Chapter 1, Problem 1.3.31P

Space Frame ABC is clamped at A, except it is free to rotate at A about the x and y axes. Cables DC and EC support the frame at C. Force P_y= - 50 lb is applied at the mid-span of AS, and a concentrated moment M_x= -20 in-lb acts at joint B.

(a) Find reactions at support A.

(b) Find cable tension Forces.

  

To determine

Reactions at support A.

Answer to Problem 1.3.31P

The correct answers are:

$A_x=5.77\text{ lb, }{A}_y=47.3\text{ lb, }{A}_z=-2.31\text{ lb, }{M}_{Az}=200\text{ lb-in}\text{.}$

Explanation of Solution

Given Information:

You have following figure with all relevant information,

and $P_y=-50\text{ lb, and }{M}_x=-20\text{ lb-in}\text{.}$

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,

Analyze the free body diagram of member ABC.

The forces and corresponding position vectors are,

Force	Position vector
$\overrightarrow{A}=(A_x,A_y,A_z)$	$\overrightarrow{r_A}=(0,0,0)$
$\overrightarrow{P}=(0,P_y,0)=(0,-50,0)$	$\overrightarrow{r_P}=(5,0,0)$
$\begin{array}{c}\overrightarrow{F_{CE}}=\frac{F_{CE}}{\sqrt{{(-10)}^2+4^2+{(14)}^2}}(-10,4,14)\\ =F_C(-0.56,0.2,0.8)\end{array}$	$\overrightarrow{r_C}=(10,4,-4)$
$\begin{array}{c}\overrightarrow{F_{CD}}=\frac{F_{CD}}{\sqrt{{(-10)}^2+6^2+{(-16)}^2}}(-10,6,-16)\\ =F_C(-0.5,0.3,-0.8)\end{array}$	$\overrightarrow{r_C}=(10,4,-4)$

Take equilibrium of forces vector form,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}=0}\\ \Rightarrow \overrightarrow{A}+\overrightarrow{P}+\overrightarrow{F_{CE}}+\overrightarrow{F_{CD}}=0\\ \Rightarrow (A_x,A_y,A_z)+(0,-50,0)+(-0.56F_{CE},0.2F_{CE},0.8F_{CE})+(-0.5F_{CD},0.3F_{CD},-0.8F_{CD})=0\end{array}$

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

  $\begin{array}{c}-0.56F_{CE}-0.5F_{CD}+A_x=0\mathrm{....}\text{(1)}\\ 0.2F_{CE}+0.3F_{CD}+A_x-50=0\mathrm{....}\text{(2)}\\ 0.79F_{CE}-0.8F_{CD}+A_x=0\mathrm{....}\text{(3)}\end{array}$

Now take equilibrium of moments about A in vector form as,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{M_A}}=0\\ \Rightarrow \overrightarrow{M}+\overrightarrow{M_x}+\overrightarrow{r_P}\times \overrightarrow{P}+\overrightarrow{r_C}\times \overrightarrow{F_{CE}}+\overrightarrow{r_C}\times \overrightarrow{F_{CD}}=0\\ \Rightarrow (0,0,M_A{}_z)+(M_x,0,0)+(5,0,0)\times (0,-50,0)\\ +(10,4,-4)\times (-0.56F_{CE},0.2F_{CE},0.8F_{CE})+(10,4,-4)\times (-0.5F_{CD},0.3F_{CD},-0.8F_{CD})=0\end{array}$

Evaluate the cross products to get,

  $\Rightarrow (0,0,M_A{}_z)+(M_x,0,0)+(0,0,-250)+(4.07F_{CE},-5.67F_{CE},4.53F_{CE})+(-2F_{CD},10F_{CD},5F_{CD})=0$

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

  $\begin{array}{c}{M}_x+4.07F_{CE}-2.02F_{CD}=0\mathrm{....}\text{(4)}\\ -5.6F_{CE}+10.1F_{CD}=0\mathrm{....}\text{(5)}\\ M_{Az}-250+4.52F_{CE}+5F_{CD}=0\mathrm{....}\text{(6)}\end{array}$

Solve equations (1-6) to get,

$\begin{array}{l}{A}_x=5.77\text{ lb, }{A}_y=47.3\text{ lb, }{A}_z=-2.31\text{ lb, }{M}_{Az}=200\text{ lb-in,}\\ F_{CE}=6.79{\text{ lb, F}}_{CD}=3.8\text{ lb}\text{.}\end{array}$

Conclusion:

Therefore the forces and moments are:

$A_x=5.77\text{ lb, }{A}_y=47.3\text{ lb, }{A}_z=-2.31\text{ lb, }{M}_{Az}=200\text{ lb-in}\text{.}$

To determine

Cable forces.

Answer to Problem 1.3.31P

The correct answers are:

$F_{CE}=6.79{\text{ lb, F}}_{CD}=3.8\text{ lb}\text{.}$

Explanation of Solution

Given Information:

You have following figure with all relevant information,

and $P_y=-50\text{ lb, and }{M}_x=-20\text{ lb-in}\text{.}$

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,

Analyze the free body diagram of member ABC.

The forces and corresponding position vectors are,

Force	Position vector
$\overrightarrow{A}=(A_x,A_y,A_z)$	$\overrightarrow{r_A}=(0,0,0)$
$\overrightarrow{P}=(0,P_y,0)=(0,-50,0)$	$\overrightarrow{r_P}=(5,0,0)$
$\begin{array}{c}\overrightarrow{F_{CE}}=\frac{F_{CE}}{\sqrt{{(-10)}^2+4^2+{(14)}^2}}(-10,4,14)\\ =F_C(-0.56,0.2,0.8)\end{array}$	$\overrightarrow{r_C}=(10,4,-4)$
$\begin{array}{c}\overrightarrow{F_{CD}}=\frac{F_{CD}}{\sqrt{{(-10)}^2+6^2+{(-16)}^2}}(-10,6,-16)\\ =F_C(-0.5,0.3,-0.8)\end{array}$	$\overrightarrow{r_C}=(10,4,-4)$

Take equilibrium of forces vector form,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{F}=0}\\ \Rightarrow \overrightarrow{A}+\overrightarrow{P}+\overrightarrow{F_{CE}}+\overrightarrow{F_{CD}}=0\\ \Rightarrow (A_x,A_y,A_z)+(0,-50,0)+(-0.56F_{CE},0.2F_{CE},0.8F_{CE})+(-0.5F_{CD},0.3F_{CD},-0.8F_{CD})=0\end{array}$

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

  $\begin{array}{c}-0.56F_{CE}-0.5F_{CD}+A_x=0\mathrm{....}\text{(1)}\\ 0.2F_{CE}+0.3F_{CD}+A_x-50=0\mathrm{....}\text{(2)}\\ 0.79F_{CE}-0.8F_{CD}+A_x=0\mathrm{....}\text{(3)}\end{array}$

Now take equilibrium of moments about A in vector form as,

  $\begin{array}{l}{\displaystyle \sum \overrightarrow{M_A}}=0\\ \Rightarrow \overrightarrow{M}+\overrightarrow{M_x}+\overrightarrow{r_P}\times \overrightarrow{P}+\overrightarrow{r_C}\times \overrightarrow{F_{CE}}+\overrightarrow{r_C}\times \overrightarrow{F_{CD}}=0\\ \Rightarrow (0,0,M_A{}_z)+(M_x,0,0)+(5,0,0)\times (0,-50,0)\\ +(10,4,-4)\times (-0.56F_{CE},0.2F_{CE},0.8F_{CE})+(10,4,-4)\times (-0.5F_{CD},0.3F_{CD},-0.8F_{CD})=0\end{array}$

Evaluate the cross products to get,

  $\Rightarrow (0,0,M_A{}_z)+(M_x,0,0)+(0,0,-250)+(4.07F_{CE},-5.67F_{CE},4.53F_{CE})+(-2F_{CD},10F_{CD},5F_{CD})=0$

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

  $\begin{array}{c}{M}_x+4.07F_{CE}-2.02F_{CD}=0\mathrm{....}\text{(4)}\\ -5.6F_{CE}+10.1F_{CD}=0\mathrm{....}\text{(5)}\\ M_{Az}-250+4.52F_{CE}+5F_{CD}=0\mathrm{....}\text{(6)}\end{array}$

Solve equations (1-6) to get,

$\begin{array}{l}{A}_x=5.77\text{ lb, }{A}_y=47.3\text{ lb, }{A}_z=-2.31\text{ lb, }{M}_{Az}=200\text{ lb-in,}\\ F_{CE}=6.79{\text{ lb, F}}_{CD}=3.8\text{ lb}\text{.}\end{array}$

Conclusion:

Therefore cable forces are:

$F_{CE}=6.79{\text{ lb, F}}_{CD}=3.8\text{ lb}\text{.}$