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Two 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.

 

 

 

 

 

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

 

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Tags: Beam Support

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