# Three Span Continuous Beam - Equal Spans, Uniformly Distributed Load

Written by Jerry Ratzlaff on . Posted in Structural Engineering

## Three Span Continuous Beam - Equal Spans, Uniformly Distributed Load formulas

 $$\large{ R_1 = V_1 = R_4 = V_4 = 0.400\;w\;L }$$ $$\large{ R_2 = R_3 = 1.100\;w\;L }$$ $$\large{ V_{2_1} = V_{3_2} = 0.500\;w\;L }$$ $$\large{ V_{2_2} = V_{3_1} = 0.600\;w\;L }$$ $$\large{ M_1 = M_5 \; }$$ at  $$\large{ \left( 0.400\;L \right) \; }$$ from  $$\large{ \left( R_1 \right) \; }$$  or   $$\large{ \left( R_4 \right) = 0.080\;w\;L^2 }$$ $$\large{ M_2 = M_4 \; }$$ at   $$\large{ \left( R_2 \right) \; }$$  or   $$\large{ \left( R_3 \right) = 0.100\;w\;L^2 }$$ $$\large{ M_3 \; }$$  (at mid center Span)  $$\large{ = 0.025\;w\;L^2 }$$ $$\large{ \Delta_{max} \; }$$ at  $$\large{ \left( 0.446\;L \right) \; }$$ from  $$\large{ \left( R_1 \right) \; }$$  or   $$\large{ \left( R_4 \right) = \frac{0.0069\;w\;L^4}{\lambda\; I} }$$

### 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