# Simple Beam - Uniform Load Partially Distributed at One End

Written by Jerry Ratzlaff on . Posted in Structural

### Simple 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