# Question: 6–3. Determine the force in each member of the truss. State if the members are in tension or compression. –Free Chegg Question Answer

6–3. Determine the force in each member of the truss. State if the members are in tension or compression.

Transcribed text From Image: Determine the force in each member of the truss. Stale if the members are in tension or compression. Prob.6-3

Answer:

## Step-by-step

### Step 1 of 2

Consider the joint A.

Write the force equilibrium along y direction.

\begin{array}{l}\\\sum {{F_y}\;{\rm{ = }}\;{\rm{0}}} \;\\\\\frac{4}{5}\left( {{F_{AC}}} \right) – \frac{{12}}{{13}}\left( {130} \right) = 0\\\\\left( {{F_{AC}}} \right) = 150\;{\rm{lb}}\left( {{\rm{compression}}} \right)\\\end{array}∑Fy​=054​(FAC​)−1312​(130)=0(FAC​)=150lb(compression)​

Here, {F_{AC}}FAC​ is the force in member ACAC .

Write the force equilibrium along x direction.

\begin{array}{l}\\\sum {{F_x}\;{\rm{ = }}\;{\rm{0}}} \;\\\\{F_{AB}} – \frac{3}{5}{F_{AC}} – \frac{5}{{13}}\left( {130} \right) = 0\\\\{F_{AB}} – \frac{3}{5}\left( {150} \right) – \frac{5}{{13}}\left( {130} \right) = 0\\\\{F_{AB}} = 140\;{\rm{lb}}\left( {{\rm{tension}}} \right)\\\end{array}∑Fx​=0FAB​−53​FAC​−135​(130)=0FAB​−53​(150)−135​(130)=0FAB​=140lb(tension)​

Here, {F_{AB}}FAB​ is the force in member ABAB .

Consider the joint B.

Write the force equilibrium along x direction.

\begin{array}{l}\\\sum {{F_x}\;{\rm{ = }}\;{\rm{0}}} \;\\\\{F_{BD}} – {F_{AB}} = 0\\\\{F_{BD}} – \left( {140} \right) = 0\\\\{F_{BD}} = 140\;{\rm{lb}}\left( {{\rm{tension}}} \right)\\\end{array}∑Fx​=0FBD​−FAB​=0FBD​−(140)=0FBD​=140lb(tension)​

Here, {F_{BD}}FBD​ is the force in member BCBC .

Write the force equilibrium in vertical direction.

\begin{array}{l}\\\sum {{F_y}\;{\rm{ = }}\;{\rm{0}}} \;\\\\{F_{BC}} = 0\;{\rm{lb}}\\\end{array}∑Fy​=0FBC​=0lb​

Force in member AC,AB, BD, BC is {\bf{150}}\;{\bf{lb}}\left( {{\bf{compression}}} \right)150lb(compression) , {\bf{140}}\;{\bf{lb}}\left( {{\bf{tension}}} \right)140lb(tension) , {\bf{140}}\;{\bf{lb}}\left( {{\bf{tension}}} \right)140lb(tension) , {\bf{0}}0 respectively.
Explanation | Common mistakes | Hint for next step

Using the equilibrium of force, forces in each members are determined.

### Step 2 of 2

Consider the following diagram at joint C.

Write the force equilibrium along y direction.

\begin{array}{l}\\\sum {{F_y}} = 0\\\\\frac{4}{5}\left( {{F_{CD}}} \right) – \frac{4}{5}\left( {{F_{AC}}} \right) = 0\\\\\frac{4}{5}\left( {{F_{CD}}} \right) – \frac{4}{5}\left( {150} \right) = 0\\\\{F_{CD}} = 150\;{\rm{lb}}\left( {{\rm{tension}}} \right)\\\end{array}∑Fy​=054​(FCD​)−54​(FAC​)=054​(FCD​)−54​(150)=0FCD​=150lb(tension)​

Here, force in member CD is {F_{CD}}FCD​ .

Write the force equilibrium along x direction.

\begin{array}{l}\\\sum {{F_x}} = 0\\\\ – {F_{CE}} + \frac{3}{5}\left( {{F_{AC}}} \right) + \frac{3}{5}\left( {{F_{CD}}} \right) = 0\\\\ – {F_{CE}} + \frac{3}{5}\left( {150} \right) + \frac{3}{5}\left( {150} \right) = 0\\\\{F_{CE}} = 180\;{\rm{lb}}\left( {{\rm{compression}}} \right)\\\end{array}∑Fx​=0−FCE​+53​(FAC​)+53​(FCD​)=0−FCE​+53​(150)+53​(150)=0FCE​=180lb(compression)​

Here, force in member CE is {F_{CE}}FCE​ .

Draw the forces at joint D.

Write the force equilibrium along y direction.

\begin{array}{l}\\\sum {{F_y}} = 0\\\\\left( {{F_{DE}}} \right) – \frac{4}{5}\left( {{F_{CD}}} \right) = 0\\\\\left( {{F_{DE}}} \right) – \frac{4}{5}\left( {150} \right) = 0\\\\{F_{DE}} = 120\;{\rm{lb}}\left( {{\rm{compression}}} \right)\\\end{array}∑Fy​=0(FDE​)−54​(FCD​)=0(FDE​)−54​(150)=0FDE​=120lb(compression)​

Here, force in member DE is {F_{DE}}FDE​ .

Write the force equilibrium along x direction.

\begin{array}{l}\\\sum {{F_x}} = 0\\\\{F_{DF}} – \frac{3}{5}\left( {{F_{CD}}} \right) – {F_{BD}} = 0\\\\{F_{DF}} = 230\;{\rm{lb}}\left( {{\rm{tension}}} \right)\\\end{array}∑Fx​=0FDF​−53​(FCD​)−FBD​=0FDF​=230lb(tension)​

Here, force in member DF is {F_{DF}}FDF​ .

Consider the forces at joint D.

Write the force equilibrium along x direction.

\begin{array}{l}\\\sum {{F_x}} = 0\\\\{F_{CE}} – \frac{3}{5}\left( {{F_{EF}}} \right) = 0\\\\\left( {180} \right) – \frac{3}{5}\left( {{F_{EF}}} \right) = 0\\\\{F_{EF}} = 300\;{\rm{lb}}\left( {{\rm{compression}}} \right)\\\end{array}∑Fx​=0FCE​−53​(FEF​)=0(180)−53​(FEF​)=0FEF​=300lb(compression)​

Here, force in member EF is {F_{EF}}FEF​ .

The force in member CD, CE,DE, DF,EF is {\bf{150}}\;{\bf{lb}}\left( {{\bf{tension}}} \right)150lb(tension) , {\bf{180}}\;{\bf{lb}}\left( {{\bf{compression}}} \right)180lb(compression) , {\bf{120}}\;{\bf{lb}}\left( {{\bf{compression}}} \right)120lb(compression) , {\bf{230}}\;{\bf{lb}}\left( {{\bf{tension}}} \right)230lb(tension) , {\bf{300}}\;{\bf{lb}}\left( {{\bf{compression}}} \right)300lb(compression) respectively.
Explanation

Using force equilibrium conditions, the force in each member is calculated.