Chapter 1, Problem 1.3.5P

Segments AB and BCD of beam ABCD are pin connected at x = 10 ft. The beam is supported by a pin support at A and roller supports at C and D; the roller at D is rotated by 30* from the x axis (see figure). A trapezoidal distributed load on BC varies in intensity from 5 lb/ft at B to 2.5 lb/ft at C. A concentrated moment is applied at joint A, and a 40-lb inclined load is applied at the mid-span or CD.

(a) Find reactions at supports A, C, and D.

(b) Find the resultant force in the pin connection at B.

(c) Repeat parts (a) and (b) if a rotational spring(k_r= 50 ft-lb/radian ) is added at A and the roller at C is removed.

  

To determine

Determine the reactions at supports A, C, and D in the following figure.

Answer to Problem 1.3.5P

The correct answers are: $A_x=12.55\text{ lb, }{A}_y=-15\text{ lb, }{C}_y=104.3\text{ lb, }{D}_x=11.45\text{ lb, }{D}_y=-19.83\text{ lb}\text{.}$

Explanation of Solution

Given Information:

The following figure is given for reference.

Calculation:

To find support reactions, draw free body diagram as shown below.

First analyze left hand side of the above figure as shown below.

Take equilibrium of moments about point A,

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ \Rightarrow -150+10B_y=0\\ \Rightarrow B_y=15\text{ lb}\end{array}$

Take equilibrium of forces in $y-$direction (horizontal direction),

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow A_y+B_y=0\\ \Rightarrow A_y=-B_y=-15\\ \Rightarrow A_y=-15\text{ lb}\end{array}$

Take equilibrium of forces in $x-$direction (horizontal direction):

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

Now, analyze the right hand side of the above figure as shown below,

Take equilibrium of moments about point B,

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ \Rightarrow -\frac{5\times 10}{2}\times \frac{10}{3}-\frac{2.5\times 10}{2}\times \frac{2\times 10}{3}+10C_y-32\times 15+20D\cos {30}^0=0\end{array}$

$10C_y+10D\sqrt{3}=\frac{1940}{3}$

Take equilibrium of forces in $y-$direction (horizontal direction),

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow -B_y-\frac{10\times 5}{2}-\frac{2.5\times 10}{2}+C_y-32+D\cos {30}^0=0\\ \Rightarrow -15-\frac{10\times 5}{2}-\frac{2.5\times 10}{2}+C_y-32+D\cos {30}^0=0\end{array}$

$C_y+\frac{1}{2}D\sqrt{3}=-\frac{169}{2}$

Solve equations and for $C_y\text{=104}\text{.33 lb, }D=-22.9\text{ lb, }{D}_y=D\cos {30}^0=-19.83\text{ lb, }{D}_x=D\sin {30}^0=-11.45\text{ lb}\text{.}$.

Take equilibrium of forces in $x-$direction (horizontal direction):

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow -B_x-24-D\sin {30}^0=0\\ \Rightarrow -B_x-24-(-22.9)\sin {30}^0=0\\ \Rightarrow B_x=-12.55\text{ lb}\end{array}$

Now solve equation

$A_x=-B_x=-(-12.55)=12.55\text{ lb}$

Conclusion:

Thus reactions forces are: $A_x=12.55\text{ lb, }{A}_y=-15\text{ lb, }{C}_y=104.3\text{ lb, }{D}_x=11.45\text{ lb, }{D}_y=-19.83\text{ lb}\text{.}$

To determine

Determine the resultant force in pin B.

Answer to Problem 1.3.5P

The correct answer is: $\text{19}\text{.56 lb}\text{.}$

Explanation of Solution

Given Information:

The following figure is given for reference.

Calculation:

To find support reactions, draw free body diagram as shown below.

First analyze left hand side of the above figure as shown below.

Take equilibrium of moments about point A,

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ \Rightarrow -150+10B_y=0\\ \Rightarrow B_y=15\text{ lb}\end{array}$

Take equilibrium of forces in $y-$direction (horizontal direction),

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow A_y+B_y=0\\ \Rightarrow A_y=-B_y=-15\\ \Rightarrow A_y=-15\text{ lb}\end{array}$

Now, analyze the right hand side of the above figure as shown below,

Take equilibrium of moments about point B,

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ \Rightarrow -\frac{5\times 10}{2}\times \frac{10}{3}-\frac{2.5\times 10}{2}\times \frac{2\times 10}{3}+10C_y-32\times 15+20D\cos {30}^0=0\end{array}$

$10C_y+10D\sqrt{3}=\frac{1940}{3}$

Take equilibrium of forces in $y-$direction (horizontal direction),

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow -B_y-\frac{10\times 5}{2}-\frac{2.5\times 10}{2}+C_y-32+D\cos {30}^0=0\\ \Rightarrow -15-\frac{10\times 5}{2}-\frac{2.5\times 10}{2}+C_y-32+D\cos {30}^0=0\end{array}$

$C_y+\frac{1}{2}D\sqrt{3}=-\frac{169}{2}$

Solve equations and for $C_y\text{=104}\text{.33 lb, }D=-22.9\text{ lb, }{D}_y=D\cos {30}^0=-19.83\text{ lb, }{D}_x=D\sin {30}^0=-11.45\text{ lb}\text{.}$.

Take equilibrium of forces in $x-$direction (horizontal direction):

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow -B_x-24-D\sin {30}^0=0\\ \Rightarrow -B_x-24-(-22.9)\sin {30}^0=0\\ \Rightarrow B_x=-12.55\text{ lb}\end{array}$

Conclusion:

Thus resultant force is: $R=\sqrt{B_x{}^2+B_y{}^2}=\sqrt{{(-12.55)}^2+{15}^2}=19.55\text{ lb}$

To determine

Determine the reactions at supports A, C, and D; and resultant force at B in the following figure.

Answer to Problem 1.3.5P

The correct answers are: $A_x=42.7\text{ lb, }{A}_y=37.2\text{ lb, }{M}_A=522\text{ lb-ft, }{D}_x=-18.67\text{ lb, }{D}_y=32.3\text{ lb, }{R}_B=56.6\text{ lb}\text{.}$

Explanation of Solution

Given Information:

The following figure is given for reference.

Calculation:

To find support reactions, draw free body diagram as shown below.

First, analyze the right hand side of the above figure as shown below,

Take equilibrium of moments about point B,

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ \Rightarrow -\frac{5\times 10}{2}\times \frac{10}{3}-\frac{2.5\times 10}{2}\times \frac{2\times 10}{3}-32\times 15+20D\cos {30}^0=0\\ \Rightarrow 10D\sqrt{3}=\frac{1940}{3}\\ \Rightarrow D=37.33\text{ lb}\end{array}$

So $D_y=D\cos {30}^0=32.32\text{ lb, }{D}_x=D\sin {30}^0=18.66\text{ lb}\text{.}$

Take equilibrium of forces in $y-$direction (horizontal direction),

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow -B_y-\frac{10\times 5}{2}-\frac{2.5\times 10}{2}-32+D\cos {30}^0=0\\ \Rightarrow -B_y-\frac{10\times 5}{2}-\frac{2.5\times 10}{2}-32+37.33\cos {30}^0=0\\ \Rightarrow B_y=-37.17\text{ lb}\end{array}$

Take equilibrium of forces in $x-$direction (horizontal direction):

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow -B_x-24-D\sin {30}^0=0\\ \Rightarrow -B_x-24-(37.33)\sin {30}^0=0\\ \Rightarrow B_x=-42.66\text{ lb}\end{array}$

Thus resultant force at B is: $R=\sqrt{B_x{}^2+B_y{}^2}=\sqrt{{(-42.66)}^2+{(-37.17)}^2}=56.58\text{ lb}$.

Now, analyze left hand side of the above figure as shown below.

Take equilibrium of moments about point A,

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ \Rightarrow M_A-150+10B_y=0\\ \Rightarrow M_A-150+10(-37.17)=0\\ \Rightarrow M_A=521.7\text{ lb-ft}\end{array}$

Take equilibrium of forces in $y-$direction (horizontal direction),

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow A_y+B_y=0\\ \Rightarrow A_y-37.17=0\\ \Rightarrow A_y=37.17\text{ lb}\end{array}$

Take equilibrium of forces in $x-$direction (horizontal direction):

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow A_x+B_x=0\\ \Rightarrow A_x=-B_x=-(-42.66)=42.67\text{ lb}\end{array}$

Conclusion:

The correct answers are: $A_x=42.7\text{ lb, }{A}_y=37.2\text{ lb, }{M}_A=522\text{ lb-ft, }{D}_x=-18.67\text{ lb, }{D}_y=32.3\text{ lb, }{R}_B=56.6\text{ lb}\text{.}$