Chapter 4.1, Problem 4.59P

(a)

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 $\underset{\_}{\text{completely}\text{constrained}}$.

(b)

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 $\underset{\_}{\text{indeterminate}}$.

(c)

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 $\underset{\_}{\text{equilibrium}}$.