Chapter 11, Problem 28P

To determine

Analyze the structure calculate the horizontal displacement of joint B.

Explanation of Solution

Given information:

The Young’s modulus E is constant and equals to $30,000\text{kips}/{\text{in}}^{\text{2}}$.

Calculation:

Sketch the free body diagram of frame as shown in Figure 1.

Find the stiffness of each member connected to joint B as follows;

$\begin{array}{c}{K}_{BA}=\frac{3}{4}\frac{I}{L}\\ =\frac{3}{4}\frac{120}{12}\\ =7.5\\ K_{BC}=\frac{I}{L}\\ =\frac{240}{24}\\ =10\end{array}$

Find the total stiffness at joint B as follows;

$\begin{array}{c}{\displaystyle \sum K_B}=K_{BA}+K_{BC}\\ =7.5+10\\ =17.5\end{array}$

Find the stiffness of each member connected to joint C as follows;

$\begin{array}{c}{K}_{CB}=\frac{I}{L}\\ =\frac{240}{12}\\ =10\\ K_{CD}=\frac{I}{L}\\ =\frac{150}{18}\\ =8.33\end{array}$

Find the total stiffness at joint C as follows;

$\begin{array}{c}{\displaystyle \sum K_C}=K_{CD}+K_{CB}\\ =8.33+10\\ =18.33\end{array}$

Find the distribution factors at joint B using the relation;

$\begin{array}{c}D{F}_{BA}=\frac{K_{BA}}{K_B}\\ =\frac{7.5}{17.5}\\ =0.43\\ D{F}_{BC}=\frac{K_{BC}}{K_B}\\ =\frac{10}{17.5}\\ =0.571\end{array}$

Find the distribution factors at joint C using the relation;

$\begin{array}{c}D{F}_{CB}=\frac{K_{CB}}{K_C}\\ =\frac{10}{18.33}\\ =0.55\\ D{F}_{CD}=\frac{K_{CD}}{K_C}\\ =\frac{8.33}{18.33}\\ =0.45\end{array}$

Refer the appendix Table A.4 to find the fixed end moments.

Find the fixed end moment at each end of the member BC as follows;

$\begin{array}{c}{\text{FEM}}_{BC}=\pm \frac{2PL}{9}\\ =\pm \frac{2\times 6\times 24}{9}\\ =\pm 32\text{kips}\cdot \text{ft}\end{array}$

Joint	A	B		C		D
Member	AB	BA	BC	CB	CD	DC
DF		0.43	0.57	0.55	0.45
FEM			-32	32
Balancing		13.76	18.24	-17.6	-14.4
CO			-8.8	9.12		-7.2
Balancing		3.78	5.02	-5.02	-4.1
CO			-2.51	2.51		-2.1
Balancing		1.08	1.43	-1.38	-1.13
CO			-0.69	0.72		-0.57
Balancing		0.3	0.39	-0.4	-0.32
CO			-0.2	0.2		-0.16
Balancing		0.09	0.11	-0.11	-0.09
		19	-19	20.06	-20.06	-10.03

Sketch the free body diagram as shown in Figure 2.

Show the members AB, BC, and CD as in Figure 3.

Taking moment about B ,

$\begin{array}{l}\sum M_B=0\\ \left(-V_{CB}\times 24\right)+\left(6\times 8\right)+\left(6\times 16\right)+20.86-19=0\\ -24V_{CB}+48+96+20.86-19=0\\ V_{CB}=6.07\text{k}\end{array}$

Summation of vertical force is equal to zero.

$\begin{array}{l}\sum F_y=0\\ -6-6+V_{CB}+V_{BC}=0\\ -6-6+6.07+V_{BC}=0\\ V_{BC}=5.93\text{kips}\end{array}$

Taking moment about A ,

$\begin{array}{l}\sum M_A=0\\ 19-\left(V_{BA}\times 12\right)=0\\ V_{BA}=1.58\text{kips}\end{array}$

Taking moment about D,

$\begin{array}{l}\sum M_D=0\\ V_{CD}\times 18-20.06-10.03=0\\ V_{CD}=1.67\text{kips}\end{array}$

Summation of horizontal force is equal to zero.

$\begin{array}{l}\sum H=0\\ -H_C+1.58+3-1.67=0\\ H_C=2.91\text{kips}\end{array}$

Find the fixed end moment at each end of the member AB as follows;

$FE{M}_{AB}=\frac{2EI}{L}\left(-3\psi \right)$

Consider the $\psi =\frac{1^{″}}{L}$

$\begin{array}{c}FE{M}_{AB}=FE{M}_{BA}=\frac{2E120}{12\times 12}\left(-3\times \frac{1}{12\times 12}\right)\\ =-1041.67\text{ft}\cdot \text{k}\\ \text{=}-\text{86}\text{.80}\text{kip}\cdot \text{ft}\end{array}$

Find the fixed end moment at each end of the member CD as follows;

$\begin{array}{c}FE{M}_{CD}=FE{M}_{DC}=\frac{2E150}{18\times 12}\left(-3\times \frac{1}{18\times 12}\right)\\ =-578.70\text{ft}\cdot \text{k}\\ \text{=}-48.22\text{kip}\cdot \text{ft}\end{array}$

Joint	A	B		C		D
Member	AB	BA	BC	CB	CD	DC
DF		0.43	0.57	0.55	0.45
FEM	-86.8	-86.8			-48.22	-48.22
Balancing	86.8	37.32	49.48	26.52	21.7
CO		43.4	13.26	24.74		10.85
Balancing		-24.36	-32.3	-13.61	-11.13
CO			-6.81	-16.15		-5.57
Balancing		2.93	3.88	8.88	7.27
CO			4.44	1.94		3.64
Balancing		-1.91	-2.53	-1.07	-0.87
CO			-0.54	-1.27		-0.44
Balancing		0.23	0.31	0.7	0.57
CO			0.35	0.16		0.29
Balancing		-0.15	-0.2	-0.09	-0.06
CO			-0.05	-0.1		-0.03
Balancing		0.02	0.03	0.06	0.04
	0	-29.32	29.32	30.75	-30.75	-39.98

Show the member AB,BC, CD, as shown in figure 4.

Consider span AB:

Take moment about B is Equal to zero.

$\begin{array}{l}\sum M_B=0\\ V_{AB}\times 12-29.32=0\\ V_{AB}=2.44\text{ kips}(\leftarrow )\end{array}$

Summation of forces along x-direction is Equal to zero.

$\begin{array}{l}+\to \sum F_x=0\\ -V_{BA}-2.44=0\\ V_{BA}=2.44\text{ kips}(\to )\end{array}$

Consider span BC:

Take moment about B is Equal to zero.

$\begin{array}{l}\sum M_B=0\\ -V_{CB}\times 24+29.32+30.7=0\\ V_{CB}=2.5\text{ kips}(\uparrow )\end{array}$

Summation of forces along y-direction is Equal to zero.

$\begin{array}{l}+\uparrow \sum F_y=0\\ V_{BC}+2.5=0\\ V_{BC}=2.5\text{ kips}(\downarrow )\end{array}$

Consider span CD.

Take moment about C is Equal to zero.

$\begin{array}{l}\sum M_C=0\\ V_{DC}\times 18-30.7-39.42=0\\ V_{DC}=3.9\text{ kips}(\leftarrow )\end{array}$

Summation of forces along x-direction is Equal to zero.

$\begin{array}{l}+\to \sum F_x=0\\ -V_{CD}-3.9=0\\ V_{CD}=3.9\text{ kips}(\to )\end{array}$

Consider the entire structure.

Summation of forces along x-direction is Equal to zero.

$\begin{array}{l}+\to \sum F_x=0\\ C_x-3.9-2.44=0\\ C_x=6.34\text{ kips}(\to )\end{array}$

Take moment about A is Equal to zero.

$\begin{array}{l}\sum M_A=0\\ -D_y\times 24-39.42+3.9\times 6+6.34\times 12=0\\ D_y=2.5\text{ kips}(\uparrow )\end{array}$

Summation of forces along y-direction is Equal to zero.

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y+2.5=0\\ A_y=2.5\text{ kips}(\downarrow )\end{array}$

Sketch the free body diagram of shear and moment as shown in Figure 5.

Sketch the deflected shape as shown in Figure 6.