Chapter 5, Problem 5.56RP

The uniform bars AB and BC each weigh 4 lb/ft. Calculate the tension in cable DE, and the magnitudes of the ball-and-socket reactions at A, B, and C.

To determine

Tension in cable DE and magnitude of reactions at A, B and C.

Answer to Problem 5.56RP

  $A=12\text{ lb, }B=12\text{ lb, }C=140.7\text{ lb, }{T}_{DE}=161.4\text{ lb}\text{.}$

Explanation of Solution

Given Information:

Weight of the uniform bars AB and BC is 4 lb/ft.

  

Calculation:

Draw free body diagram of as shown in following figure,

  

Take equilibrium of moments about x-axis passing through A in free body diagram of AB as,

  $\begin{array}{l}\sum M_x=0\\ \Rightarrow 6B_z-3W_{AB}=0\\ \Rightarrow B_z=W_{Ab}/2=4\times 6/2=12\text{ lb}\text{. }\mathrm{...}\text{(1)}\end{array}$

Take equilibrium of moments about z-axis passing through A in free body diagram of AB as,

  $\begin{array}{l}\sum M_z=0\\ \Rightarrow 6B_x=0\\ \Rightarrow B_x=0\mathrm{...}\text{(2)}\end{array}$

Take equilibrium of forces in x-direction in free body diagram of AB as,

  $\begin{array}{l}\sum F_x=0\\ \Rightarrow B_x+A_x=0\\ \Rightarrow A_x=-B_x=0\\ \Rightarrow A_x=0.\mathrm{...}\text{(3)}\end{array}$

Take equilibrium of forces in y-direction in free body diagram of AB as,

  $\begin{array}{l}\sum F_y=0\\ \Rightarrow B_y+A_y=0\mathrm{...}\text{(4)}\end{array}$

Take equilibrium of forces in z-direction in free body diagram of AB as,

  $\begin{array}{l}\sum F_z=0\\ \Rightarrow B_z+A_z-W_{AB}=0\\ \Rightarrow 12+A_z-4\times 6=0\\ \Rightarrow A_z=12\text{ lb }\mathrm{...}\text{(5)}\end{array}$

Take equilibrium of moments about C in free body diagram of BC as,

  $\begin{array}{l}\sum {\overrightarrow{M}}_C=0\\ \Rightarrow {\overrightarrow{r}}_{CD}\times \left\{{\overrightarrow{W}}_{BC}+{\overrightarrow{T}}_{DE}\right\}+{\overrightarrow{r}}_{CB}\times \left(-\overrightarrow{B}\right)=0\\ \Rightarrow \left({\overrightarrow{r}}_D-{\overrightarrow{r}}_C\right)\times \left\{{\overrightarrow{W}}_{BC}+{\overrightarrow{T}}_{DE}\right\}+\left({\overrightarrow{r}}_B-{\overrightarrow{r}}_C\right)\times \left(-\overrightarrow{B}\right)=0\\ \Rightarrow \left(6.3\widehat{i}+3\widehat{j}-12.65\widehat{i}\right)\times \left\{\left(-14\times 4\widehat{k}\right)+T_{DE}\left(\frac{6.3\widehat{i}-3\widehat{j}+4\widehat{k}}{\sqrt{{6.3}^2+{(-3)}^2+4^2}}\right)\right\}\\ +\left(6\widehat{j}-12.65\widehat{i}\right)\times \left(-B_y\widehat{j}-12\widehat{k}\right)=0\end{array}$

Above vector equation yields following scalar equations,

  $\begin{array}{c}-240+1.48T_{DE}=0\mathrm{...}\text{(6)}\\ -0.009T_{DE}+12.65B_y=0\mathrm{...}\text{(7)}\end{array}$

Solve equations 6 and 7 to get $T_{DE}=161.4\text{ lb, }{B}_y=0.12\text{ lb}\text{.}$

From equation (4) $A_y=-B_y=-0.12\text{ lb}$.

Take equilibrium of forces in free body diagram of BC as,

  $\begin{array}{l}\sum \overrightarrow{F}=0\\ \Rightarrow \left\{{\overrightarrow{W}}_{BC}+{\overrightarrow{T}}_{DE}\right\}+\left(-\overrightarrow{B}\right)+\overrightarrow{C}=0\\ \Rightarrow \left\{\left(-14\times 4\widehat{k}\right)+T_{DE}\left(\frac{6.3\widehat{i}-3\widehat{j}+4\widehat{k}}{\sqrt{{6.3}^2+{(-3)}^2+4^2}}\right)\right\}+\left(-B_y\widehat{j}-12\widehat{k}\right)+\left(C_x\widehat{i}+C_y\widehat{j}+C_z\widehat{k}\right)=0\end{array}$

Above vector equation yields following scalar equations,

  $\begin{array}{c}{C}_x+0.78T_{DE}=0\mathrm{...}\text{(8)}\\ C_y-B_y-0.37T_{DE}=0\mathrm{...}\text{(9)}\\ -68+C_z+0.5T_{DE}=0\mathrm{...}\text{(10)}\end{array}$

Substitute known parameters and solve equations (8-10) to get,

  $C_x=-126.65\text{ lb, }{C}_y=60.2\text{ lb, }{C}_z=-12.1\text{ lb}\text{. }$

Magnitude of reaction at A: $A=\sqrt{A_x{}^2+A_y{}^2+A_z{}^2}=\sqrt{0^2+{(-0.12)}^2+{12}^2}=12\text{ lb}\text{.}$

Magnitude of reaction at B: $B=\sqrt{B_x{}^2+B_y{}^2+B_z{}^2}=\sqrt{0^2+{(0.12)}^2+{12}^2}=12\text{ lb}\text{.}$

Magnitude of reaction at C:

  $C=\sqrt{C_x{}^2+C_y{}^2+C_z{}^2}=\sqrt{{(-126.6)}^2+{(60.2)}^2+{(-12.1)}^2}=140.7\text{ lb}\text{.}$

Conclusion:

Therefore, $A=12\text{ lb, }B=12\text{ lb, }C=140.7\text{ lb, }{T}_{DE}=161.4\text{ lb}\text{.}$