Chapter 11, Problem 133RP

A uniform rod of flexural rigidity EI is bent and loaded as shown. Determine (a) the horizontal deflection of point D, (b) the slope at point D.

Fig. P11.133

To determine

Find the horizontal deflection of point D.

Answer to Problem 133RP

The horizontal deflection of point D is $\underset{\_}{{\delta }_D=\frac{5P{L}^3}{3EI}\to }$.

Explanation of Solution

Given information:

The flexural rigidity of the rod is $EI$.

Calculation:

Apply a couple $M_D$ at D.

Sketch the Free Body Diagram as shown in Figure 1.

Refer to Figure 1.

Apply the Equations of Equilibrium as shown below.

Summation of forces along vertical direction is Equal to zero.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ R_{Ay}=0\end{array}$

Summation of forces along horizontal direction is Equal to zero.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ R_{Ax}+P=0\\ R_{Ax}=-P\end{array}$

Take moment about A is Equal to zero.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ M_A-M_D=0\\ M_A=M_D\end{array}$

Calculate the strain energy $\left(U\right)$ as shown below.

$\begin{array}{c}U=U_{AB}+U_{BC}+U_{CD}\\ ={\displaystyle {\int }_0^l\frac{M_{AB}^2}{2EI}dx}+{\displaystyle {\int }_0^l\frac{M_{BC}^2}{2EI}dx}+{\displaystyle {\int }_0^l\frac{M_{CD}^2}{2EI}dv}\end{array}$ (1)

Calculate the deflection of point D $\left({\delta }_D\right)$ as shown below.

Differentiate both side of the Equation (1) with respect to P as shown below.

$\begin{array}{c}{\delta }_D=\frac{\partial U}{\partial P}\\ =\frac{1}{2EI}\left({\displaystyle {\int }_0^{l}2M_{AB}\frac{\partial M_{AB}}{\partial P}dx}+{\displaystyle {\int }_0^{l}2M_{BC}\frac{\partial M_{BC}}{\partial P}dx}+{\displaystyle {\int }_0^{\frac{L}{2}}2M_{CD}\frac{\partial M_{CD}}{\partial P}dv}\right)\\ =\frac{1}{EI}\left({\displaystyle {\int }_0^l{M}_{AB}\frac{\partial M_{AB}}{\partial P}dx}+{\displaystyle {\int }_0^l{M}_{BC}\frac{\partial M_{BC}}{\partial P}dx}+{\displaystyle {\int }_0^l{M}_{CD}\frac{\partial M_{CD}}{\partial P}dv}\right)\end{array}$ (2)

Consider the member AB.

The bending moment over portion AB is $M_{AB}=M_D+Py$ (3)

Differentiate both side of the Equation (3) with respect to P as shown below.

$\frac{\partial M_{AB}}{\partial P}=y$

Differentiate both side of the Equation (3) with respect to $M_D$ as shown below.

$\frac{\partial M_{AB}}{\partial M_D}=1$

Consider the member BC.

The bending moment over portion BC is $M_{BC}=M_D+Pl$ (4)

Differentiate both side of the Equation (4) with respect to P as shown below.

$\frac{\partial M_{BC}}{\partial P}=l$

Differentiate both side of the Equation (4) with respect to $M_D$ as shown below.

$\frac{\partial M_{BC}}{\partial M_D}=1$

Consider the member CD.

The bending moment over portion CD is $M_{CD}=M_D+Py$ (5)

Differentiate both side of the Equation (5) with respect to P as shown below.

$\frac{\partial M_{CD}}{\partial P}=y$

Differentiate both side of the Equation (5) with respect to $M_D$ as shown below.

$\frac{\partial M_{CD}}{\partial M_D}=1$

Calculate the deflection of point D $\left({\delta }_D\right)$ as shown below.

Substitute $M_D+Py$ for $M_{AB}$, $y$ for $\frac{\partial M_{AB}}{\partial P}$, $M_D+Pl$ for $M_{BC}$, $l$ for $\frac{\partial M_{BC}}{\partial M_D}$, $M_D+Py$ for $M_{CD}$, and $y$ for $\frac{\partial M_{CD}}{\partial M_D}$ and apply the limits in Equation (2).

${\delta }_D=\frac{1}{EI}\left({\displaystyle {\int }_0^l\left(M_D+Py\right)ydy}+{\displaystyle {\int }_0^l\left(M_D+Pl\right)ldx}+{\displaystyle {\int }_0^l\left(M_D+Py\right)ydy}\right)$

Substitute $0$ for $M_D$.

$\begin{array}{c}{\delta }_D=\frac{1}{EI}\left({\displaystyle {\int }_0^{l}P{y}^{2}dy}+{\displaystyle {\int }_0^{l}P{l}^{2}dx}+{\displaystyle {\int }_0^{l}P{y}^{2}dy}\right)\\ =\frac{P}{EI}\left({\left(\frac{y^3}{3}\right)}_0^l+{\left(l^{2}x\right)}_0^l+{\left(\frac{y^3}{3}\right)}_0^l\right)\\ =\frac{P{l}^3}{3EI}\left(1+3+1\right)\\ =\frac{5P{l}^3}{3EI}\end{array}$

Therefore, the horizontal deflection of point D is $\underset{\_}{{\delta }_D=\frac{5P{L}^3}{3EI}\to }$.

To determine

Find the slope at point D.

Answer to Problem 133RP

The slope at point D is $\underset{\_}{{\theta }_D=\frac{2P{l}^2}{EI}}$.

Explanation of Solution

Given information:

The flexural rigidity of the rod is $EI$.

Calculation:

Calculate the slope at D $\left({\theta }_D\right)$ as shown below.

Differentiate both side of the Equation (1) with respect to $M_D$ as shown below.

$\begin{array}{c}{\theta }_D=\frac{\partial U}{\partial M_D}\\ =\frac{1}{2EI}\left({\displaystyle {\int }_0^{l}2M_{AB}\frac{\partial M_{AB}}{\partial M_D}dx}+{\displaystyle {\int }_0^{l}2M_{BC}\frac{\partial M_{BC}}{\partial M_D}dx}+{\displaystyle {\int }_0^{\frac{L}{2}}2M_{CD}\frac{\partial M_{CD}}{\partial M_D}dv}\right)\\ =\frac{1}{EI}\left({\displaystyle {\int }_0^l{M}_{AB}\frac{\partial M_{AB}}{\partial M_D}dx}+{\displaystyle {\int }_0^l{M}_{BC}\frac{\partial M_{BC}}{\partial M_D}dx}+{\displaystyle {\int }_0^l{M}_{CD}\frac{\partial M_{CD}}{\partial M_D}dv}\right)\end{array}$

Substitute $M_D+Py$ for $M_{AB}$, $1$ for $\frac{\partial M_{AB}}{\partial P}$, $M_D+Pl$ for $M_{BC}$, $1$ for $\frac{\partial M_{BC}}{\partial M_D}$, $M_D+Py$ for $M_{CD}$, and $1$ for $\frac{\partial M_{CD}}{\partial M_D}$ and apply the limits in Equation (2).

${\theta }_D=\frac{1}{EI}\left({\displaystyle {\int }_0^l\left(M_D+Py\right)\left(1\right)dy}+{\displaystyle {\int }_0^l\left(M_D+Pl\right)\left(1\right)dx}+{\displaystyle {\int }_0^l\left(M_D+Py\right)\left(1\right)dy}\right)$

Substitute $0$ for $M_D$.

$\begin{array}{c}{\theta }_D=\frac{1}{EI}\left({\displaystyle {\int }_0^{l}Pydy}+{\displaystyle {\int }_0^{l}Pldx}+{\displaystyle {\int }_0^{l}Pydy}\right)\\ =\frac{P}{EI}\left({\left(\frac{y^2}{2}\right)}_0^l+{\left(lx\right)}_0^l+{\left(\frac{y^2}{2}\right)}_0^l\right)\\ =\frac{P{l}^2}{2EI}\left(1+2+1\right)\\ =\frac{2P{l}^2}{EI}\end{array}$

Therefore, the slope of point D is $\underset{\_}{{\theta }_D=\frac{2P{l}^2}{EI}}$.