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Beam Fixed at Both Ends - Uniformly Distributed Load

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.

 

febe 1A

beam fixed at both ends - Uniformly Distributed Load formulas

\( R \;=\; V \;=\;   \dfrac{ w \cdot L }{ 2  }\) 

\( V_x \;=\;   w  \cdot   \left( \dfrac{ L }{ 2 } - x \right)  \) 

\( M_{max}  \; (at \;ends )  \;=\;   \dfrac{ w\cdot L^2 }{ 12 } \) 

\( M_1  \; (at\; center )  \;=\;  \dfrac{ w \cdot L^2 }{ 24 } \)

\( M_x  \;=\;    \dfrac{w}{12}  \cdot  ( 6 \cdot L \cdot x - L^2 - 6 \cdot x^2 )   \)

\( M_{max}  \; (at \;center )  \;=\;  \dfrac{ w \cdot L^4 }{ 384 \cdot \lambda \cdot I }  \)

\( \Delta_x  \;=\;    \dfrac{ w \cdot x^2 }{ 24 \cdot \lambda \cdot I }  \cdot  ( L - x ) ^2   \)

\( x  \; (points \;of \;contraflexure )  \;=\;   (\sqrt{3} - 3 ) \cdot  L   \)

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 \(in\) \(mm\)
\( x \) = Horizontal Distance from Reaction to Point on Beam \(in\) \(mm\)
\( w \) = Load per Unit Length \(lbf\;/\;in\) \(N\;/\;m\)
\( L \) = Span Length of the Bending Member \(in\) \(mm\)
\( \lambda  \)   (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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