Chapter 10.1, Problem 10.6P

A spring of constant 15 kN/m connects points C and F of the linkage shown. Neglecting the weight of the spring and linkage, determine the force in the spring and the vertical motion of point G when a vertical downward 120-N force is applied (a) at point E, (b) at points E and F.

Fig. P10.5 and P10.6

To determine

Find the force in the spring and the vertical motion of point G when a vertical load of $120\text{N}$ force is applied at point E.

Answer to Problem 10.6P

The force in the spring is $\underset{\_}{120\text{N}\left(\text{C}\right)}$.

The vertical motion of point G is $\underset{\_}{\text{16}\text{mm}(\downarrow )}$.

Explanation of Solution

Given information:

The spring constant is $k=15\text{kN/m}$.

Calculation:

Show the free-body diagram of the spring assembly as in Figure 1.

Write the relation of the deflections at point G, H, F, E, D with C as follows;

$y_G=4y_C;\delta y_G=4\delta y_C$

$\begin{array}{l}{y}_H=4y_C;\delta y_H=4\delta y_C\\ y_F=3y_C;\delta y_F=3\delta y_C\\ y_D=2y_C;\delta y_D=2\delta y_C\\ y_E=2y_C;\delta y_E=2\delta y_C\end{array}$

The deflection $\Delta$ of the spring is;

$\begin{array}{c}\Delta =y_F-y_C\\ =3y_C-y_C\\ =2y_C\end{array}$

Assume the spring force Q is in tension.

Find the force in the spring Q using the relation.

$Q=+k\Delta$

Here, the spring constant is k.

Substitute $15\text{kN/m}$ for k and $2y_C$ for $\Delta$.

$\begin{array}{c}Q=15\left(2y_C\right)\\ =30y_C\end{array}$ (1)

Use the virtual work principle:

$\begin{array}{l}\delta U=0\\ -C\delta y_C+Q\delta y_C-Q\delta y_F-F\delta y_F-H\delta y_H-E\delta y_E=0\end{array}$

Here, $E=120\text{N};F=0;H=0;C=0$

Substitute 0 for C, $3\delta y_C$ for $\delta y_F$, 0 for F, 0 for H, $4\delta y_C$ for $\delta y_H$, 120 N for E, and $2\delta y_C$ for $\delta y_E$.

$\begin{array}{l}-\left(0\right)\delta y_C+Q\delta y_C-Q\left(3\delta y_C\right)-\left(0\right)\left(3\delta y_C\right)-\left(0\right)\left(4\delta y_C\right)-120\left(2\delta y_C\right)=0\\ -0+Q-3Q-0-0-240=0\\ 2Q=-240\\ Q=-120N\end{array}$

$Q=120\text{N}\left(\text{C}\right)$

The spring force Q is in compression. The assumption is incorrect.

Therefore, the force in the spring is $\underset{\_}{120\text{N}\left(\text{C}\right)}$.

Substitute –120 N for Q in Equation (1).

$\begin{array}{l}-120=30y_C\\ y_C=-4\text{mm}\end{array}$

Find the vertical motion $\left(y_G\right)$ of point G using the relation.

$y_G=4y_C$

Substitute –4 mm for $y_C$.

$\begin{array}{c}{y}_G=4\left(-4\right)\\ =-16\text{mm}\\ =\text{16}\text{mm}(\downarrow )\end{array}$

Therefore, the vertical motion of point G is $\underset{\_}{\text{16}\text{mm}(\downarrow )}$.

To determine

Find the force in the spring and the vertical motion of point G when a vertical load of 120-N force is applied at point E and F.

Answer to Problem 10.6P

The force in the spring is $\underset{\_}{300\text{N}\left(\text{C}\right)}$.

The vertical motion of point G is $\underset{\_}{\text{40}\text{mm}(\downarrow )}$.

Explanation of Solution

Given information:

The spring constant is $k=15\text{kN/m}$.

Calculation:

Use the virtual work principle:

$\begin{array}{l}\delta U=0\\ -C\delta y_C+Q\delta y_C-Q\delta y_F-F\delta y_F-H\delta y_H-E\delta y_E=0\end{array}$

Here, $E=120\text{N};F=120\text{N};H=0;C=0$

Substitute 0 for C, $3\delta y_C$ for $\delta y_F$, 120 N for F, 0 for H, $4\delta y_C$ for $\delta y_H$, 120 N for E, and $2\delta y_C$ for $\delta y_E$.

$\begin{array}{l}-\left(0\right)\delta y_C+Q\delta y_C-Q\left(3\delta y_C\right)-\left(120\right)\left(3\delta y_C\right)-\left(0\right)\left(4\delta y_C\right)-\left(120\right)\left(2\delta y_C\right)=0\\ -0+Q-3Q-360-0-240=0\\ 2Q=-600\\ Q=-300N\end{array}$

$Q=300\text{N}\left(\text{C}\right)$

The spring force Q is in compression. The assumption is incorrect.

Therefore, the force in the spring is $\underset{\_}{300\text{N}\left(\text{C}\right)}$.

Substitute –300 N for Q in Equation (1).

$\begin{array}{l}-300=30y_C\\ y_C=-10\text{mm}\end{array}$

Find the vertical motion $\left(y_G\right)$ of point G using the relation.

$y_G=4y_C$

Substitute –10 mm for $y_C$.

$\begin{array}{c}{y}_G=4\left(-10\right)\\ =-40\text{mm}\\ =\text{40}\text{mm}(\downarrow )\end{array}$

Therefore, the vertical motion of point G is $\underset{\_}{\text{40}\text{mm}(\downarrow )}$.