# Simple Beam - Uniformly Distributed Load and Variable End Moments

Written by Jerry Ratzlaff on . Posted in Structural

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

$$\large{ R_1 = V_1 = \frac { wL } { 2 } + \frac { M_1 - M_2 } { L } }$$

$$\large{ R_2 = V_2 = \frac { wL } { 2 } - \frac { M_1 - M_2 } { L } }$$

$$\large{ V_x = w \left( \frac { L } { 2 } - x \right) + \frac { M_1 - M_2 } { L } }$$

$$\large{ a }$$  (inflection points)  $$\large{ = \sqrt{ \frac { L^2 } { 4 } - \left( \frac { M_1 + M_2 } { w } \right) + \left( \frac { M_1 + M_2 } { wL } \right)^2 } }$$

$$\large{ M_x = \frac { wx } { 2 } \left( L - x \right) + \left( \frac { M_1 - M_2 } { L } \right) x -M_1 }$$

$$\large{ M_3 }$$  at  $$\large{ \left( x = \frac { L } { 2 } + \frac { M_1 - M_2 } { wL } \right) = \frac { wL^2 } { 8 } - \frac { M_1 + M_2 } { 2 } + \frac { \left( M_1 - M_2 \right)^2 } { 2wL^2 } }$$

$$\large{ \Delta_x = \frac { wx } { 48 \lambda I } \left[ x^3 - \left( 2L + \frac { 4M_1 } { wL } - \frac { 4M_2 } { wL } \right) x^2 + \frac { 12M_1 } { w } + L^3 + \frac { 8M_1 L } { w } - \frac { 4M_2 L } { w } \right] }$$

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