Chapter 7.2, Problem 7.60P

Knowing that P = Q = 150 lb, determine (a) the distance a for which the maximum absolute value of the bending moment in beam AB is as small as possible, (b) the corresponding value of |M|_max. (See the hint for Prob. 7.55.)

Fig. P7.60

To determine

The distance a from the ends of the beam to the points where the cables should be attached if the maximum absolute value of the bending moment in the beam AB is the smallest.

Answer to Problem 7.60P

The distance a from the ends of the beam to the points where the cables should be attached if the maximum absolute value of the bending moment in the beam AB is the smallest is $52.0\text{in}$.

Explanation of Solution

Refer Figures 1.

Write an expression to calculate the counter clockwise moment at point A.

$\Sigma M_A=0$                                       

Here, $\Sigma M_A$ is the moment about point A.

Write an expression to calculate the counter clockwise moment at point A.

$\Sigma M_C=0$                                       

Here, $\Sigma M_C$ is the moment about point A.

Write an expression to calculate the counter clockwise moment at point A.

$\Sigma M_D=0$                                        

Here, $\Sigma M_D$ is the moment about point A.

Conclusion:

Refer Figure 1:

Calculate the moment about point A.

$\begin{array}{c}\Sigma M_A=Da-\left(P\right)\left(AC\right)-\left(Q\right)\left(AB\right)\\ =0\end{array}$

Here, $P$ is the weight at point A and a is the distance between points A and D, Q is the weight at Q.

Rearrange the equation to calculate the D.

$D=\frac{\left(P\right)\left(AC\right)+\left(Q\right)\left(AB\right)}{a}$

Substitute $150\text{lb}$ for $P$, $30\text{in}$ for $AC$, $150\text{lb}$ for $Q$ and $60\text{in}$ for $AB$ to find $D$.

$\begin{array}{c}D=\frac{\left(150\text{lb}\right)\left(30\text{in}\right)+\left(150\text{lb}\right)\left(60\text{in}\right)}{a}\\ =\frac{13,500\text{lb}\cdot \text{in}}{a}\end{array}$

Refer figure 2.

Calculate the moment about point C.

$\begin{array}{c}\Sigma M_C=-M_C-\left(P\right)\left(AC\right)+\left(D\right)\left(a-AC\right)\\ =0\end{array}$

Rearrange the equation to calculate the $M_C$.

$\begin{array}{c}{M}_C=-\left(P\right)\left(AC\right)+\left(D\right)\left(a-AC\right)\\ =0\end{array}$

Substitute $150\text{lb}$ for $P$, $30\text{in}$ for $AC$, $\frac{13,500\text{lb}\cdot \text{in}}{a}$ for $D$ and $60\text{in}$ for $AB$ to find $D$.

$\begin{array}{c}{M}_C=-\left(150\text{lb}\right)\left(30\text{in}\right)+\left(\frac{13,500\text{lb}\cdot \text{in}}{a}\right)\left(a-30\text{in}\right)\\ =9,000\text{lb}\cdot \text{in}\left(1-\frac{45}{a}\right)\end{array}$ (I)

Refer figure 2.

Calculate the moment about point D.

$\begin{array}{c}\Sigma M_D=-M_D-\left(DB\right)\left(AB-a\right)\\ =0\end{array}$

Rearrange the equation to calculate the $M_D$.

$M_D=-\left(DB\right)\left(AB-a\right)$

Substitute $150\text{lb}$ for $DB$ and $150\text{lb}$ for $Q$ to find $M_D$.

$M_D=-\left(150\text{lb}\right)\left(60\text{in}-a\right)$ (II)

The magnitude of the maximum moment is equal to the magnitude of the minimum moment.

$\begin{array}{c}{M}_{\max }=\left|M_{\min }\right|\\ M_C=\left|M_D\right|\end{array}$

Substitute (I) and (II) in above equation to find a.

$\left(9,000\text{lb}\cdot \text{in}\right)\left(1-\frac{45}{a}\right)=\left(150\text{lb}\right)\left(60\text{in}-a\right)$

Rearrange the equation to find a.

$\begin{array}{c}a=\sqrt{60\text{in}-60\text{ in}+2700{\text{in}}^2}\\ =52.0\text{in}\end{array}$

Thus, the distance a from the ends of the beam to the points where the cables should be attached if the maximum absolute value of the bending moment in the beam AB is the smallest is $52.0\text{in}$.

To determine

The value of $M_{\max }$

Answer to Problem 7.60P

The value of $M_{\max }$ is $1,206\text{lb}\cdot \text{in}$.

Explanation of Solution

Refer figure 4.

The magnitude of the maximum moment is equal to the magnitude of the minimum moment.

$\begin{array}{c}{M}_{\max }=\left|M_{\min }\right|\\ =\left|M_D\right|\end{array}$

Conclusion:

Substitute $52\text{in}$ for $a$ in equation (II) to calculate the value of $M_{\max }$ .

$\begin{array}{c}{M}_{\max }=\left(150\text{lb}\right)\left(60\text{in}-\left(52\text{in}\right)\right)\\ =1206\text{lb}\cdot \text{in}\end{array}$

Thus, the value of $M_{\max }$ is $1,206\text{lb}\cdot \text{in}$.