Chapter 9, Problem 9.5.18P

A beam ABCD consisting of a simple span BD and an overhang AB\s loaded by a force P acting at the end of the bracket CEF (see figure),

1. Determine the deflection at the end of the over h a tig.

Under what conditions is this deflection upward? Under what conditions is it downward?

  

To determine

Deflection at A.

Answer to Problem 9.5.18P

The deflection at A is ${\delta }_A=P{L}^2(10L-9a)/324EI\text{ (Positive upward)}$.

Explanation of Solution

Given Information:

The following figure is given along with relevant information,

  

Calculation:

Consider the following diagram,

  

Transfer the load P from F to C and draw reaction forces as shown in following figure,

  

Take equilibrium of forces in horizontal direction as,

  $\begin{array}{l}{\displaystyle \sum F_x}=0\\ \Rightarrow B_x=0\end{array}$

Take equilibrium of moments about D as,

  $\begin{array}{l}{\displaystyle \sum M_D}=0\\ \Rightarrow -B_{y}L+Pa+2LP/3=0\\ \Rightarrow B_y=\frac{Pa+2LP/3}{L}\end{array}$

Take equilibrium of forces in vertical direction as,

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ \Rightarrow B_y+D_y=P\\ \Rightarrow D_y=P-B_y=P-\frac{Pa+2LP/3}{L}\end{array}$

The bending moment at distance x from point A is given by,

  $M=\{\begin{array}{l}0x\le L/2\\ B_y(x-L/2)x\le L/2+L/3\\ B_y(x-L/2)-P(x-L/2-L/3)-Pax\le L/2+L/3+2L/3\end{array}$

The deflection and bending moment is related by following differential equation

  $\frac{d^{2}v}{d{x}^2}=\frac{M}{EI}\mathrm{....}\text{(1)}$

Integrate differential equation (1) with respect to x by putting expression for M to get angle of rotations, as,

  $\theta =\frac{dv}{dx}=\frac{1}{EI}\{\begin{array}{l}{C}_{1}x\le L/2\\ B_y{(}^x/2+C_{2}x\le L/2+L/3\\ B_y{(}^x/2-P{(}^x/2-Pax+C_{3}x\le L/2+L/3+2L/3\end{array}$

Integrate angle of rotation with respect to x get deflections as,

  $\delta =v=\frac{1}{EI}\{\begin{array}{l}{C}_{1}x+C_{4}x\le L/2\\ B_y{(}^x/6+C_{2}x+C_{5}x\le 5L/6\\ B_y{(}^x/6-P{(}^x/6-Pa{x}^2/2+C_{3}x+C_{6}x\le 3L/2\end{array}$

The following conditions are used to evaluate integration constants,

  $\begin{array}{l}\underset{x\to {(L/2)}^{-}}{\lim }\theta =\underset{x\to {(L/2)}^{+}}{\lim }\theta \mathrm{....}\text{(1)}\\ \underset{x\to {(5L/6)}^{-}}{\lim }\theta =\underset{x\to {(5L/6)}^{+}}{\lim }\theta \mathrm{....}\text{(2)}\\ \underset{x\to {(L/2)}^{-}}{\lim }\delta =\underset{x\to {(L/2)}^{+}}{\lim }\delta \mathrm{....}\text{(3)}\\ \underset{x\to {(5L/6)}^{-}}{\lim }\delta =\underset{x\to {(5L/6)}^{+}}{\lim }\delta \mathrm{....}\text{(4)}\\ \delta (L/2)=0\mathrm{....}\text{(5)}\\ \delta (3L/2)=0\mathrm{....}\text{(6)}\end{array}$

Then substitute values of constants and $x=0$ in $\delta$ to get

  ${\delta }_A=P{L}^2(10L-9a)/324EI\text{ (Positive upward)}$

Conclusion:

Therefore the deflection at A is ${\delta }_A=P{L}^2(10L-9a)/324EI\text{ (Positive upward)}$ .

To determine

Condition for upward and downward deflection at A .

Answer to Problem 9.5.18P

The deflection at A is upward when $a/L<10/9$and downward when $a/L>10/9$

Explanation of Solution

Given Information:

The following figure is given along with relevant information,

  

Calculation:

Consider the following diagram,

  

Transfer the load P from F to C and draw reaction forces as shown in following figure,

  

Take equilibrium of forces in horizontal direction as,

  $\begin{array}{l}{\displaystyle \sum F_x}=0\\ \Rightarrow B_x=0\end{array}$

Take equilibrium of moments about D as,

  $\begin{array}{l}{\displaystyle \sum M_D}=0\\ \Rightarrow -B_{y}L+Pa+2LP/3=0\\ \Rightarrow B_y=\frac{Pa+2LP/3}{L}\end{array}$

Take equilibrium of forces in vertical direction as,

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ \Rightarrow B_y+D_y=0\\ \Rightarrow D_y=-B_y=-\frac{Pa+2LP/3}{L}\end{array}$

The bending moment at distance x from point A is given by,

  $M=\{\begin{array}{l}0x\le L/2\\ B_y(x-L/2)x\le L/2+L/3\\ B_y(x-L/2)-P(x-L/2-L/3)-Pax\le L/2+L/3+2L/3\end{array}$

The deflection and bending moment is related by following differential equation

  $\frac{d^{2}v}{d{x}^2}=\frac{M}{EI}\mathrm{....}\text{(1)}$

Integrate differential equation (1) with respect to x by putting expression for M to get angle of rotations, as,

  $\theta =\frac{dv}{dx}=\frac{1}{EI}\{\begin{array}{l}{C}_{1}x\le L/2\\ B_y{(}^x/2+C_{2}x\le L/2+L/3\\ B_y{(}^x/2-P{(}^x/2-Pax+C_{3}x\le L/2+L/3+2L/3\end{array}$

Integrate angle of rotation with respect to x get deflections as,

  $\delta =v=\frac{1}{EI}\{\begin{array}{l}{C}_{1}x+C_{4}x\le L/2\\ B_y{(}^x/6+C_{2}x+C_{5}x\le 5L/6\\ B_y{(}^x/6-P{(}^x/6-Pa{x}^2/2+C_{3}x+C_{6}x\le 3L/2\end{array}$

The following conditions are used to evaluate integration constants,

  $\begin{array}{l}\underset{x\to {(L/2)}^{-}}{\lim }\theta =\underset{x\to {(L/2)}^{+}}{\lim }\theta \mathrm{....}\text{(1)}\\ \underset{x\to {(5L/6)}^{-}}{\lim }\theta =\underset{x\to {(5L/6)}^{+}}{\lim }\theta \mathrm{....}\text{(2)}\\ \underset{x\to {(L/2)}^{-}}{\lim }\delta =\underset{x\to {(L/2)}^{+}}{\lim }\delta \mathrm{....}\text{(3)}\\ \underset{x\to {(5L/6)}^{-}}{\lim }\delta =\underset{x\to {(5L/6)}^{+}}{\lim }\delta \mathrm{....}\text{(4)}\\ \delta (L/2)=0\mathrm{....}\text{(5)}\\ \delta (3L/2)=0\mathrm{....}\text{(6)}\end{array}$

Then substitute values of constants and $x=0$ in $\delta$ to get

  ${\delta }_A=P{L}^2(10L-9a)/324EI\text{ (Positive upward)}$

Now, the deflection at A is upward when $a/L<10/9$ and downward when $a/L>10/9$

Conclusion:

Therefore the deflection at A is upward when $a/L<10/9$ and downward when $a/L>10/9$