Simple Beam - 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.

 

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Simple Beam - Uniformly Distributed Load formulas

\( R \;=\; V_{max} \;=\; w \; L\;/\;2 \)

\( V_x  \;=\; w  \; [\; (L\;/\;2)  - x \;]  \)

\( M_{max} \; \left(at \;center \right)  \;=\; w \; L^2\;/\;8  \)

\( M_x  \;=\; (w \; x\;/\;2) \;  ( L  - x )   \)

\( \Delta_{max} \; \left(at \;center \right)  \;=\; 5 \;w \;L^4\;/\;384\; \lambda \;I \)

\( \Delta_x  \;=\; (w\; x\;/\;24\; \lambda \;I )  \; ( L^3 - 2\;L\;x^2 + x^3 )  \)

S B Uniformly Distributed Load - Solve for R

\(\large{ R = V_{max} = \frac{w \; L}{2}  }\)

load per unit length, w
span length, L

 

S B Uniformly Distributed Load - Solve for Vx

\(\large{ V_x =  w  \; \left(   \frac{L}{2}  - x    \right)     }\)

load per unit length, w
span length, L
distance from reaction, x

S B Uniformly Distributed Load - Solve for Mmax

\(\large{ M_{max} \; \left(at \;center \right) =  \frac{w \; L^2}{8}  }\)

load per unit length, w
span length, L

S B Uniformly Distributed Load - Solve for Mx

\(\large{ M_x =  \frac{w \; x}{2} \;  \left(   L  - x    \right)   }\)

load per unit length, w
distance from reaction, x
span length, L

S B Uniformly Distributed Load - Solve for Δmax

\(\large{ \Delta_{max} \; \left(at \;center \right) =  \frac{5 \;w \;L^4}{384\; \lambda \;I}  }\)

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

S B Uniformly Distributed Load - Solve for Δx

\(\large{ \Delta_x = \frac{w\; x}{24\; \lambda \;I}  \; \left(   L^3 - 2\;L\;x^2 + x^3    \right)  }\)

load per unit length, w
distance from reaction, x
modulus of elasticity, λ
second moment of area, I
span length, L

Symbol English Metric
\( R \) = reaction load at bearing point \(lbf\) \(N\)
\( V \) = maximum shear force \(lbf\) \(N\)
\( M \) = maximum bending moment \(lbf-in\) \(N-mm\)
\( \Delta \) = deflection or deformation \(in\) \(mm\)
\( w \) = load per unit length \(lbf\;/\;in\) \(N\;/\;m\)
\( L \) = span length of the bending member \(in\) \(mm\)
\( x \) = horizontal distance from reaction to point on beam \(in\) \(mm\)
\( \lambda  \)   (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