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Chapter 8

Q. 8.3

Draw the influence lines for the vertical reactions at supports A, C, and E, the shear just to the right of support C, and the bending moment at point B of the beam shown in Fig. 8.5 (\mathrm{a}).

Step-by-Step

Verified Solution

The bean is composed of two rigid parts, A D and D E, connected by an intermal hinge at D. To avoid solving simultaneous equations in determining the expressions for the reactions, we will apply the equations of equilibrium and condition in such an order that each equation involves only one unknown.

Influence Line for \boldsymbol E_{y}, We will apply the equation of condition, \sum M_{D}^{D E}=0, to determine the expression for E_{y}, First, we place, the unit load at a variable position x to the left of the hinge D- that is, on the rigid part A D of the beam-to obtain

\begin{array}{rl} +\curvearrowleft \sum M_{D}^{D E}=0 & \\ E_{y}(20)=0 & \\ E_{y}=0 & 0 \leq x \leq 40  \mathrm{ft} \end{array}

Next, the unit load is located to the right of hinge D—that is, on the rigid part D E of the beam—to obtain

\begin{array}{l} \quad+\curvearrowleft \sum M_{D}^{D E}=0 \\ -1(x-40)+E_{y}(20)=0 \\ E_{y}=\frac{1(x-40)}{20}=\frac{x}{20}-2 \quad 40  \mathrm{ft} \leq x \leq 60  \mathrm{ft} \end{array}

Thus, the equations of the influence line for E_{y} are

E_{y}=\left\{\begin{array}{lr} 0 & 0 \leq x \leq 40  \mathrm{ft} \\ \frac{x}{20}-2 & 40  \mathrm{ft} \leq x \leq 60  \mathrm{ft} \end{array}\right.

The influence line for E_{y}, is shown in Fig. 8.5(\mathrm{c}).

Influence Line for \boldsymbol C_{y}. Applying the equilibrium equation:

\begin{aligned} +\curvearrowleft \sum M_{A} &=0 \\ -1(x)+C_{y}(20)+E_{y}(60) &=0 \\ C_{y} &=\frac{x}{20}-3 E_{y} \end{aligned}

By substituting the expressions for E_{y}, we obtain

C_{y}=\left\{\begin{array}{cr} \frac{x}{20}-0=\frac{x}{20} & 0 \leq x \leq 40  \mathrm{ft} \\ \frac{x}{20}-3\left(\frac{x}{20}-20\right)=6-\frac{x}{10} & 40  \mathrm{ft} \leq x \leq 60  \mathrm{ft} \end{array}\right.

The influence line for C_{y}, which is obtained by plotting these equations, is shown in Fig. 8.5(d).

Influence Line for \boldsymbol A_{y}.

\begin{array}{c} +\uparrow \sum F_{y}=0 \\ A_{y}-1+C_{y}+E_{y}=0 \\ A_{y}=1-C_{y}-E_{y} \end{array}

By substituting the expressions for C_{y} and E_{y}, we obtain the following equations of the influence line for A_{y} :

A_{y}=\left\{\begin{array}{rr} 1-\frac{x}{20}-0=1-\frac{x}{20} & 0 \leq x \leq 40  \mathrm{ft} \\ 1-\left(6-\frac{x}{20}\right)-\left(\frac{x}{20}-2\right)=\frac{x}{20}-3 & 40  \mathrm{ft} \leq x \leq 60  \mathrm{ft} \end{array}\right.

The influence line for A, is shown in Fig. 8.5 (c).

Influence Line for Shear at Just to the Right of \boldsymbol C, \boldsymbol S_{C, R}.

s_{c, R}=\left\{\begin{array}{ll} -E_{y} & 0 \leq x<20  \mathrm{ft} \\ 1-E_{y} & 20  \mathrm{ft}<x \leq 60   \mathrm{ft} \end{array}\right.

By substituting the expressions for E_{\text {Y }}, we obtain

s_{c, R}=\left\{\begin{array}{cc} 0 & 0 \leq x<20  \mathrm{ft} \\ 1-0=1 & 20  \mathrm{ft}<x \leq 40 \mathrm{ft} \\ 1-\left(\frac{x}{20}-2\right)=3-\frac{x}{20} & 40  \mathrm{ft} \leq x \leq 60  \mathrm{ft} \end{array}\right.

The influence line for S_{C, R} is shown in Fig. 8.5(f).

Influence Line for \boldsymbol M_{B}.

M_{B}=\left\{\begin{array}{lr} 10 A_{y}-1(10-x) & 0 \leq x \leq 10  \mathrm{ft} \\ 10 A_{y} & 10  \mathrm{ft} \leq x \leq 60  \mathrm{ft} \end{array}\right.

By substituting the expressions for A_{Y}, we obtain

M_{B}=\left\{\begin{array}{rr} 10\left(1-\frac{x}{20}\right)-1(10-x)=\frac{x}{2} & 0 \leq x \leq 10  \mathrm{ft} \\ 10\left(1-\frac{x}{20}\right)=10-\frac{x}{2} & 10  \mathrm{ft} \leq x \leq 40  \mathrm{ft} \\ 10\left(\frac{x}{20}-3\right)=\frac{x}{2}-30 & 4  \mathrm{ft} \leq x \leq 60  \mathrm{ft} \end{array}\right.

The influence line for M_{B} is shown in Fig. 8.5(\mathrm{~g}).