Chapter 9, Problem 9.5.23P

The overhanging beam A BCD supports two concentrated loads P and Q (see figure),

1. For what ratio PIQ will the deflection at point B be zero?
2. For what ratio will the deflection at point D be zero?
3. If Q is replaced by a uniform load with intensity q (on the overhang), repeat parts (a) and (b), but find ratio Pl(qa).

  

To determine

Ratio P/Q for which deflection at B is zero .

Answer to Problem 9.5.23P

Ratio P/Q for which deflection at B is zero is $P/Q=9a/4L$.

Explanation of Solution

Given Information:

The following figure is given along with relevant information,

  

The deflection at B is zero.

Calculation:

Consider the following free body diagram,

  

Take equilibrium of forces in horizontal direction as,

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

Take equilibrium of forces in vertical direction as,

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ \Rightarrow C_y=P+Q\end{array}$

Take equilibrium of moments about A as,

  $\begin{array}{l}{\displaystyle \sum M_A}=0\\ \Rightarrow M_A-PL/2+C_{y}L-Q(L+a)=0\\ \Rightarrow M_A=PL/2-C_{y}L+Q(L+a)\\ \Rightarrow M_A=PL/2-(P+Q)L+Q(L+a)\end{array}$

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

  $M=\{\begin{array}{l}-M_{A}x\le L/2\\ -M_A-P(x-L/2)x\le L\\ -M_A-P(x-L/2)+C_y(x-L)x\le L+a\end{array}$

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,

  $\theta =\frac{dv}{dx}=\frac{1}{EI}\{\begin{array}{l}-M_{A}x+C_{1}x\le L/2\\ -M_{A}x-P{(}^x/2+C_{2}x\le L\\ -M_{A}x-P{(}^x/2+C_y{(}^x/2+C_{3}x\le L+a\end{array}$

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

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

The following conditions are used to evaluate integration constants,

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

Since deflection at B is zero, hence

  ${\delta }_B=\delta (L/2)=0$

Now substitute values of constants and solve the above equation to get

  $P/Q=9a/4L$ .

Conclusion:

Therefore, the ratio P/Q for which deflection at B is zero is $P/Q=9a/4L$ .

To determine

Ratio P/Q for which deflection at D is zero .

Answer to Problem 9.5.23P

Ratio P/Q for which deflection at D is zero is $P/Q=8a(3L+a)/9L^2$.

Explanation of Solution

Given Information:

The following figure is given along with relevant information,

  

The deflection at D is zero.

Calculation:

Consider the following free body diagram,

  

Take equilibrium of forces in horizontal direction as,

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

Take equilibrium of forces in vertical direction as,

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ \Rightarrow C_y=P+Q\end{array}$

Take equilibrium of moments about A as,

  $\begin{array}{l}{\displaystyle \sum M_A}=0\\ \Rightarrow M_A-PL/2+C_{y}L-Q(L+a)=0\\ \Rightarrow M_A=PL/2-C_{y}L+Q(L+a)\\ \Rightarrow M_A=PL/2-(P+Q)L+Q(L+a)\end{array}$

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

  $M=\{\begin{array}{l}-M_{A}x\le L/2\\ -M_A-P(x-L/2)x\le L\\ -M_A-P(x-L/2)+C_y(x-L)x\le L+a\end{array}$

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,

  $\theta =\frac{dv}{dx}=\frac{1}{EI}\{\begin{array}{l}-M_{A}x+C_{1}x\le L/2\\ -M_{A}x-P{(}^x/2+C_{2}x\le L\\ -M_{A}x-P{(}^x/2+C_y{(}^x/2+C_{3}x\le L+a\end{array}$

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

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

The following conditions are used to evaluate integration constants,

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

Since deflection at D is zero, hence

  ${\delta }_D=\delta (L+a)=0$

Now substitute values of constants and solve the above equation to get

  $P/Q=8a(3L+a)/9L^2$ .

Conclusion:

Therefore, the ratio P/Q for which deflection at D is zero is $P/Q=8a(3L+a)/9L^2$ .

To determine

Ratio P/Q for which deflection at B and D is zero .

Answer to Problem 9.5.23P

  $P/qa=9a/8L\text{ for }{\delta }_B=0\text{ and }P/qa=a(4L+a)/3L^2\text{ for }{\delta }_D=0$.

Explanation of Solution

Given Information:

The following figure is given along with relevant information,

  

The deflection at B and D is zero.

Calculation:

Consider the following free body diagram,

  

Take equilibrium of forces in horizontal direction as,

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

Take equilibrium of forces in vertical direction as,

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ \Rightarrow C_y=P+qa\end{array}$

Take equilibrium of moments about A as,

  $\begin{array}{l}{\displaystyle \sum M_A}=0\\ \Rightarrow M_A-PL/2+C_{y}L-qa(L+a/2)=0\\ \Rightarrow M_A=-[-PL/2+C_{y}L-qa(L+a/2)]\end{array}$

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

  $M=\{\begin{array}{l}-M_{A}x\le L/2\\ -M_A-P(x-L/2)x\le L\\ -M_A-P(x-L/2)+C_y(x-L)-q{(}^x/2x\le L+a\end{array}$

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,

  $\theta =\frac{dv}{dx}=\frac{1}{EI}\{\begin{array}{l}-M_{A}x+C_{1}x\le L/2\\ -M_{A}x-P{(}^x/2+C_{2}x\le L\\ -M_{A}x-P{(}^x/2+C_y{(}^x/2-q{(}^x/6+C_{3}x\le L+a\end{array}$

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

  $\delta =v=\frac{1}{EI}\{\begin{array}{l}-M_A{x}^2/2+C_{1}x+C_{4}x\le L/2\\ -M_A{x}^2/2-P{(}^x/6+C_{2}x+C_{5}x\le L\\ -M_A{x}^2/2-P{(}^x/6+C_y{(}^x/6-q{(}^x/24+C_{3}x+C_{6}x\le L+a\end{array}$

The following conditions are used to evaluate integration constants,

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

Substitute values of constants to get the expression for deflection.

Since deflection at B is zero, hence

  ${\delta }_B=\delta (L/2)=0$

Solve the above equation to get

  $P/qa=9a/8L$ .

For deflection at D is zero,

  ${\delta }_D=\delta (L+a)=0$

Solve the above equation to get

  $P/qa=a(4L+a)/3L^2$ .

Conclusion:

Therefore, the ratio P/Q $P/qa=9a/8L\text{ for }{\delta }_B=0\text{ and }P/qa=a(4L+a)/3L^2\text{ for }{\delta }_D=0$