Overhanging Beam - Point Load on Beam End

Written by Jerry Ratzlaff on . Posted in Structural

ob 5AOverhanging Beam - Point Load on Beam End Formula

\(\large{ R_1 = V_1 = \frac{Pa}{L}    }\)

\(\large{ R_2 = V_1 + V_2 =  \frac{ P }{L}   \left( L + a  \right)     }\)

\(\large{ V_2  =   P   }\)

\(\large{ M_{max}  \; }\)  at   \(\large{  \left( R_2 \right)    =  Pa   }\)

\(\large{ M_x    \; }\)  (between supports)    \(\large{  =  \frac{Pax}{L}  }\) 

\(\large{ M_{x_1}    \; }\)  (for overhang)    \(\large{  =    P  \left(  a - x_1 \right)    }\)   

\(\large{ \Delta_x    \; }\)  (between supports)    \(\large{  =    \frac{ -Pax }{6 \lambda IL}   \left( L^2 - x^2  \right)    }\)

\(\large{ \Delta_{x_1}    \; }\)  (overhang)    \(\large{  =   \frac{ Px_1 }{6 \lambda I }   \left( 2aL + 3ax_1 - x_{1}{^2}  \right)   }\)

\(\large{ \Delta_{max}    \; }\)  for overhang at   \(\large{  \left(x_1 = a \right)   =   \frac{ Pa^2 }{3 \lambda I }   \left( L + a  \right)   }\)

\(\large{ \Delta_{max}    \; }\)  between supports at   \(\large{  \left( x = \frac{L}{\sqrt{3}}   \right)    =   \frac{ -PaL^2 }{9 \sqrt{3}  \lambda I }  =  0.06415  \frac{ PaL^2 }{ \lambda I }   }\)

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