Simple Beam - Uniform Load Partially Distributed at One End

Written by Jerry Ratzlaff on . Posted in Structural

sb 4DSimple Beam - Uniform Load Partially Distributed at One End Formula

(Eq. 1)  \(\large{ R_1 = V_1 = \frac {w a} {2L}    \left(  2L - a  \right)     }\)

(Eq. 2)  \(\large{ R_2 = V_2 = \frac { w a^2 } {2L}  }\)

(Eq. 3)  \(\large{ V_x    \; }\)  when \(\large{ \left(  x < a \right)  =     R_1 - wx  }\)       

(Eq. 4)  \(\large{ M_{max} \; }\)  at \(\large{ \left(  x = \frac {R_1}{w}  \right)  =  \frac { R_{1}{^2} } { 2w }    }\)

(Eq. 5)  \(\large{ M_x \; }\)  when \(\large{ \left(  x < a \right)  =     R_1  x -  \frac {wx^2}{2}  }\)

(Eq. 6)  \(\large{ M_x \; }\)  when \(\large{ \left(  x > a \right)  =     R_2   \left(  L - x  \right)   }\)

(Eq. 7)  \(\large{ \Delta_x  \; }\)  when \(\large{ \left(  x < a \right)  =   \frac { w x} { 24 \lambda I L}   \left[ a^2    \left(  2L - a  \right)^2 - 2ax^2    \left(  2L - a  \right)  + Lx^3  \right]    }\)

(Eq. 8)  \(\large{ \Delta_x  \; }\)  when \(\large{ \left(  x > a \right)   =  \frac { w a^2   \left(  L - x  \right)  } { 24 \lambda I L}   \left( 4xL - 2x^2 - a^2     \right)    }\)

Where:

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

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

\(\large{ a }\) = width of UDL

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

\(\large{ R }\) = reaction load at bearing point

\(\large{ V }\) = maximum shear force

\(\large{ w }\) = load per unit length

\(\large{ x }\) = 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