Examples II - Slope Deflection Equation - Displacement Method, Strength of Materials | Strength of Material Notes - Agricultural Engg - Agricultural Engineering PDF Download

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 Example

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

Fig.14.1

Step 1: Fixed end Moments

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

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

Step 2: Slope-Deflection Equaitons

Fig.14.2

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

Since axial deformation is neglected, δ_B = δ_C = δ

For span AB,

\[{M_{AB}} = {M_{FAB}} + {{2EI} \over {{L_{AB}}}}\left( {2{\theta _A} + {\theta _B} - {{3\delta } \over {{L_{AB}}}}} \right) = 0.4EI{\theta _B} - 0.24EI\delta \]       (14.1)

\[{M_{BA}} = {M_{FBA}} + {{2EI} \over {{L_{AB}}}}\left( {{\theta _A} + 2{\theta _B} - {{3\delta } \over {{L_{AB}}}}} \right) = 0.8EI{\theta _B} - 0.24EI\delta \]       (14.2)

For span BC,

\[{M_{BC}} = {M_{FBC}} + {{2EI} \over {{L_{BC}}}}\left( {2{\theta _B} + {\theta _C} - {{3\delta } \over {{L_{BC}}}}} \right) =-62.5 + 0.2EI\left( {2{\theta _B} + {\theta _C}} \right)\]            (14.3)

\[{M_{CB}} = {M_{FCB}} + {{2EI} \over {{L_{BC}}}}\left( {2{\theta _C} + {\theta _B} - {{3\delta } \over {{L_{BC}}}}} \right) = 62.5 + 0.2EI\left( {{\theta _B} + 2{\theta _C}} \right)\]            (14.4)

For span CD,

\[{M_{CD}} = {M_{FCD}} + {{2EI} \over {{L_{CD}}}}\left( {2{\theta _C} + {\theta _D} - {{3\delta } \over {{L_{CD}}}}} \right) = 0.8EI{\theta _C} - 0.24EI\delta \]        (14.5)

\[{M_{DC}} = {M_{FDC}} + {{2EI} \over {{L_{CD}}}}\left( {{\theta _C} + 2{\theta _D} - {{3\delta } \over {{L_{CD}}}}} \right) = 0.4EI{\theta _C} - 0.24EI\delta \]        (14.6)

Step 3: Equilibrium Equaitons

At B,

\[{M_{BA}} + {M_{BC}} = 0 \Rightarrow 0.8EI{\theta _B} - 0.24EI\delta-62.5+0.2EI\left( {2{\theta _B} + {\theta _C}} \right) = 0\]      

\[\Rightarrow 1.2EI{\theta _B} + 0.2EI{\theta _C}-0.24EI\delta-62.5=0\]                 (14.7)

At C,

\[{M_{CB}} + {M_{CD}} = 0 \Rightarrow 62.5 + 0.2EI\left( {{\theta _B} + 2{\theta _C}} \right) + 0.8EI{\theta _C}--0.24EI\delta= 0\]        

\[\Rightarrow 0.2EI{\theta _B} + 1.2EI{\theta _C}-0.24EI\delta+62.5=0\]                (14.8)

Step 3: Additional Shear Equation

Free body diagram of each member are shown bellow.

Fig.14.3

Summation of force in horizontal direction is zero \[ \Rightarrow \sum {{F_x} = 0}\]

Here, horizontal forces are, external horizontal force of 10 kN and shear forces V_A and V_B respectively at support A and D.

From the above FBD, V_A and V_B may be expressed as,

\[{V_A} = {{{M_{AB}} + {M_{BA}}} \over {{L_{AB}}}} = {{0.4EI{\theta _B} - 0.24EI\delta+0.8EI{\theta _B}-0.24EI\delta }\over 5}={{1.2EI{\theta _B}-0.48EI\delta } \over 5}\]

\[{V_B} = {{{M_{AB}} + {M_{BA}}} \over {{L_{AB}}}} = {{0.8EI{\theta _C} - 0.24EI\delta+0.4EI{\theta _C}-0.24EI\delta }\over 5}={{1.2EI{\theta _C}-0.48EI\delta } \over 5}\]

Now,

\[\sum {{F_x} = 0}\Rightarrow {V_A} + {V_B} + 10 = 0\]

\[{{1.2EI{\theta _B} - 0.48EI\delta } \over 5} + {{1.2EI{\theta _C} - 0.48EI\delta } \over 5} + 10 = 0\]

\[1.2EI{\theta _B} + 1.2EI{\theta _C}-0.96EI\delta+50= 0\]                 (14.9)

Solving equations (7) – (9), we have,

\[{\theta _B} = {{78.125} \over {EI}}\]  ,   \[{\theta _C} =-{{46.875} \over {EI}}\]  ,    \[\delta=-{{91.1458} \over {EI}}\]

Step 4: End Moment calculation

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

\[{M_{AB}} = 0.4EI{\theta _B} - 0.24EI\delta= 9.375{\rm{ kNm}}\]

\[{M_{BA}} = 0.8EI{\theta _B} - 0.24EI\delta= 40.625{\rm{ kNm}}\]

\[{M_{BC}} =-62.5 + 0.2EI\left( {2{\theta _B} + {\theta _C}} \right) =-40.625{\rm{kNm}}\]

\[{M_{CB}} = 62.5 + 0.2EI\left( {{\theta _B} + 2{\theta _C}} \right) =59.375{\rm{ kNm}}\]

\[{M_{CD}} = 0.8EI{\theta _C} - 0.24EI\delta=-59.375{\rm{kNm}}\]

\[{M_{DC}} = 0.4EI{\theta _C} - 0.24EI\delta=-40.625{\rm{ kNm}}\]

 Fig. 14.4. Bending Moment Diagram.