Three Member Frame - Pin/Pin Side Uniformly Distributed Load

Written by Jerry Ratzlaff on . Posted in Structural

3fpbe 5Three Member Frame - Pin/Pin Side Uniformly Distributed Load Formula

\(\large{ e  = \frac{h}{L}  }\)

\(\large{ \beta = \frac{I_h}{I_v}  }\)

\(\large{ R_A  = R_D = \frac{ w\;h^2 }{2\;L}  }\)

\(\large{ H_A  = \frac{w\;h}{8}  \;  \left( \frac{ 11\;\beta\;e \;+\; 18 }{ 2\;\beta\;e \;+\; 3 }  \right)  }\)

\(\large{ H_D  = \frac{w\;h}{8}  \;  \left( \frac{ 5\;\beta\;e \;+\; 6 }{ 2\;\beta\;e \;+\; 3 }  \right)  }\) 

\(\large{ M_B  = \frac{3\;w\;h^2}{8}  \;  \left( \frac{ \beta\;e \;+\; 2 }{ 2\;\beta\;e \;+\; 3 }  \right)  }\)

\(\large{ M_C  = \frac{w\;h^2}{8}  \;  \left( \frac{ 5\; \beta\;e \;+\; 6 }{ 2\;\beta\;e \;+\; 3 }  \right)  }\)

Where:

\(\large{ h }\) = height of frame

\(\large{ H }\) =  horizontal reaction load at bearing point

\(\large{ w }\) = load per unit length

\(\large{ M }\) = maximum bending moment

\(\large{ A, B, C, D, E }\) = points of intersection on frame

\(\large{ R }\) = reaction load at bearing point

\(\large{ I }\) = second moment of area (moment of inertia)

\(\large{ I_h }\) = horizontal second moment of area (moment of inertia)

\(\large{ I_v }\) = vertical second moment of area (moment of inertia)

\(\large{ L }\) = span length of the bending member

 

Tags: Equations for Frame Support