOLUTION

The overhanging span is replaced with its equivalent moment equal to 18kNm (12 x 1.5 =

18). This moment is positive in accordance with our adopted sign convention since the 12kN

load tends to rotate joint C in a clockwise direction.

Notice that the fixed-end moments for spans AB and BC will be the same as for Example

1.1.

Thus

FEMAB = 24kNm; FEMBA = -24kNm;

FEMBC = 45kNm; FEMCB = -45kNm.

Considering the overhanging span as a fixed cantilever, the fixed-end moment at support C

will be

FEMCD = 12 x 1.5 = 18kNm.

Slope-deflection equation

,244

2 B

AB

EIM

OR 245.0 BAB EIM (1)

,244

4 B

BA

EIM

OR 24 BBA EIM (2)

,4526

2 CBBC

EIM OR 4533.067.0 CB EIEI (3)

,4526

2 BCCB

EIM OR 4533.067.0 BC EIEI (4)

kNmMCD 18 (5)

Next, we consider the equilibrium conditions of the free bodies of the joints (Fig.1.9).

Fig.1.9 Free-body diagram of the joints

From Fig.1.9, we write:

MBA + MBC = 0 (6)

MBA

MBC C B

MCB

MCD

MBA



MCB + MCD = 0 (7)

Next, we substitute the values of the end moments from equations 2 to 5 into equations 6 and

7 and solve the resulting simultaneous equations for the end rotations, as follows:

04533.067.024 CBB EIEIEI or 2133.067.1 CB EIEI (8)

and 0184533.067.0 BC EIEI or 2767.033.0 CB EIEI (9)

Solving equations 8 and 9 simultaneously yields:

EI

C

51.51 and

EIB

75.22

Substituting these values of displacements C and B into the slope-deflection equations will

yield the end (or support) moments.

Thus:

kNmEI

EIM AB 63.122475.22

5.0

kNmEI

EIMBA 75.462475.22

kNmEI

EIEI

EIMBC 75.464551.51

33.075.22

67.0

kNmEI

EIEI

EIMBC 184575.22

33.051.51

67.0

and kNmMCD 18

If it is required to obtain the support reactions, this can be conveniently carried out by

separately obtaining the simply supported beam reactions and the reactions due to support

moments, and then algebraically adding them together. This is carried out as shown in

Fig.1.10



(a)

(b)

(c)

Fig.1.10 (a) Simply supported beam reactions; (b) Reactions due to support moments;

(c) Total (Final) support reactions

Note that in applying the end moments obtained from the slope-deflection equation at the

beam support points, the adopted sign convention (i.e., counterclockwise moments are

positive) is observed but in using these moments for further calculations (such as obtaining

support reactions or span moments) the beam sign convention (i.e., sagging moments are

positive and shears upward to the left of the beam or downward to the right of the beam is

positive) must be followed.

Checks To check the correctness of the final support reactions obtained in Fig.1.10(c), the

equilibrium of the entire beam under vertical forces is considered as follows:

Fy = 0; 27.47 + 79.32 +37.21 - (18 x 4) - 60 12 = 0

144 - 144 = 0 (Satisfied)

To check the correctness of the moment at support A, we can take the sum of the moments of

all the active and reactive forces about A as follows:

18kN/m

37.21kN

46.75kNm 12.63kNm

1.5m

12kN 30kN 30kN 36kN 36kN

6m

12kN

60kN

4m

79.32kN

12.63kNm

27.47kN

4.79kN 4.79kN 8.53kN 8.53kN

18kNm 46.75kNm



MA = - 12(1.5 + 6 + 4) + 37.21(6 + 4) - 60(3 + 4) + (79.32 x 4) - (18 x 4 x 2)

= - 138 + 372.1 - 420 + 317.3 - 144

= - 12.6kNm (Satisfied)

STRUCTURES SUBJECT TO SUPPORT SETTLEMENT

If any of the supports settles, its effect on each span can be taken care of in the slope-

deflection equation. The known support settlement () is used to obtain the chord rotation ,

which is then appropriately substituted into the slope-deflection equation(1.7) and the rest of

the analysis steps remains as before. This is illustrated using the following example.

Example 1.3

Determine the support moments for the beam of Fig.1.11 if under the given loading support

B sinks by 5mm. Take E = 210 x 106 kN/m

2; I = 360 x 10

-6m

4


SOLUTION

Fixed-end moments

kNmFEM AB 4812

436 2

kNmFEMBA 48

kNmFEMBC 458

660

kNmFEMCB 45

36kN/m

A

LAB = 4m

B 3m

C D

60kN

LBC = 6m 2m

EI = Const

30kN



kNmFEMCD 60230

The chord rotations are obtained as follows:

.00125.04000

5rad

LAB

AB

Note that the minus sign is because the chord rotation for member AB is clockwise as

support B sinks (See sign convention).

.000833.06000

5rad

LBC

BC

Slope-deflection equations

4800125.0322

BA

AB

ABL

EIM or

48001875.05.0 EIEIM BAB (1)

4800125.0324

2 ABBA

EIM or

48001875.0 EIEIM BBA (2)

45000833.0326

2 CBBC

EIM or

45000833.033.067.0 EIEIEIM CBBC (3)

45000833.0326

2 BCCB

EIM

or

45000833.033.067.0 EIEIEIM BCBC (4)

kNmMCD 60 (5)

Joint equilibrium equations (See Fig.1.12)

Fig.1.12 Free-body diagrams of the joints

0 BCBA MM (6)

0 CDCB MM (7)

MBA

B C MBC

MCB

MCD



Substitution of eqns (2) to (5) into eqns (6) and (7) gives the following.

For eqn.(6):

045000833.033.067.048001875.0 EIEIEIEIEI CBB

or 300104.033.067.1 EIEIEI CB

Dividing throughout by EI, we have:

00104.03

33.067.1 EI

CB (8)

For eqn.(7):

06045000833.033.067.0 BC EIEI

Dividing throughout by EI, we obtain:

000833.015

67.033.0 EI

CB (9)

Solving eqns.(8) and (9) simultaneously, we obtain:

,782.25

00171.0EI

C 00096.0891.6

EI

B

But 266 756001036010210 kNmEI

00137.0000341.000171.0 C or .1037.1 3rad

and

000869.000096.00000911.0 B or rad41069.8

Final end moments:

4875600001875.0000869.0756005.0 ABM or kNmM AB 9.156

4875600001875.0000869.075600 BAM or kNmMBA 28

4575600000833.000137.07560033.0000869.07560067.0 BCM or

kNmMBC 28

4575600000833.0000869.07560033.000137.07560067.0 CBM or

kNmMCB 60



We now consider an example of a beam for which spans have different second moments of

area I.

Example 1.4

Evaluate the member-end moments of the beam shown in Fig.1.13


Fixed-end moments

kNmFEM AB 10012

548 2

kNmFEMBA 100

kNmFEMBC 508

580

kNmFEMCB 50


100

5

22 BAB

IEM or 1008.0 BAB EIM (1)

1002

5

22 BBA

IEM or 1006.1 BBA EIM (2)

5025

2 CBBC

EIM or 504.08.0 CBBC EIEIM (3)

5025

2 BCCB

EIM or 504.08.0 BCCB EIEIM (4)

Joint equilibrium equations: (See Fig.1.14)

w = 36kN/m

5m

B

5m

2I I 2.5m

80kN

C A




0 BCBA MM (5)

and 0CBM (6)

Substituting the values of MBA and MBC from eqns (2), (3) and (4) into eqns (5) and (6), we

obtain the following:

0504.08.01006.1 CBB EIEIEI or 504.04.2 CB EIEI (7)

and 0504.08.0 BC EIEI (8)

Solving eqns (7) and (8) simultaneously, we obtain:

36.11BEI and 84.56CEI

Final end moments

kNmM AB 09.10910036.118.0

kNmMBA 82.8110036.116.1

kNmMBC 82.815084.564.036.118.0

05036.114.084.568.0 CBM (as expected)

1.4. FRAMES WITH NO LATERAL TRANSLATION OF JOINTS

The procedure for the analysis of frames whose joint translations are prevented is similar to

that for the analysis of continuous beams. The following example illustrates such a case.

MBA

C B

MCB =0

MBC



Example 1.5

Determine the support moments of the frame shown in Fig.1.15

Fig.1.15 Given frame and loading

SOLUTION

Observe that joints B and C can rotate but none of the joints can translate.

Fixed-end moments

kNmFEM AB 89.86

42102

2

kNmFEMBA 44.46

24102

2

kNmFEMBC 209

36302

2

kNmFEMCB 409

63302

2

Stiffnesses

46

5.1 IIKAB

39

3 IIKBC

3

4

AB

BC

K

K

Therefore, the relative values of the stiffness are KKBC 4 and KKAB 3

We now use these relative stiffness values in the slope-deflection equations.

30kN

2m

A

10kN

4m

B

1.5I 1.5I

3I

6m

9m D

6m

C

3m




89.832 BAB EKM or 89.86 BAB EKM (1)

44.434 BBA EKM or 44.412 BBA EKM (2)

204244 CBBC EKEKM or 20816 CBBC EKEKM (3)

404244 BCCB EKEKM or 40816 BCCB EKEKM (4)

CCD EKM 34 or CCD EKM 12 (5)

CDC EKM 32 or CDC EKM 6 (6)

Joint equilibrium equations

0 BCBA MM (7)

0 CDCB MM (8)

Substituting eqns (2) to (5) into eqns (7) and (8), we have:

02081644.412 CBB EKEKEK (9)

and 01240816 CBC EKEKEK (10)

Simultaneous solution of eqns (9) and (10) yields:

048.1BEK and 728.1CEK

Final end moments (or support moments)

kNmM AB 60.289.8048.16

kNmMBA 02.1744.4048.112

kNmMBC 06.1720728.18048.116

kNmMCB 74.2040048.18728.116

kNmMCD 74.20728.112

kNmMDC 37.10728.16

Example 1.6

Determine the member-end moments for the frame of Fig.1.16

100kN

0.75I

B 1.5I

0.75I

1.5I A

C

6m 4m 4m

15kN/m

75kN

D

E

3m

3m




SOLUTION

Here only joints B and D are free to rotate. The structure therefore has only two degrees of

freedom, which are the unknown joint rotations B and D.

Fixed-end moments

kNmFEM AB 1008

8100

;

;100kNmFEMBA

0 CBBC FEMFEM ;

;4512

615 2

kNmFEMBD

;45kNmFEMDB

;25.568

675kNmFEMDE

.25.56 kNmFEMED

Chord rotations

There is no settlement of any of the supports so the chord rotation of each of the four

members is zero, i.e., 0 DEBDBCAB .


100375.01008

5.12

BBAB EI

EIM (1)

10075.010028

5.12

BBBA EI

EIM (2)

BBBC EIEI

M 5.026

75.02

(3)

BBCB EIEI

M 25.06

75.02

(4)

455.04526

5.12

DBDBBD EIEI

EIM (5)



455.04526

5.12

BDBDDB EIEI

EIM (6)

25.565.025.5626

75.02

DDDE EI

EIM (7)

25.5625.025.566

75.02

DDED EI

EIM (8)


We now apply moment equilibrium equations to the free bodies of joints B and D shown in

Fig 1.17

0 BDBCBA MMM (9)

0 DEDB MM (10)


Joint rotations

To determine the unknown joint rotations B and D , we substitute slope-deflection eqns (2),

(3), (5), (6) and (7) into the joint equilibrium equations (9) and (10) to obtain the following

equations:

0455.05.010075.0 DBBB EIEIEIEI

or 555.025.2 DB EIEI (11)

and

025.565.0455.0 DBD EIEIEI

or 25.115.15.0 DB EIEI (12)

Simultaneous solution of eqns (11) and (12) yields the following:

;2.28BEI 9.16DEI

Member-end moments

MBA

B

MBC

MBD

D

MDB

MDE



The member-end moments are obtained by substituting the numerical values of BEI and

DEI into the slope-deflection equations as follows:

;58.1101002.28375.0 kNmMAB

;85.781002.2875.0 kNmMBA

;1.142.285.0 kNmMBC

;05.72.2825.0 CBM

;75.64459.165.02.28 kNmMBD

;8.47452.285.09.16 kNmMDB

;8.4725.569.165.0 kNmMDE

.48.6025.569.1625.0 kNmMED

1.5. FRAMES WITH LATERAL TRANSLATION OF JOINTS

For the analysis of frames with joint translation, some other equilibrium conditions is are

required, in addition to joint equilibrium equations. The procedure involved is illustrated by

means of the following example.

Example 1.7

Determine the member-end moments for the frame shown in Fig.1.18.


SOLUTION

2m

A

10kN

4m

B

1.5I 1.5I

3I

9m D

6m

C

3m



Recall that this frame was earlier analysed in Example 1.5 (Fig.1.15) but with the joint

translation prevented by provision of a lateral support at joint C. In this example, the support

at C has been removed to allow translation of joints to take place. A possible deformed shape

of the frame is shown in Fig.1.19. Since axial deformation is negligible, the lateral

translations of joints B and C () are equal.

Fig.1.19 Possible deformed shape of the frame

Fixed-end moments

The fixed-end moments were obtained in Example 1.5 and are as follows:

;89.8 kNmFEM AB ;44.4 kNmFEMBA

;20kNmFEMBC .40kNmFEMCB

As was done in the analysis of the frame of Example 1.5, in this example we shall use for

convenience, the relative values of the stiffness as KKK CDAB 3 and KKBC 4

Note also that the chord rotations are equal, i.e. 6

CDAB


89.8332 BAB EKM

or 89.8186 EKEKM BAB (1)

44.4326 BBA EKM

or 44.41812 EKEKM BBA (2)

20242 CBBC EKM

or 20816 CBBC EKEKM (3)

40242 BCCB EKM

A D

C B



or 40816 BCCB EKEKM (4)

3232 CCD EKM

or EKEKM CCD 1812 (5)

332 CDC EKM

or EKEKM CDC 186 (6)


0 BCBA MM (7)

0 CDCB MM (8)

Note that the joint equilibrium equations are only two but there are three unknowns: B ; C

and . A third equation is therefore necessary in order to evaluate the three unknowns. This

can be obtained by considering the equilibrium of the entire frame under horizontal forces,

i.e. Fx =0. The forces involved here are the external 10kN force and the shears at the

column bases. The two columns, together with the various forces needed to establish the third

equation, are shown in Fig.1.20.

Fig.1.20 Shears at column bases

Referring to Fig.1.20, we obtain the following equilibrium equation for horizontal forces:

Fx =0; 10 - HAB - HDC = 0 (9)

Expressions for the shears (HAB and HDC) can be obtained in terms of the end moments as

follows:

;0 BM 04106 BAABAB MMH

MBA

2m

4m

A

B

MAB

HAB

MDC

10kN

C

MCD

HDC

D

6m



or

6

40 BAAB

AB

MMH (10)

and ;0 CM

6

DCCDDC

MMH

(11)

Substituting equations (1) and (2) into eqn (10) and eqns(5) and (6) into eqn(11), and then

substituting the resulting expressions for HAB and HDC into eqn(9), we have:

06

1861812

6

56.35181289.818610

EKEKEKEKEKEKEKEK CCBB

or 592.21233 EKEKEK CB (12)

Substitution of eqns (2) and (3) into eqn(7) and eqns(4) and (5) into eqn(8) yields the

following equations (13) and (14) respectively;

56.1518828 EKEKEK CB (13)

and 4018288 EKEKEK CB (14)

Simultaneous solution of equations (12), (13) and (14) yields the joint displacements as

follows:

;0847.1BEK ;697.1CEK .0629.0EK

Final member-end moments

;51.389.80629.0180847.16 kNmM AB

;32.1644.40629.0180847.112 kNmMBA

;22.1620697.180847.116 kNmMBC

;52.21400847.18697.116 kNmMCB

;52.210629.018697.112 kNmMCD

.31.110629.018697.16 kNmMDC

CHAPTER 2 ANALYSIS OF INDETERMINATE STRUCTURES: SLOPE ...

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