Chapter 6.4, Problem 6.162P

Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal joint at C. The bearings at B and D do not exert any axial force. A couple with a magnitude of 500 lb·in. (clockwise when viewed from the positive x axis) is applied to shaft CF at F. At a time when the arm of the crosspiece attached to shaft CF is horizontal, determine (a) the magnitude of the couple that must be applied to shaft AC at A to maintain equilibrium, (b) the reactions at B, D, and E. (Hint: The sum of the couples exerted on the crosspiece must be zero.)

Fig. P6.161

*6.162 Solve Prob. 6.161 assuming that the arm of the crosspiece attached to shaft CF is vertical.

To determine

The magnitude of the couple that must be applied to shaft $AC$ at $A$ to maintain equilibrium.

Answer to Problem 6.162P

The magnitude of the couple that must be applied to shaft $AC$ at $A$ to maintain equilibrium is $433\text{lb}\text{.in}$.

Explanation of Solution

Take all vectors along the $x$ axis , $y$ axis and $y$ axis as positive.

The free body diagram of the shaft $DF$ is shown in figure 1.

Here, $C_x$ , $C_y$ and $C_z$ are the component of force $C$ at $C$ , $D_y$ and $D_z$ are the $y$ and $z$ component of force at $D$ , $E_x,E_y$ and $E_z$ are the $x,y\text{and z}$ component force at $E$ respectively.

At equilibrium total moment will be zero.

Write the expression for the equilibrium moment about $x$ axis.

${\displaystyle \sum M_x}=0$ (I)

Here, ${\displaystyle \sum M_x}$ are the equilibrium moment of force about $x$ axis.

The moment along the $x$ axis are the moment of force about $C$ and about $E$.

From free body diagram in figure1, write the complete equilibrium expression of moment about $x$ for the shaft $DF$.

$M_C-500\text{lb}\cdot \text{in=0}$ (II)

Here, $M_C$ is the moment of force about the point $C$ in the shaft $DF$.

The free body diagram of the shaft $BC$ is shown in figure 1.

Here,$x^{\prime },y^{\prime },{\text{z}}^{\prime }$ are the arbitrary axis with $x^{\prime }$ along $BC$ , $B_y{j}^{\prime }$ , $B_z{k}^{\prime }$ are the component of force at $B$ along $y^{\prime }\text{and }{\text{z}}^{\prime }$ ,$E_x{i}^{\prime },E_y{j}^{\prime }\text{and }{E}_z{k}^{\prime }$ are the component of force at $E$ along arbitrary axes.

Write the expression for the equilibrium moment about $C$ axis.

${\displaystyle \sum M_C}=0$

Here, ${\displaystyle \sum M_C}$ are the equilibrium moment of force about $C$.

Resolve $\left(500\text{lb}\cdot \text{in}\text{.}\right)$ into components along $x^{\prime }$ and $y^{\prime }$  axes.

$-M_C=-\left(500\text{lb}\cdot \text{in}\text{.}\right)\left(\cos 30°i^{\prime }+\sin 30°j^{\prime }\right)$

From free body diagram in figure2, write the complete equilibrium expression of moment about $C$  for the shaft $BC$.

$M_A{i}^{\prime }-\left(500\text{lb}\cdot \text{in}\text{.}\right)\left(\cos 30°i^{\prime }+\sin 30°j^{\prime }\right)+\left(-5\text{in}\text{.}\right)i^{\prime }\times \left(B_y{j}^{\prime }+B_z{k}^{\prime }\right)=0$

$M_A{i}^{\prime }-\left(433.35\text{lb}\cdot \text{in}\text{.}\right)i^{\prime }-\left(250\text{lb}\cdot \text{in}\text{.}\right)j^{\prime }+\left(5B_{y^{\prime }}\text{lb}\cdot \text{in}\text{.}\right)k+\left(5B_z\text{lb}\cdot \text{in}\text{.}\right)j^{\prime }=0$ (III)

Here, $M_A$ is the moment of force about the point $A$ in the shaft $DF$.

Calculation:

Rearrange equation (II) to get $M_C$.

$M_C-500\text{lb}\cdot \text{in=0}$

$M_C\text{=}500\text{lb}\cdot \text{in}\text{.}$

Equate coefficient of $i^{\prime }$ in equation (III)to zero to get $M_A$.

$\begin{array}{c}\left(M_A-\left(433\text{lb}\cdot \text{in}\text{.}\right)\right)=0\\ M_A=433\text{lb}\end{array}$

Therefore, the magnitude of the couple that must be applied to shaft $AC$ at $A$ to maintain equilibrium is $433\text{lb}\text{.in}$.

To determine

The reaction at $B,D$ and $E$.

Answer to Problem 6.162P

The reaction at $B$ is equal to $B=-\left(50\text{lb}\right)$ , reaction at $D$ is equal to $D=\left(83.3\text{lb}\right)k$ and reaction at $E$ is equal to $E=-\left(33.3\text{lb}\right)k$.

Explanation of Solution

From free body diagram in figure2, write the complete equilibrium expression of moment about $C$ for the shaft $BC$.

$M_A{i}^{\prime }-\left(500\text{lb}\cdot \text{in}\text{.}\right)\left(\cos 30°i^{\prime }+\sin 30°j^{\prime }\right)+\left(-5\text{in}\text{.}\right)i^{\prime }\times \left(B_{y^{\prime }}{j}^{\prime }+B_{z}k\right)=0$

$M_A{i}^{\prime }-\left(433\text{lb}\cdot \text{in}\text{.}\right)i^{\prime }+\left(5B_{y^{\prime }}\text{lb}\cdot \text{in}\text{.}\right)k^{\prime }-\left(5B_z\text{lb}\cdot \text{in}\text{.}\right)j^{\prime }=0$ (III)

Here, $M_A$ is the moment of force about the point $A$ in the shaft $DF$.

Since the net force at the shaft $BC$ is zero. Write the expression for the net force.

${\displaystyle \sum F=0}\Rightarrow B-C=0$ (IV)

Using free body diagram in figure1, apply the equilibrium condition for moment about $D$.

${\displaystyle \sum M_D}=0$ (V)

Here, ${\displaystyle \sum M_D}$ is the net moment at $D$.

From figure1, write the complete expression of moment $M_D$ about $D$

$\begin{array}{l}\left(500\text{lb}\cdot \text{in}\text{.}\right)i-\left(500\text{lb}\cdot \text{in}\text{.}\right)i+\left(6\text{in}\text{.}\right)i\times \left(E_{x}i+E_{y}j+E_{z}k\right)-\left(4\text{in}\text{.}\right)i\times \left(-50\text{lb}\right)k=0\\ -\left(200\text{lb}\cdot \text{in}\text{.}\right)j+\left(6\text{in}\text{.}\right)E_{y}k-\left(6\text{in}\text{.}\right)E_{z}j=0\end{array}$ (VI)

Since net force at shaft $DF$ is zero, write the expression for the net force.

${\displaystyle \sum F=0}$

$C+D_{y}j+D_{z}k+E_{x}i+E_{z}k=0$ (VII)

Calculation:

Equate coefficient of $j^{\prime }$ in equation (III)to zero to get $B_z$.

$-250\text{lb}\cdot \text{in}\text{.-}\left(5\text{in}\text{.}\right)B_z=0$

Equate coefficient of $k$ in equation (III)to zero to get $B_{y^{\prime }}$.

$B_{y^{\prime }}=0$

Therefore,

$\begin{array}{c}B=\left(0\text{lb}\right)j^{\prime }+\left(-50\text{lb}\right)k=0\\ =-\left(-50\text{lb}\right)k\end{array}$

Substitute $-50\text{lb}$ for $B$ in equation (III) to get $C$.

$\begin{array}{c}-50\text{lb}-C=0\\ C=-50\text{lb}\end{array}$

Equate coefficient of $j$ in equation (VI) to get $E_z$.

$\begin{array}{c}-\left(200\text{lb}\cdot \text{in}\text{.}\right)-\left(6\text{in}\text{.}\right)E_z=0\\ E_z=-33.3\text{lb}\end{array}$

Equate coefficient of $k$ in equation (VI) to zero to get $E_y$.

$E_y=0$

Equate coefficient of $i$ in equation (VII) to zero to get $E_x$.

$E_x=0$

Therefore, net reaction at $E$ is equal to, $E=E_{x}i+E_{y}j+E_{z}k=-\left(33.3\text{lb}\right)k$.

Substitute $-50\text{lb}k$ for $C$ , $-\left(33.3\text{lb}\right)k$ for $E$ and equate coefficient of $k$ in equation (VII) to zero to get $D_z$.

$\begin{array}{c}-50\text{lb}+D_z+-\left(33.3\text{lb}\right)=0\\ D_z=83.3\text{lb}\end{array}$

Equate coefficient of $j$ in equation (VII) to zero to get $D_y$.

$D_y=0$

Therefore, total reaction at $D$ equal to $D=83.3\text{lb}k$

Therefore, the reaction at $B$ is equal to $B=-\left(50\text{lb}\right)$ , reaction at $D$ is equal to $D=\left(83.3\text{lb}\right)k$ and reaction at $E$ is equal to $E=-\left(33.3\text{lb}\right)k$.