# Overhanging Beam - Uniformly Distributed Load on Overhang

Written by Jerry Ratzlaff on . Posted in Structural

### Overhanging Beam - Uniformly Distributed Load on Overhang Formula

$$\large{ R_1 = V_2 = \frac{w a^2 }{2L} }$$

$$\large{ R_2 = V_1 + V_2 = \frac{w a }{2L} \left( 2L + a \right) }$$

$$\large{ V_2 = w a }$$

$$\large{ V_{x _1} = w \left( a - x_1 \right) }$$

$$\large{ M_{max} \; }$$  at  $$\large{ \left( R_2 \right) = \frac{ w a^2 }{2} }$$

$$\large{ M_x \; }$$  (between supports)  $$\large{ = \frac{ w a^2 x }{2L} }$$

$$\large{ M_{x_1} \; }$$  (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  $$\large{ \left( x_1 = a \right) = \frac{ w x^3 }{24 \lambda I } \left( 4L + 3a \right) }$$

$$\large{ \Delta_{x1} \; }$$  (for overhang)  $$\large{ = \frac{ w x_1 }{24 \lambda I } \left( 4a^2 L + 6a^2 x_1 - 4a x_{1}{^2} + x_{1}{^3} \right) }$$

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