Examples of frames where joint translations (side sway) are restrained - Moment Distribution Method | Strength of Material Notes - Agricultural Engg - Agricultural Engineering PDF Download

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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.