Chapter 1, Problem 1.3.14P

A space truss is restrained at joints O, A. B. and C, as shown in the figure. Load P is applied at joint A and load IP acts downward at joint C.

(a) Find reaction force components A_x, B_y, and B. in terms of load variable P.

(b) Find the axial force in truss member AB in terms of load variable P.

  

To determine

You need to determine the force components $A_x,B_y,B_z$in the following figure.

Answer to Problem 1.3.14P

The correct answers are:

$A_x=-1.25P,B_y=0,B_z=-P$

Explanation of Solution

Given Information:

You have following figure with all relevant information,

Draw free body diagram of joints and use equilibrium of forces to determine the unknowns.

Calculation:

Draw free body diagram as shown in the following figure,

Joint C :

The forces at joint C are,

  $\begin{array}{l}\overrightarrow{P}=2P(0,-1,0),\\ \overrightarrow{F_{AC}}=\frac{1}{\sqrt{0^2+{(-0.6L)}^2+{(0.8L)}^2}}{F}_{AC}(0,-0.6L,0.8L)=F_{AC}(0,-0.6,0.8),\\ \overrightarrow{F_{CO}}=F_{CO}(0,-1,0),\\ \overrightarrow{C}=(C_x,0,0).\end{array}$

Take equilibrium of forces at joint C in vector form,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}=0}\\ \Rightarrow \overrightarrow{P}+\overrightarrow{F_{AC}}+\overrightarrow{F_{CO}}+\overrightarrow{C}=0\\ \Rightarrow 2P(0,-1,0)+F_{AC}(0,-0.6,0.8)+F_{CO}(0,-1,0)+(C_x,0,0)=0\end{array}$

The vector equation yields three equations in components form as below,

  $\begin{array}{c}{C}_x=0\mathrm{....}\text{(1)}\\ -2P-0.6F_{AC}-F_{CO}=0\mathrm{....}\text{(2)}\\ F_{AC}=0\mathrm{....}\text{(3)}\end{array}$

Solve the three equations to get $C_x=0,{\text{ F}}_{AC}=0,F_{CO}=-2P$

Joint A :

The forces at joint A are,

  $\begin{array}{l}\overrightarrow{P}=(0,0,P),\\ \overrightarrow{F_{AC}}=\frac{1}{\sqrt{0^2+{(0.6L)}^2+{(-0.8L)}^2}}{F}_{AC}(0,0.6L,-0.8L)=F_{AC}(0,0.6,-0.8),\\ \overrightarrow{F_{AB}}=\frac{1}{\sqrt{L^2+{(0)}^2+{(-0.8L)}^2}}{F}_{AB}(L,0,-0.8L)=\frac{1}{\sqrt{1.64}}{F}_{AB}(1,0,-0.8),\\ \overrightarrow{A}=(A_x,A_y,0).\end{array}$

Take equilibrium of forces at joint A in vector form,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}=0}\\ \Rightarrow \overrightarrow{P}+\overrightarrow{F_{AC}}+\overrightarrow{F_{AB}}+\overrightarrow{A}=0\\ \Rightarrow P(0,0,P)+F_{AC}(0,0.6,-0.8)+\frac{1}{\sqrt{1.64}}{F}_{AB}(1,0,-0.8)+(A_x,A_y,0)=0\end{array}$

The vector equation yields three equations in components form as below,

  $\begin{array}{c}\frac{1}{\sqrt{1.64}}{F}_{AB}+A_x=0\mathrm{....}\text{(1)}\\ 0.6F_{AC}+A_y=0\mathrm{....}\text{(2)}\\ P-0.8F_{AC}-\frac{0.8}{\sqrt{1.64}}{F}_{AB}=0\mathrm{....}\text{(3)}\end{array}$

Solve the three equations to get

  $\begin{array}{l}{A}_y=-0.6F_{AC}=0\\ F_{AB}=\sqrt{1.64}\left(P-0.8F_{AC}\right)/0.8=\sqrt{1.64}\left(P\right)/0.8=1.6P\\ A_x=-\frac{1}{\sqrt{1.64}}{F}_{AB}=-\frac{1}{\sqrt{1.64}}\sqrt{1.64}\left(P\right)/0.8=-\frac{P}{0.8}=-1.25P\end{array}$

Joint B :

The forces at joint B are,

  $\begin{array}{l}\overrightarrow{F_{OB}}=F_{OB}(-1,0,0),\\ \overrightarrow{F_{AB}}=-\frac{1}{\sqrt{L^2+{(0)}^2+{(-0.8L)}^2}}{F}_{AB}(L,0,-0.8L)=-\frac{1}{\sqrt{1.64}}{F}_{AB}(1,0,-0.8),\\ \overrightarrow{B}=(0,B_y,B_z).\end{array}$

Take equilibrium of forces at joint B in vector form,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}=0}\\ \Rightarrow \overrightarrow{F_{OB}}+\overrightarrow{F_{AB}}+\overrightarrow{B}=0\\ \Rightarrow F_{OB}(-1,0,0)-\frac{1}{\sqrt{1.64}}{F}_{AB}(1,0,-0.8)+(0,B_y,B_z)=0\end{array}$

The vector equation yields three equations in components form as below,

  $\begin{array}{c}-\frac{1}{\sqrt{1.64}}{F}_{AB}-F_{OB}=0\mathrm{....}\text{(1)}\\ B_y=0\mathrm{....}\text{(2)}\\ B_z+\frac{0.8}{\sqrt{1.64}}{F}_{AB}=0\mathrm{....}\text{(3)}\end{array}$

Solve the three equations to get

  $\begin{array}{l}{B}_y=0\\ B_z=-\frac{0.8}{\sqrt{1.64}}{F}_{AB}=-\frac{0.8}{\sqrt{1.64}}\sqrt{1.64}\left(P\right)/0.8=-P\end{array}$

Conclusion:

Thus the forces are: $A_x=-1.25P,B_y=0,B_z=-P$

  $\text{.}$

To determine

You need to determine the member force $F_{AB}$in the following figure.

Answer to Problem 1.3.14P

The correct answers are:

$F_{AB}=1.6P$

Explanation of Solution

Given Information:

You have following figure with all relevant information,

Draw free body diagram of joints and use equilibrium of forces to determine the unknowns.

Calculation:

Draw free body diagram as shown in the following figure,

Joint C :

The forces at joint C are,

  $\begin{array}{l}\overrightarrow{P}=2P(0,-1,0),\\ \overrightarrow{F_{AC}}=\frac{1}{\sqrt{0^2+{(-0.6L)}^2+{(0.8L)}^2}}{F}_{AC}(0,-0.6L,0.8L)=F_{AC}(0,-0.6,0.8),\\ \overrightarrow{F_{CO}}=F_{CO}(0,-1,0),\\ \overrightarrow{C}=(C_x,0,0).\end{array}$

Take equilibrium of forces at joint C in vector form,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}=0}\\ \Rightarrow \overrightarrow{P}+\overrightarrow{F_{AC}}+\overrightarrow{F_{CO}}+\overrightarrow{C}=0\\ \Rightarrow 2P(0,-1,0)+F_{AC}(0,-0.6,0.8)+F_{CO}(0,-1,0)+(C_x,0,0)=0\end{array}$

The vector equation yields three equations in components form as below,

  $\begin{array}{c}{C}_x=0\mathrm{....}\text{(1)}\\ -2P-0.6F_{AC}-F_{CO}=0\mathrm{....}\text{(2)}\\ F_{AC}=0\mathrm{....}\text{(3)}\end{array}$

Solve the three equations to get $C_x=0,{\text{ F}}_{AC}=0,F_{CO}=-2P$

Joint A :

The forces at joint A are,

  $\begin{array}{l}\overrightarrow{P}=(0,0,P),\\ \overrightarrow{F_{AC}}=\frac{1}{\sqrt{0^2+{(0.6L)}^2+{(-0.8L)}^2}}{F}_{AC}(0,0.6L,-0.8L)=F_{AC}(0,0.6,-0.8),\\ \overrightarrow{F_{AB}}=\frac{1}{\sqrt{L^2+{(0)}^2+{(-0.8L)}^2}}{F}_{AB}(L,0,-0.8L)=\frac{1}{\sqrt{1.64}}{F}_{AB}(1,0,-0.8),\\ \overrightarrow{A}=(A_x,A_y,0).\end{array}$

Take equilibrium of forces at joint A in vector form,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}=0}\\ \Rightarrow \overrightarrow{P}+\overrightarrow{F_{AC}}+\overrightarrow{F_{AB}}+\overrightarrow{A}=0\\ \Rightarrow P(0,0,P)+F_{AC}(0,0.6,-0.8)+\frac{1}{\sqrt{1.64}}{F}_{AB}(1,0,-0.8)+(A_x,A_y,0)=0\end{array}$

The vector equation yields three equations in components form as below,

  $\begin{array}{c}\frac{1}{\sqrt{1.64}}{F}_{AB}+A_x=0\mathrm{....}\text{(1)}\\ 0.6F_{AC}+A_y=0\mathrm{....}\text{(2)}\\ P-0.8F_{AC}-\frac{0.8}{\sqrt{1.64}}{F}_{AB}=0\mathrm{....}\text{(3)}\end{array}$

Solve the three equations to get

  $\begin{array}{l}{A}_y=-0.6F_{AC}=0\\ F_{AB}=\sqrt{1.64}\left(P-0.8F_{AC}\right)/0.8=\sqrt{1.64}\left(P\right)/0.8=1.6P\\ A_x=-\frac{1}{\sqrt{1.64}}{F}_{AB}=-\frac{1}{\sqrt{1.64}}\sqrt{1.64}\left(P\right)/0.8=-\frac{P}{0.8}=-1.25P\end{array}$

Conclusion:

Thus the forces are: $F_{AB}=1.6P$

  $\text{.}$