Chapter 1, Problem 1.3.13P

A space truss has three-dimensional pin supports at joints 0, B, and C, Load P is applied at joint A and acts toward point Q. Coordinates of all joints arc given in feet (see figure).

(a) Find reaction force components B x, B z, and Oz

(b) Find the axial force in truss member AC.

To determine

You need to determine the force components $B_x,B_z$and $O_z$in the following figure.

Answer to Problem 1.3.13P

The forces are: $O_z=-1.25P,B_x=-0.8P,B_z=2.0P.$

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 A :

The forces at joint A are,

  $\begin{array}{l}\overrightarrow{P}=\frac{1}{\sqrt{4^2+{(-3)}^2+0^2}}P(4,-3,0)=\frac{1}{5}P(4,-3,0),\\ \overrightarrow{F_{AC}}=\frac{1}{\sqrt{0^2+{(4)}^2+{(-5)}^2}}{F}_{AC}(0,4,-5)=\frac{1}{\sqrt{41}}{F}_{AC}(0,4,-5),\\ \overrightarrow{F_{AO}}=\frac{1}{\sqrt{0^2+{(0)}^2+{(-5)}^2}}{F}_{AO}(0,0,-5)=\frac{1}{5}{F}_{AO}(0,0,-5),\\ \overrightarrow{F_{AB}}=\frac{1}{\sqrt{2^2+{(0)}^2+{(-5)}^2}}{F}_{AB}(2,0,-5)=\frac{1}{\sqrt{29}}{F}_{AB}(2,0,-5).\end{array}$

Take equilibrium of forces at joint A in vector form,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}=0}\\ \Rightarrow \overrightarrow{P}+\overrightarrow{F_{AC}}+\overrightarrow{F_{AO}}+\overrightarrow{F_{AB}}=0\\ \Rightarrow \frac{1}{5}P(4,-3,0)+\frac{1}{\sqrt{41}}{F}_{AC}(0,4,-5)+\frac{1}{5}{F}_{AO}(0,0,-5)+\frac{1}{\sqrt{29}}{F}_{AB}(2,0,-5)=0\end{array}$

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

  $\begin{array}{c}4\times \frac{1}{5}P+2\times \frac{1}{\sqrt{29}}{F}_{AB}=0\mathrm{....}\text{(1)}\\ -3\times \frac{1}{5}P+4\times \frac{1}{\sqrt{41}}{F}_{AC}=0\mathrm{....}\text{(2)}\\ -5\times \frac{1}{\sqrt{41}}{F}_{AC}-5\times \frac{1}{5}{F}_{AO}-5\times \frac{1}{\sqrt{29}}{F}_{AB}=0\mathrm{....}\text{(3)}\end{array}$

Solve the three equations to get $F_{AB}=-\frac{2}{5}\sqrt{29}P,F_{AC}=\frac{3}{20}\sqrt{41}P,F_{AO}=\frac{5}{4}P.$

Joint O :

Consider the free body diagram again,

The forces at joint O are,

  $\begin{array}{l}\overrightarrow{F_{AO}}=\frac{1}{5}{F}_{AO}(0,0,5),\\ \overrightarrow{O}=(O_x,O_y,O_z).\end{array}$

Take equilibrium of forces at joint O in vector form,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}=0}\\ \Rightarrow \overrightarrow{F_{AO}}+\overrightarrow{O}=0\\ \Rightarrow \frac{1}{5}{F}_{AO}(0,0,5)+(O_x,O_y,O_z)=0\end{array}$

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

  $\begin{array}{c}{O}_X=0\mathrm{....}\text{(1)}\\ O_y=0\mathrm{....}\text{(2)}\\ 5\times \frac{1}{5}{F}_{AO}+O_z=0\mathrm{....}\text{(3)}\end{array}$

Solve the equation (3) to get ${\text{O}}_z=-F_{AO}=-1.25P.$

Joint B :

Consider the free body diagram again,

The forces at joint B are,

  $\begin{array}{l}\overrightarrow{F_{AB}}=\frac{1}{\sqrt{29}}{F}_{AB}(-2,0,5),\\ \overrightarrow{B}=(B_x,B_y,B_z).\end{array}$

Take equilibrium of forces at joint O in vector form,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}=0}\\ \Rightarrow \overrightarrow{F_{AB}}+\overrightarrow{B}=0\\ \Rightarrow \frac{1}{\sqrt{29}}{F}_{AB}(-2,0,5)+(B_x,B_y,B_z)=0\end{array}$

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

  $\begin{array}{c}-2\times \frac{1}{\sqrt{29}}{F}_{AB}+B_x=0\mathrm{....}\text{(1)}\\ B_y=0\mathrm{....}\text{(2)}\\ 5\times \frac{1}{\sqrt{29}}{F}_{AB}+B_z=0\mathrm{....}\text{(3)}\end{array}$

Solve the equation (3) to get $B_x=\frac{2}{\sqrt{29}}{F}_{AB}=\frac{2}{\sqrt{29}}\times \frac{-2}{5}\sqrt{29}P=-0.8P$and $B_z=-\frac{5}{\sqrt{29}}{F}_{AB}=-\frac{5}{\sqrt{29}}\times \frac{-2}{5}\sqrt{29}P=2P$.

Conclusion:

Therefore the forces are: $O_z=-1.25P,B_x=-0.8P,B_z=2.0P.$

To determine

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

Answer to Problem 1.3.13P

The forces are: $F_{AC}=0.96P.$

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 A :

The forces at joint A are,

  $\begin{array}{l}\overrightarrow{P}=\frac{1}{\sqrt{4^2+{(-3)}^2+0^2}}P(4,-3,0)=\frac{1}{5}P(4,-3,0),\\ \overrightarrow{F_{AC}}=\frac{1}{\sqrt{0^2+{(4)}^2+{(-5)}^2}}{F}_{AC}(0,4,-5)=\frac{1}{\sqrt{41}}{F}_{AC}(0,4,-5),\\ \overrightarrow{F_{AO}}=\frac{1}{\sqrt{0^2+{(0)}^2+{(-5)}^2}}{F}_{AO}(0,0,-5)=\frac{1}{5}{F}_{AO}(0,0,-5),\\ \overrightarrow{F_{AB}}=\frac{1}{\sqrt{2^2+{(0)}^2+{(-5)}^2}}{F}_{AB}(2,0,-5)=\frac{1}{\sqrt{29}}{F}_{AB}(2,0,-5).\end{array}$

Take equilibrium of forces at joint A in vector form,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}=0}\\ \Rightarrow \overrightarrow{P}+\overrightarrow{F_{AC}}+\overrightarrow{F_{AO}}+\overrightarrow{F_{AB}}=0\\ \Rightarrow \frac{1}{5}P(4,-3,0)+\frac{1}{\sqrt{41}}{F}_{AC}(0,4,-5)+\frac{1}{5}{F}_{AO}(0,0,-5)+\frac{1}{\sqrt{29}}{F}_{AB}(2,0,-5)=0\end{array}$

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

  $\begin{array}{c}4\times \frac{1}{5}P+2\times \frac{1}{\sqrt{29}}{F}_{AB}=0\mathrm{....}\text{(1)}\\ -3\times \frac{1}{5}P+4\times \frac{1}{\sqrt{41}}{F}_{AC}=0\mathrm{....}\text{(2)}\\ -5\times \frac{1}{\sqrt{41}}{F}_{AC}-5\times \frac{1}{5}{F}_{AO}-5\times \frac{1}{\sqrt{29}}{F}_{AB}=0\mathrm{....}\text{(3)}\end{array}$

Solve the three equations to get $F_{AB}=-\frac{2}{5}\sqrt{29}P,F_{AC}=\frac{3}{20}\sqrt{41}P,F_{AO}=\frac{5}{4}P.$

Conclusion:

Therefore the member force in AC is: $F_{AC}=\frac{3}{20}\sqrt{41}P=0.96P$.