# Simple Beam - Uniform Load Partially Distributed at Each End

Written by Jerry Ratzlaff on . Posted in Structural Engineering

## Simple Beam - Uniform Load Partially Distributed at Each End formulas

 $$\large{ R_1 = V_1 = \frac{ w_1 \;a \; \left( 2\;L \;-\; a \right) \; + \;w_2 \;c^2 }{ 2\;L } }$$ $$\large{ R_2 = V_2 = \frac{ w_2 \;c \; \left( 2\;L \;-\; c \right) \;+\; w_1\; a^2 }{ 2\;L } }$$ $$\large{ V_x \; }$$  when $$\large{ \left( x < a \right) = R_1 - w_1 \;x }$$ $$\large{ V_x \; }$$  when $$\large{ \left[ a < x < \left( a + b \right) \right] = R_1 - w_1 \;a }$$ $$\large{ V_x \; }$$  when $$\large{ \left[ x > \left( a + b \right) \right] = R_2 - w_2 \left( 1 - x \right) }$$ $$\large{ M_{max} \; }$$   at  $$\large{ \left( x = \frac{R_1}{w_1} \right) }$$  when  $$\large{ \left( R_1 < w_1 \;a \right) = \frac{ R_{1}{^2} }{ 2\;w_1 } }$$ $$\large{ M_{max} \; }$$   at  $$\large{ \left( x = L - \frac{R_2}{w_2} \right) }$$  when  $$\large{ \left( R_2 < w_2 \;c \right) = \frac{ R_{2}{^2} }{ 2\;w_2 } }$$ $$\large{ M_x \; }$$  when $$\large{ \left( w < a \right) = R_1 \;x - \frac{ w_1 \;x^2}{ 2 } }$$ $$\large{ M_x \; }$$  when $$\large{ \left[ a < x < \left( a + b \right) \right] = R_1\; x - \frac{ w_1 \;a}{ 2 } \; \left( 2\;x - a \right) }$$ $$\large{ M_x \; }$$  when $$\large{ \left[ x > \left( a + b \right) \right] = R_2 \; \left( L - x \right) - \frac{ w_2 \; \left( L\; -\; x \right)^2 }{ 2 } }$$

### Where:

$$\large{ x }$$ = horizontal distance from reaction to point on beam

$$\large{ w }$$ = load per unit length

$$\large{ M }$$ = maximum bending moment

$$\large{ V }$$ = maximum shear force

$$\large{ R }$$ = reaction load at bearing point

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

$$\large{ a, b, c }$$ = width and seperation of UDL