Step 1

We are given the force $F = \left\{ { - 200i + 400j} \right\}\;{\rm{N}}$.

We are asked to determine the force in each member of the space truss and also to determine their state that either they are in tension or in compression.

 
Step 2

In vector form, the force acting on the truss can be written as:

\[F = {F_x}i + {F_y}j + {F_z}k\]

The force acting on the member in x, y and z direction will be:

\[\begin{array}{c} {F_x} = - 200\;{\rm{N}}\\ {F_y} = 400\;{\rm{N}}\\ {F_z} = 0\;{\rm{N}} \end{array}\]

The free-body diagram at joint D can be drawn as:

Here, ${F_{AD}}$ is the force in member AD and ${F_{CD}}$ is the force in member CD.

From the above figure, the coordinates of point A, B, C and D can be written as:

\[\begin{array}{c} A\left( {0,\;0,\;0} \right)\;{\rm{m}}\\ B\left( {6,\;3.5,\;0} \right)\;{\rm{m}}\\ C\left( {0,\;3.5,\;0} \right)\;{\rm{m}}\\ D\left( {1,\;2,\;2} \right)\;{\rm{m}} \end{array}\]

The position vector of member AD, BD and CD can be written as:

\[\begin{array}{c} {r_{AD}} = \left( {1i + 2j + 2k} \right)\;{\rm{m}}\\ {r_{AC}} = \left( {0i + 3.5j + 0k} \right)\;{\rm{m}}\\ {r_{BD}} = \left( { - 5i - 1.5j + 2k} \right)\;{\rm{m}}\\ {r_{BC}} = \left( { - 6i + 0j + 0k} \right)\;{\rm{m}}\\ {r_{CD}} = \left( {1i - 1.5j + 2k} \right)\;{\rm{m}} \end{array}\]
 
Step 3

For equilibrium condition, we have:

\[\begin{array}{c} \sum F = 0\\ \left[ {F + {F_{AD}}\left( {\frac{{ - {r_{AD}}}}{{\left| {{r_{AD}}} \right|}}} \right) + {F_{BD}}\left( {\frac{{ - {r_{BD}}}}{{\left| {{r_{BD}}} \right|}}} \right) + {F_{CD}}\left( {\frac{{ - {r_{CD}}}}{{\left| {{r_{CD}}} \right|}}} \right)} \right] = 0\\ \left[ \begin{array}{l} \left\{ {\left( { - 200i + 400j} \right)\;{\rm{N}}} \right\} + \left\{ {\frac{{\left( { - 1i - 2j - 2k} \right)\;{\rm{m}}}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( { - 2} \right)}^2}} \;{\rm{m}}}}{F_{AD}}} \right\} + \\ \left\{ {\frac{{\left( { - 5i - 1.5j + 2k} \right)\;{\rm{m}}}}{{\sqrt {{{\left( { - 5} \right)}^2} + {{\left( { - 1.5} \right)}^2} + {{\left( 2 \right)}^2}} \;{\rm{m}}}}{F_{BD}}} \right\} + \left\{ {\frac{{\left( {1i - 1.5j + 2k} \right)\;{\rm{m}}}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( { - 1.5} \right)}^2} + {{\left( 2 \right)}^2}} \;{\rm{m}}}}{F_{CD}}} \right\} \end{array} \right] = 0\\ \left[ \begin{array}{l} \left( { - 200i + 400j} \right)\;{\rm{N}} + \left( { - 0.33i - 0.66j - 0.66k} \right){F_{AD}} + \\ \left( {0.89i + 0.267j - 0.356k} \right){F_{BD}} + \left( { - 0.37i + 0.557j - 0.74k} \right){F_{CD}} \end{array} \right] = 0 \end{array}\]

On applying the equilibrium condition along the x direction, we get:

\[ - \left( {200\;{\rm{N}}} \right) - 0.33{F_{AD}} + 0.89{F_{BD}} - 0.37{F_{CD}} = 0\]   … (1)

On applying the equilibrium condition along the y direction, we get:

\[\begin{array}{c} \sum {F_y} = 0\\ - \left( {400\;{\rm{N}}} \right) - 0.66{F_{AD}} + 0.267{F_{BD}} + 0.557{F_{CD}} = 0\\ - 0.66{F_{AD}} + 0.267{F_{BD}} + 0.557{F_{CD}} = 400\;{\rm{N}} \end{array}\]   … (2)

Similarly, on applying the equilibrium condition along the z direction, we get:

\[\begin{array}{c} \sum {F_z} = 0\\ - 0.66{F_{AD}} - 0.356{F_{BD}} - 0.74{F_{CD}} = 0 \end{array}\]    … (3)

On equating equation (2) and (3), we get:

\[\begin{array}{c} \left( \begin{array}{l} - 0.66{F_{AD}} + 0.267{F_{BD}} + \\ 0.557{F_{CD}} + 400\;{\rm{N}} \end{array} \right) = \left( \begin{array}{l} - 0.66{F_{AD}} - 0.356{F_{BD}} - \\ 0.74{F_{CD}} \end{array} \right)\\ - 0.623{F_{BD}} + 1.297 - 400\;{\rm{N}} = 0 \end{array}\]   ... (4)

To find the force in member BD, we need to multiply the equation (1) by 2 and subtract with the equation (3) as:

On solving the equation (1), (2) and (3), we get:

\[\begin{array}{c} 2 \times \left[ \begin{array}{l} - \left( {200\;{\rm{N}}} \right) - 0.33{F_{AD}} + \\ 0.89{F_{BD}} - 0.37{F_{CD}} \end{array} \right] - \left[ { - 0.66{F_{AD}} - 0.356{F_{BD}} - 0.74{F_{CD}}} \right] = 0\\ \left[ \begin{array}{l} - \left( {400\;{\rm{N}}} \right) - 0.66{F_{AD}} + 1.78{F_{BD}} - 0.74{F_{CD}} + \\ 0.66{F_{AD}} + 0.356{F_{BD}} + 0.74{F_{CD}} \end{array} \right] = 0\\ - \left( {400\;{\rm{N}}} \right) + 2.136{F_{BD}} = 0\\ {F_{BD}} = 187.26\;{\rm{N}}\;\left( {{\rm{Tension}}} \right) \end{array}\]

Substitute the value of ${F_{BD}}$ in equation (4), we get:

\[\begin{array}{c} - 0.623 \times \left( {187.26\;{\rm{N}}} \right) + 1.297{F_{CD}} - 400\;{\rm{N}} = 0\\ \left( { - 516.66\;{\rm{N}}} \right) = 1.297{F_{CD}}\\ {F_{CD}} = 398.35\;{\rm{N}}\;\left( {{\rm{Compression}}} \right) \end{array}\]

Substitute the value of ${F_{BD}}$ and ${F_{CD}}$ in equation (1), we get:

\[\begin{array}{c} - \left( {200\;{\rm{N}}} \right) - 0.33{F_{AD}} + 0.89 \times \left( {187.26\;{\rm{N}}} \right) - 0.37\left( {218.45\;{\rm{N}}} \right) = 0\\ - 114.16 = 0.33{F_{AD}}\\ {F_{AD}} = 345.95\;{\rm{N}}\;\left( {{\rm{Compression}}} \right) \end{array}\]
 
Step 4

The free-body diagram at joint C can be drawn as:

For equilibrium condition, we have:

\[\begin{array}{c} \sum F = 0\\ \left[ {{F_{CD}}\left( {\frac{{ - {r_{CD}}}}{{\left| {{r_{CD}}} \right|}}} \right) + {F_{BC}}\left( {\frac{{ - {r_{BC}}}}{{\left| {{r_{BC}}} \right|}}} \right) + {F_{AC}}\left( {\frac{{ - {r_{AC}}}}{{\left| {{r_{AC}}} \right|}}} \right) - {F_{EC}}k} \right] = 0\\ \left[ \begin{array}{l} \left\{ {\frac{{\left( {i - 1.5j + 2k} \right)\;{\rm{m}}}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( { - 1.5} \right)}^2} + {{\left( 2 \right)}^2}} \;{\rm{m}}}}{F_{CD}}} \right\} + \left\{ {\frac{{\left( {6i} \right)\;{\rm{m}}}}{{\sqrt {{{\left( 6 \right)}^2}} \;{\rm{m}}}}{F_{BC}}} \right\} + \\ \left\{ {\frac{{\left( { - 3.5j} \right)\;{\rm{m}}}}{{\sqrt {{{\left( { - 3.5} \right)}^2}} \;{\rm{m}}}}{F_{AC}} - {F_{EC}}k} \right\} \end{array} \right] = 0\\ \left[ {\left( {0.37i - 0.55j + 0.74k} \right){F_{CD}} + \left( i \right){F_{BC}} + \left( { - j} \right){F_{AC}} + \left( { - k} \right){F_{EC}}} \right] = 0 \end{array}\]

The force in member BC can be calculated as:

\[\begin{array}{c} \sum {F_x} = 0\\ 0.37{F_{CD}} + {F_{BC}} = 0\\ {F_{BC}} = - \left( {0.37} \right) \times \left( {398.35\;{\rm{N}}} \right)\\ {F_{BC}} = - 147.38\;{\rm{N}}\;\left( {{\rm{Tension}}} \right) \end{array}\]

The force in member AC can be calculated as:

\[\begin{array}{c} \sum {F_y} = 0\\ - 0.55{F_{CD}} - {F_{AC}} = 0\\ - 0.55 \times \left( {398.35\;{\rm{N}}} \right) = {F_{AC}}\\ {F_{AC}} = - 219.1\;{\rm{N}}\;\left( {{\rm{Tension}}} \right) \end{array}\]

The force in member EC can be calculated as:

\[\begin{array}{c} \sum {F_z} = 0\\ 0.74{F_{CD}} - {F_{EC}} = 0\\ 0.74 \times \left( {398.35\;{\rm{N}}} \right) = {F_{EC}}\\ {F_{EC}} = 294.8\;{\rm{N}}\;\left( {{\rm{Compression}}} \right) \end{array}\]