Step 1

We are given that the maximum force supported at point D is ${F_t} = 8\;{\rm{kN}}$in tension and ${F_c} = 6\;{\rm{kN}}$ in compression.

We are asked to calculate the maximum force supported by point D.

We have the distance between point A and E is ${d_1} = 4\;{\rm{m}}$.

We have the distance between point E and D is ${d_2} = 4\;{\rm{m}}$.

We have the distance between point B and C is ${d_3} = 4\;{\rm{m}}$.

 
Step 2

The free body diagram of the system is shown as:

 
Step 3

Applying the equilibrium of force along x-axis:

\[\begin{array}{c} \sum {F_x} = 0\\ {A_x} = 0 \end{array}\]
 
Step 4

Applying the moment of forces about point A:

\[{E_y}{d_1} - P\left( {{d_1} + {d_2}} \right) = 0\]
 
Step 5

Substitute the known values in the equation:

\[\begin{array}{c} {E_y}\left( {4\;{\rm{m}}} \right) - P\left( {4\;{\rm{m}} + 4\;{\rm{m}}} \right) = 0\\ {E_y}\left( {4\;{\rm{m}}} \right) = P\left( {8\;{\rm{m}}} \right)\\ {E_y} = 2P \end{array}\]
 
Step 6

Applying the equilibrium of force in vertical direction:

\[\begin{array}{c} \sum {F_y} = 0\\ {E_y} + {A_y} - P = 0 \end{array}\]
 
Step 7

Substitute the known values in the equation:

\[\begin{array}{c} 2P + {A_y} - P = 0\\ {A_y} = P - 2P\\ = - P \end{array}\]
 
Step 8

Applying the equilibrium of vertical forces at point D:

\[{F_{CD}}\sin 60^\circ - P = 0\]
 
Step 9

Substitute the known values in the equation:

\[\begin{array}{c} {F_{CD}}\left( {0.866} \right) - P = 0\\ {F_{CD}} = \frac{P}{{0.866}}\\ = 1.155P \end{array}\]
 
Step 10

Applying the equilibrium of horizontal forces at point D:

\[ - {F_{ED}} - {F_{CD}}\cos 60^\circ = 0\]
 
Step 11

Substitute the known values in the equation:

\[\begin{array}{c} - {F_{ED}} - \left( {1.155P} \right)\left( {0.5} \right) = 0\\ {F_{ED}} = - 0.5775P \end{array}\]
 
Step 12

Applying the equilibrium of vertical forces at point C:

\[\begin{array}{c} - {F_{EC}}\sin 60^\circ - {F_{CD}}\sin 60^\circ = 0\\ {F_{EC}}\sin 60^\circ = - {F_{CD}}\sin 60^\circ \\ {F_{EC}} = - {F_{CD}} \end{array}\]
 
Step 13

Substitute the known values in the equation:

\[{F_{EC}} = - 1.155P\]

Therefore, the force is maximum compression in member EC.

 
Step 14

Applying the equilibrium of forces along x-axis:

\[ - {F_{BC}} - {F_{EC}}\cos 60^\circ + {F_{CD}}\cos 60^\circ = 0\]
 
Step 15

Substitute the known values in the equation for maximum compression:

\[\begin{array}{c} - 6\;{\rm{kN}} - \left( { - 1.155P} \right)\left( {0.5} \right) + \left( {1.155P} \right)\left( {0.5} \right) = 0\\ 0.5775P + 0.5775P = 6\;{\rm{kN}}\\ 1.155P = 6\;{\rm{kN}}\\ P = 5.195\;{\rm{kN}} \end{array}\]
 
Step 16

Applying the equilibrium of horizontal forces at point B:

\[{F_{BC}} + \left( {{F_{BE}} - {F_{AB}}} \right)\cos 60^\circ = 0\]
 
Step 17

Substitute the known values in the equation:

\[\begin{array}{c} 1.155P + \left( {{F_{BE}} - \left( { - {F_{BE}}} \right)} \right)\left( {0.5} \right) = 0\\ 2{F_{BE}}\left( {0.5} \right) = - 1.155P\\ {F_{BE}} = - 1.155P \end{array}\]
 
Step 18

Applying the equilibrium of vertical forces at point B:

\[\begin{array}{c} - {F_{AB}}\sin 60^\circ - {F_{BE}}\sin 60^\circ = 0\\ {F_{AB}}\sin 60^\circ = - {F_{BE}}\sin 60^\circ \\ {F_{AB}} = - {F_{BE}} \end{array}\]...... (1)

Therefore, the force is maximum tension in member AB.

 
Step 19

Substitute the known value in the equation (1) for maximum tension:

\[\begin{array}{c} 8\;{\rm{kN}} = - \left( { - 1.155P} \right)\\ 1.155P = 8\;{\rm{kN}}\\ P = 6.926\;{\rm{kN}} \end{array}\]

For maximum allowable load, consider the minimum value of the obtained loads. Therefore, the maximum allowed load for the system is $5.195\;{\rm{kN}}$.