Chapter 7.4, Problem 7.121P

To determine

Find the distances $d_B$ and $d_D$.

Answer to Problem 7.121P

The value of distance is $\underset{\_}{d_B=1.733\text{m}}$.

The value of distance is $\underset{\_}{d_D=4.20\text{m}}$.

Explanation of Solution

Given information:

The value of distance is $d_C=3\text{m}$.

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.

Apply the following sign convention for calculating the bending moment at any section x-x while approaching from the left hand side.

• Take clockwise moment as positive and anticlockwise moment as negative

Apply the following sign convention for calculating the shear force at any section x-x while approaching from the left hand side.

Take downward force as negative and upward force as positive.

Calculation:

Show the free-body diagram of the converted beam as in Figure 1.

Find the vertical reaction at point E by taking moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ E_y\left(10\right)-10\left(7\right)-5\left(4\right)-5\left(2\right)=0\\ 10E_y-70-20-10=0\\ E_y=10\text{kN}(\uparrow )\end{array}$

Find the vertical reaction at point A by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-5-5-10+E_y=0\\ A_y-20+10=0\\ A_y=10\text{kN}(\uparrow )\end{array}$

Show the calculation of shear force as follows;

Shear force at $x=0$;

$\begin{array}{c}{V|}_{x=0\left(\text{Just}\text{left}\right)}=0\\ {V|}_{x=0\left(\text{Just}\text{right}\right)}=A_y\\ =10\text{kN}\end{array}$

Shear force at $x=2\text{m}$;

$\begin{array}{c}{V|}_{x=2\text{m}\left(\text{Just}\text{left}\right)}=10\text{kN}\\ {V|}_{x=2\text{m}\left(\text{Just}\text{right}\right)}=10-5\\ =5\text{kN}\end{array}$

Shear force at $x=4\text{m}$;

$\begin{array}{c}{V|}_{x=4\text{m}\left(\text{Just}\text{left}\right)}=5\text{kN}\\ {V|}_{x=4\text{m}\left(\text{Just}\text{right}\right)}=5-5\\ =0\end{array}$

Shear force at $x=7\text{m}$;

$\begin{array}{c}{V|}_{x=7\text{m}\left(\text{Just}\text{left}\right)}=0\\ {V|}_{x=7\text{m}\left(\text{Just}\text{right}\right)}=0-10\\ =-10\text{kN}\end{array}$

Shear force at $x=10\text{m}$;

$\begin{array}{c}{V|}_{x=10\text{m}\left(\text{Just}\text{left}\right)}=-10\text{kN}\\ {V|}_{x=10\text{m}\left(\text{Just}\text{right}\right)}=-10+E_y\\ =-10+10\\ =0\end{array}$

Tabulate the calculated shear force values as in Table 1.

Distance, x (m)	Shear force, V (kN)
0 (Just left)	0
0 (Just right)	10
2 (Just left)	10
2 (Just right)	5
4 (Just left)	5
4 (Just right)	0
7 (Just left)	0
7 (Just right)	–10
10 (Just left)	–10
10 (Just left)	0

Plot the shear force diagram as in Figure 2.

Show the calculation of bending moment as follows;

Bending moment at $x=0$;

${M|}_{x=0}=0$

Bending moment at $x=2\text{m}$;

$\begin{array}{c}{M|}_{x=2\text{m}}=10\left(2\right)\\ =20\text{kN-m}\end{array}$

Bending moment at $x=4\text{m}$;

$\begin{array}{c}{M|}_{x=4\text{m}}=10\left(4\right)-5\left(2\right)\\ =30\text{kN-m}\end{array}$

Bending moment at $x=7\text{m}$;

$\begin{array}{c}{M|}_{x=7\text{m}}=10\left(7\right)-5\left(5\right)-5\left(3\right)\\ =30\text{kN-m}\end{array}$

Bending moment at $x=10\text{m}$;

$\begin{array}{c}{M|}_{x=10\text{m}}=10\left(10\right)-5\left(8\right)-5\left(6\right)-10\left(3\right)\\ =0\end{array}$

Tabulate the calculated bending moment values as in Table 2.

Distance, x (m)	Bending moment, M (kN-m)
0	0
2	20
4	30
7	30
10	0

Plot the bending moment diagram as in Figure 3.

Show the geometry of the cable as in Figure 4.

Find the distance $h_C$ using the relation.

$d_C=h_C+1.6$

Substitute 3 m for $d_C$.

$\begin{array}{l}3=h_C+1.6\\ h_C=1.4\text{m}\end{array}$

Use the concept given in Problem (7.119);

Write the relationship between the moment and the height h as;

$\frac{h_B}{M_B}=\frac{h_C}{M_C}=\frac{h_D}{M_D}$ (1)

Refer to the Figure 3;

$\begin{array}{l}{M}_B=20\text{kN-m}\\ M_C=30\text{kN-m}\\ M_D=30\text{kN-m}\end{array}$

Substitute $20\text{kN-m}$ for $M_B$, 1.4 m for $h_C$, $30\text{kN-m}$ for $M_C$, and $30\text{kN-m}$ for $M_D$ in Equation (1).

$\begin{array}{l}\frac{h_B}{20}=\frac{1.4}{30}=\frac{h_D}{30}\\ h_B=0.933\text{m}\\ h_D=1.4\text{m}\end{array}$

Refer to Figure 3;

Find the distance $d_D$ using the relation.

$d_D=2.8+h_D$

Substitute 1.4 m for $h_D$.

$\begin{array}{c}{d}_D=2.8+1.4\\ =4.20\text{m}\end{array}$

Find the distance $d_B$ using the relation.

$d_B=0.8+h_B$

Substitute 0.933 m for $h_B$.

$\begin{array}{c}{d}_B=0.8+0.933\\ =1.733\text{m}\end{array}$

Therefore, the value of distance $\underset{\_}{d_B=1.733\text{m}}$.

Therefore, the value of distance $\underset{\_}{d_D=4.20\text{m}}$.