Analysis of Frames Without Sidesway - Slope Deflection Equation - Displacement Method, | Strength of Material Notes - Agricultural Engg - Agricultural Engineering PDF Download

Analysis of Frames Without Sidesway

The general procedure for analysis of frames without sidesway is same as for continous beams (lesson 12). This is illustrated in the following two examples.

Example 1

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

Fig. 13.2.

Step 1: Fixed end Moments

\[M{}_{FAB} =-{{5 \times 4} \over 8}=-2.5{\rm{kNm}}\]  ;   \[M{}_{FCB} = {{7.5 \times {{10}^2}} \over {12}} = 62.5{\rm{kNm}}\]

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

Step 2: Slope-Deflection Equaitons

Since A and D are fixed ends, θ_A = θ_D = 0

Since there is no support settlement, δ = 0

For span AB,

\[{M_{AB}} = {M_{FAB}} + {{2EI} \over {{L_{AB}}}}\left( {2{\theta _A} + {\theta _B} - {{3\delta } \over {{L_{AB}}}}} \right) = 0.4EI{\theta _B}\]               (13.1)

\[{M_{BA}} = {M_{FBA}} + {{2EI} \over {{L_{AB}}}}\left( {{\theta _A} + 2{\theta _B} - {{3\delta } \over {{L_{AB}}}}} \right) = 0.8EI{\theta _B}\]               (13.2)

For span BC,

\[{M_{BC}} = {M_{FBC}} + {{2EI} \over {{L_{BC}}}}\left( {2{\theta _B} + {\theta _C} - {{3\delta } \over {{L_{BC}}}}} \right)=-62.5 + 0.2EI\left( {2{\theta _B} + {\theta _C}} \right)\]               (13.3)

\[{M_{CB}} = {M_{FCB}} + {{2EI} \over {{L_{BC}}}}\left( {2{\theta _C} + {\theta _B} - {{3\delta } \over {{L_{BC}}}}} \right) = 62.5 + 0.2EI\left( {{\theta _B} + 2{\theta _C}} \right)\]               (13.4)

For span CD,

\[{M_{CD}} = {M_{FCD}} + {{2EI} \over {{L_{CD}}}}\left( {2{\theta _C} + {\theta _D} - {{3\delta } \over {{L_{CD}}}}} \right) = 0.8EI{\theta _C}\]              (13.5)

\[{M_{DC}} = {M_{FDC}} + {{2EI} \over {{L_{CD}}}}\left( {{\theta _C} + 2{\theta _D} - {{3\delta } \over {{L_{CD}}}}} \right) = 0.4EI{\theta _C}\]              (13.6)

Step 3: Equilibrium Equaitons

At B,

\[{M_{BA}} + {M_{BC}} = 0 \Rightarrow 0.8EI{\theta _B} - 62.5 + 0.2EI\left( {2{\theta _B} + {\theta _C}} \right)=0\]

\[\Rightarrow 1.2EI{\theta _B} + 0.2EI{\theta _C} - 62.5 = 0\]             (13.7)

At C,

\[{M_{CB}} + {M_{CD}} = 0 \Rightarrow 62.5 + 0.2EI\left( {{\theta _B} + 2{\theta _C}} \right) + 0.8EI{\theta _C}=0\]

\[\Rightarrow 0.2EI{\theta _B} + 1.2EI{\theta _C} + 62.5 = 0\]            (13.8)

Solving equations (7) and (8),

\[{\theta _B} = {{62.5} \over {EI}}\]  and  \[{\theta _C}=-{{62.5} \over {EI}}\]

Step 4: End Moment calculation

Substituting,  θ_B and θ_C into equations (1) – (6), we have,

\[{M_{AB}} = 0.4EI{\theta _B} = 25{\rm{ kNm}}\]

\[{M_{BA}} = 0.8EI{\theta _B} = 50{\rm{ kNm}}\]

\[{M_{BC}}=-62.5 + 0.2EI\left( {2{\theta _B} + {\theta _C}} \right) =-50{\rm{ kNm}}\]

\[{M_{CB}} = 62.5 + 0.2EI\left( {{\theta _B} + 2{\theta _C}} \right) = 50{\rm{ kNm}}{M_{CB}} = 62.5 + 0.2EI\left( {{\theta _B} + 2{\theta _C}} \right) = 50{\rm{ kNm}}\]

\[{M_{CD}} = 0.8EI{\theta _C} =-50{\rm{ kNm}}\]

\[{M_{DC}} = 0.4EI{\theta _C} =-25{\rm{ kNm}}\]

Fig. 13.3. Bending Moment Diagram.

Example 2

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

Fig.13.4.

Step 1: Fixed end Moments

\[M{}_{FAB}=-{{5 \times 4} \over 8}=-2.5{\rm{kNm}}\]  ;               \[M{}_{FBA} = {{5 \times 4} \over 8} = 2.5{\rm{kNm}}\]

\[M{}_{FBC}=-{{4 \times 4} \over 8}=-2{\rm{kNm}}\] ;                   \[M{}_{FCD}=-{{4 \times 4} \over 8}=-2{\rm{kNm}}\] 

Step 2: Slope-Deflection Equaitons

Since A and C are fixed ends, θ_A = θ_C = 0

Since there is no support settlement, δ = 0

For span AB,

\[{M_{AB}} = {M_{FAB}} + {{2EI} \over {{L_{AB}}}}\left( {2{\theta _A} + {\theta _B} - {{3\delta } \over {{L_{AB}}}}} \right) =-2.5 + 0.5EI{\theta _B}\]                 (13.9)

\[{M_{BA}} = {M_{FBA}} + {{2EI} \over {{L_{AB}}}}\left( {{\theta _A} + 2{\theta _B} - {{3\delta } \over {{L_{AB}}}}} \right) = 2.5 + EI{\theta _B}\]                       (13.10)

For span BC,

\[{M_{BC}} = {M_{FBC}} + {{2EI} \over {{L_{BC}}}}\left( {2{\theta _B} + {\theta _C} - {{3\delta } \over {{L_{BC}}}}} \right) =-2 + EI{\theta _B}\]                          (13.11)

\[{M_{CB}} = {M_{FCB}} + {{2EI} \over {{L_{BC}}}}\left( {2{\theta _C} + {\theta _B} - {{3\delta } \over {{L_{BC}}}}} \right) = 2 + 0.5EI{\theta _B}\]                     (13.12)

For span BD,

\[{M_{BD}} =-{{1.5 \times {2^2}} \over 2} =  - 3{\rm{ kNm}}\]                         (13.3)

Step 3: Equilibrium Equaitons

At B,

\[{M_{BA}} + {M_{BC}} + {M_{BD}} = 0 \Rightarrow 2.5 + EI{\theta _B} - 2 + EI{\theta _B} - 3 = 0\]

\[\Rightarrow 2EI{\theta _B} - 19.5 = 0 \Rightarrow {\theta _B} = {{1.25} \over {EI}}\]

Step 4: End Moment calculation

Substituting, θ_B into equations (9) – (12), we have,

\[{M_{AB}} =-2.5 + 0.5EI{\theta _B} =-1.875{\rm{ kNm}}\]

\[{M_{BA}} = 2.5 + EI{\theta _B} = 3.75{\rm{ kNm}}\]

\[{M_{BC}} =-2 + EI{\theta _B} =-0.75{\rm{kNm}}\]

\[{M_{CB}} = 2 + 0.5EI{\theta _B} = 2.265{\rm{ kNm}}\]

Fig.13.5. Bending Moment Diagram.