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Simple Beam - Load Increasing Uniformly to Center

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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.

 

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Simple Beam - Load Increasing Uniformly to Center formulas

\(\large{ R = V_{max} = \frac{W}{2}  }\)

\(\large{ V_x  \;  \left(  x < \frac{L}{2} \right)  =  \frac{W}{2\;L^2}  \; \left( L^2 - 4\;x^2    \right) }\)

\(\large{ M_{max}  \; \left(at \;center\right) =  \frac{W \;L}{6}  }\)

\(\large{ M_x \; \left(  x < \frac{L}{2}  \right)   =  W\;x  \; \left(  \frac{1}{2} - \frac {2\;x^2}{3\;L^2}  \right)  }\)

\(\large{ \Delta_{max} \; \left(at \;center\right) = \frac{W \;L^3} {60 \; \lambda \;I }  }\)

\(\large{ \Delta_x \; \left(  x < \frac{L}{2}  \right)  =  \frac{W\; x}{480\; \lambda \;I \;L^2}  \; \left(  5\;L^2 - 4\;x^2  \right)^2     }\)

S B - Load Increasing Unif to Center - Solve for R

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

load per unit length, w
span length, L

S B - Load Increasing Unif to Center - Solve for Vx

\(\large{ V_x  =  \frac{ \frac{w\;L}{2} }{2\;L^2}  \; \left( L^2 - 4\;x^2    \right) }\)

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

S B - Load Increasing Unif to Center - Solve for Mmax

\(\large{ M_{max}  =  \frac{ \frac{w\;L}{2} \;L}{6}  }\)

load per unit length, w
span length, L

S B - Load Increasing Unif to Center - Solve for Mx

\(\large{ M_x  =  \frac{w\;L}{2} \; x  \; \left(  \frac{1}{2} - \frac {2\;x^2}{3\;L^2}  \right)  }\)

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

S B - Load Increasing Unif to Center - Solve for Δmax

\(\large{ \Delta_{max}  = \frac{ \frac{w\;L}{2} \;L^3} {60 \; \lambda \;I }  }\)

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

S B - Load Increasing Unif to Center - Solve for Δx

\(\large{ \Delta_x  =  \frac{ \frac{w\;L}{2} \; x}{480\; \lambda \;I \;L^2}  \; \left(  5\;L^2 - 4\;x^2  \right)^2     }\)

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

Symbol English Metric
\(\large{ R }\) = reaction load at bearing point \(\large{lbf}\) \(\large{N}\)
\(\large{ V }\) = maximum shear force \(\large{lbf}\) \(\large{N}\)
\(\large{ M }\) = maximum bending moment \(\large{lbf-in}\) \(\large{N-mm}\)
\(\large{ \Delta }\) = deflection or deformation \(\large{in}\) \(\large{mm}\)
\(\large{ W }\) = total load or \(\large{ \frac{w\;L}{2} }\) \(\large{lbf}\) \(\large{N}\)
\(\large{ w }\) = highest load per unit length of UIL \(\large{\frac{lbf}{in}}\) \(\large{\frac{N}{m}}\)
\(\large{ L }\) = span length of the bending member \(\large{in}\) \(\large{mm}\)
\(\large{ x }\) = horizontal distance from reaction to point on beam \(\large{in}\) \(\large{mm}\)
\(\large{ \lambda  }\)   (Greek symbol lambda) = modulus of elasticity \(\large{\frac{lbf}{in^2}}\) \(\large{Pa}\)
\(\large{ I }\) = second moment of area (moment of inertia) \(\large{in^4}\) \(\large{mm^4}\)

 

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Tags: Beam Support Equations

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