We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program


Advertise your business, and reach millions of students around the world.


All the data tables that you may search for.


For Arabic Users, find a teacher/tutor in your City or country in the Middle East.


Find the Source, Textbook, Solution Manual that you are looking for in 1 click.


Need Help? We got you covered.

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}).


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).