Overhanging Beam - Uniformly Distributed Load

Written by Jerry Ratzlaff on . Posted in Structural

Overhanging Beam - Uniformly Distributed Loadob 1A

Uniformly Distributed Load Formula

\(\large{ R_1 = V_1 = \frac{w}{2L}    \left( L^2 - a^2  \right)     }\)

\(\large{ R_2 = V_2 + V_3 =  \frac{ w }{2L}   \left( L + a  \right)^2     }\)

\(\large{ V_2  =   wa   }\)

\(\large{ V_3  =  \frac{ w }{2L}   \left( L^2 + a^2  \right)    }\)

\(\large{ V_x    \; }\)  (between supports)    \(\large{  =     R_1 - wx  }\)  

\(\large{ V_{x_1}    \; }\)  (for overhang)    \(\large{  =    w  \left(  a - x_1 \right)    }\)    

\(\large{ M_x    \; }\)  (between supports)    \(\large{  =    \frac{ wx }{2L}   \left( L^2 - a^2 - xL  \right)    }\)

\(\large{ M_{x_1}    \; }\)  (overhang)    \(\large{  =    \frac{ w }{2}   \left( a - x_1  \right)^2    }\)

\(\large{ M_2    \; }\)  at   \(\large{ \left( R_2 \right)  =    \frac{ wa^2 }{2}     }\)

\(\large{ M_1  \; }\)  at   \(\large{ \left[  x = \frac{L}{2}    \left(   1 - \frac{a^2}{L^2}   \right)   \right]   =    \frac{ w }{8 L^2}    \left(  L + a  \right)^2     \left(  L - a  \right)^2          }\)

\(\large{ \Delta_x  \; }\)  (between supports)   \(\large{   =   \frac { w x} { 24 \lambda I L}    \left(  L^4  - 2L^2x^2  + Lx^3  - 2a^2L^2 + 2a^2x^2   \right)    }\)

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

 

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