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## Q. 8.10

Draw the influence lines for the shear in panel $CD$ and the bending moment at $D$ of the girder with floor system shown in Fig. $8.15 (\mathrm{a})$.

## Verified Solution

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

$\begin{array}{lll} 1 \mathrm{kN} \text { is at } B, & F_{y}=0 & S_{C D}=0 \\ 1 \mathrm{kN} \text { is at } C, & F_{y}=\frac{1}{4} \mathrm{kN} & S_{C D}=-\frac{1}{4} \mathrm{kN} \\ 1 \mathrm{kN} \text { is at } D, & B_{y}=\frac{2}{4}=\frac{1}{2} \mathrm{kN} & S_{C D}=\frac{1}{2} \mathrm{kN} \\ 1 \mathrm{kN} \text { is at } F, & B_{y}=0 & S_{C D}=0 \end{array}$

The influence line for $S_{C D}$ is constructed by plotting these ordinates and by connecting them with straight lines, as shown in Fig. 8.15(c). The ordinates at the ends $A$ and $H$ of the girder are then determined from the geometry of the influence line.

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

$\begin{array}{lll}1 \mathrm{kN} \text { is at } B, & F_{y}=0 & M_{D}=0 \\ 1 \mathrm{kN} \text { is at } D, & B_{y}=\frac{1}{2} \mathrm{kN} & M_{D}=\left(\frac{1}{2}\right) 8=4 \mathrm{kN} \cdot \mathrm{m} \\ 1 \mathrm{kN} \text { is at } F, & B_{y}=0 & M_{D}=0\end{array}$

The influence line for $M_{D}$ thus obtained is shown in Fig. $8.15(\mathrm{~d})$.