Simer - Space Trusses

NotationπŸ”—

`\vec{\textcolor{#dac83c}{F}}_{\textcolor{#1ba7ae}{AB}_\textcolor{#f90bec}{C}}` is a force vector `\color{#dac83c}{F}`, along member `\textcolor{#1ba7ae}{AB}`, acting at joint `\textcolor{#f90bec}{C}` (uppercase subscript).

`\vec{\textcolor{#dac83c}{F}}_{\textcolor{#1ba7ae}{AB}_\textcolor{#f90bec}{x}}` is a force vector `\color{#dac83c}{F}`, along member `\textcolor{#1ba7ae}{AB}`, acting along the axis `\textcolor{#f90bec}{x}` (lowercase subscript).

`\vec{\textcolor{#dac83c}{F}}_{\textcolor{#f90bec}{C}_\textcolor{#1ba7ae}{AB}}` is a force vector `\color{#dac83c}{F}`, acting at joint `\textcolor{#f90bec}{C}`, on member `\textcolor{#1ba7ae}{AB}`.

The space truss supports a force `\vec F = \left\{600\hat i + 450\hat j - 750\hat k\right\}\ lb`.

Determine the force in each member, and state if the members are in tension or compression.

Joint `\boldsymbol{C}`: `\boldsymbol{ \begin{bmatrix} \boldsymbol{F_{AC} = 1000\ lb\ (C)} \\ \boldsymbol{F_{DC} = -406.25\ lb\ (T)} \\ \boldsymbol{F_{BC} = 343.75\ lb\ (C)} \\ \end{bmatrix} }`πŸ”—

\begin{aligned} \vec F_{R_C} &= \vec F_{AC_C} + \vec F_{BC_C} + \vec F_{DC_C} + \vec F \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -0.6F_{AC} \\ 0 \\ 0.8F_{AC} \\ \end{bmatrix} + \begin{bmatrix} 0 \\ -0.6F_{BC} \\ 0.8F_{BC} \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 0.6F_{DC} \\ 0.8F_{DC} \\ \end{bmatrix} + \begin{bmatrix} 600 \\ 450 \\ -750 \\ \end{bmatrix} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -0.6F_{AC} + 600 \\ -0.6F_{BC} + 0.6F_{DC} + 450 \\ 0.8F_{AC} + 0.8F_{BC} + 0.8F_{DC} - 750 \\ \end{bmatrix} \\[1pt] \hline F_{AC} &= 1000\ lb\ (C) \quad\quad \vec F_{AC_C} = \begin{bmatrix} -600 \\ 0 \\ 800 \\ \end{bmatrix} \ lb \\[1pt] F_{DC} &= -406.25\ lb\ (T) \quad\quad \vec F_{DC_C} = \begin{bmatrix} 0 \\ -243.75 \\ -325 \\ \end{bmatrix} \ lb \\[1pt] F_{BC} &= 343.75\ lb\ (C) \quad\quad \vec F_{BC_C} = \begin{bmatrix} 0 \\ -206.25 \\ 275 \\ \end{bmatrix} \ lb\\[1pt] \end{aligned}

Joint `\boldsymbol{A}`: `\boldsymbol{ \begin{bmatrix} \boldsymbol{F_{AD} = 424.26\ lb\ (T)} \\ \boldsymbol{F_{AB} = 424.26\ lb\ (T)} \\ \end{bmatrix} }`πŸ”—

\begin{aligned} \vec F_{R_A} &= \vec F_{AB_A} + \vec F_{AC_A} + \vec F_{AD_A} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -0.71F_{AB} \\ 0.71F_{AB} \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 600 \\ 0 \\ -800 \\ \end{bmatrix} + \begin{bmatrix} -0.71F_{AD} \\ -0.71F_{AD} \\ 0 \\ \end{bmatrix} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -0.71F_{AB} + 600 - 0.71F_{AD} \\ 0.71F_{AB} - 0.71F_{AD} \\ -800 \\ \end{bmatrix} \\[1pt] \hline -0.71F_{AD} - 0.71F_{AB} &= -600 \color{green}\text{ //equation 1} \\[1pt] -0.71F_{AD} + 0.71F_{AB} &= 0 \color{green}\text{ //equation 2} \\[1pt] \hline F_{AD} &= 424.26\ lb\ (T) \quad\quad \vec F_{AD_A} = \begin{bmatrix} -300 \\ -300 \\ 0 \\ \end{bmatrix} \ lb \\[1pt] F_{AB} &= 424.26\ lb\ (T) \quad\quad \vec F_{AB_A} = \begin{bmatrix} -300 \\ 300 \\ 0 \\ \end{bmatrix} \ lb \\[1pt] \end{aligned}

Joint `\boldsymbol{D}`: `\boldsymbol{F_{BD} = 543.75\ lb\ (C)}`πŸ”—

\begin{aligned} \vec F_{R_D} &= \vec F_{BD_D} + \vec F_{AD_D} + \vec F_{DC_D} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} 0 \\ -F_{BD} \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 300 \\ 300 \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 243.75 \\ 325 \\ \end{bmatrix} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} 300 \\ -F_{BD} + 543.75 \\ 325 \\ \end{bmatrix} \\[1pt] \hline F_{BD} &= 543.75\ lb \ (C) \quad\quad \vec F_{BD_D} = \begin{bmatrix} 0 \\ -543.75 \\ 0 \\ \end{bmatrix} \ lb \\[1pt] \end{aligned}

The space truss supports a force `\vec F = \left\{-500\hat i + 600\hat j + 400\hat k\right\}\ lb`.

Determine the force in each member, and state if the members are in tension or compression.

Joint `\boldsymbol{C}`: `\boldsymbol{ \begin{bmatrix} \boldsymbol{F_{AC} = -833.33\ lb\ (T)} \\ \boldsymbol{F_{DC} = -333.33\ lb\ (T)} \\ \boldsymbol{F_{BC} = 666.67\ lb\ (C)} \\ \end{bmatrix} }`πŸ”—

\begin{aligned} \vec F_{R_C} &= \vec F_{AC_C} + \vec F_{BC_C} + \vec F_{DC_C} + \vec F \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -0.6F_{AC} \\ 0 \\ 0.8F_{AC} \\ \end{bmatrix} + \begin{bmatrix} 0 \\ -0.6F_{BC} \\ 0.8F_{BC} \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 0.6F_{DC} \\ 0.8F_{DC} \\ \end{bmatrix} + \begin{bmatrix} -500 \\ 600 \\ 400 \\ \end{bmatrix} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -0.6F_{AC} - 500 \\ -0.6F_{BC} + 0.6F_{DC} + 600 \\ 0.8F_{AC} + 0.8F_{BC} + 0.8F_{DC} + 400 \\ \end{bmatrix} \\[1pt] \hline F_{AC} &= -833.33\ lb\ (T) \quad\quad \vec F_{AC_C} = \begin{bmatrix} 500 \\ 0 \\ -666.67 \\ \end{bmatrix} \ lb \\[1pt] F_{DC} &= -333.33\ lb\ (T) \quad\quad \vec F_{DC_C} = \begin{bmatrix} 0 \\ -200 \\ -266.67 \\ \end{bmatrix} \ lb \\[1pt] F_{BC} &= 666.67\ lb\ (C) \quad\quad \vec F_{BC_C} = \begin{bmatrix} 0 \\ -400 \\ 533.33 \\ \end{bmatrix} \ lb\\[1pt] \end{aligned}

Joint `\boldsymbol{A}`: `\boldsymbol{ \begin{bmatrix} \boldsymbol{F_{AD} = -353.55\ lb\ (C)} \\ \boldsymbol{F_{AB} = -353.55\ lb\ (C)} \\ \end{bmatrix} }`πŸ”—

\begin{aligned} \vec F_{R_A} &= \vec F_{AB_A} + \vec F_{AC_A} + \vec F_{AD_A} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -0.71F_{AB} \\ 0.71F_{AB} \\ 0 \\ \end{bmatrix} + \begin{bmatrix} -500 \\ 0 \\ 666.67 \\ \end{bmatrix} + \begin{bmatrix} -0.71F_{AD} \\ -0.71F_{AD} \\ 0 \\ \end{bmatrix} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -0.71F_{AB} - 500 - 0.71F_{AD} \\ 0.71F_{AB} - 0.71F_{AD} \\ 666.67 \\ \end{bmatrix} \\[1pt] \hline -0.71F_{AD} - 0.71F_{AB} &= 500 \color{green}\text{ //equation 1} \\[1pt] -0.71F_{AD} + 0.71F_{AB} &= 0 \color{green}\text{ //equation 2} \\[1pt] \hline F_{AD} &= -353.55\ lb\ (C) \quad\quad \vec F_{AD_A} = \begin{bmatrix} 250 \\ 250 \\ 0 \\ \end{bmatrix} \ lb \\[1pt] F_{AB} &= -353.55\ lb\ (C) \quad\quad \vec F_{AB_A} = \begin{bmatrix} 250 \\ -250 \\ 0 \\ \end{bmatrix} \ lb \\[1pt] \end{aligned}

Joint `\boldsymbol{D}`: `\boldsymbol{F_{BD} = -50\ lb\ (T)}`πŸ”—

\begin{aligned} \vec F_{R_D} &= \vec F_{BD_D} + \vec F_{AD_D} + \vec F_{DC_D} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} 0 \\ -F_{BD} \\ 0 \\ \end{bmatrix} + \begin{bmatrix} -250 \\ -250 \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 200 \\ 266.67 \\ \end{bmatrix} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -250 \\ -F_{BD} - 50 \\ 266.67 \\ \end{bmatrix} \\[1pt] \hline F_{BD} &= -50\ lb\ (T) \quad\quad \vec F_{BD_D} = \begin{bmatrix} 0 \\ 50 \\ 0 \\ \end{bmatrix} \ lb \\[1pt] \end{aligned}

Determine the force in each member of the space truss and state if the members are in tension or compression.

The truss is supported by ball-and-socket joints at `A, B, C,\ and\ D`.

Joint `\boldsymbol{G}`: `\boldsymbol{ \begin{bmatrix} \boldsymbol{F_{DG}} \\ \boldsymbol{F_{CG}} \\ \boldsymbol{F_{EG}} \\ \end{bmatrix} = \left[ \begin{array}{r} \boldsymbol{-4.47} \\ \boldsymbol{4.47} \\ \boldsymbol{-6} \\ \end{array} \right] \ kN}`πŸ”—

\begin{aligned} \vec F_{R_G} &= \vec F_{CG_G} + \vec F_{DG_G} + \vec F_{EG_G} + \vec F_G \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -0.89F_{CG} \\ 0.45F_{CG} \\ 0 \\ \end{bmatrix} + \begin{bmatrix} -0.89F_{DG} \\ -0.45F_{DG} \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ -F_{EG} \\ \end{bmatrix} + \begin{bmatrix} 0 \\ -4 \\ -6 \\ \end{bmatrix} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -0.89F_{CG} - 0.89F_{DG} \\ 0.45F_{CG} - 0.45F_{DG} - 4 \\ -F_{EG} - 6 \\ \end{bmatrix} \\[1pt] \hline \left[ \begin{array}{rrr} -0.89 & -0.89 & 0 \\ -0.45 & 0.45 & 0 \\ 0 & 0 & -1 \\ \end{array} \right] \begin{bmatrix} F_{DG} \\ F_{CG} \\ F_{EG} \\ \end{bmatrix} &= \left[ \begin{array}{r} 0 \\ 4 \\ 6 \\ \end{array} \right] \\[1pt] \begin{bmatrix} F_{DG} \\ F_{CG} \\ F_{EG} \\ \end{bmatrix} &= \left[ \begin{array}{r} -4.47 \\ 4.47 \\ -6 \\ \end{array} \right] \ kN \\[1pt] \hline \vec F_{DG_G} &= \begin{bmatrix} 4 \\ 2 \\ 0 \\ \end{bmatrix} \ kN \quad\quad \vec F_{CG_G} = \begin{bmatrix} -4 \\ 2 \\ 0 \\ \end{bmatrix} \ kN \quad\quad \vec F_{EG_G} = \begin{bmatrix} 0 \\ 0 \\ 6 \\ \end{bmatrix} \ kN \\[1pt] \end{aligned}

Joint `\boldsymbol{E}`: `\boldsymbol{ \begin{bmatrix} \boldsymbol{F_{DE}} \\ \boldsymbol{F_{AE}} \\ \boldsymbol{F_{BE}} \\ \end{bmatrix} = \left[ \begin{array}{r} \boldsymbol{9} \\ \boldsymbol{-6.71} \\ \boldsymbol{0} \\ \end{array} \right] \ kN}`πŸ”—

\begin{aligned} \vec F_{R_E} &= \vec F_{BE_E} + \vec F_{AE_E} + \vec F_{DE_E} + \vec F_{EG_E} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -0.89F_{BE} \\ 0.45F_{BE} \\ 0 \\ \end{bmatrix} + \begin{bmatrix} -0.89F_{AE} \\ -0.45F_{AE} \\ 0 \\ \end{bmatrix} + \begin{bmatrix} -0.67F_{DE} \\ -0.33F_{DE} \\ 0.67F_{DE} \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ -6 \\ \end{bmatrix} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -0.89F_{BE} - 0.89F_{AE} - 0.67F_{DE} \\ 0.45F_{BE} - 0.45F_{AE} - 0.33F_{DE} \\ 0.67F_{DE} - 6 \\ \end{bmatrix} \\[1pt] \hline \left[ \begin{array}{rrr} -0.67 & -0.89 & -0.89 \\ -0.33 & -0.45 & 0.45 \\ 0.67 & 0 & 0 \\ \end{array} \right] \begin{bmatrix} F_{DE} \\ F_{AE} \\ F_{BE} \\ \end{bmatrix} &= \left[ \begin{array}{r} 0 \\ 0 \\ 6 \\ \end{array} \right] \\[1pt] \begin{bmatrix} F_{DE} \\ F_{AE} \\ F_{BE} \\ \end{bmatrix} &= \left[ \begin{array}{r} 9 \\ -6.71 \\ 0 \\ \end{array} \right] \ kN \\[1pt] \hline \vec F_{DE_E} &= \begin{bmatrix} -6 \\ -3 \\ 6 \\ \end{bmatrix} \ kN \quad\quad \vec F_{AE_E} = \begin{bmatrix} 6 \\ 3 \\ 0 \\ \end{bmatrix} \ kN \quad\quad \vec F_{BE_E} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} \ kN \\[1pt] \end{aligned}

Determine the force in each member of the space truss and state if the members are in tension or compression.

Hint: The support reaction at `E` acts along member `EB`. Why?

Joint `\boldsymbol{A}`: `\boldsymbol{F_{AB} = 6.46\ kN\ (T)}`πŸ”—

\begin{aligned} \vec F_{R_A} &= \vec F_{AC_A} + \vec F_{AD_A} + \vec F_{AB_A} + \vec F_A \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -0.6F_{AC} \\ -0.8F_{AC} \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 0.6F_{AD} \\ -0.8F_{AD} \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 0 \\ -0.37F_{AB} \\ 0.93F_{AB} \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ -6 \\ \end{bmatrix} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -0.6F_{AC} + 0.6F_{AD} \\ -0.8F_{AC} - 0.8F_{AD} - 0.37F_{AB} \\ 0.93F_{AB} - 6 \\ \end{bmatrix} \\[1pt] \hline 0.6F_{AD} - 0.6F_{AC} &= 0 \color{green}\text{ //equation 1} \\[1pt] -0.8F_{AD} - 0.8F_{AC} &= 2.4 \color{green}\text{ //equation 2} \\[1pt] F_{AB} &= 6.46\ kN\ (T) \quad\quad \vec F_{AB_A} = \begin{bmatrix} 0 \\ -2.4 \\ 6 \\ \end{bmatrix} \ kN \\[1pt] \end{aligned}

`\boldsymbol{ \begin{bmatrix} \boldsymbol{F_{AD}} \\ \boldsymbol{F_{AC}} \\ \end{bmatrix} = \left[ \begin{array}{r} \boldsymbol{-1.5} \\ \boldsymbol{-1.5} \\ \end{array} \right] \ kN\ (C)}`πŸ”—

Solve the two equations.

\begin{aligned} \left[ \begin{array}{rr} 0.6 & -0.6 \\ -0.8 & -0.8 \\ \end{array} \right] \begin{bmatrix} F_{AD} \\ F_{AC} \\ \end{bmatrix} &= \left[ \begin{array}{r} 0 \\ 2.4 \\ \end{array} \right] \\[1pt] \begin{bmatrix} F_{AD} \\ F_{AC} \\ \end{bmatrix} &= \left[ \begin{array}{r} -1.5 \\ -1.5 \\ \end{array} \right] \ kN\ (C) \\[1pt] \hline \vec F_{AD_A} &= \begin{bmatrix} -0.9 \\ 1.2 \\ 0 \\ \end{bmatrix} \ kN \quad\quad \vec F_{AC_A} = \begin{bmatrix} 0.9 \\ 1.2 \\ 0 \\ \end{bmatrix} \ kN \\[1pt] \end{aligned}

Joint `\boldsymbol{B}`: `\boldsymbol{ \begin{bmatrix} \boldsymbol{F_{BC} = -3.7\ kN (C)} \\ \boldsymbol{F_{BD} = -3.7\ kN \ (C)} \\ \boldsymbol{F_{BE} = 4.8\ kN \ (T)} \\ \end{bmatrix} }`πŸ”—

\begin{aligned} \vec F_{R_B} &= \vec F_{BD_B} + \vec F_{BC_B} + \vec F_{BE_B} + \vec F_{AB_B} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} 0.49F_{BD} \\ -0.32F_{BD} \\ -0.81F_{BD} \\ \end{bmatrix} + \begin{bmatrix} -0.49F_{BC} \\ -0.32F_{BC} \\ -0.81F_{BC} \\ \end{bmatrix} + \begin{bmatrix} 0 \\ -F_{BE} \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 2.4 \\ -6 \\ \end{bmatrix} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} 0.49F_{BD} - 0.49F_{BC} \\ -0.32F_{BD} - 0.32F_{BC} - F_{BE} + 2.4 \\ -0.81F_{BD} - 0.81F_{BC} - 6 \\ \end{bmatrix} \\[1pt] \hline \left[ \begin{array}{rrr} -0.49 & 0.49 & 0 \\ -0.32 & -0.32 & -1 \\ -0.81 & -0.81 & 0 \\ \end{array} \right] \begin{bmatrix} F_{BC} \\ F_{BD} \\ F_{BE} \\ \end{bmatrix} &= \left[ \begin{array}{r} 0 \\ -2.4 \\ 6 \\ \end{array} \right] \\[1pt] \begin{bmatrix} F_{BC} \\ F_{BD} \\ F_{BE} \\ \end{bmatrix} &= \left[ \begin{array}{r} -3.7 \\ -3.7 \\ 4.8 \\ \end{array} \right] \ kN \\[1pt] \hline \vec F_{BD_B} &= \begin{bmatrix} -1.8 \\ 1.2 \\ 3 \\ \end{bmatrix} \quad\quad \vec F_{BC_B} = \begin{bmatrix} 1.8 \\ 1.2 \\ 3 \\ \end{bmatrix} \quad\quad \vec F_{BE_B} = \begin{bmatrix} 0 \\ -4.8 \\ 0 \\ \end{bmatrix} \\[1pt] \end{aligned}

Determine the force developed in each member of the space truss and state if the members are in tension or compression.

The crate has a weight of `150\ lb`.

Joint `\boldsymbol{C}`: ForcesπŸ”—

First, find all force vectors that act at joint `C`.

`` \vec F_{crate} = \begin{bmatrix} 0 \\ 0 \\ -150 \\ \end{bmatrix} \quad\quad \vec F_{AC_C} = \begin{bmatrix} 0.35F_{AC} \\ -0.71F_{AC} \\ -0.61F_{AC} \\ \end{bmatrix} \quad\quad \vec F_{BC_C} = \begin{bmatrix} -0.35F_{BC} \\ -0.71F_{BC} \\ -0.61F_{BC} \\ \end{bmatrix} \quad\quad \vec F_{CD_C} = \begin{bmatrix} 0 \\ -F_{CD} \\ 0 \\ \end{bmatrix} ``

`\boldsymbol{ \begin{bmatrix} \boldsymbol{F_{BC} = -122.47\ lb\ (C)} \\ \boldsymbol{F_{AC} = -122.47\ lb\ (C)} \\ \boldsymbol{F_{CD} = 173.21\ lb\ (T)} \\ \end{bmatrix} }`πŸ”—

Joint `C` must be in equilibrium.

\begin{aligned} \vec F_{R_C} &= \vec F_{AC_C} + \vec F_{BC_C} + \vec F_{CD_C} + \vec F_{crate} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} 0.35F_{AC} \\ -0.71F_{AC} \\ -0.61F_{AC} \\ \end{bmatrix} + \begin{bmatrix} -0.35F_{BC} \\ -0.71F_{BC} \\ -0.61F_{BC} \\ \end{bmatrix} + \begin{bmatrix} 0 \\ -F_{CD} \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ -150 \\ \end{bmatrix} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} 0.35F_{AC} - 0.35F_{BC} \\ -0.71F_{AC} - 0.71F_{BC} - F_{CD} \\ -0.61F_{AC} - 0.61F_{BC} - 150 \\ \end{bmatrix} \\[1pt] \hline \left[ \begin{array}{rrr} -0.35 & 0.35 & 0 \\ -0.71 & -0.71 & -1 \\ -0.61 & -0.61 & 0 \\ \end{array} \right] \begin{bmatrix} F_{BC} \\ F_{AC} \\ F_{CD} \\ \end{bmatrix} &= \left[ \begin{array}{r} 0 \\ 0 \\ 150 \\ \end{array} \right] \\[1pt] \begin{bmatrix} F_{BC} \\ F_{AC} \\ F_{CD} \\ \end{bmatrix} &= \left[ \begin{array}{r} -122.47 \\ -122.47 \\ 173.21 \\ \end{array} \right] \ lb \\[1pt] \hline \vec F_{BC_C} &= \begin{bmatrix} 43.3 \\ 86.6 \\ 75 \\ \end{bmatrix} \ lb \quad\quad \vec F_{AC_C} = \begin{bmatrix} -43.3 \\ 86.6 \\ 75 \\ \end{bmatrix} \ lb \quad\quad \vec F_{CD_C} = \begin{bmatrix} 0 \\ -173.21 \\ 0 \\ \end{bmatrix} \ lb \\[1pt] \end{aligned}

Joint `\boldsymbol{B}`: ForcesπŸ”—

`` \vec F_{AB_B} = \begin{bmatrix} -F_{AB} \\ 0 \\ 0 \\ \end{bmatrix} \quad\quad \vec F_{BD_B} = \begin{bmatrix} 0.5F_{BD} \\ 0 \\ 0.87F_{BD} \\ \end{bmatrix} \quad\quad \vec F_{BC_B} = -1 \times \vec F_{BC_C} = \begin{bmatrix} -43.3 \\ -86.6 \\ -75 \\ \end{bmatrix} ``

`\boldsymbol{ \begin{bmatrix} \boldsymbol{F_{BD} = 86.6\ lb\ (T)} \\ \boldsymbol{F_{AB} = 0\ lb} \\ \end{bmatrix} }`πŸ”—

Joint `B` must be in equilibrium.

\begin{aligned} \vec F_{R_B} &= \vec F_{AB_B} + \vec F_{BD_B} + \vec F_{BC_B} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -F_{AB} \\ 0 \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 0.5F_{BD} \\ 0 \\ 0.87F_{BD} \\ \end{bmatrix} + \begin{bmatrix} -43.3 \\ -86.6 \\ -75 \\ \end{bmatrix} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -F_{AB} + 0.5F_{BD} - 43.3 \\ -86.6 \\ 0.87F_{BD} - 75 \\ \end{bmatrix} \\[1pt] \hline F_{BD} &= 86.6\ lb\ (T) \\[1pt] F_{AB} &= 0\ lb \\[1pt] \hline \vec F_{BD_B} &= \begin{bmatrix} 43.3 \\ 0 \\ 75 \\ \end{bmatrix} \ lb \quad\quad \vec F_{AB_B} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} \ lb \\[1pt] \end{aligned}

Joint `\boldsymbol{D}`: ForcesπŸ”—

`` \vec F_{AD_D} = \begin{bmatrix} -0.5F_{AD} \\ 0 \\ 0.87F_{AD} \\ \end{bmatrix} \quad\quad \vec F_{BD_D} = -1 \times \vec F_{BD_B} = \begin{bmatrix} -43.3 \\ 0 \\ -75 \\ \end{bmatrix} \quad\quad \vec F_{CD_D} = -1 \times \vec F_{CD_C} = \begin{bmatrix} 0 \\ 173.21 \\ 0 \\ \end{bmatrix} ``

`\boldsymbol{F_{AD} = -86.6\ lb\ (T)}`πŸ”—

Joint `D` must be in equilibrium.

\begin{aligned} \vec F_{R_D} &= \vec F_{AD_D} + \vec F_{BD_D} + \vec F_{CD_D} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -0.5F_{AD} \\ 0 \\ 0.87F_{AD} \\ \end{bmatrix} + \begin{bmatrix} -43.3 \\ 0 \\ -75 \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 173.21 \\ 0 \\ \end{bmatrix} \\[1pt] \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} &= \begin{bmatrix} -0.5F_{AD} - 43.3 \\ 173.21 \\ 0.87F_{AD} - 75 \\ \end{bmatrix} \\[1pt] \hline F_{AD} &= -86.6\ lb\ (T) \\[1pt] \hline \vec F_{AD_D} &= \begin{bmatrix} 43.3 \\ 0 \\ -75 \\ \end{bmatrix} \ lb \\[1pt] \end{aligned}