Overhanging Beam - Uniformly Distributed Load on Overhang

Written by Jerry Ratzlaff on . Posted in Structural

ob 2Aformulas that use Overhanging Beam - Uniformly Distributed Load on Overhang

\(\large{ R_1 = V_2 =  \frac{w\; a^2 }{2\;L}      }\)   
\(\large{ R_2 = V_1 + V_2 =  \frac{w\; a }{2\;L} \;  \left( 2\;L + a  \right)   }\)   
\(\large{ V_2 =  w \;a    }\)   
\(\large{ V_{x _1} =  w \;  \left( a - x_1  \right)   }\)  
\(\large{ M_{max} \; }\)  at  \(\large{ \left( R_2 \right)  =  \frac{ w \;a^2 }{2}      }\)  
\(\large{ M_x \; }\)  (between supports)  \(\large{  =  \frac{ w \;a^2 \;x }{2\;L}      }\)  
\(\large{ M_{x_1} \; }\)  (for overhang)  \(\large{  =  \frac{ w }{2} \; \left( a - x_1 \right)^2        }\)  
\(\large{ \Delta_x \; }\)  (between supports)  \(\large{  =  \frac{ - \;w \;a^2\; x }{12\; \lambda\; I \;L} \; \left( L^2 - x^2 \right)        }\)  
\(\large{ \Delta_{max} \; }\)  between supports at  \(\large{ \left( x = \frac{L}{\sqrt{3}} \right)     =  \frac{ - \;w\; a^2 \;L^2 }{18 \; \sqrt{3} \; \lambda\; I } =  0.3208\;  \frac{ w \;a^2 \; L^2 }{\lambda\; I}        }\)  
\(\large{ \Delta_{max} \; }\)  for overhang at  \(\large{ \left( x_1 = a \right)   =  \frac{  w\; x^3 }{24\; \lambda\; I }  \; \left(  4\;L + 3\;a   \right)       }\)  
\(\large{ \Delta_{x1} \; }\)  (for overhang)  \(\large{   =  \frac{  w\; x_1 }{24\; \lambda\; I } \;  \left(  4\;a^2 \;L + 6\;a^2\; x_1 - 4\;a \;x_{1}{^2}  + x_{1}{^3}  \right)       }\)  

Where:

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

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

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

\(\large{ R }\) = 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