# Simple Beam - Uniformly Distributed Load and Variable End Moments

Written by Jerry Ratzlaff on . Posted in Structural Engineering

## Simple Beam - Uniformly Distributed Load and Variable End Moments formulas

 $$\large{ R_1 = V_1 = \frac { w\;L } { 2 } + \frac { M_1\; - \;M_2 } { L } }$$ $$\large{ R_2 = V_2 = \frac { w\;L } { 2 } - \frac { M_1\; - \;M_2 } { L } }$$ $$\large{ V_x = w \; \left( \frac { L } { 2 } - x \right) + \frac { M_1 \;-\; M_2 } { L } }$$ $$\large{ a }$$  (inflection points)  $$\large{ = \sqrt{ \frac { L^2 } { 4 } - \left( \frac { M_1 \;+ \;M_2 } { w } \right) + \left( \frac { M_1 \;+\; M_2 } { w\;L } \right)^2 } }$$ $$\large{ M_x = \frac { w\;x } { 2 } \; \left( L - x \right) + \left( \frac { M_1\; - \;M_2 } { L } \right) x -M_1 }$$ $$\large{ M_3 }$$  at  $$\large{ \left( x = \frac { L } { 2 } + \frac { M_1\; - \;M_2 } { w\;L } \right) = \frac { w\;L^2 } { 8 } - \frac { M_1 \;+ \;M_2 } { 2 } + \frac { \left( M_1 \;- \;M_2 \right)^2 } { 2\;w\;L^2 } }$$ $$\large{ \Delta_x = \frac { w\;x } { 48\; \lambda\; I } \; \left[ x^3 - \; \left( 2\;L + \frac { 4\;M_1 } { w\;L } - \frac { 4\;M_2 } { w\;L } \right) x^2 + \frac { 12\;M_1 } { w } + L^3 + \frac { 8\;M_1 \;L } { w } - \frac { 4\;M_2\; L } { w } \right] }$$

### 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

$$\large{ V }$$ = 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