# Two Span Continuous Beam - Equal Spans, Concentrated Load at Center of One Span

Written by Jerry Ratzlaff on . Posted in Structural

### Two Span Continuous Beam - Equal Spans, Concentrated Load at Center of One Span Formula

$$\large{ R_1 = V_1 = \frac{13P}{32} }$$

$$\large{ R_2 = V_2 + V_3 = \frac{11P}{16} }$$

$$\large{ R_3 = V_3 = \frac{3P}{32} }$$

$$\large{ V_2 = \frac{19P}{32} }$$

$$\large{ M_{max} \; }$$ (at point of load)   $$\large{ = \frac{13PL}{64} }$$

$$\large{ M_{max} \; }$$  at support   $$\large{ \left( R_2 \right) = \frac{3PL}{32} }$$

$$\large{ \Delta_{max} \; \left( 0.408L \right) }$$  from  $$\large{ \left( R_1 \right) = 0.015 \frac{PL^3}{\lambda I} }$$

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