Chapter 9, Problem 38P

To determine

Find the reactions and all bar forces for the truss.

Explanation of Solution

Given information:

EI is constant.

The Young’s modulus E of the beam is $30,000\text{kips/in}{\text{.}}^2$.

The area of all the bars are $2{\text{in}}^2$.

Calculation:

Consider the force in the member AC and the horizontal reaction at support B are the redundant.

Remove the redundant to get the released structure.

Sketch the released structure with external loading as shown in Figure 1.

Refer Figure 1.

Consider the force in the member AC as zero.

Find the value of the angle $\theta$ as follows:

$\begin{array}{l}\tan \theta =\frac{15}{20}\\ \theta ={\tan }^{-1}\left(\frac{15}{20}\right)\\ \theta =36.869°\end{array}$

Find the support reaction as follows:

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ -60\times 15+B_{y1}\times 20=0\\ B_{y1}=\frac{900}{20}\\ B_{y1}=45\text{kips}\end{array}$

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_{y1}+B_{y1}=0\\ A_{y1}=-B_{y1}\\ A_{y1}=-45\text{kips}\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_{x1}+60=0\\ A_{x1}=-60\text{kips}\end{array}$

Consider support A.

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_{y1}+F_{AD1}=0\\ F_{AD1}=-A_{y1}\\ F_{AD1}=-\left(-45\right)\text{kips}\\ F_{AD1}=45\text{kips}\left(\text{T}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_{x1}+F_{AB1}=0\\ F_{AB1}=-A_{x1}\\ F_{AB1}=-\left(-60\right)\text{kips}\\ F_{AB1}=60\text{kips}\left(\text{T}\right)\end{array}$

Consider joint C.

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{CD1}+60=0\\ F_{CD1}=60\text{kips}\left(\text{T}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{BC1}=0\end{array}$

Consider joint D,

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{CD1}+F_{BD1}\cos \theta =0\\ F_{CD1}+F_{BD1}\cos \left(36.869°\right)=0\\ 60+0.8F_{BD1}=0\\ F_{BD1}=75\text{kips}\left(\text{T}\right)\end{array}$

Sketch the P system for ${\Delta }_{10}$ and ${\Delta }_{20}$ as shown in Figure 2.

Sketch the released structure with unit member force in member AC as shown in Figure 3.

Refer Figure 4.

Consider the force in the member AC is 1 kips.

The support reactions at A and B are zero as no external load acts on the truss.

Consider support A.

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{AC}\cos \theta +F_{AB}=0\\ 1\cos \left(36.869°\right)+F_{AB}=0\\ F_{AB}=-0.8\text{kips}\\ F_{AB}=0.8\text{kips}\left(\text{C}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{AC}\sin \theta +F_{AD}=0\\ 1\sin \left(36.869°\right)+F_{AD}=0\\ F_{AD}=-0.6\text{kips}\\ F_{AD}=0.6\text{kips}\left(\text{C}\right)\end{array}$

Consider joint C.

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ -F_{CA}\cos \theta -F_{CD}=0\\ -\cos \left(36.869°\right)-F_{CD}=0\\ F_{CD}=-0.8\text{kips}\\ F_{CD}=0.8\text{kips}\left(\text{C}\right)\end{array}$

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -F_{CA}\sin \theta -F_{CB}=0\\ -\sin \left(36.869°\right)-F_{CB}=0\\ F_{CB}=-0.6\text{kips}\\ F_{CB}=0.6\text{kips}\left(\text{C}\right)\end{array}$

Consider joint D,

Apply Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{CD}+F_{BD}\cos \theta =0\\ -0.8+F_{BD}\cos \left(36.869°\right)=0\\ -0.8+0.8F_{BD}=0\\ F_{BD}=1\text{kips}\left(\text{T}\right)\end{array}$

Sketch the Q system for ${\Delta }_{10}$ as shown in Figure 4.

Sketch the Q system for ${\Delta }_{20},{\delta }_{21},{\delta }_{22}$ and P system for ${\delta }_{22}$ as shown in Figure 5.

Refer Figure 5.

Find the support reaction at A as follows:

Applying Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x-1=0\\ A_x=1\text{kips}\end{array}$

There is no external load at joint D and C and no vertical reaction at A and B.

The force in all the members of the truss except AB is zero.

Consider joint A.

Applying Equation of Equilibrium,

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x+F_{AB2}=0\\ 1+F_{AB2}=0\\ F_{AB2}=-1\text{kips}\\ F_{AB2}=-1\text{kips}\end{array}$

Find the deflection ${\Delta }_{10}$ as follows:

$\begin{array}{c}{\Delta }_{10}=\frac{F_Q{F}_{P}L}{AE}\\ =\left[\begin{array}{l}\frac{\left(-0.8\right)\times 60\times \left(20\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}\right)}{2\times 30,000}\times 2\\ +\frac{\left(-0.6\right)\times 45\times \left(15\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}\right)}{2\times 30,000}+\frac{1\times \left(-75\right)\times \left(25\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}\right)}{4\times 30,000}\end{array}\right]\\ =-0.6525\text{in}\text{.}\end{array}$

Find the deflection ${\Delta }_{20}$ as follows:

$\begin{array}{c}{\Delta }_{20}=\frac{F_Q{F}_{P}L}{AE}\\ =\frac{-1\times 60\times \left(20\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}\right)}{2\times 30,000}\\ =-0.24\text{in}\text{.}\end{array}$

Find the deflection ${\delta }_{11}$ as follows:

$\begin{array}{c}{\delta }_{11}=\frac{F_Q^{2}L}{AE}\\ =\left[\begin{array}{l}\frac{{\left(-0.8\right)}^2\times \left(20\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}\right)}{2\times 30,000}\times 2+\frac{{\left(-0.6\right)}^2\times \left(15\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}\right)}{2\times 30,000}\\ +\frac{1^2\times \left(25\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}\right)}{2\times 30,000}+\frac{1^2\times \left(25\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}\right)}{4\times 30,000}\end{array}\right]\\ =0.0148\text{in}\text{.}\end{array}$

Find the deflection ${\delta }_{21}$ as follows:

$\begin{array}{c}{\delta }_{21}=\frac{F_Q{F}_{P}L}{AE}\\ =\frac{\left(-1\right)\left(-0.8\right)\times \left(20\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}\right)}{2\times 30,000}\\ =0.0032\text{in}\text{.}\end{array}$

Find the deflection ${\delta }_{12}$ as follows:

$\begin{array}{c}{\delta }_{12}=\frac{F_Q{F}_{P}L}{AE}\\ =\frac{\left(-0.8\right)\left(-1\right)\times \left(20\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}\right)}{2\times 30,000}\\ =0.0032\text{in}\text{.}\end{array}$

Find the deflection ${\delta }_{22}$ as follows:

$\begin{array}{c}{\delta }_{22}=\frac{F_Q{F}_{P}L}{AE}\\ =\frac{\left(-1\right)\left(-1\right)\times \left(20\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}\right)}{2\times 30,000}\\ =0.004\text{in}\text{.}\end{array}$

Show the compatibility Equation for ${\Delta }_{1,ACREL}$ as follows:

$\begin{array}{l}{\Delta }_{1,ACREL}=0\\ {\Delta }_{10}+{\delta }_{11}{X}_1+{\delta }_{12}{X}_2=0\\ -0.6525+0.0148X_1+0.0032X_2=0\\ 0.0148X_1+0.0032X_2=0.6525\end{array}$        (1)

Show the compatibility Equation for ${\Delta }_{BX}$ as follows:

$\begin{array}{l}{\Delta }_{BX}=0\\ {\Delta }_{20}+{\delta }_{21}{X}_1+{\delta }_{22}{X}_2=0\\ -0.24+0.0032X_1+0.004X_2=0\\ 0.0032X_1+0.004X_2=0.24\end{array}$        (2)

Solve Equation (1) and (2).

$X_1=37.62\text{kips}$

$X_2=29.9\text{kips}$

Find the reaction and bar forces in the members of the truss as shown in 6 and 7.

Apply the sum shown in Figure 6 to get the final reactions and member forces shown in Figure 7.