Chapter 2, Problem 2.2.20P

A framework ABC consists of two rigid bars AB and BC. Each having a length b (see the first part of the figure part a). The bars have pin connections at A, B, and C and are joined by a spring of stiffness k. The spring is attached at the midpoints of the bars. The framework has a pin support at A and a roller support al C, and the bars are at an angle a to the horizontal.

When a vertical load P is applied at joint B (see the second part of the figure part a.) the roller support C moves to the right, the spring is stretched, and the angle of the bars decreases from a to the angle ??.

(a) Determine the angle 0 and the increase S in the distance between points A and C. Also find reactions at A and C. (Use the following data: b = 200 mm. ft = 3.2 kN/m. a = 45°. and

P = 50 N.)

(b) Repeat part (a) if a translational spring k_t= kll is added at C and a rotational spring k_r= kb-l2 is added at A (see figure pan b).

  

To determine

The angle $\theta$ and the increase $\delta$ in the distance between points A and C, also, find the reactions at A and C.

Answer to Problem 2.2.20P

The angle $\theta ={35.1}^o$

The distance between points A and C, $\delta =44.6\text{mm}$

The reaction at support C $R_C=25N$

The reaction at support A $R_A=25N$

Explanation of Solution

Given information:

The length of the rigid bars b = 200 mm

The stiffness of the spring k = 3.2 kN/m

The angle of bars from the horizontal = $\alpha ={45}^o$

The valueof load P = 50 N

  

Figure: Initial and displaced positions of the framework.

Calculation:

Let us consider the structure in its displaced position. Let us use the free body diagrams of left-hand bars and right-hand bars.

From the free body diagram of left-hand bar,

   $\begin{array}{l}{\displaystyle \sum M_B=0}\\ H_{A}h+k\frac{\delta }{2}\left(\frac{h}{2}\right)-R_A\left(\frac{L_2}{2}\right)+k_r\left(\alpha -\theta \right)=0\end{array}$

Reaction at support A,

   $R_A=\frac{2}{L_2}\left(k_1\delta h+k\frac{\delta }{2}\left(\frac{h}{2}\right)\right)+k_r\left(\alpha -\theta \right)$

From the free body diagram of right-hand bar,

   $\begin{array}{l}{\displaystyle \sum M_B=0}\\ -k\frac{\delta }{2}\left(\frac{h}{2}\right)-k_1\delta h+R_c\left(\frac{L_2}{2}\right)=0\end{array}$

   $R_C=\frac{2}{L_2}\left(k\frac{\delta }{2}\left(\frac{h}{2}\right)+k_1\delta h\right)$

From overall FBD of the beams,

   $\begin{array}{l}{\displaystyle \sum F_H}=0\\ H_A-k_1\delta =0\end{array}$

   $H_A=k_1\delta$

   $\begin{array}{l}{\displaystyle \sum F_V}=0\\ R_A+R_C=P\end{array}$

   $\begin{array}{l}{\displaystyle \sum M_A}=0\\ k_r\left(\alpha -\theta \right)-P\frac{L_2}{2}+R_c{L}_2=0\end{array}$

   $R_c=\frac{1}{L_2}\left(P\frac{L_2}{2}-k_r\left(\alpha -\theta \right)\right)$

We have two expressions for R_c, equating both of the expressions and then substitute expressions for L₂, k_r, k₁, h and $\delta$ .

   $\frac{1}{L_2}\left(P\frac{L_2}{2}-k_r\left(\alpha -\theta \right)\right)=\frac{2}{L_2}\left(k\frac{\delta }{2}\left(\frac{h}{2}\right)+k_1\delta h\right)$

Now,

   $\frac{1}{L_2}\left(P\frac{L_2}{2}-k_r\left(\alpha -\theta \right)\right)=\left(\frac{2}{L_2}\left(k\frac{2b\cos \theta -\cos \alpha }{2}\times \frac{b\sin \theta }{2}\right)+k_1\left(\left(2b\left(\cos \theta \right)-\cos \alpha \right)\right)\left(b\sin \theta \right)\right)=0$

Let us substitute the numerical values of variables and calculate the values of angle $\theta$ and the increase $\delta$ in the distance.

   $\begin{array}{l}b=200\text{mm}\\ \text{k = 3}\text{.2 kN/m}\\ \alpha \text{=}{\text{45}}^o\\ P=50N\\ k_1=0\\ k_r=0\end{array}$

   $\begin{array}{l}{L}_2=2b\cos \left(\theta \right)\\ L_1=2b\cos \left(\alpha \right)\\ \delta =L_2-L_1\\ \delta =2b\left(\cos \theta -\cos \alpha \right)\\ h=b\sin \theta \end{array}$

   $\begin{array}{l}\frac{1}{2\times 0.2\cos \theta }\left(50\times \frac{2\times 0.2\times \cos \theta }{2}-0\times \left(45-\theta \right)\right)-\\ \left(\frac{1}{2\times 0.2\times \cos \theta }\left(0\times \frac{2\times 0.2\cos \theta -\cos 45}{2}\times \frac{b\sin \theta }{2}\right)+0\times \left(\left(2b\left(\cos \theta \right)-\cos \alpha \right)\right)\left(b\sin \theta \right)\right)=0\end{array}$

Solving the above equation gives,

   $\theta ={35.1}^o$ and

   $\delta =44.6\text{mm}$

Now, let us compute the reactions at point C and A.

The reaction at point C

   $\begin{array}{l}{R}_C=\frac{2}{L_2}\left(k\frac{\delta }{2}\times \frac{h}{2}+k_1\delta h\right)\\ R_C=\frac{1}{L_C}\left(P\frac{L_2}{2}-k_r\left(\alpha -\theta \right)\right)\\ R_C=25N\end{array}$

And the reaction at point A is

   $\begin{array}{l}{R}_A=\frac{2}{L_2}\left(k_1\delta h+k\frac{\delta }{2}\left(\frac{h}{2}\right)+k_r\left(\alpha -\theta \right)\right)\\ R_A=25N\end{array}$

Conclusion: Thus, the angle $\theta ={35.1}^o$

The distance between points A and C, $\delta =44.6\text{mm}\text{.}$

The reaction at support C $R_C=25N.$

The reaction at support A $R_A=25N.$

To determine

The angle $\theta$ and the increase $\delta$ in the distance between points A and C, also, find the reactions at A and C.

Answer to Problem 2.2.20P

The angle $\theta ={43.3}^o$

The distance between points A and C, $\delta =8.19\text{mm}$

The reaction at support C $R_C=18.5N$

The reaction at support A $R_A=31.5N$

The moment reaction at point A is $M_A=1.8989\text{N}\text{.m}$

Explanation of Solution

Given information:

The length of the rigid bars b = 200 mm

The stiffness of the spring k = 3.2 kN/m

The angle of bars from the horizontal = $\alpha ={45}^o$

The value of load P = 50 N

  

Figure: Initial and displaced positions of the framework.

Calculation:

Let us consider the structure in its displaced position. Let us use the free body diagrams of left-hand bars and right-hand bars.

From the free body diagram of left-hand bar,

   $\begin{array}{l}{\displaystyle \sum M_B=0}\\ H_{A}h+k\frac{\delta }{2}\left(\frac{h}{2}\right)-R_A\left(\frac{L_2}{2}\right)+k_r\left(\alpha -\theta \right)=0\end{array}$

Reaction at support A,

   $R_A=\frac{2}{L_2}\left(k_1\delta h+k\frac{\delta }{2}\left(\frac{h}{2}\right)\right)+k_r\left(\alpha -\theta \right)$

From the free body diagram of right-hand bar,

   $\begin{array}{l}{\displaystyle \sum M_B=0}\\ -k\frac{\delta }{2}\left(\frac{h}{2}\right)-k_1\delta h+R_c\left(\frac{L_2}{2}\right)=0\end{array}$

   $R_C=\frac{2}{L_2}\left(k\frac{\delta }{2}\left(\frac{h}{2}\right)+k_1\delta h\right)$

From overall FBD of the beams,

   $\begin{array}{l}{\displaystyle \sum F_H}=0\\ H_A-k_1\delta =0\end{array}$

   $H_A=k_1\delta$

   $\begin{array}{l}{\displaystyle \sum F_V}=0\\ R_A+R_C=P\end{array}$

   $\begin{array}{l}{\displaystyle \sum M_A}=0\\ k_r\left(\alpha -\theta \right)-P\frac{L_2}{2}+R_c{L}_2=0\end{array}$

   $R_c=\frac{1}{L_2}\left(P\frac{L_2}{2}-k_r\left(\alpha -\theta \right)\right)$

We have two expressions for R_c, equating both of the expressions and then substitute expressions for L₂, k_r, k₁, h and $\delta$ .

   $\frac{1}{L_2}\left(P\frac{L_2}{2}-k_r\left(\alpha -\theta \right)\right)=\frac{2}{L_2}\left(k\frac{\delta }{2}\left(\frac{h}{2}\right)+k_1\delta h\right)$

Now,

   $\frac{1}{L_2}\left(P\frac{L_2}{2}-k_r\left(\alpha -\theta \right)\right)=\left(\frac{2}{L_2}\left(k\frac{2b\cos \theta -\cos \alpha }{2}\times \frac{b\sin \theta }{2}\right)+k_1\left(\left(2b\left(\cos \theta \right)-\cos \alpha \right)\right)\left(b\sin \theta \right)\right)=0$

Let us substitute the numerical values of variables and calculate the values of angle $\theta$ and the increase $\delta$ in the distance.

   $\begin{array}{l}b=200\text{mm}\\ \text{k = 3}\text{.2 kN/m}\\ \alpha \text{=}{\text{45}}^o\\ P=50N\\ k_1=\frac{k}{2}\\ k_r=\frac{k}{2}{b}^2\end{array}$

   $\begin{array}{l}{L}_2=2b\cos \left(\theta \right)\\ L_1=2b\cos \left(\alpha \right)\\ \delta =L_2-L_1\\ \delta =2b\left(\cos \theta -\cos \alpha \right)\\ h=b\sin \theta \end{array}$

   $\begin{array}{l}\frac{1}{2\times 0.2\cos \theta }\left(50\times \frac{2\times 0.2\times \cos \theta }{2}-\frac{3.2\times {10}^3\times {\left(0.2\right)}^2}{2}\times \left(45-\theta \right)\right)-\\ \left(\begin{array}{l}\frac{1}{2\times 0.2\times \cos \theta }\left(3.2\times \frac{2\times 0.2\cos \theta -\cos 45}{2}\times \frac{0.2\times \sin \theta }{2}\right)\\ +\frac{3.2\times {10}^3}{2}\times \left(\left(2\times 0.2\times \left(\cos \theta \right)-\cos 45\right)\right)\left(0.2\times \sin \theta \right)\end{array}\right)=0\end{array}$

Solving the above equation gives,

   $\theta ={43.3}^o$ and

   $\delta =8.19\text{mm}$

Now, let us compute the reactions at point C and A.

The reaction at point C,

   $\begin{array}{l}{R}_C=\frac{2}{L_2}\left(k\frac{\delta }{2}\times \frac{h}{2}+k_1\delta h\right)\\ R_C=\frac{1}{L_2}\left(P\frac{L_2}{2}-k_r\left(\alpha -\theta \right)\right)\\ R_C=18.5N\end{array}$

And the reaction at point A is

   $\begin{array}{l}{R}_A=\frac{2}{L_2}\left(k_1\delta h+k\frac{\delta }{2}\left(\frac{h}{2}\right)+k_r\left(\alpha -\theta \right)\right)\\ R_A=31.5N\end{array}$

Now, moment reaction at point A is,

   $\begin{array}{l}{M}_A=k_r\left(\alpha -\theta \right)\\ M_A=\frac{3.2\times {10}^3}{2}\times {\left(0.2\right)}^2\times \left(45-43.3\right)\times \frac{\pi }{180}\end{array}$

   $M_A=1.8989\text{N}\text{.m}$

Conclusion: Thus, the angle $\theta ={43.3}^o$

The distance between points A and C, $\delta =8.19\text{mm}\text{.}$

The reaction at support C $R_C=18.5N.$

The reaction at support A $R_A=31.5N.$

The moment reaction at point A is $M_A=1.8989\text{N}\text{.m}\text{.}$