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## Q. 8.5

Draw the influence lines for the horizontal and vertical reactions at supports $A$ and $B$ and the shear at hinge $E$ of the three-hinged bridge frame shown in Fig. $8.7 (\mathrm{a})$.

## Verified Solution

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

\begin{aligned} +\curvearrowleft \sum M_{B} &=0 \\ -A_{y}(10)+1(15-x) &=0 \\ A_{y} &=\frac{1(15-x)}{10}=1.5-\frac{x}{10} \end{aligned}

The influence line for $A_{y}$, is shown in Fig. $8.7$ (c).

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

\begin{aligned} +\uparrow \sum F_{y} &=0 \\ A_{y}-1+B_{y} &=0 \\ B_{y} &=1-A_{y}=1-\left(1.5-\frac{x}{10}\right)=\frac{x}{10}-0.5 \end{aligned}

The influence line for $B_{y}$, is shown in Fig. $8.7(\mathrm{~d})$.

Influence Line for $\boldsymbol A_{x}$. We will use the equation of condition $\sum M_{E}^{C E}=0$ to determine the expressions for $A_{x}$ First, we place the unit load to the left of hinge $E$ —that is, on the rigid part $C E$ of the frame—to obtain

\begin{aligned} +\curvearrowleft \sum M_{E}^{C E} &=0 \\ A_{x}(3)-A_{y}(5)+1(10-x) &=0 \\ A_{x} &=\frac{5}{3} A_{y}-\frac{1}{3}(10-x)=\frac{5}{3}\left(1.5-\frac{x}{10}\right)-\frac{1}{3}(10-x) \\ &=\frac{x-5}{6} \quad 0 \leq x \leq 10 \mathrm{~m} \end{aligned}

Next, the unit load is located to the right of hinge $E$—that is, on the rigid part $E G$ of the frame—to obtain

$\begin{array}{l} +\curvearrowleft \sum M_{E}^{C E}=0 \\ A_{x}(3)-A_{y}(5)=0 \\ A_{x}=\frac{5}{3} A_{y}=\frac{5}{3}\left(1.5-\frac{x}{10}\right)=\frac{15-x}{6} \quad 10 \mathrm{~m} \leq x \leq 20 \mathrm{~m} \end{array}$

Thus, the equations of the influence line for $A_{x}$ are

$A_{x}=\left\{\begin{array}{lr} \frac{x-5}{6} & 0 \leq x \leq 10 \mathrm{~m} \\ \frac{15-x}{6} & 10 \mathrm{~m} \leq x \leq 20 \mathrm{~m} \end{array}\right.$

The influence line for $A_{x}$ is shown in Fig. $8.7(\mathrm{c})$.

Inflence Line for $\boldsymbol B_{x}$.

$\begin{array}{r} +\rightarrow \sum F_{x}=0 \\ A_{x}-B_{x}=0 \\ B_{x}=A_{x} \end{array}$

which indicates that the influence line for $B_{x}$ is the same as that for $A_{x}$ (Fig. $8.7$ (e)).

Influence Line for $\boldsymbol S_{E}$.

$S_{E}=\left\{\begin{array}{cr} -B_{y}=-\frac{x}{10}+0.5 & 0 \leq x<10 \mathrm{~m} \\ A_{y}=1.5-\frac{x}{10} & 10 \mathrm{~m}

The influence line for $S_{E}$ is shown in Fig. $8.7(f)$.