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

$$\large{ R_1 = V_1 = \frac{w \;a}{2\;L} \; \left( 2\;L - a \right) }$$

$$\large{ R_2 = V_2 = \frac{ w \;a^2 }{2\;L} }$$

$$\large{ V_x \; }$$  when $$\large{ \left( x < a \right) = R_1 - w\;x }$$

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

$$\large{ M_x \; }$$  when $$\large{ \left( x < a \right) = R_1 \; x - \frac {w\;x^2}{2} }$$

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

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

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

Where:

$$\large{ \Delta }$$ = deflection or deformation

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

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

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

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

$$\large{ \lambda }$$   (Greek symbol lambda) = modulus of elasticity

$$\large{ I }$$ = moment of inertia

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

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

$$\large{ a }$$ = width of UDL