LESSON 17. Displacement Method: Moment Distribution Method – 3

17.1 Introduction: In this lesson the application of the Moment Distribution Method in continusous beam is illustrated via two examples.

Example 1

Draw the bending moment diagram for the following continuous beam. All spans have constant EI.

Module 2 Lesson 17 Fig.17.1Fig. 17.1.

From lesson 15.1.3 we have,

\[{k_{BA}} = {{4E{I_{BA}}} \over {{L_{BA}}}} = {{4EI} \over 5}\]  and  \[{k_{BC}} = {{3E{I_{BC}}} \over {{L_{BC}}}} = {{3EI} \over 5}\]

Distribution factors for BA and BC are,

\[D{F_{BA}} = {4 \over 7}\]  and  \[D{F_{BC}} = {3 \over 7}\]

End A is fixed and therefore no moment will be carrid over to B from A. Carry over factors for other joints,

\[{C_{BA}} = {1 \over 2}\] ,  \[{C_{BC}} = {1 \over 2}\]  ,  \[{C_{CB}} = {1 \over 2}\]

Fixed end moments are,

\[M{}_{FAB}=-{{3 \times {5^2}} \over {12}}=-6.25{\rm{ kNm}}\] ;  \[M{}_{FBA} = {{3 \times {5^2}} \over {12}} = 6.25{\rm{ kNm}}\]

\[M{}_{FBC}=-{{10 \times 2 \times {3^2}} \over {{5^2}}}=-7.2{\rm{ kNm}}\] ;  \[M{}_{FCB} = {{10 \times 3 \times {2^2}} \over {{5^2}}} = 4.8{\rm{ kNm}}\]

Calculations are performed in the following table.

Module 2 Lesson 17 Table17.1

Module 2 Lesson 17 Fig.17.2Fig. 17.2: Bending moment diagram (in kNm).

Example 2

Replace the fixed support at A by a hinge in the continuous beam shown in Example 1 and determine the  bending moments.

Module 2 Lesson 17 Fig.17.3

Fig. 17.3 .

From lesson 15.1.3 we have,

\[{k_{BA}} = {{4E{I_{BA}}} \over {{L_{BA}}}} = {{4EI} \over 5}\]  and  \[{k_{BC}} = {{3E{I_{BC}}} \over {{L_{BC}}}} = {{3EI} \over 5}\]

Distribution factors for BA and BC are,

\[D{F_{BA}} = {4 \over 7}\]  and  \[D{F_{BA}} = {3 \over 7}\]

Carry over factors,

\[{C_{AB}} = {1 \over 2}\] ,  \[{C_{BA}} = {1 \over 2}\] ,  \[{C_{BC}} = {1 \over 2}\] , \[{C_{CB}} = {1 \over 2}\]

Fixed end moments are,

\[M{}_{FAB}=-{{3 \times {5^2}} \over {12}}=-6.25{\rm{ kNm}}\] ;  \[M{}_{FBA} = {{3 \times {5^2}} \over {12}} = 6.25{\rm{ kNm}}\]

\[M{}_{FBC}=-{{10 \times 2 \times {3^2}} \over {{5^2}}}=-7.2{\rm{ kNm}}\] ;  \[M{}_{FCB}=-{{10 \times 3 \times {2^2}} \over {{5^2}}} = 4.8{\rm{ kNm}}\]

Calculations are performed in the following table.

Module 2 Lesson 17 Table17.2

 Module 2 Lesson 17 Fig.17.4

Fig. 17.4. Bending moment diagram (in kNm).

 

Suggested Readings

Hbbeler, R. C. (2002). Structural Analysis, Pearson Education (Singapore) Pte. Ltd.,Delhi.

Jain, A.K., Punmia, B.C., Jain, A.K., (2004). Theory of Structures. Twelfth Edition, Laxmi Publications. 

Menon, D.,  (2008), Structural Analysis, Narosa Publishing House Pvt. Ltd., New Delhi.

Hsieh, Y.Y., (1987), Elementry Theory of Structures , Third Ddition, Prentrice Hall.

Last modified: Saturday, 21 September 2013, 5:17 AM