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Chapter 8

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}).

Step-by-Step

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}<x \leq 20  \mathrm{~m} \end{array}\right.

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