Beam Fixed at Both Ends - 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.

 

febe 1A

beam fixed at both ends - Uniformly Distributed Load formulas

\( R \;=\; V \;=\; w\; L\;/\;2  \) 

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

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

\( M_1  \; (at\; center )  \;=\; w\; L^2\;/\;24  \)

\( M_x  \;=\; (w\;/\;12)  \; ( 6\;L\;x - L^2 - 6\;x^2 )   \)

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

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

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

B F at B E - Uniformly Distributed Load - Solve for R

\(\large{ R =  \frac{w\; L}{2}  }\) 

load per unit length, w
span length, L

B F at B E - Uniformly Distributed Load - Solve for Vx

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

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

B F at B E - Uniformly Distributed Load - Solve for MmaxE

\(\large{ M_{max}  =  \frac{w\; L^2}{12}  }\) 

load per unit length, w
span length, L

B F at B E - Uniformly Distributed Load - Solve for M1

\(\large{ M_1  = \frac{w\; L^2}{24}  }\)

load per unit length, w
span length, L

B F at B E - Uniformly Distributed Load - Solve for Mx

\(\large{ M_x  = \frac{w}{12}  \;  \left( 6\;L\;x - L^2 - 6\;x^2  \right)   }\)

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

B F at B E - Uniformly Distributed Load - Solve for MmaxC

\(\large{ M_{max}  =  \frac{w\; L^4}{384\; \lambda\; I}     }\)

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

B F at B E - Uniformly Distributed Load - Solve for Δx

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

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

B F at B E - Uniformly Distributed Load - Solve for x

\(\large{ x  =  \left(\sqrt{3} - 3  \right)  L    }\)

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