Four Span Continuous Beam - Equal Spans, Uniform Load on Two Spans

Written by Jerry Ratzlaff on . Posted in Structural

cb5s 3AFour Span Continuous Beam - Equal Spans, Uniform Load on Two Spans  Formula

 

 

 

 

 

 

 

 

 

 

 

 

\(\large{ R_1 = V_1   = 0.446wL    }\)

\(\large{ R_2  = 0.572wL    }\)

\(\large{ R_3  = 0.464wL    }\)

\(\large{ R_4  = 0.572wL    }\)

\(\large{ R_5  = -0.054wL    }\)

\(\large{ V_{2_1}   = 0.0554wL    }\)

\(\large{ V_{2_2} = V_{3_1}  = 0.018wL    }\)

\(\large{ V_{3_2}   = 0.482wL    }\)

\(\large{ V_{4_1}   = 0.518wL    }\)

\(\large{ V_{4_2} = V_5  = 0.054wL    }\)

\(\large{ M_1  \;   \left( 0.446L \right)  \; }\) from  \(\large{ \left( R_1 \right) = 0.0996wL^2   }\)

\(\large{ M_2  \; }\) at  \(\large{ \left( R_2 \right)  \;   = 0.0536wL^2    }\)

\(\large{ M_3  \; }\) at  \(\large{ \left( R_3 \right)  \;   = -0.0357wL^2    }\)

\(\large{ M_4  \;   \left( 0.518L \right)  \; }\) from  \(\large{ \left( R_4 \right) = 0.805wL^2   }\)

\(\large{ M_5  \; }\) at  \(\large{ \left( R_4 \right)  \;   = -0.0536wL^2    }\)

\(\large{ \Delta_{max}  \; }\) at  \(\large{  \left(  0.477L \right)  \; }\) from  \(\large{ \left( R_1 \right)  \;   =  \frac{0.0097wL^4}{\lambda I}    }\)

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