Chapter 4.1, Problem 4.59P

Eight identical 500 × 750-mm rectangular plates, each of mass = 40 kg, are held in a vertical plane as shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether (a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determine or indeterminate, (c) the equilibrium of the plate is maintained the position shown. Also, wherever possible, compute the reactions.

To determine

Find whether the plate is completely, partially, or improperly constrained.

Answer to Problem 4.59P

The plate in figure 1 is $\underset{\_}{\text{completely}\text{constrained}}$.

The plate figure 2 is $\underset{\_}{\text{completely}\text{constrained}}$.

The plate figure 3 is $\underset{\_}{\text{completely}\text{constrained}}$.

The plate figure 4 is $\underset{\_}{\text{improperly}\text{constrained}}$.

The plate figure 5 is $\underset{\_}{\text{partially}\text{constrained}}$.

The plate figure 6 is $\underset{\_}{\text{completely}\text{constrained}}$.

The plate figure 7 is $\underset{\_}{\text{partially}\text{constrained}}$.

The plate figure 8  is $\underset{\_}{\text{completely}\text{constrained}}$.

Explanation of Solution

Given information:

The size of the identical plates is $500\text{mm}\times 750\text{mm}$.

Number of plates is 8.

The mass of each plate is $m=40\text{kg}$.

Calculation:

Find the weight (W) of the plate using the relation.

$W=mg$

Here, the acceleration due to gravity is g.

Consider the acceleration due to gravity as $g=9.81{\text{m/s}}^2$.

Substitute 40 kg for m and $9.81{\text{m/s}}^2$ for g.

$\begin{array}{c}W=40\times 9.81\\ =392.4\text{N}\end{array}$

Figure 1:

Show the free-body diagram of the Figure 1.

The three reactions in the plate behave like non-concurrent and non-parallel force system.

The plate in figure 1 is $\underset{\_}{\text{completely}\text{constrained}}$.

Figure 2:

Show the free-body diagram of the Figure 2.

The three reactions in the plate behave like non-concurrent and non-parallel force system.

The plate figure 2 is $\underset{\_}{\text{completely}\text{constrained}}$.

Figure 3:

Show the free-body diagram of the Figure 3.

The four reactions in the plate behave like non-concurrent and non-parallel force system.

The plate figure 3 is $\underset{\_}{\text{completely}\text{constrained}}$.

Figure 4:

Show the free-body diagram of the Figure 4.

The three reactions in the plate behave like concurrent force system.

The plate figure 4 is $\underset{\_}{\text{improperly}\text{constrained}}$.

Figure 5:

Show the free-body diagram of the Figure 5.

The two reactions in the plate behave like concurrent force system.

The plate figure 5 is $\underset{\_}{\text{partially}\text{constrained}}$.

Figure 6:

Show the free-body diagram of the Figure 6.

The three reactions in the plate behave like non-concurrent and non-parallel force system.

The plate figure 6 is $\underset{\_}{\text{completely}\text{constrained}}$.

Figure 7:

Show the free-body diagram of the Figure 7.

The two reactions in the plate behave like concurrent force system.

The plate figure 7 is $\underset{\_}{\text{partially}\text{constrained}}$.

Figure 8:

Show the free-body diagram of the Figure 8.

The four reactions in the plate behave like non-concurrent and non-parallel force system.

The plate figure 8  is $\underset{\_}{\text{completely}\text{constrained}}$.

To determine

Find whether the reactions are statically determinate or indeterminate.

Answer to Problem 4.59P

The reactions in figure 1 is $\underset{\_}{\text{determinate}}$.

The reactions in figure 2 is $\underset{\_}{\text{determinate}}$.

The reactions in figure 3 is $\underset{\_}{\text{indeterminate}}$.

The reactions in figure 4 is $\underset{\_}{\text{indeterminate}}$.

The reactions in figure 5 is $\underset{\_}{\text{determinate}}$.

The reactions in figure 6 is $\underset{\_}{\text{determinate}}$.

The reactions in figure 7 is $\underset{\_}{\text{indeterminate}}$.

The reactions in figure 8 is $\underset{\_}{\text{indeterminate}}$.

Explanation of Solution

Refer Figure 1:

The equilibrium equations are;

$\begin{array}{l}{\displaystyle \sum F_x=0;}\\ {\displaystyle \sum F_y=0;}\\ {\displaystyle \sum M=0;}\end{array}$

$\text{number}\text{of}\text{unknowns}\le \text{equilibrium}\text{equations}$.

The equilibrium equations are enough to determine the unknown reactions.

The reactions in figure 1 is $\underset{\_}{\text{determinate}}$.

Refer Figure 2:

The equilibrium equations are;

$\begin{array}{l}{\displaystyle \sum F_x=0;}\\ {\displaystyle \sum F_y=0;}\\ {\displaystyle \sum M=0;}\end{array}$

$\text{number}\text{of}\text{unknowns}\le \text{equilibrium}\text{equations}$.

The equilibrium equations are enough to determine the unknown reactions.

The reactions in figure 2 is $\underset{\_}{\text{determinate}}$.

Refer Figure 3:

The equilibrium equations are;

$\begin{array}{l}{\displaystyle \sum F_x=0;}\\ {\displaystyle \sum F_y=0;}\\ {\displaystyle \sum M=0;}\end{array}$

$\text{number}\text{of}\text{unknowns}>\text{equilibrium}\text{equations}$.

The equilibrium equations are not enough to determine the unknown reactions.

The reactions in figure 3 is $\underset{\_}{\text{indeterminate}}$.

Refer Figure 4:

The equilibrium equations are;

$\begin{array}{l}{\displaystyle \sum F_x=0;}\\ {\displaystyle \sum F_y=0;}\\ {\displaystyle \sum M=0;}\end{array}$

$\text{number}\text{of}\text{unknowns}\le \text{equilibrium}\text{equations}$.

The equilibrium equations are enough to determine the unknown reactions.

But the plate is improperly constrained and the plate is not in equilibrium. $\left({\displaystyle \sum M_D\ne 0}\right)$

The reactions in figure 4 is $\underset{\_}{\text{indeterminate}}$.

Refer Figure 5:

The equilibrium equations are;

$\begin{array}{l}{\displaystyle \sum F_x=0;}\\ {\displaystyle \sum F_y=0;}\\ {\displaystyle \sum M=0;}\end{array}$

$\text{number}\text{of}\text{unknowns}\le \text{equilibrium}\text{equations}$.

The equilibrium equations are enough to determine the unknown reactions.

The reactions in figure 5 is $\underset{\_}{\text{determinate}}$.

Refer Figure 6:

The equilibrium equations are;

$\begin{array}{l}{\displaystyle \sum F_x=0;}\\ {\displaystyle \sum F_y=0;}\\ {\displaystyle \sum M=0;}\end{array}$

$\text{number}\text{of}\text{unknowns}\le \text{equilibrium}\text{equations}$.

The equilibrium equations are enough to determine the unknown reactions.

The reactions in figure 6 is $\underset{\_}{\text{determinate}}$.

Refer Figure 7:

The equilibrium equations are;

$\begin{array}{l}{\displaystyle \sum F_x=0;}\\ {\displaystyle \sum F_y=0;}\\ {\displaystyle \sum M=0;}\end{array}$

$\text{number}\text{of}\text{unknowns}\le \text{equilibrium}\text{equations}$.

The equilibrium equations are enough to determine the unknown reactions.

But the plate is improperly constrained and the plate is not in equilibrium. $\left({\displaystyle \sum F_y\ne 0}\right)$.

The reactions in figure 7 is $\underset{\_}{\text{indeterminate}}$.

Refer Figure 8:

The equilibrium equations are;

$\begin{array}{l}{\displaystyle \sum F_x=0;}\\ {\displaystyle \sum F_y=0;}\\ {\displaystyle \sum M=0;}\end{array}$

$\text{number}\text{of}\text{unknowns}>\text{equilibrium}\text{equations}$.

The equilibrium equations are not enough to determine the unknown reactions.

The reactions in figure 8 is $\underset{\_}{\text{indeterminate}}$.

To determine

Find whether the equilibrium of the plate is maintained.

Answer to Problem 4.59P

The reactions in the plate 1 are $\underset{\_}{C=196.2\text{N}(\uparrow );A=196.2\text{N}(\uparrow )}$.

The plate 1 is in $\underset{\_}{\text{equilibrium}}$.

The reactions in the plate 2 are $\underset{\_}{C=196.2\text{N}(\uparrow );D=196.2\text{N}(\uparrow )}$.

The plate 2 is in $\underset{\_}{\text{equilibrium}}$.

The reactions in the plate 3 are $\underset{\_}{D_x=294.3\text{N}(\leftarrow );A_x=294.3\text{N}(\to )}$.

The plate 3 is in $\underset{\_}{\text{equilibrium}}$.

The plate 4 is in $\underset{\_}{\text{not}\text{equilibrium}}$.

The reactions in the plate 5 are $\underset{\_}{C=196.2\text{N}(\uparrow );D=196.2\text{N}(\uparrow )}$.

The plate 5 is in $\underset{\_}{\text{equilibrium}}$.

The reactions in the plate 6 are $\underset{\_}{B=294.3\text{N}(\to );D=491\text{N}\left(\nwarrow 53.1°\right)}$.

The plate 6 is in $\underset{\_}{\text{equilibrium}}$.

The plate 7 is in $\underset{\_}{\text{not}\text{equilibrium}}$.

The reactions in the plate 8 are $\underset{\_}{B=196.2\text{N}(\uparrow );D_y=196.2\text{N}(\uparrow )}$.

The plate 8 is in $\underset{\_}{\text{equilibrium}}$.

Explanation of Solution

Refer Figure 1:

The equilibrium equations are;

$\begin{array}{l}{\displaystyle \sum F_x=0;}\\ {\displaystyle \sum F_y=0;}\\ {\displaystyle \sum M=0;}\end{array}$

Take moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ C\left(750\right)-392.4\left(\frac{750}{2}\right)=0\\ C=196.2\text{N}(\uparrow )\end{array}$

Resolve the horizontal component of forces.

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

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-392.4+C_y=0\\ A_y-392.4+196.2=0\\ A_y=196.2\text{N}(\uparrow )\end{array}$

Therefore, the reactions in the plate 1 are $\underset{\_}{C=196.2\text{N}(\uparrow );A=196.2\text{N}(\uparrow )}$.

The plate 1 is in $\underset{\_}{\text{equilibrium}}$.

Refer Figure 2:

The equilibrium equations are;

$\begin{array}{l}{\displaystyle \sum F_x=0;}\\ {\displaystyle \sum F_y=0;}\\ {\displaystyle \sum M=0;}\end{array}$

Take moment about point B.

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ -D\left(750\right)+392.4\left(\frac{750}{2}\right)=0\\ D=196.2\text{N}(\uparrow )\end{array}$

Resolve the horizontal component of forces.

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

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ D-392.4+C=0\\ 196.2-392.4+C=0\\ C=196.2\text{N}(\uparrow )\end{array}$

Therefore, the reactions in the plate 2 are $\underset{\_}{C=196.2\text{N}(\uparrow );D=196.2\text{N}(\uparrow )}$.

The plate 2 is in $\underset{\_}{\text{equilibrium}}$.

Refer Figure 3:

The equilibrium equations are;

$\begin{array}{l}{\displaystyle \sum F_x=0;}\\ {\displaystyle \sum F_y=0;}\\ {\displaystyle \sum M=0;}\end{array}$

Take moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ D_x\left(500\right)-392.4\left(\frac{750}{2}\right)=0\\ D_x=294.3\text{N}(\leftarrow )\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -D_x+A_x=0\\ -294.3+A_x=0\\ A_x=294.3\text{N}(\to )\end{array}$

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y+D_y=392.4\text{N}\end{array}$

Therefore, the reactions in the plate 3 are $\underset{\_}{D_x=294.3\text{N}(\leftarrow );A_x=294.3\text{N}(\to )}$.

The plate 3 is in $\underset{\_}{\text{equilibrium}}$.

Refer Figure 4:

The equilibrium equations are;

$\begin{array}{l}{\displaystyle \sum F_x=0;}\\ {\displaystyle \sum F_y=0;}\\ {\displaystyle \sum M=0;}\end{array}$

The moment about point D is not equal to zero.

The plate 4 is in $\underset{\_}{\text{not}\text{equilibrium}}$.

Refer Figure 5:

The equilibrium equations are;

$\begin{array}{l}{\displaystyle \sum F_x=0;}\\ {\displaystyle \sum F_y=0;}\\ {\displaystyle \sum M=0;}\end{array}$

Take moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ C\left(750\right)-392.4\left(\frac{750}{2}\right)=0\\ C=196.2\text{N}(\uparrow )\end{array}$

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ D-392.4+C=0\\ D-392.4+196.2=0\\ D=196.2\text{N}(\uparrow )\end{array}$

Therefore, the reactions in the plate 5 are $\underset{\_}{C=196.2\text{N}(\uparrow );D=196.2\text{N}(\uparrow )}$.

The plate 5 is in $\underset{\_}{\text{equilibrium}}$.

Refer Figure 6:

The equilibrium equations are;

$\begin{array}{l}{\displaystyle \sum F_x=0;}\\ {\displaystyle \sum F_y=0;}\\ {\displaystyle \sum M=0;}\end{array}$

Take moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ D_x\left(500\right)-392.4\left(\frac{750}{2}\right)=0\\ D_x=294.3\text{N}(\leftarrow )\end{array}$

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ D_y-392.4=0\\ D_y=392.4\text{N}(\uparrow )\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -D_x+B=0\\ -294.3+B=0\\ B=294.3\text{N}(\to )\end{array}$

Find the resultant force at D;

$\begin{array}{c}D=\sqrt{D_x^2+D_y^2}\\ =\sqrt{{\left(-294.3\right)}^2+{392.4}^2}\\ =491\text{N}\end{array}$

Find the angle $\theta$ at which the resultant passes from the horizontal axis.

$\begin{array}{c}\tan \theta =\frac{D_y}{D_x}\\ =\frac{392.4}{294.3}\\ \theta =53.1°\end{array}$

Therefore, the reactions in the plate 6 are $\underset{\_}{B=294.3\text{N}(\to );D=491\text{N}\left(\nwarrow 53.1°\right)}$.

The plate 6 is in $\underset{\_}{\text{equilibrium}}$.

Refer Figure 7:

The equilibrium equations are;

$\begin{array}{l}{\displaystyle \sum F_x=0;}\\ {\displaystyle \sum F_y=0;}\\ {\displaystyle \sum M=0;}\end{array}$

The plate 7 is in $\underset{\_}{\text{not}\text{equilibrium}}$.

Refer Figure 8:

The equilibrium equations are;

$\begin{array}{l}{\displaystyle \sum F_x=0;}\\ {\displaystyle \sum F_y=0;}\\ {\displaystyle \sum M=0;}\end{array}$

Take moment about point D.

$\begin{array}{l}{\displaystyle \sum M_D=0}\\ B\left(750\right)-392.4\left(\frac{750}{2}\right)=0\\ B=196.2\text{N}(\uparrow )\end{array}$

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ D_y-392.4+B=0\\ D_y-392.4+196.2=0\\ D_y=196.2\text{N}(\uparrow )\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -D_x+C=0\end{array}$

Therefore, the reactions in the plate 8 are $\underset{\_}{B=196.2\text{N}(\uparrow );D_y=196.2\text{N}(\uparrow )}$.

The plate 8 is in $\underset{\_}{\text{equilibrium}}$.