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

Written by Jerry Ratzlaff on . Posted in Structural

cb5s 2Aformulas that use Four Span Continuous Beam - Equal Spans, Uniform Load on Three Spans

 

 

 

 

 

 

 

 

 

 

\(\large{ R_1 = V_1   = 0.380\;w\;L    }\)   
\(\large{ R_2  = 1.223\;w\;L    }\)   
\(\large{ R_3  = 0.357\;w\;L    }\)   
\(\large{ R_4  = 0.598\;w\;L    }\)  
\(\large{ R_5 = V_5   = 0.442\;w\;L    }\)  
\(\large{ V_{2_1}   = 0.620\;w\;L    }\)  
\(\large{ V_{2_2}   = 0.603\;w\;L    }\)  
\(\large{ V_{3_1}   = 0.397\;w\;L    }\)  
\(\large{ V_{3_2} =  V_{4_1}    = 0.040\;w\;L    }\)  
\(\large{ V_{4_2}  = 0.558\;w\;L    }\)  
\(\large{ M_1  \;   \left( 0.380\;L \right)  \; }\) from  \(\large{ \left( R_1 \right) = 0.072\;w\;L^2   }\)  
\(\large{ M_2  \; }\) at  \(\large{ \left( R_2 \right)  \;   = -\;0.1205\;w\;L^2    }\)  
\(\large{ M_3  \;   \left( 0.603\;L \right)  \; }\) from  \(\large{ \left( R_2 \right) = 0.611\;w\;L^2   }\)  
\(\large{ M_4  \; }\) at  \(\large{ \left( R_3 \right)  \;   = -\;0.0179\;w\;L^2    }\)  
\(\large{ M_5  \; }\) at  \(\large{ \left( R_4 \right)  \;   = -\;0.058\;w\;L^2    }\)  
\(\large{ M_6  \;   \left( 0.442\;L \right)  \; }\) from  \(\large{ \left( R_5 \right) = 0.0977\;w\;L^2   }\)  
\(\large{ \Delta_{max}  \; }\) at  \(\large{  \left(  0.475\;L \right)  \; }\) from  \(\large{ \left( R_5 \right)  \;   =  \frac{0.0094\;w\;L^4}{\lambda\; I}    }\)  

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