Two Member Frame - Fixed/Pin Side Uniformly Distributed Load

Written by Jerry Ratzlaff on . Posted in Structural

2ffp 4ATwo Member Frame - Fixed/Pin Side Uniformly Distributed Load Formula

\(\large{ e  = \frac{h}{L}  }\)

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

\(\large{ R_A  = R_C =  \frac{w\;h}{4}  \;  \left(  \frac{\beta\;e^2 }{3\; \beta\;e \;+\; 4 } \right) }\)

\(\large{ H_A  =  \frac{w\;h}{2}  \;  \left(  \frac{3\; \beta\;e \;+\; 5 }{3\; \beta\;e \;+\; 4 } \right) }\)

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

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

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

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

\(\large{ A, B, C }\) = 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