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## Q. 8.6

Draw the influence lines for the vertical reactions at supports $B$ and $D$ and the shear and bending moment at point $C$ of the beam shown in Fig. $8.9 (\mathrm{a})$.

## Verified Solution

Influence Line for $\boldsymbol B_{y}$. To determine the general shape of the influence line for $B_{y}$, we remove the roller support at $B$ from the given beam (Fig. 8.9(a)) to obtain the released beam shown in Fig. 8.9(b). Next, point $B$ of the released beam is given a small displacement, $\Delta$, in the positive direction of $B_{y}$, and a deflected shape of the beam is drawn, as shown by the dashed line in the figure. Note that the deflected shape is consistent with the support conditions of the released structure; that is, the right end of the released beam, which is attached to the hinged support $D$, does not displace. The shape of the influence line is the same as the deflected shape of the released structure, as shown in Fig. 8.9(b).

To obtain the numerical value of the influence-line ordinate at $B$, we place a $1 \mathrm{kN}$ load at point $B$ on the original beam (Fig. 8.9(b)) and apply an equilibrium equation to obtain $B_{y}$,

$+\curvearrowleft \sum M_{D}=0 \quad 1(9)-B_{y}(9)=0 \quad B_{y}=1 \mathrm{kN}$

Thus the value of the influence-line ordinate at $B$ is $1 \mathrm{kN} / \mathrm{kN}$. The value of the ordinate at $A$ can now be determined from the geometry of the influence line (Fig. 8.9(b)). Observing that the triangles $A A^{\prime} D$ and $B B D^{\prime}$ are similar, we write

$A A^{\prime}=\left(\frac{1}{9}\right)(12)=\frac{4}{3} \mathrm{kN} / \mathrm{kN}$

The influence line for $B_{y}$ thus obtained is shown in Fig. $8.9(\mathrm{~b})$.

Influence Line for $\boldsymbol D_{y}$. The influence line for $D_{y}$ is constructed in a similar manner and is shown in Fig. $8.9$ (c).

Influence Line for $\boldsymbol S_{C}$. To determine the general shape of the influence line for the shear at point $C$, we cut the given beam at $C$ to obtain the released structure shown in Fig. 8.9(d). Next, the released structure is given a small relative displacement in the positive direction of $S_{C}$ by moving end $C$ of the portion $A C$ downward by $\Delta_{1}$ and end $C$ of the portion $C D$ upward by $\Delta_{2}$ to obtain the deflected shape shown in Fig. $8.9$ (d). The shape of the influence line is the same as the deflected shape of the released structure, as shown in the figure.

To obtain the numerical values of the influence-line ordinates at $C$, we place the $1 \mathrm{kN}$ load first just to the left of $C$ and then just to the right of $C$, as shown by the solid and dashed arrows, respectively, in Fig. 8.9(d). The reactions $B_{y}$ and $D_{y}$ are then determined by applying the equilibrium equations:

$\begin{array}{lll} +\curvearrowleft \sum M_{D}=0 & -B_{y}(9)+1(6)=0 & B_{y}=\frac{2}{3} \mathrm{kN} \uparrow \\ +\uparrow \sum F_{y}=0 & \left(\frac{2}{3}\right)-1+D_{y}=0 & D_{y}=\frac{1}{3} \mathrm{kN} \uparrow \end{array}$

Note that the magnitudes of $B_{y}$ and $D_{y}$ could, alternatively, have been obtained from the influence lines for these reactions constructed previously. It can be seen from Fig. $8.9(\mathrm{~b})$ and (c) that the ordinates at $C$ (or just to the left or right of $C$ ) of the influence lines for $B_{y}$ and $D_{y}$ are indeed $2 / 3$ and $1 / 3$, respectively. When the unit load is at just to the left of $C$ (see Fig. $8.9$ (d)), the shear at $C$ is

$S_{c}=-D_{y}=-\frac{1}{3} \mathrm{kN}$

When the unit load is at just to the right of $C$, the shear at $C$ is

$S_{c}=B_{y}=\frac{2}{3} \mathrm{kN}$

Thus, the values of the influence-line ordinates at $C$ are $-1 / 3 \mathrm{kN} / \mathrm{kN}$ (just to the left of $C$ ) and $2 / 3 \mathrm{kN} / \mathrm{kN}$ (just to the right of $C$ ), as shown in the figure. The ordinate of the influence line at $A$ can now be obtained from the geometry of the influence line (Fig. 8.9(d)). Observing that the triangles $A A^{\prime} B$ and $B C C^{\prime}$ are similar, we obtain the ordinate at $A, A A^{\prime}=1 / 3 \mathrm{kN} / \mathrm{kN}$. The influence line for $S_{C}$ thus obtained is shown in Fig. $8.9$ (d).

Influence Line for $\boldsymbol M_{c}$. To obtain the general shape of the influence line for the bending moment at $C$, we insert a hinge at $C$ in the given beam to obtain the released structure shown in Fig. 8.9(e). Next, a small rotation $\theta$, in the positive direction of $M_{C}$, is introduced at $C$ in the released structure by rotating the portion $A C$ counterclockwise and the portion $C D$ clockwise to obtain the deflected shape shown in Fig. $8.9(\mathrm{e})$. The shape of the influence line is the same as the deflected shape of the released structure, as shown in the figure.

To obtain the numerical value of the influence-line ordinate at $C$, we place a $1 \mathrm{kN}$ load at $C$ on the original beam (Fig. 8.9(e)). By applying, in order, the equilibrium equations $\sum M_{D}=0$ and $\sum F_{y}=0$, we compute the reactions $B_{y}=2 / 3 \mathrm{kN}$ and $D_{y}=1 / 3 \mathrm{kN}$, after which the bending moment at $C$ is determined as

$M_{c}=\left(\frac{2}{3}\right)(3)=2 \mathrm{kN} \cdot \mathrm{m}$

Thus, the value of the influence-line ordinate at $C$ is $2 \mathrm{kN} \cdot \mathrm{m} / \mathrm{kN}$. Finally, to complete the influence line, we determine the ordinate at $A$ by considering the geometry of the influence line. From Fig. $8.9(\mathrm{e})$, we observe that because the triangles $A A^{\prime} B$ and $B C C^{\prime}$ are similar, the ordinate at $A$ is $A A^{\prime}=-2 \mathrm{kN} \cdot \mathrm{m} / \mathrm{kN}$. The influence line for $M_{C}$ thus obtained is shown in Fig. 8.9(e).