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Chapter 8

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}).

Step-by-Step

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).