Chapter 4.1, Problem 4.60P

(a)

To determine

whether the bracket is completely, partially, or improperly constrained.

Answer to Problem 4.60P

The bracket in case 1 is $\underset{\_}{\text{completely}\text{constrained}}$.

The bracket in case 2 is $\underset{\_}{\text{improperly}\text{constrained}}$.

The bracket in case 3 is $\underset{\_}{\text{partially}\text{constrained}}$.

The bracket in case 4 is $\underset{\_}{\text{completely}\text{constrained}}$.

The bracket in case 5 is $\underset{\_}{\text{completely}\text{constrained}}$.

The bracket in case 6 is $\underset{\_}{\text{completely}\text{constrained}}$.

The bracket in case 7 is $\underset{\_}{\text{completely}\text{constrained}}$.

The bracket in case 8 is $\underset{\_}{\text{improperly}\text{constrained}}$.

Explanation of Solution

Given information:

The magnitude of the force $P=100\text{lb}$.

Calculation:

Assumption:

Apply the sign convention for calculating the equations of equilibrium as shown below.

• For the horizontal forces equilibrium condition, take the force acting towards right side as positive $\left(\to +\right)$ and the force acting towards left side as negative $\left(\leftarrow -\right)$.
• For the vertical forces equilibrium condition, take the upward force as positive $\left(\uparrow +\right)$ and the downward force as negative $\left(\downarrow -\right)$.
• For moment equilibrium condition, take the clockwise moment as negative and counterclockwise moment as positive.

Case 1:

Draw the free body diagram of the bracket in case 1 as in Figure (1).

Refer to Figure (1).

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

Hence, the bracket in case 1 is $\underset{\_}{\text{completely}\text{constrained}}$.

Case 2:

Draw the free body diagram of the bracket in case 2 as in Figure (2).

Refer to Figure (2).

The four reactions in the bracket behave like concurrent and force system through A.

Hence, the bracket in case 2 is $\underset{\_}{\text{improperly}\text{constrained}}$.

Case 3:

Draw the free body diagram of the bracket in case 3 as in Figure (3).

Refer to Figure (3).

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

Hence, the bracket in case 3 is $\underset{\_}{\text{partially}\text{constrained}}$.

Case 4:

Draw the free body diagram of the bracket in case 4 as in Figure (4).

Refer to Figure (4).

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

Hence, the bracket in case 4 is $\underset{\_}{\text{completely}\text{constrained}}$.

Case 5:

Draw the free body diagram of the bracket in case 5 as in Figure (5).

Refer to Figure (5).

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

Hence, the bracket in case 5 is $\underset{\_}{\text{completely}\text{constrained}}$.

Case 6:

Draw the free body diagram of the bracket in case 6 as in Figure (6).

Refer to Figure (6).

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

Hence, the bracket in case 6 is $\underset{\_}{\text{completely}\text{constrained}}$.

Case 7:

Draw the free body diagram of the bracket in case 7 as in Figure (7).

Refer to Figure (7).

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

Hence, the bracket in case 7 is $\underset{\_}{\text{partially}\text{constrained}}$.

Case 8:

Draw the free body diagram of the bracket in case 8 as in Figure (8).

Refer to Figure (8).

The three reactions in the bracket behave like concurrent and force system through A.

Hence, the bracket in case 8  is $\underset{\_}{\text{completely}\text{constrained}}$.

(b)

To determine

whether the reactions are statically determinate or indeterminate.

Answer to Problem 4.60P

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

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

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

The reactions in case 4 is $\underset{\_}{\text{determinate}}$.

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

The reactions in case 6 is $\underset{\_}{\text{indeterminate}}$.

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

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

Explanation of Solution

Given information:

The magnitude of the force $P=100\text{lb}$.

Calculation:

The equilibrium equations are;

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

Case 1:

$\text{Number}\text{of}\text{unknowns}\le \text{Equilibrium}\text{equations}$.

The equilibrium equations are enough to determine the unknown reactions.

Hence, the reactions in case 1 is $\underset{\_}{\text{determinate}}$.

Case 2:

$\text{Number}\text{of}\text{unknowns}>\text{Equilibrium}\text{equations}$.

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

Hence, the reactions in case 2 is $\underset{\_}{\text{indeterminate}}$.

Case 3:

$\text{Number}\text{of}\text{unknowns}>\text{Equilibrium}\text{equations}$.

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

Hence, the reactions in case 3 is $\underset{\_}{\text{indeterminate}}$.

Case 4:

$\text{Number}\text{of}\text{unknowns}\le \text{Equilibrium}\text{equations}$.

The equilibrium equations are enough to determine the unknown reactions.

Hence, the reactions in case 4 is $\underset{\_}{\text{indeterminate}}$.

Case 5:

$\text{Number}\text{of}\text{unknowns}>\text{Equilibrium}\text{equations}$.

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

Hence, the reactions in case 5 is $\underset{\_}{\text{indeterminate}}$.

Case 6:

$\text{Number}\text{of}\text{unknowns}>\text{Equilibrium}\text{equations}$.

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

Hence, the reactions in case 6 is $\underset{\_}{\text{indeterminate}}$.

Case 7:

$\text{Number}\text{of}\text{unknowns}\le \text{Equilibrium}\text{equations}$.

The equilibrium equations are enough to determine the unknown reactions.

Hence, the reactions in case 7 is $\underset{\_}{\text{determinate}}$.

Case 8:

$\text{Number}\text{of}\text{unknowns}>\text{Equilibrium}\text{equations}$.

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

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

(c)

To determine

whether the equilibrium of the bracket is maintained.

Answer to Problem 4.60P

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

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

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

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

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

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

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

The bracket in case 8 is $\underset{\_}{\text{not in equilibrium}}$.

Explanation of Solution

Given information:

The magnitude of the force $P=100\text{lb}$.

Assumption:

Apply the sign convention for calculating the equations of equilibrium as below.

• For the horizontal forces equilibrium condition, take the force acting towards right side as positive $\left(\to +\right)$ and the force acting towards left side as negative $\left(\leftarrow -\right)$.
• For the vertical forces equilibrium condition, take the upward force as positive $\left(\uparrow +\right)$ and downward force as negative $\left(\downarrow -\right)$.
• For moment equilibrium condition, take the clockwise moment as negative and counter clockwise moment as positive.

Calculation:

Case 1:

Refer to Figure 1.

Apply the Equations of Equilibrium as shown below.

Apply the Equilibrium equation of moment at point A and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ -100\times 2-B_x\times 3=0\\ 3B_x=-200\\ B_x=-66.67\text{lb}\end{array}$

Hence, the reaction at B is $\underset{\_}{66.67\text{lb}\leftarrow }$.

Apply the Equilibrium equation of horizontal forces at point A and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -66.67+A_x=0\\ A_x=66.67\text{lb}\end{array}$

Apply the Equilibrium equation of vertical forces at point A and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -100+A_y=0\\ A_y=100\text{lb}\end{array}$

Calculate the reaction at A using the formula;

$A=\sqrt{A_x^2+A_y^2}$

Substitute 66.67 lb for $A_x$ and 100 lb for $A_y$.

$\begin{array}{c}A=\sqrt{{66.67}^2+{100}^2}\\ =\sqrt{14,444.8889}\\ =120.19\text{lb}\end{array}$

Determine the angle $\left(\alpha \right)$ using the formula:

$\tan \alpha =\frac{A_y}{A_x}$

Substitute 66.67 lb for $A_x$ and 100 lb for $A_y$.

$\begin{array}{l}\tan \alpha =\frac{100}{66.67}\\ \alpha ={\tan }^{-1}\left(1.5\right)\\ \alpha =56.3°\end{array}$

Hence, the reaction at a is $\underset{\_}{120.2\text{lb}\nearrow \text{56}\text{.3}°}$.

Hence, the bracket 1 is in $\underset{\_}{\text{equilibrium}}$.

Case 2:

Refer to Figure 2.

The Equilibrium of moment ${\displaystyle \sum M_A}\ne 0$.

Hence, the bracket 2 is $\underset{\_}{\text{not in equilibrium}}$.

Case 3:

Refer to Figure 3.

Apply the Equations of Equilibrium as shown below.

Apply the Equilibrium equation of moment at point A and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ -100\times 2+C_y\times 4=0\\ 4C_y=200\\ C_y=50\text{lb}\end{array}$

Hence, the reaction at C is $\underset{\_}{50\text{lb}\uparrow }$.

Apply the Equilibrium equation of vertical forces at point A and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ 50-100+A=0\\ A=50\text{lb}\end{array}$

Hence, the reaction at A is $\underset{\_}{50\text{lb}\uparrow }$.

Hence, the bracket 3 is in $\underset{\_}{\text{equilibrium}}$.

Case 4:

Refer to Figure 4.

Calculate the angle of force B to the horizontal $\left(\theta \right)$ using the formula:

$\begin{array}{l}\tan \theta =\frac{3}{4}\\ \theta ={\tan }^{-1}\left(\frac{3}{4}\right)\\ \theta =36.9°\end{array}$

Apply the Equations of Equilibrium as shown below.

Apply the Equilibrium equation of moment at point A and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ -100\times 2+B\cos 36.9°\times 3=0\\ 2.4B=200\\ B=83.3\text{lb}\end{array}$

Hence, the reaction at B is $\underset{\_}{83.3\text{lb}\nwarrow 36.9°}$.

Apply the Equilibrium equation of horizontal forces at point A and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ C_x-83.3\cos 36.9=0\\ C_x=66.6\text{lb}\end{array}$

Hence, the reaction at C is $\underset{\_}{B=66.6\text{lb}\to }$.

Apply the Equilibrium equation of vertical forces at point A and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -100+A+83.3\sin 36.9°=0\\ A=50\text{lb}\end{array}$

Hence, the reaction at A is $\underset{\_}{50\text{lb}\uparrow }$.

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

Case 5:

Refer to Figure 5.

Apply the Equations of Equilibrium as shown below.

Apply the Equilibrium equation of moment at point C and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum M_C}=0\\ -A_y\times 4+100\times 2=0\\ A_y=50\text{lb}\uparrow \end{array}$

Hence, the bracket 5 is in $\underset{\_}{\text{equilibrium}}$.

Case 6:

Refer to Figure 6.

Apply the Equations of Equilibrium as shown below.

Apply the Equilibrium equation of moment at point A and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ -100\times 2-B_x\times 3=0\\ 3B_x=-200\\ B_x=66.67\text{lb}\leftarrow \end{array}$

Hence, the reaction at B is $\underset{\_}{66.67\text{lb}\leftarrow }$.

Apply the Equilibrium equation of horizontal forces at point A and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x-66.67=0\\ A_x=66.67\text{lb}\to \end{array}$

Hence, the reaction at A is $\underset{\_}{66.6\text{lb}\to }$.

Apply the Equilibrium equation of vertical forces at point A and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y+B_y-100=0\\ A_y+B_y=100\text{lb}\end{array}$

Hence, the bracket 6 is in $\underset{\_}{\text{equilibrium}}$.

Refer In case 7:

Refer to Figure 7.

Apply the Equations of Equilibrium as shown below.

Apply the Equilibrium equation of moment at point A and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ -100\times 2+C\times 4=0\\ 4C=200\\ C=50\text{lb}\end{array}$

Hence, the reaction at C is $\underset{\_}{50\text{lb}\uparrow }$.

Apply the Equilibrium equation of vertical forces at point A and equate to zero for equilibrium.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ 50-100+A=0\\ A=50\text{lb}\end{array}$

Hence, the reaction at A is $\underset{\_}{50\text{lb}\uparrow }$.

Hence, the bracket 7 is in $\underset{\_}{\text{not}\text{equilibrium}}$.

Case 8:

Refer to Figure 8.

The Equilibrium of moment ${\displaystyle \sum M_A}\ne 0$.

Therefore, the bracket 8 is $\underset{\_}{\text{not in equilibrium}}$.