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## Q. 8.1

Draw the influence lines for the vertical reactions at supports $A$ and $C$, and the shear and bending moment at point $B$, of the simply supported beam shown in Fig. $8.3 (\mathrm{a})$. ## Verified Solution

The free-body diagram of the beam is shown in Fig. 8.3(b). This diagram shows the beam subjected to a moving $1 \mathrm{k}$ load, whose position is defined by the coordinate $x$ measured from the left end $A$ of the beam. The two vertical reactions, $A_{y}$ and $C_{y}$, are assumed to be positive in the upward direction, as indicated on the free-body diagram.

Influence Line for $\boldsymbol A_{y}$. To determine the expression for $A_{y}$, we apply the equilibrium equation:

\begin{aligned} +\curvearrowleft \sum M_{C} &=0 \\-A_{y}(20)+1(20-x) &=0 \\A_{y} &=\frac{1(20-x)}{20}=1-\frac{x}{20}\end{aligned}

The influence line for $A_{y}$, which is obtained by plotting this equation, is shown in Fig. 8.3(c). Note that the ordinates of the influence line are expressed in the units obtained by dividing the units of the response function, $A_{y}$, by the units of the unit load-that is, $k / k$

Influence Line for $\boldsymbol C_{y}$.

\begin{aligned} +\curvearrowleft \sum M_{A} &=0 \\ -1(x)+C_{y}(20) &=0 \\ C_{y} &=\frac{1(x)}{20}=\frac{x}{20} \end{aligned}

The influence line for $C_{y}$, which is obtained by plotting this equation, is shown in Fig. 8.3(d).

Influence Line for $\boldsymbol S_{B}$. First, we place the unit load at a variable position $x$ to the left of point $B$—that is, on the segment $A B$ of the beam—and determine the shear at $B$ by using the free body of the portion $B C$ of the beam, which is to the right of $B$ :

$S_{B}=-C_{y} \quad 0 \leq x<12 \mathrm{ft}$

Next, the unit load is located to the right of $B$—that is, on the segment $B C$ of the beam—and we use the free body of the portion $A B$, which is to the left of $B$, to determine $S_{B}$ :

$S_{B}=A_{y} \quad 12 \mathrm{ft}

Thus, the equations of the influence line for $S_{B}$ are

$S_{B}=\left\{\begin{array}{cc} -C_{y}=-\frac{x}{20} & 0 \leq x<12 \mathrm{ft} \\ A_{y}=1-\frac{x}{20} & 12 \mathrm{ft}

The influence line for $S_{B}$ is shown in Fig. 8.3(e).

Influence Line for $\boldsymbol M_{B}$. First, we place the unit load at a position $x$ to the left of $B$ and determine the bending moment at $B$ by using the free body of the portion of the beam to the right of $B$ :

$M_{B}=8 C_{y} \quad 0 \leq x \leq 12 \mathrm{ft}$

Next, the unit load is located to the right of $B$, and we use the free body of the portion of the beam to the left of $B$ to determine $M_{B}$ :

$M_{B}=12 A_{y} \quad 12 \mathrm{ft} \leq x \leq 20 \mathrm{ft}$

Thus, the equations of the influence line for $M_{B}$ are

$M_{B}=\left\{\begin{array}{cc} 8 C_{y}=\frac{2 x}{5} & 0 \leq x \leq 12 \mathrm{ft} \\ 12 A_{y}=12-\frac{3 x}{5} & 12 \mathrm{ft} \leq x \leq 20 \mathrm{ft} \end{array}\right.$

The influence line for $M_{B}$ is shown in Fig. 8.3(f).