Chapter 6.3, Problem 6.120P

6.119 through 6.121 Each of the frames shown consists of two L-shaped members connected by two rigid links. For each frame, determine the reactions at the supports and indicate whether the frame is rigid.

Fig. P6.120

To determine

The reactions at the frame and the rigidness of the frame.

Answer to Problem 6.120P

The reactions at the frame for figure (a) is $B=\underset{\_}{2.06P}$ at $\underset{\_}{14.04°}$ above the negative $x$ axis, $A=\underset{\_}{2.06P}$ at $\underset{\_}{14.04°}$ above the positive $x$ axis and the rigidness of the frame in figure (a) rigid, for figure (b) the reaction is $P=0$ and the system is not rigid, and for figure (c) the reactions are $B=\underset{\_}{1.031P}$ at $\underset{\_}{14.04°}$ above the positive $x$ axis, $A=\underset{\_}{1.250P}$ at $\underset{\_}{36.9°}$ above the negative $x$ axis and the system is rigid.

Explanation of Solution

The following figure gives the free body diagram of the member in figure P6.120(a).

Write the equation to find the moment of force.

$M=Fa$

Here, $F$ is the force acting and $a$ is the perpendicular distance from the force to the point about which the moment is calculated.

Write the equation to find the total moment about the point $A$.

$\Sigma M_A=\Sigma \left(Fa\right)$

Write the equations for equilibrium for the free body diagram in figure 1.

$\Sigma M_A=0$ (I)

$\Sigma F_y=0$ (II)

$\Sigma F_x=0$ (III)

Here, $M_A$ is the torque about the point $A$, $F_x$ is the force in the $x$ direction, and $F_y$ is the force in the $y$ direction.

The following figure gives the free body diagram of the member in figure P6.120(b).

Write the equations for equilibrium for the free body diagram in figure 2.

$\Sigma M_A=0$ (IV)

Here, $M_A$ is the torque about the point $A$.

The following figure gives the free body diagram of the member in figure P6.120(c).

Write the equations for equilibrium for the free body diagram in figure 3.

$\Sigma M_A=0$ (V)

$\Sigma F_y=0$ (VI)

$\Sigma F_x=0$ (VII)

Here, $M_A$ is the torque about the point $A$, $F_x$ is the force in the $x$ direction, and $F_y$ is the force in the $y$ direction.

Write the expression to find the magnitude of the vector from its components.

$C=\sqrt{C_x^2+C_y^2}$ (VIII)

Here,$C_x$ is the $x$ component of the vector $C$ and $C_y$ is the $y$ component of the vector $C$.

Write the equation to find the angle of orientation of the vector $C$.

$\theta ={\text{tan}}^{-1}\left(\frac{C_y}{C_x}\right)$ (IX)

Conclusion:

Solve equation (I) using figure 1.

$-2aP-\frac{a}{4}\frac{4}{\sqrt{17}}B+5a\frac{1}{\sqrt{17}}B=0$

Rewrite the above equation to find $B$.

$B=2.06P$

Solve equation (II) using figure 1.

$A_x-\frac{4}{\sqrt{17}}B=0$

Substitute $2.06P$ for $B$ in the above equation and rearrange it to find $A_x$.

$\begin{array}{c}{A}_x=\frac{4}{\sqrt{17}}\left(2.06P\right)\\ =1.99P\\ =2P\end{array}$

Solve equation (III) using figure 1.

$A_y-\frac{1}{\sqrt{17}}B=0$

Substitute $2.06P$ for $B$ and rewrite the above equation to find $A_y$.

$\begin{array}{c}{A}_y=\frac{1}{\sqrt{17}}\left(2.06P\right)\\ =\frac{P}{2}\end{array}$

Rewrite equation (VIII) in terms of the vector $A$.

$A=\sqrt{A_x^2+A_y^2}$

Substitute $2P$ for $A_x$, and $\frac{P}{2}$ for $A_y$ in the above equation.

$\begin{array}{c}A=\sqrt{{\left(2P\right)}^2+{\left(\frac{P}{2}\right)}^2}\\ =2.061P\end{array}$

Rewrite equation (IX) in terms of the vector $A$.

$\theta ={\text{tan}}^{-1}\left(\frac{A_y}{A_x}\right)$

Substitute $2P$ for $A_x$, and $\frac{P}{2}$ for $A_y$ in the above equation.

$\begin{array}{c}\theta ={\text{tan}}^{-1}\left(\frac{\frac{P}{2}}{2P}\right)\\ =14.036°\\ =14.04°\end{array}$

Solve equation (IV) using figure 2.

$\begin{array}{c}2aP=0\\ P=0\end{array}$

Solve equation (V) using figure 3.

$5a\frac{1}{\sqrt{17}}B+\frac{3a}{4}\frac{4}{\sqrt{17}}B-2aP=0$

Rewrite the above equation to find $B$.

$\begin{array}{c}B=\frac{\sqrt{17}}{4}P\\ =1.0307P\end{array}$

Solve equation (VI) using figure 3.

$A_x+\frac{4}{\sqrt{17}}B=0$

Substitute $\frac{\sqrt{17}P}{4}$ for $B$ in the above equation and rearrange it to find $A_x$.

$\begin{array}{c}{A}_x=-\frac{4}{\sqrt{17}}\left(\frac{\sqrt{17}P}{4}\right)\\ =-P\end{array}$

Solve equation (VII) using figure 3.

$A_y-P+\frac{1}{\sqrt{17}}B=0$

Substitute $\frac{\sqrt{17}P}{4}$ for $B$ and rewrite the above equation to find $A_y$.

$\begin{array}{c}{A}_y=P-\frac{P}{4}\\ =\frac{3P}{4}\end{array}$

Rewrite equation (VIII) in terms of the vector $A$.

$A=\sqrt{A_x^2+A_y^2}$

Substitute $-P$ for $A_x$, and $\frac{3P}{4}$ for $A_y$ in the above equation.

$\begin{array}{c}A=\sqrt{{\left(-P\right)}^2+{\left(\frac{3P}{4}\right)}^2}\\ =1.250P\end{array}$

Rewrite equation (IX) in terms of the vector $A$.

$\theta ={\text{tan}}^{-1}\left(\frac{A_y}{A_x}\right)$

Substitute $-P$ for $A_x$, and $\frac{3P}{4}$ for $A_y$ in the above equation.

$\begin{array}{c}\theta ={\text{tan}}^{-1}\left(\frac{\frac{3P}{4}}{-P}\right)\\ =-36.86°\end{array}$

Therefore, the reactions at the frame for figure (a) is $B=\underset{\_}{2.06P}$ at $\underset{\_}{14.04°}$ above the negative $x$ axis, $A=\underset{\_}{2.06P}$ at $\underset{\_}{14.04°}$ above the positive $x$ axis and the rigidness of the frame in figure (a) rigid, for figure (b) the reaction is $P=0$ and the system is not rigid, and for figure (c) the reactions are $B=\underset{\_}{1.031P}$ at $\underset{\_}{14.04°}$ above the positive $x$ axis, $A=\underset{\_}{1.250P}$ at $\underset{\_}{36.9°}$ above the negative $x$ axis and the system is rigid.