# Three 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.

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

$$\large{ R_1 = V_1 = R_4 = V_4 \;\;=\;\; 0.400\;w\;L }$$

$$\large{ R_2 = R_3 \;\;=\;\; 1.100\;w\;L }$$

$$\large{ V_{2_1} = V_{3_2} \;\;=\;\; 0.500\;w\;L }$$

$$\large{ V_{2_2} = V_{3_1} \;\;=\;\; 0.600\;w\;L }$$

$$\large{ M_1 = M_5 \; \left(at\; 0.400\;L \; from \; R_1 \;or\; R_4 \right) \;\;=\;\; 0.080\;w\;L^2 }$$

$$\large{ M_2 = M_4 \; \left(at\; R_2 \;or\; R_3 \right) \;\;=\;\; 0.100\;w\;L^2 }$$

$$\large{ M_3 \; \left(at \;mid \;center \;span \right) \;\;=\;\; 0.025\;w\;L^2 }$$

$$\large{ \Delta_{max} \; \left(at\; 0.446\;L \; from \; R_1 \;or\; R_4 \right) \;\;=\;\; \frac{0.0069\;w\;L^4}{\lambda\; I} }$$

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

$$\large{ R_1 = 0.400 \; w \; L }$$

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

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

$$\large{ R_2 = 1.100\;w\;L }$$

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

### 3 S C B - E S, Unif Dist Load - Solve for V21

$$\large{ V_{2_1} = 0.500\;w\;L }$$

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

### 3 S C B - E S, Unif Dist Load - Solve for V22

$$\large{ V_{2_2} = 0.600\;w\;L }$$

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

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

$$\large{ M_1 = 0.080\;w\;L^2 }$$

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

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

$$\large{ M_2 = 0.100\;w\;L^2 }$$

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

### 3 S C B - E S, Unif Dist Load - Solve for M3

$$\large{ M_3 = 0.025\;w\;L^2 }$$

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

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

$$\large{ \Delta_{max} = \frac{0.0069\;w\;L^4}{\lambda\; I} }$$

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

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