Chapter 2, Problem 2.2.13P

Two rigid bars are connected to each other by two linearly elastic springs. Before loads are applied, the lengths or the springs are such, that the bars are parallel and the springs are without stress.

(a) Derive a formula for the displacement E₄at point 4 when the load P is applied at joint 3 and moment PL is applied at joint 1. as shown in the figure part a. (Assume that the bars rotate through very small angles under the action of load P.)

(b) Repeat part (a) if a rotational spring, k_r= kL², is now added at joint 6. What is the ratio of the deflection d₄ in the figure part a to that in the figure part b ?

  

To determine

The formula for the displacement at point $4$ .

Answer to Problem 2.2.13P

The formula for the displacement at point $4$ is ${\delta }_4=\frac{26P}{3K}$ .

Explanation of Solution

Given information:

The load applied at point $3$ is $P$ .

The moment art point $P$ is $PL$ .

The following figure shows the free body diagram of the bar:

  

Figure-(1)

Write the expression for the moment equilibrium about support $1$ .

  $\begin{array}{l}\left(F_{S_2}L\right)-\left(F_{S_1}\frac{2L}{3}\right)-P\left(\frac{L}{3}+\frac{2L}{3}\right)=0\\ F_{S_2}=\left(\frac{2F_{S_1}}{3}+2P\right)\end{array}$.....(I)

Here, the reaction at forces in spring $1$ is $F_{S_1}$ , the reaction force in spring $2$ is $F_{S_2}$ , the length of the bar is $L$ and the load applied on the load is $P$ .

Write the expression for the moment equilibrium about support $6$ .

  $\begin{array}{l}-F_{S_1}\left(\frac{L}{3}+L\right)+F_{S_2}L=0\\ F_{S_1}=\frac{3}{4}{F}_{S_2}\end{array}$ .....(II)

  

Figure-(2)

Write the expression for deflection ${\delta }_5$ using Figure 1.

  ${\delta }_5=\frac{3{\delta }_4}{4}$ .....(III)

Here, the deflection at point $4$ is ${\delta }_4$ .

  

Figure-(3)

Write the expression for deflection ${\delta }_2$ using Figure-(3).

  ${\delta }_2=\frac{2}{3}{\delta }_3$ .....(IV)

Here, the deflection at point $3$ is ${\delta }_3$ .

Write the expression for the spring stiffness.

  $\delta =\frac{P}{K}$ .....(V)

Here, the spring constant is $K$ .

Write the expression for the elongation of spring $1$ .

  $\left({\delta }_4-{\delta }_2\right)=\frac{F_{S_1}}{K}$ .....(VI)

Write the expression for the elongation of spring $2$ .

  $\left({\delta }_5-{\delta }_3\right)=\frac{-F_{S_2}}{2K}$.....(VII)

Calculation:

Substitute $\frac{3}{4}{F}_{S_2}$ for $F_{S_1}$ , in Equation (I).

  $\begin{array}{l}{F}_{S_2}=\frac{2}{3}\left(\frac{3}{4}{F}_{S_2}\right)+2P\\ F_{S_2}-0.5F_{S_2}=2P\\ F_{S_2}=4P\end{array}$

Substitute $4P$ for $F_{S_2}$ , in Equation (II).

  $\begin{array}{l}{F}_{S_1}=\frac{3}{4}\cdot 4P\\ F_{S_1}=3P\end{array}$

Substitute $\frac{2}{3}{\delta }_3$ for ${\delta }_2$ , $3P$ for $F_{S_1}$ in Equation (VI).

  $\left({\delta }_4-\frac{2{\delta }_3}{3}\right)=\frac{3P}{K}$

Substitute $\frac{3{\delta }_4}{4}$ for ${\delta }_5$ , $4P$ for $F_{S_2}$ in Equation (VII).

  $\begin{array}{l}\left(\frac{3{\delta }_4}{4}-{\delta }_3\right)=\frac{-4P}{2K}\\ {\delta }_4-\frac{{\delta }_4}{2}=\frac{4P}{3K}+\frac{3P}{K}\\ \frac{{\delta }_4}{2}=\frac{13P}{3K}\\ {\delta }_4=\frac{26P}{3K}\end{array}$.....(VIII)

Conclusion:

The formula for the displacement at point $4$ is ${\delta }_4=\frac{26P}{3K}$ .

To determine

The ratio of deflection at point $4$ .

Answer to Problem 2.2.13P

The ratio of deflection at point 4 is $3.75$ .

Explanation of Solution

Given Information:

The spring constant for rotational spring is $k_r=k{L}^2$ .

The following figure shows the free body diagram of the bar having rotational spring:

  

Figure-(4)

Write the expression for the moment equilibrium about support $1$ .

  $\begin{array}{l}{R}_1\times 2L+PL-PL+k{L}^2{\theta }_6=0\\ R_1=\frac{-kL{\theta }_6}{2}\end{array}$.....(IX)

Here, angle of deflection of beam is ${\theta }_6$ .

Write the expression for the net elongation of spring $1$ from Figure-(4).

  ${\delta }_{spring1}=\left({\delta }_4-{\delta }_2\right)$ .....(X)

Write the expression for the displacement at point $4$ from Figure-(4)

  ${\delta }_4={\theta }_6\frac{4L}{3}$.....(XI)

Write the expression for the displacement at point $2$ from Figure-(4)

  ${\delta }_2={\theta }_1\times \frac{2L}{3}$.....(XII)

Write the expression for the net elongation of spring $2$ from Figure-(4)

  ${\delta }_{spring2}=\left({\delta }_3-{\delta }_5\right)$.....(XIII)

Write the expression for displacement at point 3.

  ${\delta }_3={\theta }_1\times L$ .....(XIV)

Write the expression for displacement at point 5.

  ${\delta }_5={\theta }_6\times L$ .....(XV)

The figure below shows the free body diagram of the bar.

  

Figure-(5)

Write the expression for the vertical force equilibrium.

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ P=kL\left(2{\theta }_1-2{\theta }_6-\frac{{\theta }_6}{2}-\frac{4}{3}{\theta }_6+\frac{2}{3}{\theta }_1\right)\\ 16{\theta }_1-23{\theta }_6=\frac{6P}{kL}\end{array}$ .....(XVI)

Write the expression for the moment equilibrium about point 2.

  $\begin{array}{l}{\displaystyle \sum M_2}=0\\ \frac{2kL}{3}\left(2{\theta }_1-{\theta }_6\right)=\frac{4}{3}P\\ 2{\theta }_1-{\theta }_6=\frac{4P}{kL}\end{array}$ ….. (XVII)

Write the expression for the angle of deflection of the beam ${\theta }_6$ .

  $\begin{array}{l}15{\theta }_6=\frac{26P}{kL}\\ {\theta }_6=\frac{26P}{15kL}\end{array}$.....(XVIII)

The following figure shows the forces acting on point $2$ :

  

Figure 6

Write the expression for the ratio of deflection of part $a$ to the deflection of part $b$ .

  $\frac{{\left({\delta }_4\right)}_a}{{\left({\delta }_4\right)}_b}$.....(XIX)

Calculation:

Substitute ${\theta }_6\frac{4L}{3}$ for ${\delta }_4$ ,and ${\theta }_1\times \frac{2L}{3}$ for ${\delta }_2$ in Equation (X).

  ${\delta }_{spring1}=L\left(\frac{4}{3}{\theta }_6-\frac{2}{3}{\theta }_1\right)$

Substitute ${\theta }_1\times L$ for ${\delta }_3$ ,and ${\theta }_6\times L$ for ${\delta }_5$ in Equation (XIII).

  ${\delta }_{spring2}=L\left({\theta }_1-{\theta }_2\right)$

Substitute $\frac{26P}{15kL}$ for ${\theta }_6$ in Equation (VIII).

  $\begin{array}{l}{\delta }_4=\frac{4}{3}L\times \frac{26P}{15kL}\\ {\delta }_4=\left(\frac{104P}{45k}\right)\end{array}$

Substitute $\left(\frac{104P}{45k}\right)$ for ${\left({\delta }_4\right)}_a$ and $\frac{104P}{45k}$ for ${\left({\delta }_4\right)}_b$ .

  $\begin{array}{l}\frac{{\left({\delta }_4\right)}_a}{{\left({\delta }_4\right)}_b}=\left(\frac{\frac{26P}{3k}}{\frac{104P}{45k}}\right)\\ \frac{{\left({\delta }_4\right)}_a}{{\left({\delta }_4\right)}_b}=\frac{15}{4}\\ \frac{{\left({\delta }_4\right)}_a}{{\left({\delta }_4\right)}_b}=3.75\end{array}$

Conclusion:

The ratio of deflection at point $4$ is $3.75$ .