# Overhanging Beam - Point Load on Beam End

Written by Jerry Ratzlaff on . Posted in Structural

## Overhanging Beam - Point Load on Beam End ### Point Load on Beam End Formula

$$\large{ R_1 = V_1 = \frac{Pa}{L} }$$

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

$$\large{ V_2 = P }$$

$$\large{ M_{max} \; }$$  at   $$\large{ \left( R_2 \right) = Pa }$$

$$\large{ M_x \; }$$  (between supports)    $$\large{ = \frac{Pax}{L} }$$

$$\large{ M_{x_1} \; }$$  (for overhang)    $$\large{ = P \left( a - x_1 \right) }$$

$$\large{ \Delta_x \; }$$  (between supports)    $$\large{ = \frac{ -Pax }{6 \lambda IL} \left( L^2 - x^2 \right) }$$

$$\large{ \Delta_{x_1} \; }$$  (overhang)    $$\large{ = \frac{ Px_1 }{6 \lambda I } \left( 2aL + 3ax_1 - x_{1}{^2} \right) }$$

$$\large{ \Delta_{max} \; }$$  for overhang at   $$\large{ \left(x_1 = a \right) = \frac{ Pa^2 }{3 \lambda I } \left( L + a \right) }$$

$$\large{ \Delta_{max} \; }$$  between supports at   $$\large{ \left( x = \frac{L}{\sqrt{3}} \right) = \frac{ -PaL^2 }{9 \sqrt{3} \lambda I } = 0.06415 \frac{ PaL^2 }{ \lambda I } }$$

Where:

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

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

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

$$\large{ P }$$ = total concentrated load

$$\large{ R }$$ = 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