Three Span Continuous Beam - Equal Spans, Uniformly Distributed Load

on . Posted in Structural Engineering

cb4s 1A

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

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

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

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

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

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

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

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

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

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
\( R \) = 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) \(in^4\) \(mm^4\)

 

Piping Designer Logo 1

Tags: Beam Support