Chapter 5, Problem 5.49P

Determine the reactions at ball-and-socket joints D, E, and F of the space truss shown. Express the answers in vector form.

To determine

Reaction at D, E and F.

Answer to Problem 5.49P

  ${\overrightarrow{R}}_D=-100\widehat{i}+100\widehat{j}+5900\widehat{k}\text{ lb, }{\overrightarrow{R}}_E=8000\widehat{j}\text{ lb, }{\overrightarrow{R}}_F=-8000\widehat{j}\text{ lb}\text{.}$

Explanation of Solution

Given Information:

The follwing truss with ball and socket joint at D, E and F is given,

  

Calculation:

Draw free body diagram of as shown in following figure,

  

Forces and corresponding position vectors are listed in the following table,

Force	Position vector
$\overrightarrow{A}=-6000\widehat{k}$	${\overrightarrow{r}}_A=4.5\widehat{i}+6\widehat{j}$
${\overrightarrow{F}}_{DA}=F_{DA}\widehat{j}$	${\overrightarrow{r}}_D=4.5\widehat{i}$
${\overrightarrow{F}}_{DB}=F_{DB}\left[\frac{-4.5\widehat{i}+6\widehat{j}}{\sqrt{{(-4.5)}^2+6^2}}\right]$	${\overrightarrow{r}}_D=4.5\widehat{i}$
${\overrightarrow{F}}_{DC}=F_{DC}\left[\frac{-4.5\widehat{i}+6\widehat{j}+4.5\widehat{k}}{\sqrt{{(-4.5)}^2+6^2+{4.5}^2}}\right]$	${\overrightarrow{r}}_D=4.5\widehat{i}$
${\overrightarrow{F}}_{EB}=F_{EB}\widehat{j}$	${\overrightarrow{r}}_E=\overrightarrow{0}$
${\overrightarrow{F}}_{EC}=F_{EC}\left[\frac{6\widehat{j}+4.5\widehat{k}}{\sqrt{6^2+{4.5}^2}}\right]$	${\overrightarrow{r}}_E=\overrightarrow{0}$
${\overrightarrow{F}}_{FC}=F_{FC}\widehat{j}$	${\overrightarrow{r}}_F=4.5\widehat{k}$

Take equilibrium of moments about origin as,

  $\begin{array}{l}\sum \overrightarrow{M}=0\\ \Rightarrow {\overrightarrow{r}}_A\times \overrightarrow{A}+{\overrightarrow{r}}_D\times \left\{{\overrightarrow{F}}_{DA}+{\overrightarrow{F}}_{DB}+{\overrightarrow{F}}_{DC}\right\}+{\overrightarrow{r}}_E\times \left\{{\overrightarrow{F}}_{EC}+{\overrightarrow{F}}_{EB}\right\}+{\overrightarrow{r}}_F\times {\overrightarrow{F}}_{FC}=0\\ \Rightarrow \left(4.5\widehat{i}+6\widehat{j}\right)\times \left(-6000\widehat{k}\right)\\ +\left(4.5\widehat{i}\right)\times \left\{\left(F_{DA}\widehat{j}\right)+\left(F_{DB}\left[\frac{-4.5\widehat{i}+6\widehat{j}}{\sqrt{{(-4.5)}^2+6^2}}\right]\right)+\left(F_{DC}\left[\frac{-4.5\widehat{i}+6\widehat{j}+4.5\widehat{k}}{\sqrt{{(-4.5)}^2+6^2+{4.5}^2}}\right]\right)\right\}\\ +\left(\overrightarrow{0}\right)\times \left\{\left(F_{EB}\widehat{j}\right)+\left(F_{EC}\left[\frac{6\widehat{j}+4.5\widehat{k}}{\sqrt{6^2+{4.5}^2}}\right]\right)\right\}+\left(4.5\widehat{k}\right)\times \left(F_{FC}\widehat{j}\right)=0\end{array}$

Above vector equations gives following scalars equations,

  $\begin{array}{c}-36000-4.5F_{FC}=0\mathrm{...}\text{(1)}\\ 27000-2.21F_{DC}=0\mathrm{...}\text{(2)}\\ 4.5F_{DA}+3.5F_{DB}+2.97F_{DC}=0\mathrm{...}\text{(3)}\end{array}$

Take equilibrium of forces as,

  $\begin{array}{l}\sum \overrightarrow{F}=0\\ \Rightarrow \overrightarrow{A}+\left\{{\overrightarrow{F}}_{DA}+{\overrightarrow{F}}_{DB}+{\overrightarrow{F}}_{DC}\right\}+\left\{{\overrightarrow{F}}_{EC}+{\overrightarrow{F}}_{EB}\right\}+{\overrightarrow{F}}_{FC}=0\\ \Rightarrow \left(-6000\widehat{k}\right)\\ +\left\{\left(F_{DA}\widehat{j}\right)+\left(F_{DB}\left[\frac{-4.5\widehat{i}+6\widehat{j}}{\sqrt{{(-4.5)}^2+6^2}}\right]\right)+\left(F_{DC}\left[\frac{-4.5\widehat{i}+6\widehat{j}+4.5\widehat{k}}{\sqrt{{(-4.5)}^2+6^2+{4.5}^2}}\right]\right)\right\}\\ +\left\{\left(F_{EB}\widehat{j}\right)+\left(F_{EC}\left[\frac{6\widehat{j}+4.5\widehat{k}}{\sqrt{6^2+{4.5}^2}}\right]\right)\right\}+\left(F_{FC}\widehat{j}\right)=0\end{array}$

Above vector equations gives following scalars equations,

  $\begin{array}{c}-0.58F_{DB}-0.49F_{DC}=0\mathrm{...}\text{(4)}\\ F_{DA}+0.78F_{DB}+0.66F_{DC}+0.78F_{EC}+F_{EB}+F_{FC}=0\mathrm{...}\text{(5)}\\ -6000+0.49F_{DC}+0.58F_{EC}=0\mathrm{...}\text{(6)}\end{array}$

Solve equations (1-6) to get

  $F_{DA}=-13\text{ lb, }{F}_{DB}=-10000\text{ lb, }{F}_{DC}=12000\text{ lb, }{F}_{EC}=0\text{, }{F}_{EB}=8000\text{ lb, }{F}_{FC}=-8000\text{ lb}\text{. }$

Calculate reaction at D as,

  ${\overrightarrow{R}}_D={\overrightarrow{F}}_{DA}+{\overrightarrow{F}}_{DB}+{\overrightarrow{F}}_{DC}=-100\widehat{i}+100\widehat{j}+5900\widehat{k}\text{ lb}$

Calculate reaction at E as,

  ${\overrightarrow{R}}_E={\overrightarrow{F}}_{EC}+{\overrightarrow{F}}_{EB}=8000\widehat{j}\text{ lb}$

Calculate reaction at F as,

  ${\overrightarrow{R}}_F={\overrightarrow{F}}_{FC}=-8000\widehat{j}\text{ lb}\text{.}$

Conclusion:

Therefore, ${\overrightarrow{R}}_D=-100\widehat{i}+100\widehat{j}+5900\widehat{k}\text{ lb, }{\overrightarrow{R}}_E=8000\widehat{j}\text{ lb, }{\overrightarrow{R}}_F=-8000\widehat{j}\text{ lb}\text{.}$