LESSON 18. Displacement Method: Moment Distribution Method – 4

18.1 Introduction : In this lesson the application of the Moment Distribution Method in frames where joint translations (side sway) are restrained is illustrated via two examples.

18.1.1 Example 1

Draw the bending moment diagram for the follwing frame. EI is constant for all members.

Module 2 Lesson 18 Fig.18.1

 Fig. 18.1.

\[{k_{BA}}={{4E{I_{BA}}} \over {{L_{BA}}}}={{4EI} \over 5}\] ,  \[{k_{BC}}={{4E{I_{BC}}} \over {{L_{BC}}}}={{2EI} \over 5}\]  and  \[{k_{CD}}={{4E{I_{CD}}} \over {{L_{CD}}}}={{4EI} \over 5}\]

Distribution factors for BA and BC are,

\[D{F_{BA}}={2 \over 3}\] ,  \[D{F_{BC}}={1 \over 3}\] ,  \[D{F_{CB}}={1 \over 3}\]  and  \[D{F_{CD}}={2 \over 3}\]

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

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

Fixed end moments are,

\[M{}_{FBC}=-{{7.5 \times {{10}^2}} \over {12}}=-62.5{\rm{kNm}}\] ;   \[M{}_{FCB}={{7.5 \times {{10}^2}} \over {12}}=62.5{\rm{kNm}}\]

\[M{}_{FAB} = M{}_{FBA} = M{}_{FCD} = M{}_{FDC}=0\] .

Calculations are performed in the following table.

Module 2 Lesson 18 Table 18.1

Module 2 Lesson 18 Fig.18.2

 Fig. 18.2. Bending moment diagram (in kNm).

Example 2

Draw the bending moment diagram for the following rigid frame.

Module 2 Lesson 18 Fig.18.3 Fig. 18.3.

\[{k_{BA}} = {{3E{I_{BA}}} \over {{L_{BA}}}} = {{3EI} \over 3} = EI\] ,  \[{k_{BC}} = {{4E{I_{BC}}} \over {{L_{BC}}}} = {{4EI} \over 5}\]  and  \[{k_{BD}} = {{4E{I_{BC}}} \over {{L_{BC}}}} = {{8EI} \over 4} = 2EI\]

Distribution factors for BA and BC are,

\[D{F_{BA}} = {5 \over {19}}\] ,  \[D{F_{BC}} = {4 \over {19}}\]  and  \[D{F_{BD}} = {{10} \over {19}}\]

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

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

Fixed end moments are,

\[M{}_{FAB}=-{{3 \times {3^2}} \over {12}}=-2.25{\rm{ kNm}}\] ;  \[M{}_{FBA} = {{3 \times {3^2}} \over {12}} = 2.25{\rm{ kNm}}\]

\[M{}_{FBC}=-{{15 \times 2 \times {3^2}} \over {{5^2}}}=-10.8{\rm{ kNm}}\] ;  \[M{}_{FCB} = {{15 \times 3 \times {2^2}} \over {{5^2}}} = 7.2{\rm{ kNm}}\]

Calculations are performed in the following table.

Module 2 Lesson 18 Table 18.2

Module 2 Lesson 18 Fig.18.4

 Fig. 18.4. Bending moment diagram.

 

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, 4:55 AM