Products
Rewards 
from HOLOOLY

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

HOLOOLY 
BUSINESS MANAGER

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

HOLOOLY 
TABLES

All the data tables that you may search for.

HOLOOLY 
ARABIA

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

HOLOOLY 
TEXTBOOKS

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

HOLOOLY 
HELP DESK

Need Help? We got you covered.

Chapter 8

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}).

Step-by-Step

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

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).