# Two Member Frame - Fixed/Pin Top Point Load

Written by Jerry Ratzlaff on . Posted in Structural

### Two Member Frame - Fixed/Pin Top Point Load Formula

$$\large{ e = \frac{h}{L} }$$

$$\large{ \beta = \frac{I_h}{I_v} }$$

$$\large{ R_A = \frac{P\;x}{L} \; \left[ 1+ \; \frac{2}{L^2} \; \left( \frac{L^2\;-\;x^2}{ 3\;\beta\;e \;+\; 4} \right) \right] }$$

$$\large{ R_D = \frac{P\;\left( L\;-\;x \right)}{L} \; \left[ 1- \; \frac{2\;x}{L^2} \; \left( \frac{L\;+\;x}{ 3\;\beta\;e \;+\; 4} \right) \right] }$$

$$\large{ H_A = H_D = \frac{3\;P\;x}{h\;L^2} \; \left( \frac{L^2\;-\;x^2}{ 3\;\beta\;e \;+\; 4} \right) }$$

$$\large{ M_A = \frac{P\;x}{L^2} \; \left( \frac{L^2\;-\;x^2}{ 3\;\beta\;e \;+\; 4} \right) }$$

$$\large{ M_B = \frac{2\;P\;x}{L^2} \; \left( \frac{L^2\;-\;x^2}{ 3\;\beta\;e \;+\; 4} \right) }$$

$$\large{ M_C = \frac{P\;a \; \left( L\;-\;x \right) }{L} \; \left[ 1- \; \frac{2\;x}{L^2} \; \left( \frac{L\;-\;x}{ 3\;\beta\;e \;+\; 4} \right) \right] }$$

Where:

$$\large{ h }$$ = height of frame

$$\large{ x }$$ =  horizontal distance from reaction point

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

$$\large{ M }$$ = maximum bending moment

$$\large{ A, B, C, D }$$ = 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

$$\large{ P }$$ = total concentrated load