Chapter 6.3, Problem 6.94P

(a)

To determine

Find the components of the reaction at E.

Answer to Problem 6.94P

The reactions at E are $\underset{\_}{E_x=3\text{kips}(\leftarrow )}$ and $\underset{\_}{E_y=1.5\text{kips}(\uparrow )}$.

Explanation of Solution

Given information:

Consider the diameter of the pipe is $d=3\text{ft}$. Then,

The radius of the pipe is $r=1.5\text{ft}$.

The length of the pipe between the support is $l=16\text{ft}$.

The weight of the pipe is $w=500\text{lb/ft}$.

Consider the surfaces are frictionless.

Assumption:

Apply the sign convention for calculating the equations of equilibrium as below.

• For the horizontal forces equilibrium condition, take the force acting towards right side as positive $\left(\to +\right)$ and the force acting towards left side as negative $\left(\leftarrow -\right)$.
• For the vertical forces equilibrium condition, take the upward force as positive $\left(\uparrow +\right)$ and downward force as negative $\left(\downarrow -\right)$.
• For moment equilibrium condition, take the clockwise moment as negative and counter clockwise moment as positive.

Calculation:

Show the free body diagram of pipe as shown in Figure 1.

Refer Figure 1.

Calculate the total weight of the pipe as follows:

$\begin{array}{c}{W}_y=w\times l\\ =500\text{lb/ft}\times \text{16}\text{ft}\\ =\text{8000}\text{lb}\times \frac{1\text{kips}}{{10}^3\text{lb}}\\ =8\text{kips}\end{array}$

Refer the force triangle.

Consider the reaction from the contact surfaces are denoted by B and D.

$\begin{array}{l}\frac{B_y}{3}=\frac{D_y}{5}=\frac{W_z}{4}\\ \frac{B_y}{3}=\frac{D_y}{5}=\frac{8\text{kips}}{4}\end{array}$

Calculate the reaction B from the contact surface as follows:

$\begin{array}{l}\frac{B_y}{3}=\frac{8\text{kips}}{4}\\ B_y=\frac{8\text{kips}\times \text{3}}{4}\\ B_y=6\text{kips}\end{array}$

Calculate the reaction D from the contact surface as follows:

$\begin{array}{c}\frac{D_z}{5}=\frac{W_y}{4}\\ D_y=\frac{8\text{kips}\times \text{5}}{4}\\ D_y=10\text{kips}\end{array}$

Show the pipe as shown in Figure 2.

Refer Figure 2.

Show the relation between the OB, OC, and CD as follows:

$\begin{array}{l}\text{Horizontal projection of }\overrightarrow{\text{BO}}+\overrightarrow{\text{OD}}=\text{Horizontal projection of }\overrightarrow{\text{CD}}\\ \left(r+\frac{3}{5}r\right)=\frac{4}{5}\left(CD\right)\\ \frac{8}{5}r=\frac{4}{5}\left(CD\right)\\ CD=2r\end{array}$

Substitute $1.5\text{ft}$ for $r$.

$\begin{array}{c}CD=2\times 1.5\text{ft}\\ \text{=3}\text{ft}\end{array}$

The distance CD and CB are Equal. Then,

$CD=CB=\text{3}\text{ft}$

Show the free body diagram of member ABC as shown in Figure 3.

Refer Figure 3.

Consider the horizontal and vertical reaction at A is denoted by $A_x$ and $A_y$.

Consider the height of the member AC is denoted by h.

Consider the sum of the moment at A is zero. Then,

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ C_{x}h-\left(6\text{kips}\right)\left(h-3\right)=0\\ C_x=\frac{h-3}{h}\times \left(\text{6}\text{kips}\right)\end{array}$ (1)

Substitute $6\text{in}\text{.}$ for h in Equation (1).

$\begin{array}{c}{C}_x=\frac{6-3}{6}\times \left(\text{6}\text{kips}\right)\\ =3\text{kips}\end{array}$ (2)

Show the free body diagram for the member CDE as shown in Figure 4.

Refer Figure 4.

Consider the horizontal and vertical reaction at E are denoted by $E_x$ and $E_y$.

Consider the sum of the moment about the point E is zero. Then,

$\begin{array}{l}{\displaystyle \sum M_E=0}\\ \left(10\text{kips}\times \text{7}\text{ft}\right)-\left(3\text{kips}\times \text{6}\text{ft}\right)-\left(C_y\times 8\text{ft}\right)=0\\ C_y\times 8\text{ft}=52\text{kips}\cdot \text{ft}\\ {\text{C}}_y=\frac{52\text{kips}\cdot \text{ft}}{8\text{ft}}\\ {\text{C}}_y=6.5\text{ft}(\uparrow )\end{array}$ (3)

The sum of the forces in the horizontal direction is zero. Then,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -3\text{kips}+\frac{3}{5}\times 10\text{kips}+E_x=0\\ E_x=3\text{kips}-6\text{kips}\\ E_x=-3\text{kips}\end{array}$

$E_x=3\text{kips}(\leftarrow )$

The sum of the forces in the vertical direction is zero. Then,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ C_y-\frac{4}{5}\times 10\text{kips}+E_y=0\\ 6.5\text{kips}-8\text{kips}+E_y=0\\ E_y=1.5\text{kips}(\uparrow )\end{array}$

Thus, the reactions at E are $\underset{\_}{E_x=3\text{kips}(\leftarrow )}$ and $\underset{\_}{E_y=1.5\text{kips}(\uparrow )}$.

(b)

To determine

Find the force exerted at C on the member CDE.

Answer to Problem 6.94P

The force exerted at C on the member CDE are $\underset{\_}{{\text{C}}_x=3\text{kips}(\leftarrow )}$ and $\underset{\_}{{\text{C}}_y=6.5\text{ft}(\uparrow )}$.

Explanation of Solution

Given information:

Assumption:

Apply the sign convention for calculating the equations of equilibrium as below.

• For the horizontal forces equilibrium condition, take the force acting towards right side as positive $\left(\to +\right)$ and the force acting towards left side as negative $\left(\leftarrow -\right)$.
• For the vertical forces equilibrium condition, take the upward force as positive $\left(\uparrow +\right)$ and downward force as negative $\left(\downarrow -\right)$.
• For moment equilibrium condition, take the clockwise moment as negative and counter clockwise moment as positive.

Calculation:

Refer Part (a).

Refer Equation (2) and (3).

The force exerted at C on the member CDE are ${\text{C}}_x=3\text{kips}(\leftarrow )$ and ${\text{C}}_y=6.5\text{ft}(\uparrow )$.

Thus, the force exerted at C on the member CDE are $\underset{\_}{{\text{C}}_x=3\text{kips}(\leftarrow )}$ and $\underset{\_}{{\text{C}}_y=6.5\text{ft}(\uparrow )}$.