Chapter 4.1, Problem 4.60P

A truss can be supported in the eight different ways shown A connections consist of smooth pins, rollers, or short links. For case, answer the questions listed in Prob. 4.59, and, wherever possible, compute the reactions, assuming that the magnitude force P is 12 kips.

Fig. P4.60

To determine

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

Answer to Problem 4.60P

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

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

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

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

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

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

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

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

Explanation of Solution

Given information:

The magnitude of the force P is 12 kips.

Calculation:

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 concurrent force system.

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

Figure 3:

Show the free-body diagram of the Figure 3.

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

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

Figure 5:

Show the free-body diagram of the Figure 5.

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

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

Figure 6:

Show the free-body diagram of the Figure 6.

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

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

Figure 7:

Show the free-body diagram of the Figure 7.

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

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

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 4.60P

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

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

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

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

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

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

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

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.

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

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

Refer Figure 3:

The equilibrium equations are;

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

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

The equilibrium equations are enough to determine the unknown reactions.

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

Refer Figure 4:

The equilibrium equations are;

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

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

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

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.

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

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

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.

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

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 4.60P

The reactions in the plate 1 are $\underset{\_}{B=8\text{kips}(\leftarrow );A=14.42\text{kips}\left(\nearrow 56.3°\right)}$.

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

The plate 2 is in $\underset{\_}{\text{not}\text{equilibrium}}$.

The reactions in the plate 3 are $\underset{\_}{C=6\text{kips}(\uparrow );A=6\text{kips}(\uparrow )}$.

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

The reactions in the plate 4 are $\underset{\_}{B_x=8\text{kips}(\leftarrow );A_x=8\text{kips}(\to )}$.

The plate 4 is in $\underset{\_}{\text{equilibrium}}$.

The plate 5 is in $\underset{\_}{\text{not}\text{equilibrium}}$.

The reactions in the plate 6 are $\underset{\_}{C=6\text{kips}(\uparrow );A=6\text{kips}(\uparrow )}$.

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

The reactions in the plate 7 are $\underset{\_}{A=6\text{kips}(\uparrow );B=10\text{kips}\left(\nwarrow 36.9°\right);C=8\text{kips}(\to )}$.

The plate 7 is in $\underset{\_}{\text{equilibrium}}$.

The reactions in the plate 8 are $\underset{\_}{A_y=6\text{kips}(\uparrow )}$.

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.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ B\left(9\right)-12\left(6\right)=0\\ B=8\text{kips}(\leftarrow )\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -B+A_x=0\\ -8+A_x=0\\ A_x=8\text{kips}(\to )\end{array}$

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-12=0\\ A_y=12\text{kips}(\uparrow )\end{array}$

Find the resultant force at A;

$\begin{array}{c}A=\sqrt{A_x^2+A_y^2}\\ =\sqrt{8^2+{12}^2}\\ =14.42\text{kips}\end{array}$

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

$\begin{array}{c}\tan \theta =\frac{A_y}{A_x}\\ =\frac{12}{8}\\ \theta =56.3°\end{array}$

Therefore, the reactions in the plate 1 are $\underset{\_}{B=8\text{kips}(\leftarrow );A=14.42\text{kips}\left(\nearrow 56.3°\right)}$.

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

Refer Figure 2:

The equilibrium equations are;

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

The moment about point A is not equal to zero.

The plate 2 is in $\underset{\_}{\text{not}\text{equilibrium}}$.

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.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ C\left(12\right)-12\left(6\right)=0\\ C=6\text{kips}(\uparrow )\end{array}$

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.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y+C-12=0\\ A_y+6-12=0\\ A_y=6\text{kips}(\uparrow )\end{array}$

Therefore, the reactions in the plate 3 are $\underset{\_}{C=6\text{kips}(\uparrow );A=6\text{kips}(\uparrow )}$.

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

Refer Figure 4:

The equilibrium equations are;

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

Take moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ B_x\left(9\right)-12\left(6\right)=0\\ B_x=8\text{kips}(\leftarrow )\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -B_x+A_x=0\\ -8+A_x=0\\ A_x=8\text{kips}(\to )\end{array}$

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y+B_y-12=0\end{array}$

Therefore, the reactions in the plate 4 are $\underset{\_}{B_x=8\text{kips}(\leftarrow );A_x=8\text{kips}(\to )}$.

The plate 4 is in $\underset{\_}{\text{equilibrium}}$.

Refer Figure 5:

The equilibrium equations are;

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

The moment about point A is not equal to zero.

The plate 5 is in $\underset{\_}{\text{not}\text{equilibrium}}$.

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.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ C\left(12\right)-12\left(6\right)=0\\ C=6\text{kips}(\uparrow )\end{array}$

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A-12+C=0\\ A-12+6=0\\ A=6\text{kips}(\uparrow )\end{array}$

Therefore, the reactions in the plate 6 are $\underset{\_}{C=6\text{kips}(\uparrow );A=6\text{kips}(\uparrow )}$.

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

Refer Figure 7:

The equilibrium equations are;

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

Find the angle $\theta$.

$\begin{array}{c}\tan \theta =\frac{\text{opposite}\text{side}}{\text{adjacent}\text{side}}\\ =\frac{9}{12}\\ \theta =36.9°\end{array}$

Take moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ B\cos 36.9°\left(9\right)-12\left(6\right)=0\\ B=10\text{kips}\left(\nwarrow 36.9°\right)\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -B_x+C=0\\ -8+C=0\\ C=8\text{kips}(\to )\end{array}$

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A-12+10\sin 36.9°=0\\ A=6\text{kips}(\uparrow )\end{array}$

Therefore, the reactions in the plate 7 are $\underset{\_}{A=6\text{kips}(\uparrow );B=10\text{kips}\left(\nwarrow 36.9°\right);C=8\text{kips}(\to )}$.

The plate 7 is in $\underset{\_}{\text{equilibrium}}$.

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 C.

$\begin{array}{l}{\displaystyle \sum M_C=0}\\ 12\left(6\right)-A_y\left(12\right)=0\\ A_y=6\text{kips}(\uparrow )\end{array}$

Therefore, the reactions in the plate 8 are $\underset{\_}{A_y=6\text{kips}(\uparrow )}$.

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