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## Q. 8.11

Draw the influence lines for the reaction at support $A$, the shear in panel $CD$, and the bending moment at $D$ of the girder with floor system shown in Fig. $8.16 (\mathrm{a})$. ## Verified Solution

Influence Line for $\boldsymbol A_{y}$. To determine the influence line for the reaction $A_{y}$, we place a $1 \mathrm{k}$ load successively at the panel points $A, B$, and $C$. For each position of the unit load, the magnitude of $A_{y}$, is computed by applying the equation of condition $\sum M_{F}^{AF}=0$. Thus, when

$\begin{array}{rrr}1 \mathrm{k} \text { is at } A, & & A_{y}=1 \mathrm{k} \\ 1 \mathrm{k} \text { is at } B, & +\curvearrowleft \sum M_{F}^{A F}=0 & \\ & -A_{y}(15)+1(5)=0 & A_{y}=\frac{1}{3} \mathrm{k} \\ 1 \mathrm{k} \text { is at } C, & +\curvearrowleft \sum M_{F}^{A F}=0 & \\ & -A_{y}(15)=0 & \\ & & A_{y}=0\end{array}$

The influence line for $A_{y}$ thus obtained is shown in Fig. 8.16(c).

Influence Line for $\boldsymbol S_{C D}$. We place the $1 \mathrm{k}$ load successively at each of the five panel points and determine the influence-line ordinates as follows. When

$\begin{array}{lll}1 \mathrm{k} \text { is at } A, & A_{y}=1 \mathrm{k} & S_{C D}=0 \\ 1 \mathrm{k} \text { is at } B, & A_{y}=\frac{1}{3} \mathrm{k} & S_{C D}=\left(\frac{1}{3}\right)-1=-\frac{2}{3} \mathrm{k} \\ 1 \mathrm{k} \text { is at } C, & A_{y}=0 & S_{C D}=-1 \mathrm{k} \\ 1 \mathrm{k} \text { is at } D, & A_{y}=0 & S_{C D}=0 \\ 1 \mathrm{k} \text { is at } E, & A_{y}=0 & S_{C D}=0\end{array}$

The influence line for $S_{C D}$ thus obtained is shown in Fig. 8.16(d).

Influence Line for $\boldsymbol M_{D}$. We place the $1 \mathrm{k}$ load successively at each of the five panel points and determine the influence-line ordinates as follows. When

$\begin{array}{lll}1 \mathrm{k} \text { is at } A, & A_{y}=1 \mathrm{k} & M_{D}=0 \\ 1 \mathrm{k} \text { is at } B, & A_{y}=\frac{1}{3} \mathrm{k} & M_{D}=\left(\frac{1}{3}\right) 30-1(20)=-10 \mathrm{k}-\mathrm{ft} \\ 1 \mathrm{k} \text { is at } C, & A_{y}=0 & M_{D}=-1(10)=-10 \mathrm{k}-\mathrm{tt} \\ 1 \mathrm{k} \text { is at } D, & A_{y}=0 & M_{D}=0 \\ 1 \mathrm{k} \text { is at } E, & A_{y}=0 & M_{D}=0\end{array}$

The influence line for $M_{D}$ is shown in Fig. 8.16(e).