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Three Span Continuous Beam - Equal Spans, Uniformly Distributed Load

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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

\( R_1 \;=\; V_1 \;=\; R_4 \;=\; V_4  \;=\;  0.400\cdot w\cdot L    \) 

\( R_2 \;=\; R_3   \;=\;   1.100\cdot w\cdot L    \) 

\( V_{2_1} \;=\; V_{3_2}  \;=\;  0.500\cdot w\cdot L    \) 

\( V_{2_2} \;=\; V_{3_1}  \;=\;  0.600\cdot w\cdot L    \)

\( M_1 \;=\; M_5 \; (at\; 0.400\;L \; from  \; R_1  \;or\; R_4 )  \;=\;   0.080\cdot w\cdot L^2    \)

\( M_2 \;=\; M_4 \;  (at\; R_2  \;or\;  R_3 )   \;=\;  0.100\cdot w\cdot L^2    \)

\( M_3  \; (at \;mid \;center \;span )  \;=\;  0.025\cdot w\cdot L^2    \)

\( \Delta_{max}  \; (at\; 0.446\;L \; from  \; R_1  \;or\; R_4 )  \;=\;   \dfrac{ 0.0069\cdot w\cdot L^4 }{ \lambda\cdot I }  \)

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) \(in^4\) \(mm^4\)

 

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