Chapter 12, Problem 12.1P

To determine

The forces in each member of the truss.

Answer to Problem 12.1P

The forces in each member of the truss are shown below.

The force in member AE is, $2.08\text{k}\left(\text{Compression}\right)$.

The force in member BF is, $2.08\text{k}\left(\text{Tension}\right)$.

The force in member EF is, $1.67\text{k}\left(\text{Compression}\right)$.

The force in member AB is, $1.67\text{k}\left(\text{Tension}\right)$.

The force in member CE is, $2.08\left(\text{Compression}\right)$.

The force in member BD is, $2.08\text{k}\left(\text{Tension}\right)$.

The force in member DE is, $1.67\text{k}\left(\text{Compression}\right)$.

The force in member BC is, $1.67\text{k}\left(\text{Tension}\right)$.

The force in member AF is, $4.25\text{k}\left(\text{Compression}\right)$.

The force in member BE is, $2.5\text{k}\left(\text{Compression}\right)$.

The force in member CD is, $5.25\text{k}\left(\text{Compression}\right)$.

Explanation of Solution

Check the determinacy of the truss as shown below.

   $m+r-2j=0$

Here, number of members are $m$, number of reactions are $r$ and number of joints are $j$.

Substitute $11$ for $m$, $3$ for $r$ and $6$ for $j$.

   $\begin{array}{l}11+3-2\times 6>0\\ 2>0\end{array}$

Hence, the given truss is statically indeterminate by $2$ degree.

The following diagram shows the free body diagram of the truss.

Figure-(1)

Write the equation for moment about A.

   $\begin{array}{l}\Sigma M_A=0\\ \left(C_y\times 16\text{ft}\right)-\left(4\text{k}\times 16\text{ft}\right)-\left(5\text{k}\times 8\text{ft}\right)=0\\ C_y\times 16\text{ft}=104\text{k}\cdot \text{ft}\\ C_y=6.5\text{k}\end{array}$

Here, vertical reaction at C is $C_y$.

Write the equation for sum of vertical forces.

   $\begin{array}{l}\Sigma F_y=0\\ A_y+C_y-3\text{k}-4\text{k}-5\text{k}=0\\ A_y+C_y=12\text{k}\end{array}$ ...... (I)

Here, vertical reaction at A is $A_y$.

Substitute $6.5\text{k}$ for $C_y$ in Equation (I).

   $\begin{array}{l}6.5\text{k}+A_y=12\text{k}\\ A_y=\left(12-6.5\right)\text{k}\\ A_y=5.5\text{k}\end{array}$

Consider the force in diagonal members FB is in tension and AE is in compression.

   $F_{FB}=F_{AE}$

Here, force in member AE and FB are $F_{AE}$ and $F_{FB}$.

Use the method of section and cut the section as shown in figure below and calculate the forces in the members.

Figure-(2)

Calculate the angles as shown below.

   $\begin{array}{c}\cos \theta =\frac{3}{5}\\ \sin \theta =\frac{4}{5}\end{array}$

Here, the angle between the members FA and FB is $\theta$.

Consider the joint F.

Write the Equation for sum of vertical forces.

   $\begin{array}{l}\Sigma F_y=0\\ 5.5\text{k}-3\text{k}-F_{FB}\cos \theta -F_{AE}\cos \theta =0\end{array}$

Substitute $\frac{3}{5}$ for $\cos \theta$.

   $\begin{array}{l}5.5\text{k}-3\text{k}-2F_{FB}\times \frac{3}{5}=0\\ 2F_{FB}\times \frac{3}{5}=2.5\text{k}\\ F_{FB}=2.08\text{k}\left(\text{Tension}\right)\\ F_{AE}=2.08\text{k}\left(\text{Compression}\right)\end{array}$

Write the equation for the sum of horizontal forces.

   $\begin{array}{l}\Sigma F_x=0\\ F_{FE}+F_{FB}\sin \theta =0\end{array}$

Here, force in member FE is $F_{FE}$.

Substitute $2.08\text{k}$ for $F_{FB}$ and $\frac{4}{5}$ for $\sin \theta$.

   $\begin{array}{l}{F}_{FE}+\left(2.08\text{k}\times \frac{4}{5}\right)=0\\ F_{FE}=1.67\text{k}\left(\text{Compression}\right)\end{array}$

Consider joint A.

Figure-(3)

Calculate the angles as shown below.

   $\begin{array}{c}\cos \theta =\frac{4}{5}\\ \sin \theta =\frac{3}{5}\end{array}$

Write the equation for sum of vertical forces.

   $\begin{array}{l}\Sigma F_y=0\\ F_{FA}+F_{AE}\sin \theta -5.5\text{k}=0\end{array}$

Substitute $2.08\text{k}$ for $F_{AE}$ and $\frac{3}{5}$ for $\sin \theta$.

   $\begin{array}{l}{F}_{FA}+\left(2.08\text{k}\times \frac{3}{5}\right)-5.5\text{k}=0\\ F_{FA}=4.25\text{k}\left(\text{Compression}\right)\end{array}$

Here, force in member FA is $F_{FA}$.

Write the Equation for sum of horizontal forces.

   $\begin{array}{l}\Sigma F_x=0\\ F_{AB}-F_{AE}\cos \theta =0\end{array}$

Substitute $2.08\text{k}$ for $F_{AE}$ and $\frac{4}{5}$ for $\cos \theta$.

   $\begin{array}{l}{F}_{AB}-2.08\text{k}\times \frac{4}{5}=0\\ F_{AB}=1.67\text{k}\left(\text{Tension}\right)\end{array}$

Here, force in member AB is $F_{AB}$.

Consider the force in diagonal members BD is in tension and EC is in compression.

   $F_{BD}=F_{EC}$

Here, force in member BD and EC are $F_{BD}$ and $F_{EC}$.

Use the method of section method and cut the section as shown in figure below and calculate the forces in the members.

Figure-(4)

Calculate the angles as shown below.

   $\begin{array}{c}\cos \theta =\frac{4}{5}\\ \sin \theta =\frac{3}{5}\end{array}$

Write the Equation for sum of vertical forces.

   $\begin{array}{l}\Sigma F_y=0\\ 5.5-3-5+F_{BD}\sin \theta +F_{EC}\sin \theta =0\\ 2F_{BD}\sin \theta =2.5\end{array}$

Substitute $\frac{3}{5}$ for $\sin \theta$.

   $\begin{array}{l}2F_{BD}\times \frac{3}{5}=2.5\\ F_{BD}=2.08\text{k}\left(\text{Tension}\right)\\ F_{CE}=2.08\left(\text{Compression}\right)\end{array}$

Here, force in the member BD and CE are $F_{BD}$ and $F_{CE}$.

Write the equation for moment about B.

   $\left(F_{ED}\times 6+F_{EC}\cos \theta \times 6+4\times 8-6.5\times 8\right)\text{k}\cdot \text{ft}=0$

Here, force in the member ED is $F_{ED}$.

Substitute $\frac{4}{5}$ for $\cos \theta$ and $2.08\text{k}$ for $F_{EC}$.

   $\begin{array}{l}\left(F_{ED}\times 6+2.08\times \frac{4}{5}\times 6+4\times 8-6.5\times 8\right)\text{k}\cdot \text{ft}=0\\ F_{ED}\times 6\text{ft}=10.01\text{k}\cdot \text{ft}\\ F_{ED}=1.67\text{k}\left(\text{Compression}\right)\end{array}$

Write the Equation for moment about E.

   $\left(F_{BC}\times 6+F_{BD}\cos \theta \times 6-6.5\times 8+4\times 8\right)=0$

Here, force in the member BC is $F_{BC}$.

Substitute $\frac{4}{5}$ for $\cos \theta$ and $2.08\text{k}$ for $F_{BD}$.

   $\begin{array}{l}\left(F_{BC}\times 6+2.08\text{k}\times \frac{4}{5}\times 6-6.5\times 8+4\times 8\right)\text{k}\cdot \text{ft}=0\\ F_{BC}\times 6\text{ft}=10.01\text{k}\cdot \text{ft}\\ F_{BC}=1.67\text{k}\left(\text{Tension}\right)\end{array}$

Write the Equation for sum of vertical forces.

   $\begin{array}{l}\Sigma F_y=0\\ F_{BE}+F_{FB}\sin \theta +F_{BD}\sin \theta =0\end{array}$

Substitute $\frac{3}{5}$ for $\sin \theta$, $2.08\text{k}$ for $F_{FB}$ and $2.08\text{k}$ for $F_{BD}$.

   $\begin{array}{l}{F}_{BE}+2.08\text{k}\times \frac{3}{5}+2.08\text{k}\times \frac{3}{5}=0\\ F_{BE}+1.25\text{k}+1.25\text{k}=0\\ F_{BE}=2.5\text{k}\left(\text{Compression}\right)\end{array}$

Here, force in the member BE is $F_{BE}$.

Consider the joint C.

Angles will be same as calculated at joint B.

Figure-(5)

Write the Equation for sum of vertical forces.

   $\begin{array}{l}\Sigma F_y=0\\ F_{CD}+F_{EC}\sin \theta -6.5\text{k}=0\end{array}$

Here, force in the member CD is $F_{CD}$.

Substitute $\frac{3}{5}$ for $\sin \theta$, and $2.08\text{k}$ for $F_{EC}$.

   $\begin{array}{l}{F}_{CD}+\left(2.08\text{k}\times \frac{3}{5}\right)-6.5\text{k}=0\\ F_{CD}=5.25\text{k}\left(\text{Compression}\right)\end{array}$

Conclusion:

Therefore, the forces in each member of the truss are shown below.

The force in member AE is, $2.08\text{k}\left(\text{Compression}\right)$.

The force in member BF is, $2.08\text{k}\left(\text{Tension}\right)$.

The force in member EF is, $1.67\text{k}\left(\text{Compression}\right)$.

The force in member AB is, $1.67\text{k}\left(\text{Tension}\right)$.

The force in member CE is, $2.08\left(\text{Compression}\right)$.

The force in member BD is, $2.08\text{k}\left(\text{Tension}\right)$.

The force in member DE is, $1.67\text{k}\left(\text{Compression}\right)$.

The force in member BC is, $1.67\text{k}\left(\text{Tension}\right)$.

The force in member AF is, $4.25\text{k}\left(\text{Compression}\right)$.

The force in member BE is, $2.5\text{k}\left(\text{Compression}\right)$.

The force in member CD is, $5.25\text{k}\left(\text{Compression}\right)$.