Four Span Continuous Beam - Equal Spans, Uniformly Distributed Load

Written by Jerry Ratzlaff on . Posted in Structural

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Four Span Continuous Beam - Equal Spans, Uniformly Distributed Load formulas

 

\(\large{ R_1 = V_1 = R_5 = V_5    = 0.393\;w\;L    }\)   
\(\large{ R_2 = R_4   = 1.143\;w\;L    }\)   
\(\large{ R_3  = 0.928\;w\;L    }\)   
\(\large{ V_{2_1} =  V_{4_2}    = 0.607\;w\;L    }\)  
\(\large{ V_{2_2} =  V_{4_1}    = 0.536\;w\;L    }\)  
\(\large{ V_{3_1} =  V_{3_2}    = 0.464\;w\;L    }\)  
\(\large{ M_1  \; }\) at  \(\large{  \left( 0.393\;L \right)  \; }\) from  \(\large{ \left( R_1 \right) = M_6 \; }\)  at   \(\large{  \left( 0.393\;L \right)  \; }\) from  \(\large{ \left( R_5 \right)  \;   = 0.0772\;w\;L^2    }\)  
\(\large{ M_2 \; }\) at   \(\large{ \left( R_2 \right)  \;  = -\;0.1071\;w\;L^2    }\)  
\(\large{ M_3  \; }\) at  \(\large{  \left( 0.536\;L \right)  \; }\) from  \(\large{ \left( R_2 \right) = M_5 \; }\)  at   \(\large{  \left( 0.536\;L \right)  \; }\) from  \(\large{ \left( R_4 \right)  \;   = 0.0364\;w\;L^2    }\)  
\(\large{ M_4 \; }\) at   \(\large{ \left( R_3 \right)  \;  = -\;0.0714\;w\;L^2    }\)  
\(\large{ \Delta_{max}  \; }\) at  \(\large{  \left(  0.440\;L \right)  \; }\) from  \(\large{ \left( R_1 \right)  \; }\)  and   \(\large{  \left( R_5 \right)    =  \frac{0.0065\;w\;L^4}{\lambda\; I}    }\)  

Where:

\(\large{ I }\) = moment of inertia

\(\large{ L }\) = span length of the bending member

\(\large{ M }\) = maximum bending moment

\(\large{ P }\) = total concentrated load

\(\large{ R }\) = reaction load at bearing point

\(\large{ V }\) = shear force

\(\large{ w }\) = load per unit length

\(\large{ W }\) = total load from a uniform distribution

\(\large{ x }\) = horizontal distance from reaction to point on beam

\(\large{ \lambda  }\)   (Greek symbol lambda) = modulus of elasticity

\(\large{ \Delta }\) = deflection or deformation

 

Tags: Equations for Beam Support