Step 1

We are given that the compressive force applied to system is ${F_{AB}} = 800\;{\rm{lb}}$ and the maximum tensile force of system is $T = 1500\;{\rm{lb}}$. The value of constant is $a = 10\;{\rm{ft}}$.

We are asked to calculate the maximum load that can be supported by truss.

We have the distance between point B and D is ${d_1} = \frac{3}{4}a$.

We have the perpendicular distance between point D and C is ${d_2} = \frac{1}{4}a$.

We have the distance between point A and D is ${d_3} = a$.

We have the horizontal distance between point D and C is ${d_4} = a$.

 
Step 2

The free body diagram of the system is shown as:

 
Step 3

To calculate angle $BAD$ we use the formula:

\[\tan {\theta _1} = \frac{{{d_1} + {d_2}}}{{{d_3}}}\]
 
Step 4

Substitute the known values in the formula:

\[\begin{array}{c} \tan {\theta _1} = \frac{{\frac{3}{4}a + \frac{1}{4}a}}{a}\\ {\theta _1} = {\tan ^{ - 1}}\left( 1 \right)\\ = 45^\circ \end{array}\]
 
Step 5

To calculate angle $DAD'$ we use the formula:

\[\tan {\theta _2} = \frac{{{d_2}}}{{{d_3}}}\]
 
Step 6

Substitute the known values in the formula:

\[\begin{array}{c} \tan {\theta _2} = \frac{{\frac{1}{4}a}}{a}\\ {\theta _2} = {\tan ^{ - 1}}\left( {\frac{1}{4}} \right)\\ = 14.03^\circ \end{array}\]
 
Step 7

Applying the equation of moment of forces about point A:

\[{C_y}\left( {{d_3} + {d_4}} \right) - P{d_3} = 0\]
 
Step 8

Substitute the known values in the equation:

\[\begin{array}{c} {C_y}\left( {a + a} \right) - P\left( a \right) = 0\\ {C_y}\left( {2a} \right) - Pa = 0\\ {C_y}\left( {2a} \right) = Pa\\ {C_y} = \frac{P}{2} \end{array}\]
 
Step 9

Applying the equilibrium of force of equation along y-axis:

\[\begin{array}{c} \sum {F_y} = 0\\ {A_y} + {C_y} - P = 0 \end{array}\]
 
Step 10

Substitute the known value in the equation:

\[\begin{array}{c} {A_y} + \frac{P}{2} - P = 0\\ {A_y} = P - \frac{P}{2}\\ = \frac{P}{2} \end{array}\]
 
Step 11

Applying the equilibrium of force of equation along x-axis:

\[\begin{array}{c} \sum {F_x} = 0\\ - {F_{AB}}\cos {\theta _1} + {F_{AD}}\cos {\theta _2} = 0 \end{array}\]
 
Step 12

Substitute the known values in the equation:

\[\begin{array}{c} \left( { - 800\;{\rm{lb}}} \right)\cos 45^\circ + {F_{AD}}\cos 14.03^\circ = 0\\ {F_{AD}}\left( {0.970} \right) = \left( {800\;{\rm{lb}}} \right)\left( {0.707} \right)\\ {F_{AD}} = 583.1\;{\rm{lb}} \end{array}\]
 
Step 13

Applying the forces of equation along horizontal direction:

\[\begin{array}{c} \sum {F_y} = 0\\ - {F_{AB}}\sin {\theta _1} + {F_{AD}}\sin {\theta _2} + \frac{P}{2} = 0 \end{array}\]
 
Step 14

Substitute the known values in the equation:

\[\begin{array}{c} - \left( {800\;{\rm{lb}}} \right)\sin 45^\circ + \left( {583.1\;{\rm{lb}}} \right)\sin 14.03^\circ + \frac{P}{2} = 0\\ \frac{P}{2} = \left( {800\;{\rm{lb}}} \right)\left( {0.707} \right) - \left( {583.1\;{\rm{lb}}} \right)\left( {0.242} \right)\\ P = 2\left( {565.5\;{\rm{lb}} - 141.1\;{\rm{lb}}} \right)\\ = 848.8\;{\rm{lb}} \end{array}\]
 
Step 15

Applying the forces at point D along horizontal direction:

\[{F_{AD}}\sin {\theta _2} + {F_{CD}}\sin {\theta _2} + P - {F_{BD}} = 0\]
 
Step 16

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} \left( {583.1\;{\rm{lb}}} \right)\sin 14.03^\circ + \\ \left( {583.1\;{\rm{lb}}} \right)\sin 14.03^\circ + \\ 848.8\;{\rm{lb}} - {F_{BD}} \end{array} \right\} = 0\\ {F_{BD}} = \left( {583.1\;{\rm{lb}}} \right)\left( {0.242} \right) + \left( {583.1\;{\rm{lb}}} \right)\left( {0.242} \right) + 848.8\;{\rm{lb}}\\ {\rm{ = 141}}{\rm{.1}}\;{\rm{lb}} + {\rm{141}}{\rm{.1}}\;{\rm{lb}} + {\rm{848}}{\rm{.8}}\;{\rm{lb}}\\ {\rm{ = 1131}}\;{\rm{lb}} \end{array}\]

The force obtained in member BD is less than 1500 lb as the members AD, DC, and BD can support a maximum tensile force of 1500 lb.

Therefore, the greatest load the truss can support is 848.8 lb.