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

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Example
Calculate the end momens and joint rotations for the continous beam shwon bellow. The beam has constant EI for both the span.

Fig.12.1

Step 1: Fixed end Moments
\[M{}_{FAB}=-{{3 \times {5^2}} \over {12}}=-6.25{\rm{ kNm}}\] ;      \[M{}_{FBA} = {{3 \times {5^2}} \over {12}} = 6.25{\rm{ kNm}}\]
\[M{}_{FBC}=-{{10 \times 2 \times {3^2}} \over {{5^2}}}=-7.2{\rm{ kNm}}\] ;    \[M{}_{FCB}=-{{10 \times 3 \times {2^2}} \over {{5^2}}} = 4.8{\rm{ kNm}}\]
Step 2: Slope-Deflection Equaitons
For span AB,
\[{M_{AB}} ={M_{FAB}} + {{2EI} \over {{L_{BC}}}}\left( {2{\theta _A} + {\theta _B} - {{3\delta } \over {{L_{BC}}}}} \right)=-6.25{\rm{ }} + {{2EI} \over 5}{\theta _B} =-6.25{\rm{ }} + 0.4EI{\theta _B}\]        (1)

\[{M_{BA}} ={M_{FBA}} + {{2EI} \over {{L_{BC}}}}\left( {2{\theta _B} + {\theta _C} - {{3\delta } \over {{L_{BC}}}}} \right) = 6.25 + {{4EI} \over 5}{\theta _B} = 6.25 + 0.8EI{\theta _B}\]         (2)

For span BC,

\[{M_{BC}} = {M_{FBC}} + {{2EI} \over {{L_{BC}}}}\left( {2{\theta _B} + {\theta _C} - {{3\delta } \over {{L_{BC}}}}} \right)=-7.2 + {{2EI} \over 5}\left( {2{\theta _B} + {\theta _C}} \right)=-7.2 + 0.4EI\left( {2{\theta _B} + {\theta _C}} \right)\]           (3)

\[{M_{CB}} = {M_{FCB}} + {{2EI} \over {{L_{BC}}}}\left( {2{\theta _C} + {\theta _B} - {{3\delta } \over {{L_{BC}}}}} \right) = 4.8 + {{2EI} \over 5}\left( {2{\theta _C} + {\theta _B}} \right) = 4.8 + 0.4EI\left( {2{\theta _C} + {\theta _B}} \right)\]           (4)

Step 3: Equilibrium Equaitons

At B,

\[{M_{BA}} + {M_{BC}} = 0 \Rightarrow 6.25 + 0.8EI{\theta _B} - 7.2 + 0.4EI\left( {2{\theta _B} + {\theta _C}} \right)=0\]

\[\Rightarrow 1.6EI{\theta _B} + 0.4EI{\theta _C} - 0.95 = 0\]             (5)

At C,

\[{M_{CB}} = 0 \Rightarrow 0.4EI{\theta _B} + 0.8EI{\theta _C} + 4.8 = 0\]          (6)

Solving (1) and (2), we have,

\[{\theta _B} = {{2.3929} \over {EI}}\] ;   \[{\theta _C}=-{{7.1964} \over {EI}}\]

Step 4: End Moment calculation

Substituting, θ_B  and θ_C into equations (1) – (4), we have,

\[{M_{AB}}=-6.25{\rm{ }} + 0.4EI{\theta _B} =-5.29\]

\[{M_{BA}} = 6.25 + 0.8EI{\theta _B}=8.16\]

\[{M_{BC}}=-7.2 + 0.4EI\left( {2{\theta _B} + {\theta _C}} \right) =-8.16\]