Step 1

We have given a space truss which is supported at point C, D, and E. A force ${{\rm{F}}_1} = \left\{ { - 500{\rm{k}}} \right\}\;{\rm{lb}}$ is acting point A and a force ${{\rm{F}}_2} = \left\{ {400{\rm{j}}} \right\}\;{\rm{lb}}$ is acting at point B.

We are asked to find the magnitude and nature of forces in each member of the truss.

 
Step 2

Draw a free-body diagram of the complete truss.

Here, ${E_y}$ is the reaction force at point E in the y-axis direction. ${D_x}$, ${D_y}$, and ${D_z}$ are the reactions at point D in the x, y, and z-axis direction. ${C_y}$, and ${C_z}$ are the reactions at point C in the y and z-axis direction.

Apply equilibrium equation for the moment about z-axis:

\[\begin{array}{c} \sum {{M_z}} = 0\\ - \left( {{C_y} \times 3\;{\rm{ft}}} \right) - \left( {400\;{\rm{lb}} \times 3\;{\rm{ft}}} \right) = 0\\ {C_y} = - 400\;{\rm{lb}} \end{array}\]

Apply equilibrium equation of forces along the x-axis direction:

\[\begin{array}{c} \sum {{F_x}} = 0\\ {D_x} = 0 \end{array}\]

Apply equilibrium equation for the moment about y-axis:

\[\begin{array}{c} \sum {{M_y}} = 0\\ {C_z} \times 3\;{\rm{ft}} = 0\\ {C_z} = 0 \end{array}\]

Draw a free-body diagram of joint F.

Here, ${F_{FD}}$ is the force in member FD, ${F_{FC}}$ is the force in member FC, ${F_{FE}}$ is the force in member FE, and ${F_{BF}}$ is the force in member BF.

Apply equilibrium equation for the forces along the y-axis direction;

\[\begin{array}{c} \sum {{F_y}} = 0\\ {F_{BF}} = 0 \end{array}\]

Draw a free-body diagram of joint B.

Here, ${F_{BC}}$ is the force in member BC, ${F_{BE}}$ is the force in member BE, ${F_{AB}}$ is the force in member AB.

Apply equilibrium equation for the forces along the z-axis direction;

\[\begin{array}{c} \sum {{F_z}} = 0\\ {F_{BC}} \times \frac{3}{5} = 0\\ {F_{BC}} = 0 \end{array}\]

Apply equilibrium equation for the forces along the y-axis direction;

\[\begin{array}{c} \sum {{F_y}} = 0\\ \left( {{F_{BC}} \times \frac{4}{5}} \right) - \left( {{F_{BE}} \times \frac{4}{5}} \right) + \left( {400\;{\rm{lb}}} \right) = 0 \end{array}\]

Substitute the value of ${F_{BC}}$ in the above equation:

\[\begin{array}{c} \left\{ {\left( 0 \right) \times \frac{4}{5}} \right\} - \left\{ {{F_{BE}} \times \frac{4}{5}} \right\} + \left\{ {400\;{\rm{lb}}} \right\} = 0\\ {F_{BE}} \times \frac{4}{5} = 400\;{\rm{lb}}\\ {F_{BE}} = 500\;{\rm{lb}}\;\left( {{\rm{Tension}}} \right) \end{array}\]

So, the member BE is under tension.

Apply equilibrium equation for the forces along the x-axis direction;

\[\begin{array}{c} \sum {{F_x}} = 0\\ - {F_{AB}} + {F_{BE}} \times \frac{3}{5} = 0 \end{array}\]

Substitute the value of ${F_{BE}}$ in the above equation:

\[\begin{array}{c} - {F_{AB}} + \left\{ {\left( {500\;{\rm{lb}}} \right) \times \frac{3}{5}} \right\} = 0\\ {F_{AB}} = 300\;\,{\rm{lb}}\;\left( {{\rm{Compression}}} \right) \end{array}\]

So, the member AB is under compression.

Draw a free-body diagram of joint A.

Here, ${F_{AC}}$ is the force in member AC, ${F_{AD}}$ is the force in member AD, ${F_{AE}}$ is the force in member AE, and ${F_{AB}}$ is the force in member AB.

Apply equilibrium equation for the forces along the x-axis direction;

\[\begin{array}{c} \sum {{F_x}} = 0\\ {F_{AB}} - {F_{AC}} \times \frac{{3\,{\rm{ft}}}}{{\sqrt {{3^2} + {3^2} + {4^2}} \;{\rm{ft}}}} = 0\\ {F_{AB}} - {F_{AC}} \times \frac{3}{{\sqrt {34} }} = 0 \end{array}\]

Substitute the value of ${F_{AB}}$ in the above equation:

\[\begin{array}{c} \left( {300\;{\rm{lb}}} \right) - \left\{ {{F_{AC}} \times \frac{3}{{\sqrt {34} }}} \right\} = 0\\ {F_{AC}} = 583.1\;{\rm{lb}}\;\left( {{\rm{Tension}}} \right) \end{array}\]

So, the member AC is under tension.

Apply equilibrium equation for the forces along the z-axis direction;

\[\begin{array}{c} \sum {{F_z}} = 0\\ {F_{AD}} \times \frac{3}{5} + {F_{AC}} \times \frac{{3\;{\rm{ft}}}}{{\sqrt {{3^2} + {3^2} + {4^2}} \;{\rm{ft}}}} - 500\;{\rm{lb}} = 0\\ {F_{AD}} \times \frac{3}{5} + {F_{AC}} \times \frac{3}{{\sqrt {34} }} = 500\;{\rm{lb}} \end{array}\]

Substitute the value of ${F_{AC}}$ in the above equation:

\[\begin{array}{c} \left\{ {{F_{AD}} \times \frac{3}{5}} \right\} + \left\{ {\left( {583.1\;{\rm{lb}}} \right) \times \frac{3}{{\sqrt {34} }}} \right\} = 500\;{\rm{lb}}\\ {F_{AD}} \times \frac{3}{5} = 200\;{\rm{lb}}\\ {F_{AD}} = 333.3\;{\rm{lb}}\;\left( {{\rm{Tension}}} \right) \end{array}\]

So, the member AD is under tension.

Apply equilibrium equation for the forces along the y-axis direction;

\[\begin{array}{c} \sum {{F_y}} = 0\\ {F_{AE}} - {F_{AD}} \times \frac{4}{5} - {F_{AC}} \times \frac{{4\;{\rm{ft}}}}{{\sqrt {{3^2} + {3^2} + {4^2}} \;{\rm{ft}}}} = 0\\ {F_{AE}} - {F_{AD}} \times \frac{4}{5} - {F_{AC}} \times \frac{4}{{\sqrt {34} }} = 0 \end{array}\]

Substitute the value of ${F_{AC}}$ and ${F_{AD}}$ in the above equation:

\[\begin{array}{c} {F_{AE}} - \left\{ {\left( {333.3\;{\rm{lb}}} \right) \times \frac{4}{5}} \right\} - \left\{ {\left( {583.1\;{\rm{lb}}} \right) \times \frac{4}{{\sqrt {34} }}} \right\} = 0\\ {F_{AE}} = 666.7\;{\rm{lb }}\left( {{\rm{Compression}}} \right) \end{array}\]

So, the member AE is under compression.

Draw a free-body diagram of joint E.

Here, ${F_{DE}}$ is the force in member DE, ${F_{EF}}$ is the force in member EF, ${F_{BE}}$ is the force in member BE, and ${F_{AE}}$ is the force in member AE.

Apply equilibrium equation for the forces along the z-axis direction;

\[\begin{array}{c} \sum {{F_z}} = 0\\ - {F_{DE}} = 0\\ {F_{DE}} = 0 \end{array}\]

Apply equilibrium equation for the forces along the x-axis direction;

\[\begin{array}{c} \sum {{F_x}} = 0\\ {F_{EF}} - {F_{BE}} \times \frac{3}{5} = 0\\ {F_{EF}} = {F_{BE}} \times \frac{3}{5} \end{array}\]

Substitute the value of ${F_{BE}}$ in the above equation:

\[\begin{array}{c} {F_{EF}} = \left( {500\;{\rm{lb}}} \right) \times \frac{3}{5}\\ = 300\;{\rm{lb}}\;\left( {{\rm{Compression}}} \right) \end{array}\]

So, the member EF is under compression.

 
Step 3

Draw a free-body diagram of joint C.

Here, ${F_{CD}}$ is the force in member CD, ${F_{CF}}$ is the force in member CF, ${F_{AC}}$ is the force in member AC, and ${F_{BC}}$ is the force in member BC.

Apply equilibrium equation for the forces along the x-axis direction;

\[\begin{array}{c} \sum {{F_x}} = 0\\ {F_{AC}} \times \frac{{3\,{\rm{ft}}}}{{\sqrt {{3^2} + {3^2} + {4^2}} \;{\rm{ft}}}} - {F_{CD}} = 0\\ {F_{CD}} = {F_{AC}} \times \frac{3}{{\sqrt {34} }} \end{array}\]

Substitute the value of ${F_{AC}}$ in the above equation:

\[\begin{array}{c} {F_{CD}} = \left( {583.1\;{\rm{lb}}} \right) \times \frac{3}{{\sqrt {34} }}\\ {F_{CD}} = 300.0\;{\rm{lb}}\;\left( {{\rm{Compression}}} \right) \end{array}\]

So, the member CD is under compression.

Apply equilibrium equation for the forces along the z-axis direction;

\[\begin{array}{c} \sum {{F_z}} = 0\\ {F_{CF}} - {F_{AC}} \times \frac{{3\;{\rm{ft}}}}{{\sqrt {{3^2} + {3^2} + {4^2}} \;{\rm{ft}}}} + {C_z} = 0\\ {F_{CF}} - {F_{AC}} \times \frac{3}{{\sqrt {34} }} + {C_z} = 0 \end{array}\]

Substitute the value of ${F_{AC}}$ and ${C_z}$ in the above equation:

\[\begin{array}{c} {F_{CF}} - \left\{ {\left( {583.1\;{\rm{lb}}} \right) \times \frac{3}{{\sqrt {34} }}} \right\} + \left( 0 \right) = 0\\ {F_{CF}} = 300.0\;{\rm{lb}}\;\left( {{\rm{Compression}}} \right) \end{array}\]

So, the member CF is under compression.

Draw a free-body diagram of joint F.

Apply equilibrium equation of forces along the x-axis direction:

\[\begin{array}{c} \sum {{F_x}} = 0\\ {F_{DF}} \times \frac{3}{{\sqrt {18} }} - {F_{EF}} = 0 \end{array}\]

Substitute the value of ${F_{EF}}$ in the above equation:

\[\begin{array}{c} \left\{ {{F_{DF}} \times \frac{3}{{\sqrt {18} }}} \right\} - \left( {300\;{\rm{lb}}} \right) = 0\\ {F_{DF}} \times \frac{3}{{\sqrt {18} }} = 300\;{\rm{lb}}\\ {F_{DF}} = 424.26\;{\rm{lb}}\;\left( {{\rm{Tension}}} \right) \end{array}\]

So, the member DF is under tension.