Beam Fixed at One End - 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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Beam Fixed at One End - Uniformly Distributed Load formulas

\( R_1 \;=\; V_1 \;=\; 3\; w \;L\;/\;8  \) 

\( R_2 \;=\; V_2 max \;=\; 5\; w\; L\;/\;8  \) 

\( V_x \;=\; R_1 - w\;x  \) 

\( M_{max}   \;=\; w\; L^2\;/\;8  \)

\( M_1 \; \left(at\; x  = \frac {3\;L}{8}  \right)  \;=\;   9\;w\;L^2 \;/\;128   \)

\( M_x   \;=\; (R_1\; x) - (w\; x^2\;/\;2)  \)

\( \Delta_{max}  \;=\; [\; at \; x = \frac{L}{16}\; ( 1 + \sqrt{33} \;) \;or\; x = 0.4215\;L \;]  \;=\; w\; L^4\;/\;185\; \lambda\; I  \)

\( \Delta_x \;=\; (w \;x\;/\;48\; \lambda\; I)   \; (   L^3 - 3\;L\;x^2 + 2\;x^3 )  \)

B F at O E - Uniformly Distributed Load - Solve for R1

\(\large{ R_1 = \frac {3\; w \;L} {8}  }\) 

load per unit length, w
span length, L

B F at O E - Uniformly Distributed Load - Solve for R2

\(\large{ R_2 =  \frac {5\; w\; L} {8}  }\)

load per unit length, w
span length, L

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

\(\large{ V_x =   \frac {3\; w \;L} {8}  - w\;x    }\) 

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

B F at O E - Uniformly Distributed Load - Solve for Mmax

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

load per unit length, w
span length, L

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

\(\large{ M_1 =   \frac  { 9\;w\;L^2 } {128}   }\)

load per unit length, w
span length, L

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

\(\large{ M_x   =  \frac {3\; w \;L} {8}  \; x - \frac {w\; x^2} {2}  }\)

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

B F at O E - Uniformly Distributed Load - Solve for Δmax

\(\large{ \Delta_{max}   =  \frac {w\; L^4} {185\; \lambda\; I}    }\)

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

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

\(\large{ \Delta_x = \frac {w \;x} {48\; \lambda\; I}  \;  \left(   L^3 - 3\;L\;x^2 + 2\;x^3    \right)     }\)

load per unit length, w
dist 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