Examples of frames where joint translation (sidesway) is allowed - Moment Distribution Method - Disp | Strength of Material Notes - Agricultural Engg - Agricultural Engineering PDF Download

Example 1

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

Fig. 19.1.

The above problem can be represented as the superposition of two sub-problems as shwon in Figure 18.2.

Fig. 19.2.

In the first sub-problem (Figure 18.2a), sidesway is prevented and therefore can be analyzed by moment distribtuin mehod as discussed in the previous lesson. Suppose C_x be the horizontal reaction at C. Now in the second sub-problem apply an arbitrary sidesway d as shown in Figure 18.2b. Calculate horizontal force F due to arbitrary sidesway d. Then the beam end moment in the orizinal structure is obtained as,

\[M{}_{original} = M{}_{sub - problem1} + kM{}_{sub - problem2}\]

Where, \[k={{{C_x}} \over F}\]

Step1: Solution of sub-problem 1 (sidesway restrained)

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

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

\[C_{BA}={1 \over 2}\] ,  \[C_{BC}={1 \over 2}\] ,  \[C_{CB}={1 \over 2}\]  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. 19.3.

From the free body diagram of AB and CD (Figure 18.3),

\[{A_x}={{{M_{AB}} + {M_{BA}}} \over {{L_{AB}}}}={{25 + 50} \over 5}=75{\rm{ kN}}\]

\[{D_x}={{{M_{CD}} + {M_{DC}}} \over {{L_{CD}}}}={{ - 25 - 50} \over 5}=75{\rm{ kN}}\]

Also,

\[{C_x}=10 + {A_x} + {B_x}=10{\rm{ - 75}} + 7{\rm{5}}={\rm{1}}0{\rm{ kN}}\]

Step 2: Solution of sub-problem 2 (for an arbitrary sidesway)

Let an arbitray sway δ = 100 / EI is applied at C. Fixed end moments due to δ are,

\[M{}_{FAB}=M{}_{FBA}=-{{6EI\delta } \over {{L_{AB}}}}=-{{6EI} \over 5} \times {{100} \over {EI}}=-120{\rm{ kNm}}\]

\[M{}_{FCD}=M{}_{FDC}=-{{6EI\delta } \over {{L_{CD}}}}=-{{6EI} \over 5} \times {{100} \over {EI}}=-120{\rm{ kNm}}\]

Moment distribution calculations,

Horizontal reactions at A and D are,

\[{A_x}={{{M_{AB}} + {M_{BA}}} \over {{L_{AB}}}}={{ - 85.75 - 51.45} \over 5}=-27.44{\rm{ kN}}\]

\[{D_x}={{{M_{CD}} + {M_{DC}}} \over {{L_{CD}}}}={{ - 85.75 - 51.45} \over 5}=-27.44{\rm{ kN}}\]

Therefore, total horizontal force due to lateral sway δ = 100 / EI  an be ditermined by the following equilibrium equation,

\[F + {A_x} + {D_x}=0 \Rightarrow F = 54.88{\rm{ kN}}\] 

Hence, \[k = {C_x}/F=10/54.88\]

Final moment now can be obtained as,

\[M{}_{original}=M{}_{sub - problem1} + kM{}_{sub - problem2}\]

Therefore,

\[M{}_{AB} = M{}_{AB1} + kM{}_{AB2}=25 + \left( {10/54.88} \right) \times ( - 85.75) = 9.375{\rm{ kNm}}\]

\[M{}_{BA} = M{}_{BA1} + kM{}_{BA2}=50 + \left( {10/54.88} \right) \times ( - 51.45)=40.625{\rm{ kNm}}\]

\[M{}_{BC}=M{}_{BC1} + kM{}_{BC2}=-50 + \left( {10/54.88} \right) \times (51.45)=-40.625{\rm{ kNm}}\]

\[M{}_{CB}=M{}_{CB1} + kM{}_{CB2}=50 + \left( {10/54.88} \right) \times (51.45)=59.375{\rm{ kNm}}\]

\[M{}_{CD}=M{}_{CD1} + kM{}_{CD2}=-50 + \left( {10/54.88} \right) \times ( - 51.45)=-59.375{\rm{ kNm}}\]

\[M{}_{DC}=M{}_{DC1} + kM{}_{DC2}=-25 + \left( {10/54.88} \right) \times ( - 85.75)=-40.625{\rm{ kNm}}\]

Figure 19.4: Bending moment Diagram.