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

Written by Jerry Ratzlaff on . Posted in Structural

## formulas that use Four Span Continuous Beam - Equal Spans, Uniformly Distributed Load

 $$\large{ R_1 = V_1 = R_5 = V_5 = 0.393\;w\;L }$$ $$\large{ R_2 = R_4 = 1.143\;w\;L }$$ $$\large{ R_3 = 0.928\;w\;L }$$ $$\large{ V_{2_1} = V_{4_2} = 0.607\;w\;L }$$ $$\large{ V_{2_2} = V_{4_1} = 0.536\;w\;L }$$ $$\large{ V_{3_1} = V_{3_2} = 0.464\;w\;L }$$ $$\large{ M_1 \; }$$ at  $$\large{ \left( 0.393\;L \right) \; }$$ from  $$\large{ \left( R_1 \right) = M_6 \; }$$  at   $$\large{ \left( 0.393\;L \right) \; }$$ from  $$\large{ \left( R_5 \right) \; = 0.0772\;w\;L^2 }$$ $$\large{ M_2 \; }$$ at   $$\large{ \left( R_2 \right) \; = -\;0.1071\;w\;L^2 }$$ $$\large{ M_3 \; }$$ at  $$\large{ \left( 0.536\;L \right) \; }$$ from  $$\large{ \left( R_2 \right) = M_5 \; }$$  at   $$\large{ \left( 0.536\;L \right) \; }$$ from  $$\large{ \left( R_4 \right) \; = 0.0364\;w\;L^2 }$$ $$\large{ M_4 \; }$$ at   $$\large{ \left( R_3 \right) \; = -\;0.0714\;w\;L^2 }$$ $$\large{ \Delta_{max} \; }$$ at  $$\large{ \left( 0.440\;L \right) \; }$$ from  $$\large{ \left( R_1 \right) \; }$$  and   $$\large{ \left( R_5 \right) = \frac{0.0065\;w\;L^4}{\lambda\; I} }$$

### 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