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

Example 2

Draw the bending moment diagram for the following rigid frame. 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.  Fig. 18.4. Bending moment diagram.