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## Q. 8.9

Draw the influence lines for the shear in panel $BC$ and the bending moment at $B$ of the girder with floor system shown in Fig. $8.14 (\mathrm{a})$.

## Verified Solution

Influence Line for $\boldsymbol S_{B C}$. To determine the influence line for the shear in panel $B C$, we place a $1 \mathrm{k}$ load successively at the panel points $A, B, C$, and $D$. For each position of the unit load, the appropriate support reaction is first determined by proportions, and the shear in panel $B C$ is computed. Thus, when

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

The influence line for $S_{B C}$ is constructed by plotting these ordinates and by connecting them with straight lines, as shown in Fig. 8.14(c).

Influence Line for $\boldsymbol M_{B}$. To determine the influence line for the bending moment at panel point $B$, we place the $1 \mathrm{k}$ load successively at the panel points $A, B$, and $D$. For each position of the unit load, the bending moment at $B$ is determined as follows: When

$\begin{array}{lll} 1 \mathrm{k} \text { is at } A, & D_{y}=0 & M_{B}=0 \\ 1 \mathrm{k} \text { is at } B, & A_{y}=\frac{2}{3} \mathrm{k} & M_{B}=\left(\frac{2}{3}\right) 18=12 \mathrm{k}-\mathrm{ft} \\ 1 \mathrm{k} \text { is at } D, & A_{y}=0 & M_{B}=0 \end{array}$

The influence line for $M_{B}$ thus obtained is shown in Fig. 8.14(d). 