m2l13 lesson 13 the three-moment equations-ii

Module 2

Analysis of Statically Indeterminate

Structures by the Matrix Force Method

Version 2 CE IIT, Kharagpur

Lesson 13

The Three-Moment Equations-Ii


Instructional Objectives After reading this chapter the student will be able to 1. Derive three-moment equations for a continuous beam with yielding supports. 2. Write compatibility equations of a continuous beam in terms of three moments. 3. Compute reactions in statically indeterminate beams using three-moment equations. 4. Analyse continuous beams having different moments of inertia in different spans and undergoing support settlements using three-moment equations. 13.1 Introduction In the last lesson, three-moment equations were developed for continuous beams with unyielding supports. As discussed earlier, the support may settle by unequal amount during the lifetime of the structure. Such future unequal settlement induces extra stresses in statically indeterminate beams. Hence, one needs to consider these settlements in the analysis. The three-moment equations developed in the pervious lesson could be easily extended to account for the support yielding. In the next section three-moment equations are derived considering the support settlements. In the end, few problems are solved to illustrate the method. 13.2 Derivation of Three-Moment Equation Consider a two span of a continuous beam loaded as shown in Fig.13.1. Let ,

and be the support moments at left, center and right supports respectively. As stated in the previous lesson, the moments are taken to be positive when they cause tension at the bottom fibers. and denote moment of inertia of left and right span respectively and and denote left and right spans respectively. Let

LM

CM RM

LI RI

Ll Rl

CL δδ , and Rδ be the support settlements of left, centre and right supports respectively. CL δδ , and Rδ are taken as negative if the settlement is downwards. The tangent to the elastic curve at support makes an angle

CCLθ at left support and CRθ at the right support as shown in Fig. 13.1.

From the figure it is observed that,


CRCL θθ = (13.1) The rotations CLβ and CRβ due to external loads and support moments are

calculated from the EIM diagram .They are (see lesson 12)

L

LC

L

LL

LL

LLCL EI

lMEI

lMlEIxA

36++=β (13.2a)

R

RC

R

RR

RR

RRCR EI

lMEI

lMlEIxA

36++=β (13.2b)

The rotations of the chord and ' from the original position is given by ''CL 'RC

L

CLCL l

δδα

−= (13.3a)

R

CRCR l

δδα

−= (13.3b)

From Fig. 13.1, one could write,


CLCLCL βαθ −= (13.4a)

CRCRCR αβθ −= (13.4b)

Thus, from equations (13.1) and (13.4), one could write,

CRCRCLCL αββα −=− (13.5)

Substituting the values of CLCRCL βαα ,, and CRβ in the above equation,

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+

⎭⎬⎫

⎩⎨⎧

++⎟⎟⎠

⎞⎜⎜⎝

⎛

R

CR

L

CL

LL

LL

RR

RR

R

RR

R

R

L

LC

L

LL l

El

ElIxA

lIxA

IlM

Il

IlM

IlM

δδδδ66662

This may be written as

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ −+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+

⎭⎬⎫

⎩⎨⎧

++⎟⎟⎠

⎞⎜⎜⎝

⎛

R

RC

L

LC

LL

LL

RR

RR

R

RR

R

R

L

LC

L

LL ll

ElIxA

lIxA

Il

MIl

Il

MIl

Mδδδδ

666

2

(13.6) The above equation relates the redundant support moments at three successive spans with the applied loading on the adjacent spans and the support settlements. Example 13.1 Draw the bending moment diagram of a continuous beam shown in Fig.13.2a by three moment equations. The support

BCB settles by 5mm below A

and . Also evaluate reactions atC A , B and C .Assume EI to be constant for all members and , 200 GPaE = 6 48 10 mmI = ×


Assume an imaginary span having infinitely large moment of inertia and arbitrary span L′ left of A as shown in Fig.13.2b .Also it is observed that moment at C is zero.


The given problem is statically indeterminate to the second degree. The moments and ,the redundants need to be evaluated. Applying three moment equation to the span A’AB,

AM BM

0== CL δδ and mR

3105 −×−=δ

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×−−+−

××−=⎟

⎠⎞

⎜⎝⎛+

⎭⎬⎫

⎩⎨⎧ +∞

+⎟⎠⎞

⎜⎝⎛∞

−

4105(006

)4(28644'2''

3

EII

MI

LMLM BAA

410562448

3−××−−=+ EIMM BA (1)

Note that, 233

1269 kNm106.1

101010810200 ×=

××××=

−

EI

Thus,

4105106.162448

33

−××××−−=+ BA MM

3648 −=+ BA MM (2)

Again applying three moment equation to span the other equations is obtained. For this case,

ABC0=Lδ , (negative as the settlement is

downwards) and mC

3105 −×−=δ0=Rδ .

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×−

×−−

×××

−−=⎭⎬⎫

⎩⎨⎧ ++

⎭⎬⎫

⎩⎨⎧ −−

4105

41056

42667.106244424 33

EIIII

MI

M BA

41010106.163224164

33 ××××+−−=+ BA MM

32164 −=+ BA MM (3)

Solving equations (2) and (3),

1.0 kN.mBM = − 4.0 kN.mAM = − (4)

Now, reactions are calculated from equations of static equilibrium (see Fig.13.2c).


Thus,

( )( )( )( )↑=

↑=

↑=

↑=

kN75.3kN25.4kN25.1

kN75.2

C

BR

BL

A

RRR

R

The reactions at B,


kN5.5=+= BLBRB RRR (5) The area of each segment of the shear force diagram for the given continuous beam is also indicated in the above diagram. This could be used to verify the previously computed moments. For example, the area of the shear force diagram between A and B is .This must be equal to the change in the bending moment between A and D, which is indeed the case (

5.5 kN.m4 1.5 5.5 kN.m− − = ). Thus,

moments previously calculated are correct. Example 13.2 A continuous beam is supported on springs at supports ABCD B and as shown in Fig.13.3a. The loading is also shown in the figure. The stiffness of

springs is

C

20EIkB = and

30EIkC = .Evaluate support reactions and draw bending

moment diagram. Assume EI to be constant.


In the given problem it is required to evaluate bending moments at supports B and . By inspection it is observed that the support moments at C A and are zero. Since the continuous beam is supported on springs at

DB andC , the

support settles. Let BR and be the reactions at CR B and respectively. Then

the support settlement at

C

B and C are B

B

kR and

C

C

kR

respectively. Both the

settlements are negative and in other words they move downwards. Thus,

0=Aδ ,EI

RBB

20−=δ ,

EIRC

C30−

=δ and 0=Dδ (1)


Now applying three moment equations to span (see Fig.13.2b) ABC

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡ +−

+−

−×××

−×

××−=

⎭⎬⎫

⎩⎨⎧+

⎭⎬⎫

⎩⎨⎧ ++

⎭⎬⎫

⎩⎨⎧

4

3020

420

64

22064

233.21644424 EIR

EIR

EIR

EIII

MII

MI

MCB

BCBA

Simplifying,

CBCB RRMM 4560124416 −+−=+ (2) Again applying three moment equation to adjacent spans , CDBC and

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−

++−

−×

×××+××−−=

⎭⎬⎫

⎩⎨⎧ ++

⎭⎬⎫

⎩⎨⎧

EIREI

REIR

EIIII

MI

M C

BC

CB 430

4

2030

64

)13236296(604424

BCCB RRMM 309090164 −+−=+ (3)

In equation (2) and (3) express and in terms of and (see Fig.13.2c)

BR CR BM CM

( )( )

( )( )

( )( )↑+=

↑−=

↑−+=

↑−+=

↑−=

↑+=

CD

CCR

CBCL

BCBR

BBL

BA

MR

MR

MMR

MMR

MRMR

25.06

25.02

25.025.05

25.025.05

25.0825.08

(4)

Note that initially all reactions are assumed to act in the positive direction (i.e. upwards) .Now,

CBBRBLB MMRRR 25.05.013 +−=+=

CBCRCLC MMRRR 5.025.07 −+=+= (5) Now substituting the values of and in equations (2) and (3), BR CR

( ) ( )CBCBCB MMMMMM 5.025.074525.05.01360124416 −+−+−+−=+


Or,

3415.3325.57 =− CB MM (6) And from equation 3,

( ) ( )CBCBCB MMMMMM 25.05.013305.025.079090164 +−−−++−=+ Simplifying,

1505.685.33 =+− CB MM (7) Solving equations (6) and (7)

7.147 kN.m10.138 kN.m

C

B

MM

==

(8)

Substituting the values of and in (4),reactions are obtained. BM CM

( ) ( )( ) ( )( ) ( )( ) ( )

10.535 kN 5.465 kN

4.252 kN 5.748 kN

0.213 kN 7.787 kN

9.717 kN 5.961 kN

A BL

BR CL

CR D

B C

R R

R R

R R

R and R

= ↑ = ↑

= ↑ = ↑

= ↑ =

= ↑ =

↑

↑

The shear force and bending moment diagram are shown in Fig. 13.2d.


Example 13.3 Sketch the deflected shape of the continuous beam ABC of example 13.1. The redundant moments and for this problem have already been computed in problem 13.1.They are,

AM BM

1.0 kN.mBM = − 4.0 kN.mAM = −

The computed reactions are also shown in Fig.13.2c.Now to sketch the deformed shape of the beam it is required to compute rotations at B and C. These joints rotations are computed from equations (13.2) and (13.3). For calculating Aθ , consider span A’AB

ARARA αβθ −=

⎟⎠⎞

⎜⎝⎛ −

−++=436

AB

R

RA

R

RB

RR

RR

EIlM

EIlM

lEIxA δδ

⎟⎠⎞

⎜⎝⎛ −

−××

×+

×××

+××

××=

43106.14

6106.14

4106.1286

333ABAB MM δδ

( ) ( )

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×+

×××−

+××

×−+

××××

=−

4105

3106.144

6106.141

4106.1286 3

333

0= (1)

For calculating BLθ , consider span ABC

BLBLBL βαθ −=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+⎟⎟

⎠

⎞⎜⎜⎝

⎛++−=

L

BA

L

LB

L

LA

LL

LL

lEIlM

EIlM

lEIxA δδ

36

( ) ( )⎟⎟⎠

⎞⎜⎜⎝

⎛ ×+⎟

⎠⎞

⎜⎝⎛

×××−

+××

×−+

×××

−=4105

3106.141

6106.144

4106.128 3

333

31025.1 −×= radians (2)

For BRθ consider span ABC


( )⎟⎟⎠

⎞⎜⎜⎝

⎛ ×+−⎟

⎠⎞

⎜⎝⎛

×××−

+××

×=

41050

3106.141

4106.1267.10 3

33BRθ

radians (3) 31025.1 −×−=

( )

⎟⎠⎞

⎜⎝⎛ −

−⎟⎠⎞

⎜⎝⎛

×××−

+××

×−=

43106.141

4106.1267.10

33CB

Cδδ

θ

radians. (4) 31075.3 −×−=

The deflected shape of the beam is shown in Fig. 13.4.


Summary The continuous beams with unyielding supports are analysed using three-moment equations in the last lesson. In this lesson, the three-moment-equations developed in the previous lesson are extended to account for the support settlements. The three-moment equations are derived for the case of a continuous beam having different moment of inertia in different spans. Few examples are derived to illustrate the procedure of analysing continuous beams undergoing support settlements using three-moment equations.


m2l13 lesson 13 the three-moment equations-ii

Documents