Overhanging Beam - Uniformly Distributed Load on Overhang

Written by Jerry Ratzlaff on . Posted in Structural

Overhanging Beam - Uniformly Distributed Load on Overhangob 2A

Uniformly Distributed Load on Overhang Formula

\(\large{ R_1 = V_2 =  \frac{w a^2 }{2L}      }\)

\(\large{ R_2 = V_1 + V_2 =  \frac{w a }{2L}   \left( 2L + a  \right)   }\)

\(\large{ V_2 =  w a    }\)

\(\large{ V_{x _1} =  w   \left( a - x_1  \right)   }\)

\(\large{ M_{max} \; }\)  at  \(\large{ \left( R_2 \right)  =  \frac{ w a^2 }{2}      }\)

\(\large{ M_x \; }\)  (between supports)  \(\large{  =  \frac{ w a^2 x }{2L}      }\)

\(\large{ M_{x_1} \; }\)  (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  \(\large{ \left( x_1 = a \right)   =  \frac{  w x^3 }{24 \lambda I }   \left(  4L + 3a   \right)       }\)

\(\large{ \Delta_{x1} \; }\)  (for overhang)  \(\large{   =  \frac{  w x_1 }{24 \lambda I }   \left(  4a^2 L + 6a^2 x_1 - 4a x_{1}{^2}  + x_{1}{^3}  \right)       }\)

 

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