Four Span Continuous Beam - Equal Spans, Uniformly Distributed Load

Written by Jerry Ratzlaff on . Posted in Structural

cb5s 1AFour Span Continuous Beam - Equal Spans, Uniformly Distributed Load Formula

 

 

 

 

 

 

 

 

 

 

 

 

\(\large{ R_1 = V_1 = R_5 = V_5    = 0.393wL    }\)

\(\large{ R_2 = R_4   = 1.143wL    }\)

\(\large{ R_3  = 0.928wL    }\)

\(\large{ V_{2_1} =  V_{4_2}    = 0.607wL    }\)

\(\large{ V_{2_2} =  V_{4_1}    = 0.536wL    }\)

\(\large{ V_{3_1} =  V_{3_2}    = 0.464wL    }\)

\(\large{ M_1  \; }\) at  \(\large{  \left( 0.393L \right)  \; }\) from  \(\large{ \left( R_1 \right) = M_6 \; }\)  at   \(\large{  \left( 0.393L \right)  \; }\) from  \(\large{ \left( R_5 \right)  \;   = 0.0772wL^2    }\)

\(\large{ M_2 \; }\) at   \(\large{ \left( R_2 \right)  \;  = -0.1071wL^2    }\)

\(\large{ M_3  \; }\) at  \(\large{  \left( 0.536L \right)  \; }\) from  \(\large{ \left( R_2 \right) = M_5 \; }\)  at   \(\large{  \left( 0.536L \right)  \; }\) from  \(\large{ \left( R_4 \right)  \;   = 0.0364wL^2    }\)

\(\large{ M_4 \; }\) at   \(\large{ \left( R_3 \right)  \;  = -0.0714wL^2    }\)

\(\large{ \Delta_{max}  \; }\) at  \(\large{  \left(  0.440L \right)  \; }\) from  \(\large{ \left( R_1 \right)  \; }\)  and   \(\large{  \left( R_5 \right)    =  \frac{0.0065wL^4}{\lambda I}    }\)

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