# Overhanging Beam - Uniformly Distributed Load on Overhang

Written by Jerry Ratzlaff on . Posted in Structural Engineering

## Overhanging Beam - Uniformly Distributed Load on Overhang formulas

 $$\large{ R_1 = V_2 = \frac{w\; a^2 }{2\;L} }$$ $$\large{ R_2 = V_1 + V_2 = \frac{w\; a }{2\;L} \; \left( 2\;L + a \right) }$$ $$\large{ V_2 = w \;a }$$ $$\large{ V_{x _1} = w \; \left( a - x_1 \right) }$$ $$\large{ M_{max} \; }$$  at  $$\large{ \left( R_2 \right) = \frac{ w \;a^2 }{2} }$$ $$\large{ M_x \; }$$  (between supports)  $$\large{ = \frac{ w \;a^2 \;x }{2\;L} }$$ $$\large{ M_{x_1} \; }$$  (for overhang)  $$\large{ = \frac{ w }{2} \; \left( a - x_1 \right)^2 }$$ $$\large{ \Delta_x \; }$$  (between supports)  $$\large{ = \frac{ - \;w \;a^2\; x }{12\; \lambda\; I \;L} \; \left( L^2 - x^2 \right) }$$ $$\large{ \Delta_{max} \; }$$  between supports at  $$\large{ \left( x = \frac{L}{\sqrt{3}} \right) = \frac{ - \;w\; a^2 \;L^2 }{18 \; \sqrt{3} \; \lambda\; I } = 0.3208\; \frac{ w \;a^2 \; L^2 }{\lambda\; I} }$$ $$\large{ \Delta_{max} \; }$$  for overhang at  $$\large{ \left( x_1 = a \right) = \frac{ w\; x^3 }{24\; \lambda\; I } \; \left( 4\;L + 3\;a \right) }$$ $$\large{ \Delta_{x1} \; }$$  (for overhang)  $$\large{ = \frac{ w\; x_1 }{24\; \lambda\; I } \; \left( 4\;a^2 \;L + 6\;a^2\; x_1 - 4\;a \;x_{1}{^2} + x_{1}{^3} \right) }$$

### 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

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

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

$$\large{ W }$$ = total load from a uniform distribution

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

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

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