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

on . Posted in Structural Engineering

### diagram Symbols

• Bending moment diagram (BMD)  -  Used to determine the bending moment at a given point of a structural element.  The diagram can help determine the type, size, and material of a member in a structure so that a given set of loads can be supported without structural failure.
• Free body diagram (FBD)  -  Used to visualize the applied forces, moments, and resulting reactions on a structure in a given condition.
• Shear force diagram (SFD)  -  Used to determine the shear force at a given point of a structural element.  The diagram can help determine the type, size, and material of a member in a structure so that a given set of loads can be supported without structural failure.
• Uniformly distributed load (UDL)  -  A load that is distributed evenly across the entire length of the support area.

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

$$R_1 \;=\; V_1 \;=\; R_3 \;=\; V_4 \;=\; 3\;w \;L\;/\;8$$

$$R_2 \;=\; 10\;w\;L\;/\;8$$

$$V_2 = V_{max} \;=\; 5\;w\;L\;/\;8$$

$$M_1 \;=\; w\;L^2\;/\;8$$

$$M_2 \; (at\; \frac{3\;L}{8} ) \;=\; 9\;w\;L^2\;/\;128$$

$$\Delta_{max} \; ( 0.4215 \;L \;from\; R_1 \;\And\; R_3 ) \;=\; w\;L^4\;/\;185\; \lambda\; I$$

### 2 S C B - E S, Unif Dist Load - Solve for R1

$$\large{ R_1 = \frac{3\;w \;L}{8} }$$

 load per unit length, w span length under consideration, L

### 2 S C B - E S, Unif Dist Load - Solve for R2

$$\large{ R_2 = \frac{10\;w\;L}{8} }$$

 load per unit length, w span length under consideration, L

### 2 S C B - E S, Unif Dist Load - Solve for V2

$$\large{ V_2 = \frac{5\;w\;L}{8} }$$

 load per unit length, w span length under consideration, L

### 2 S C B - E S, Unif Dist Load - Solve for M1

$$\large{ M_1 = \frac{w\;L^2}{8} }$$

 load per unit length, w span length under consideration, L

### 2 S C B - E S, Unif Dist Load - Solve for M2

$$\large{ M_2 = \frac{9\;w\;L^2}{128} }$$

 load per unit length, w span length under consideration, L

### 2 S C B - E S, Unif Dist Load - Solve for Δmax

$$\large{ \Delta_{max} = \frac{w\;L^4}{185\; \lambda\; I} }$$

 load per unit length, w span length under consideration, L modulus of elasticity, λ second moment of area, I

Symbol English Metric
$$R$$ = reaction load at bearing point $$lbf$$ $$N$$
$$V$$ = maximum shear force $$lbf$$ $$N$$
$$M$$ = maximum bending moment $$lbf-ft$$ $$N-m$$
$$\Delta$$ = deflection or deformation $$in$$ $$mm$$
$$w$$ = load per unit length $$lbf\;/\;in$$ $$N\;/\;m$$
$$L$$ = span length under consideration $$in$$ $$mm$$
$$\lambda$$   (Greek symbol lambda) = modulus of elasticity $$lbf\;/\;in^2$$ $$Pa$$
$$I$$ = second moment of area (moment of inertia) $$in^4$$ $$mm^4$$

Tags: Beam Support