Chapter 6, Problem 14P

To determine

Calculate the location of the 40-kN load for the given condition.

Calculate the maximum tension in the cable and also find the reaction at support A and D.

Answer to Problem 14P

The maximum tension in the cable is $\underset{\_}{117.82\text{kN}}$.

The vertical reaction at support A is $\underset{\_}{A_y=28.57\text{kN}}$.

The vertical reaction at support D is $\underset{\_}{D_y=11.43\text{kN}}$.

Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reactions, forces, and moments using the three equations of equilibrium as shown below.

• For summation of forces along x-direction is equal to zero $\left(\sum F_x=0\right)$, consider the forces acting towards right side as positive $\left(\to +\right)$ and the forces acting towards left side as negative $\left(\leftarrow -\right)$.
• For summation of forces along y-direction is equal to zero $\left(\sum F_y=0\right)$, consider the upward force as positive $\left(\uparrow +\right)$ and the downward force as negative $\left(\downarrow -\right)$.
• For summation of moment about a point is equal to zero $\left(\sum M_{\text{at}\text{a}\text{point}}=0\right)$, consider the clockwise moment as positive and the counter clockwise moment as negative.

Calculation:

Sketch the free body diagram as shown in Figure 1.

Using Equilibrium Equation:

Take moment about point F.

$\begin{array}{l}\sum M_F=0\\ -R_{EC}\times 10+40\times x=0\\ -10R_{EC}+40x=0\\ R_{EC}=4x\end{array}$

Consider the equilibrium of forces along y direction.

$\begin{array}{l}\sum F_y=0\\ R_{FB}+R_{EC}-40=0\\ R_{FB}+4x-40=0\\ R_{FB}=4\left(10-x\right)\end{array}$

Calculate the vertical load by using the general cable theorem in which the load in the cable is assumed to be acting in the simply supported beam with same loads and span.

Sketch the free body diagram as shown in Figure 2.

Take moment about point A.

$\begin{array}{l}\sum M_A=0\\ D_y\times 40-R_{EC}\times 30-R_{FB}\times 20=0\\ 40D_y-4x\times 30-\left(40-4x\right)\times 20=0\\ 40D_y-120x-800+80x=0\\ D_y=\frac{800+40x}{40}\\ D_y=\left(x+20\right)\end{array}$

Consider the equilibrium of forces along y direction.

$\begin{array}{l}\sum F_y=0\\ A_y+D_y-R_{FB}-R_{EC}=0\\ A_y+D_y-\left(40-4x\right)-4x=0\\ A_y+D_y=40\\ A_y+\left(x+20\right)=40\\ A_y=20-x\end{array}$

Take moments about point C in the beam of general cable theorem:

$\begin{array}{c}\sum M_C=0\\ M_C=D_y\times 10\\ =\left(x+20\right)\times 10\\ =10x+200\end{array}$

Take moments about point B in the beam of general cable theorem:

$\begin{array}{c}\sum M_B=0\\ M_B=A_y\times 20\\ =\left(20-x\right)\times 20\\ =400-20x\end{array}$

Take moments about point B in the cable using reaction diagram.

$\begin{array}{l}\sum M_B=0\\ D_x\times 3=M_B\\ D_x\times 3=400-20x\\ D_x=\frac{400-20x}{3}\end{array}$        (1)

Take moments about point C in the cable using reaction diagram.

$\begin{array}{l}\sum M_C=0\\ D_x\times 2=M_C\\ D_x=10x+200\\ D_x=5x+100\end{array}$        (2)

Solve the equation (1) and (2) to get the values.

$\begin{array}{l}\frac{400-20x}{3}=5x+100\\ 400-20x=15x+300\\ x=2.857\text{m}\end{array}$

Thus, the location of load is $2.857\text{m}$ from joint B.

Substitute the value of x in Equation (2).

$\begin{array}{c}{D}_x=5\left(2.857\right)+100\\ =14.285+100\\ =114.285\text{kN}\end{array}$

Since there is no other horizontal reaction influencing the cable reaction, the horizontal reaction at D equals the horizontal reaction at A.

Thus, the horizontal reaction at the support A is $114.285\text{kN}$.

Take moments about the point A in the cable using the cable reaction diagram:

$\begin{array}{l}\sum M_A=0\\ 40\left(20+2.857\right)-D_x\times 4-D_y\left(40\right)=0\\ 914.28-4\left(114.285\right)-40D_y=0\\ 914.28-457.14-40D_y=0\\ 457.14-40D_y=0\\ D_y=11.428\text{kN}\end{array}$

Consider the equilibrium force along y direction:

$\begin{array}{l}\sum F_y=0\\ A_y+D_y-40=0\\ A_y+11.43-40=0\\ A_y=28.57\text{kN}\end{array}$

Thus, the vertical reaction at point A is $\underset{\_}{28.57\text{kN}}$.

Thus, the vertical reaction at point D is $\underset{\_}{11.43\text{kN}}$.

Calculate the maximum tension in the cable by using the following:

$\begin{array}{c}{T}_{\max }=\sqrt{{\left(A_x\right)}^2+{\left(A_y\right)}^2}\\ =\sqrt{{\left(114.3\right)}^2+{\left(28.57\right)}^2}\\ =\sqrt{13880.73}\\ =117.82\text{kN}\end{array}$

Thus, the maximum tension in the cable is $\underset{\_}{117.82\text{kN}}$.