Chapter 9.3, Problem 44P

For the beam and loading shown, determine (a) the equation of the elastic curve, (b) the deflection at point B, (c) the deflection at point D.

Fig. P9.44

To determine

Find the equation of the elastic curve.

Answer to Problem 44P

The equation of the elastic curve is,

$\underset{\_}{y=\frac{w}{24EI}\left\{-x^4+{\langle x-\frac{L}{2}\rangle }^4-{\langle x-L\rangle }^4+L{x}^3+3L{\langle x-L\rangle }^3-\frac{L^{3}x}{16}\right\}}$.

Explanation of Solution

Show the free-body diagram of the beam AD as in Figure 1.

Determine the vertical reaction at point C by taking moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ -w\left(\frac{L}{2}\right)\left(\frac{L}{4}\right)+C_y\left(L\right)-w\left(\frac{L}{2}\right)\left(\frac{5L}{4}\right)=0\\ -\frac{w{L}^2}{8}+C_{y}L-\frac{5w{L}^2}{8}=0\\ C_{y}L=\frac{6w{L}^2}{8}\\ C_y=\frac{3wL}{4}\end{array}$

Write the singularity equation for load intensity as follows;

$\frac{dV}{dx}=-w\left(x\right)=-w+w{\langle x-\frac{L}{2}\rangle }^0-w{\langle x-L\rangle }^0$

Integrate the equation to find the shear force.

$\begin{array}{c}V=A_y-wx+w{\langle x-\frac{L}{2}\rangle }^1-w{\langle x-L\rangle }^1+C_y{\langle x-L\rangle }^0\\ =A_y-wx+w{\langle x-\frac{L}{2}\rangle }^1-w{\langle x-L\rangle }^1+\frac{3wL}{4}{\langle x-L\rangle }^0\end{array}$

By definition, the change in bending moment with respect to change in distance is shear force. It is expressed as follows:

$\frac{dM}{dx}=V=A_y-wx+w{\langle x-\frac{L}{2}\rangle }^1-w{\langle x-L\rangle }^1+\frac{3wL}{4}{\langle x-L\rangle }^0$

Integrate the equation to find the bending moment.

$M=A_{y}x-\frac{w{x}^2}{2}+\frac{w}{2}{\langle x-\frac{L}{2}\rangle }^2-\frac{w}{2}{\langle x-L\rangle }^2+\frac{3wL}{4}{\langle x-L\rangle }^1$ (1)

Write the second order differential equation as follows;

$\frac{d^{2}y}{d{x}^2}=\frac{M\left(x\right)}{EI}$

Here, the moment at the corresponding section is $M\left(x\right)$, the modulus of elasticity of the material is E, and the moment of inertia of the section is I.

Substitute $\left(A_{y}x-\frac{w{x}^2}{2}+\frac{w}{2}{\langle x-\frac{L}{2}\rangle }^2-\frac{w}{2}{\langle x-L\rangle }^2+\frac{3wL}{4}{\langle x-L\rangle }^1\right)$ for $M\left(x\right)$.

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}=\frac{A_{y}x-\frac{w{x}^2}{2}+\frac{w}{2}{\langle x-\frac{L}{2}\rangle }^2-\frac{w}{2}{\langle x-L\rangle }^2+\frac{3wL}{4}{\langle x-L\rangle }^1}{EI}\\ EI\frac{d^{2}y}{d{x}^2}=A_{y}x-\frac{w{x}^2}{2}+\frac{w}{2}{\langle x-\frac{L}{2}\rangle }^2-\frac{w}{2}{\langle x-L\rangle }^2+\frac{3wL}{4}{\langle x-L\rangle }^1\end{array}$

Integrate the equation with respect to x;

$EI\frac{dy}{dx}=\frac{A_y{x}^2}{2}-\frac{w{x}^3}{6}+\frac{w}{6}{\langle x-\frac{L}{2}\rangle }^3-\frac{w}{6}{\langle x-L\rangle }^3+\frac{3wL}{8}{\langle x-L\rangle }^2+C_1$ (2)

Integrate the Equation (2) with respect to x.

$EIy=\frac{A_y{x}^3}{6}-\frac{w{x}^4}{24}+\frac{w}{24}{\langle x-\frac{L}{2}\rangle }^4-\frac{w}{24}{\langle x-L\rangle }^4+\frac{3wL}{24}{\langle x-L\rangle }^3+C_{1}x+C_2$ (3)

Boundary condition 1:

At the point D; $x=\frac{3L}{2};M=0$.

Substitute $\frac{3L}{2}$ for x and 0 for M in Equation (1).

$\begin{array}{c}0=A_y\left(\frac{3L}{2}\right)-\frac{w{\left(\frac{3L}{2}\right)}^2}{2}+\frac{w}{2}{\langle \frac{3L}{2}-\frac{L}{2}\rangle }^2-\frac{w}{2}{\langle \frac{3L}{2}-L\rangle }^2+\frac{3wL}{4}{\langle \frac{3L}{2}-L\rangle }^1\\ =\frac{3A_{y}L}{2}-\frac{9w{L}^2}{8}+\frac{w{L}^2}{2}-\frac{w{L}^2}{8}+\frac{3w{L}^2}{8}\\ =\frac{3A_{y}L}{2}-\frac{9w{L}^2}{8}+\frac{4w{L}^2}{8}-\frac{w{L}^2}{8}+\frac{3w{L}^2}{8}\\ A_y=\frac{wL}{4}\end{array}$

Boundary condition 2:

At the point A; $x=0;y=0$.

Substitute $\frac{wL}{4}$ for $A_y$, 0 for x and 0 for y in Equation (3).

$\begin{array}{l}EI\left(0\right)=\frac{\left(\frac{wL}{4}\right){\left(0\right)}^3}{6}-\frac{w{\left(0\right)}^4}{24}+\frac{w}{24}{\langle 0-\frac{L}{2}\rangle }^4-\frac{w}{24}{\langle 0-L\rangle }^4+\frac{3wL}{24}{\langle 0-L\rangle }^3+C_1\left(0\right)+C_2\\ 0=0-0+0-0+0+0+C_2\\ C_2=0\end{array}$

Boundary condition 3:

At the point C; $x=L;y=0$.

Substitute $\frac{wL}{4}$ for $A_y$, L for x, 0 for y, and 0 for $C_2$ in Equation (3).

$\begin{array}{l}EI\left(0\right)=\frac{\left(\frac{wL}{4}\right)L^3}{6}-\frac{w{L}^4}{24}+\frac{w}{24}{\langle L-\frac{L}{2}\rangle }^4-\frac{w}{24}{\langle L-L\rangle }^4+\frac{3wL}{24}{\langle L-L\rangle }^3+C_{1}L+0\\ 0=\frac{w{L}^4}{24}-\frac{w{L}^4}{24}+\frac{w{L}^4}{384}-0+0+C_{1}L+0\\ C_1=-\frac{w{L}^3}{384}\end{array}$

Substitute $\frac{wL}{4}$ for $A_y$, $-\frac{w{L}^3}{384}$ for $C_1$, and 0 for $C_2$ in Equation (3).

$\begin{array}{c}EIy=\frac{\left(\frac{wL}{4}\right)x^3}{6}-\frac{w{x}^4}{24}+\frac{w}{24}{\langle x-\frac{L}{2}\rangle }^4-\frac{w}{24}{\langle x-L\rangle }^4+\frac{3wL}{24}{\langle x-L\rangle }^3-\frac{w{L}^3}{384}x+0\\ =\frac{wL{x}^3}{24}-\frac{w{x}^4}{24}+\frac{w}{24}{\langle x-\frac{L}{2}\rangle }^4-\frac{w}{24}{\langle x-L\rangle }^4+\frac{3wL}{24}{\langle x-L\rangle }^3-\frac{w{L}^{3}x}{384}\\ y=\frac{w}{24EI}\left\{-x^4+{\langle x-\frac{L}{2}\rangle }^4-{\langle x-L\rangle }^4+L{x}^3+3L{\langle x-L\rangle }^3-\frac{L^{3}x}{16}\right\}\end{array}$ (4)

Therefore, the equation of the elastic curve is

$\underset{\_}{y=\frac{w}{24EI}\left\{-x^4+{\langle x-\frac{L}{2}\rangle }^4-{\langle x-L\rangle }^4+L{x}^3+3L{\langle x-L\rangle }^3-\frac{L^{3}x}{16}\right\}}$.

To determine

Find the deflection at point B of the beam.

Answer to Problem 44P

The deflection at mid-point B of the beam is $\underset{\_}{y_B=\frac{w{L}^4}{768EI}(\downarrow )}$.

Explanation of Solution

Refer to part (a);

Refer to Equation (4);

At point B; $x=\frac{L}{2};y=y_B$.

Substitute $y_B$ for y and $\frac{L}{2}$ for x in Equation (4).

$\begin{array}{c}{y}_B=\frac{w}{24EI}\left\{-{\left(\frac{L}{2}\right)}^4+{\langle \frac{L}{2}-\frac{L}{2}\rangle }^4-{\langle \frac{L}{2}-L\rangle }^4+L{\left(\frac{L}{2}\right)}^3+3L{\langle \frac{L}{2}-L\rangle }^3-\frac{L^3\left(\frac{L}{2}\right)}{16}\right\}\\ =\frac{w}{24EI}\left\{-{\frac{L}{16}}^4+0-0+\frac{L^4}{8}+0-\frac{L^4}{32}\right\}\\ =\frac{w}{24EI}\left\{-{\frac{2L}{32}}^4+\frac{4L^4}{32}-\frac{L^4}{32}\right\}\\ =-\frac{w{L}^4}{768EI}\end{array}$

$y_B=\frac{w{L}^4}{768EI}(\downarrow )$

Therefore, the deflection at point B of the beam is $\underset{\_}{y_B=\frac{w{L}^4}{768EI}(\downarrow )}$.

To determine

Find the deflection at point D of the beam.

Answer to Problem 44P

The deflection at mid-point D of the beam is $\underset{\_}{y_D=\frac{5w{L}^4}{256EI}(\downarrow )}$.

Explanation of Solution

Refer to part (a); Equation (4);

At point D; $x=\frac{3L}{4};y=y_D$.

Substitute $y_D$ for y and $\frac{3L}{2}$  for x in Equation (4).

$\begin{array}{c}{y}_D=\frac{w}{24EI}\left\{-{\left(\frac{3L}{2}\right)}^4+{\langle \frac{3L}{2}-\frac{L}{2}\rangle }^4-{\langle \frac{3L}{2}-L\rangle }^4+L{\left(\frac{3L}{2}\right)}^3+3L{\langle \frac{3L}{2}-L\rangle }^3-\frac{L^3\left(\frac{3L}{2}\right)}{16}\right\}\\ =\frac{w}{24EI}\left\{-\frac{81L^4}{16}+L^4-\frac{L^4}{16}+\frac{27L^4}{8}+\frac{3L^4}{8}-\frac{3L^4}{32}\right\}\\ =\frac{w}{24EI}\left\{-\frac{162L^4}{32}+\frac{32L^4}{32}-\frac{2L^4}{32}+\frac{108L^4}{32}+\frac{12L^4}{32}-\frac{3L^4}{32}\right\}\\ =-\frac{15w{L}^4}{768EI}\end{array}$

$y_D=\frac{5w{L}^4}{256EI}(\downarrow )$

Therefore, the deflection at point D of the beam is $\underset{\_}{y_D=\frac{5w{L}^4}{256EI}(\downarrow )}$.