Chapter 11, Problem 1P

To determine

Find the reactions and plot the shear force and bending moment diagrams.

Mention the points of inflection and mention the maximum shear and moment in each span.

Explanation of Solution

Given information:

EI is constant.

Calculation:

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

$\begin{array}{c}{K}_{BA}=\frac{I}{L_{BA}}\\ =\frac{I}{24}\\ K_{BC}=\frac{I}{L_{BC}}\\ =\frac{I}{12}\end{array}$

Find the total stiffness at joint B as follows;

$\begin{array}{c}{\displaystyle \sum K_B}=K_{BA}+K_{BC}\\ =\frac{I}{24}+\frac{I}{12}\\ =\frac{I}{8}\end{array}$

Find the distribution factors at joint B using the relation;

$\begin{array}{c}D{F}_{BA}=\frac{K_{BA}}{{\displaystyle \sum K_B}}\\ =\frac{\frac{I}{24}}{\frac{I}{8}}\\ =\frac{1}{3}\end{array}$

$\begin{array}{c}D{F}_{BC}=\frac{K_{BC}}{{\displaystyle \sum K_B}}\\ =\frac{\frac{I}{12}}{\frac{I}{8}}\\ =\frac{2}{3}\end{array}$

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

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

$\begin{array}{c}{\text{FEM}}_{AB}=-\frac{P{L}_{AB}}{8}\\ =-\frac{30\times 24}{8}\\ =-90\text{kips-ft}\end{array}$

$\begin{array}{c}{\text{FEM}}_{BA}=+\frac{P{L}_{BA}}{8}\\ =+\frac{30\times 24}{8}\\ =+90\text{kips-ft}\end{array}$

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

$\begin{array}{c}{\text{FEM}}_{BC}=-\frac{w{L}_{BC}^2}{12}\\ =-\frac{3.5\times {12}^2}{12}\\ =-42\text{kips-ft}\end{array}$

$\begin{array}{c}{\text{FEM}}_{CB}=+\frac{w{L}_{CB}^2}{12}\\ =+\frac{3.5\times {12}^2}{12}\\ =+42\text{kips-ft}\end{array}$

Show the moment distribution computations as in Figure 1.

Show the members AB and BC as in Figure 2.

Refer Figure (2),

Consider member AB:

Take moment about point A;

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ V_{BA}\left(24\right)-74-30\left(12\right)+98=0\\ 24V_{BA}+24-360=0\\ V_{BA}=14\text{kips}(\uparrow )\end{array}$

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ V_{AB}-30+V_{BA}=0\\ V_{AB}-30+14=0\\ V_{AB}=16\text{kips}(\uparrow )\end{array}$

Consider member BC:

Take moment about point B;

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ V_{CB}\left(12\right)-26-3.5\left(12\right)\left(\frac{12}{2}\right)+74=0\\ 12V_{CB}+48-252=0\\ V_{CB}=17\text{kips}(\uparrow )\end{array}$

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ V_{BC}-3.5\left(12\right)+V_{CB}=0\\ V_{BC}-42+17=0\\ V_{BC}=25\text{kips}(\uparrow )\end{array}$

Show the free body diagram of the shear force and bending moment as in Figure 3.