Three Member Frame - Fixed/Fixed Side Uniformly Distributed Load

Written by Jerry Ratzlaff on . Posted in Structural

3fbe 3Three Member Frame - Fixed/Fixed Side Uniformly Distributed Load Formula

\(\large{ e  = \frac{h}{L}  }\)

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

\(\large{ R_A = R_D  =  \frac{ w\;h\; \beta\; e^2 }{ 6\; \beta \;e\;+\;1 }   }\)

\(\large{ H_A =  \frac{w\;h}{4}  \;  \left(  \frac{8\; \beta\;e\;+\;17 }{2\; \left( \beta\;e\;+\;2 \right) }  -  \frac{4\; \beta\;e\;+\;3 }{6\; \beta\;e\;+\;1 }  \right)   }\)

\(\large{ H_D =  \frac{w\;h}{4}  \;  \left(  \frac{4\; \beta\;e\;+\;3 }{6\; \beta\;e\;+\;1 }  -  \frac{1 }{2\; \left( \beta\;e\;+\;2 \right) }  \right)   }\)

\(\large{ M_A =  \frac{w\;h^2}{4}  \;  \left(  \frac{4\; \beta\;e\;+\;1 }{6\; \beta\;e\;+\;1 }  -  \frac{\beta \;e \;+\; 3 }{6\; \left( \beta\;e\;+\;2 \right) }  \right)   }\)

\(\large{ M_B =  \frac{w\;h^2 \; \beta \;e}{4}  \;  \left(  \frac{ 6 }{6\; \beta\;e\;+\;1 }  -  \frac{ 1 }{6\; \left( \beta\;e\;+\;2 \right) }  \right)   }\)

\(\large{ M_C =  \frac{w\;h^2 \; \beta \;e}{4}  \;  \left(  \frac{ 2 }{6\; \beta\;e\;+\;1 }  -  \frac{ 1 }{6\; \left( \beta\;e\;+\;2 \right) }  \right)   }\)

\(\large{ M_D =  \frac{w\;h^2}{4}  \;  \left(  \frac{4\; \beta\;e\;+\;1 }{6\; \beta\;e\;+\;1 }  -  \frac{\beta \;e \;+\; 3 }{6\; \left( \beta\;e\;+\;2 \right) }  \right)   }\)

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