How to Use the Method of Joints?

Sample Solution

The following section shows how to use the method of joints:

Joint A

We first start analyzing the free-body diagram of a joint with only two unknown forces. The strategy is to use the equilibrium equations for each free-body component; For 2D trusses, we can only use two. One joint that meets the condition is joint \(A\).

Let's identify the forces acting on it: the reaction components: \(A_h\) and \(A_v\), and joint connection forces \(F_{AB}\) and \(F_{AC}\). In terms of the joint force's direction, we assume it. Let's say it acts away from the joint. We can solve the joint connection forces by applying the equilibrium equations as follows:

\(\sum{F_v}=0]\space{\uparrow_+}\)

\(6.33+F_{AB}\sin60=0\)

\(F_{AB}=-7.31kN\)

\(\sum{F_h}=0]\space{\leftarrow_+}\)

\(20+7.31\cos60-F_{AC}=0\)

\(F_{AC}=23.66kN\)

As you can see, we have solved for the joint connection force \(F_{AB}\) and \(F_{AC}\). If the answer is negative, we have the wrong assumption, and the correct direction is its opposite. In this case, \(F_{AB}\) should be towards the joint.

Member AB

Now that we have forces \(F_{AB}\) and \(F_{AC}\), we can analyze the adjacent free body diagrams, such as member \(AB\) and \(AC\), and determine the internal bar force for each member.

Let's look at member \(AB\). Using Newton's Third Law, \(F_{AB}\) on the joint must exert a force of equal magnitude and opposite direction on the connecting end of member \(AB\). Now, for the whole member to be in equilibrium, it must have a force \(F_{BA}\) that is equal in magnitude but opposite in direction to \(F_{AB}\) at the other end of the member. 

Since \(AB\) is a straight member, the bar force is the computed \(7.31kN\). At this point, we now know one of the internal forces of the truss member.

Part of the analysis is also to know whether this axial force is either tension \(T\) or compression \(C\). In this case, the two forces \(F_{AB}\) and \(F_{BA}\) cause member \(AB\) to shorten; So, the internal bar force in member \(AB\) is a compressive axial force. If these two end forces cause the member to elongate, it is a tensile force.

After analyzing the member, we repeat the process of transferring the force from the member to the next adjacent joint (with at most two unknowns) and repeat what we did in the first step.

Full Solution

This section shows the complete solution of the internal bar forces using the method of joints:

Joint B

From Member AB, let's transfer the bar force on its end and proceed with Joint B. We have two unknowns: \(F_{BC}\) and \(F_{BD}\). Let's solve these:

\(\sum{F_v}=0]\space{\uparrow_+}\)

\(7.31\sin60-F_{BC}\sin60=0\)

\(F_{BC}=7.31kN (T)\)

\(\sum{F_h}=0]\space{\leftarrow_+}\)

\(7.31\cos60+7.31\cos60+20-F_{BD}=0\)

\(F_{BD}=27.31kN (C)\)

After solving for these bar forces, we would transfer these joint connection forces to their adjacent member and analyze them again; however, since our example consists of straight members, we can already interpret the bar force from the joints; hence, there is no need to analyze each member separately. Starting at this point, we would only investigate the joints after solving the joint unknowns.

Joint C

Now that we know bar force \(F_{CA}=F_{AC}\) and \(F_{CB}=F_{BC}\), let's move to adjacent Joint C. We have two unknowns: \(F_{CD}\) and \(F_{CE}\). Let's solve these:

\(\sum{F_v}=0]\space{\uparrow_+}\)

\(7.31\sin60+F_{CD}\sin60-40=0\)

\(F_{CD}=38.88kN (T)\)

\(\sum{F_h}=0]\space{\leftarrow_+}\)

\(F_{CE}+38.88\cos60-23.66-7.31\cos60=0\)

\(F_{CE}=7.88kN (T)\)

Joint G

At this point, we have solved six bar forces out of eleven. From Joint C, we can continue to its adjacent joints; however, to demonstrate that we can start anywhere, let's go to Joint G. We have two unknowns: \(F_{GF}\) and \(F_{GE}\). Let's solve these:

\(\sum{F_v}=0]\space{\uparrow_+}\)

\(F_{GF} \sin 60-10=0\)

\(F_{GF}=11.55kN (T)\)

\(\sum{F_h}=0]\space{\leftarrow_+}\)

\(F_{GE}-11.55\cos 60=0\)

\(F_{GE}=5.78kN (C)\)

Joint F

Now that we know bar force \(F_{FG}=F_{GF}\), let's move to adjacent Joint F. We have two unknowns: \(F_{FD}\) and \(F_{FE}\). Let's solve these:

\(\sum{F_v}=0]\space{\uparrow_+}\)

\(F_{FE}\sin60-11.55\sin60=0\)

\(F_{FE}=11.55kN (C)\)

\(\sum{F_h}=0]\space{\leftarrow_+}\)

\(11.55\cos60+11.55\cos60-F_{FD}=0\)

\(F_{FD}=11.55kN (T)\)

Joint E

Now that we know bar force \(F_{EF}=F_{FE}\) and \(F_{EG}=F_{GE}\), let's move to adjacent Joint E. We have only one unknown: \(F_{DE}\). Let's solve it:

\(\sum{F_v}=0]\space{\uparrow_+}\)

\(F_{ED}\sin60+43.67-11.55\sin60=0\)

\(F_{ED}=-38.88 (C)\)

Check Your Results

When we do the method of joints, typically, the number of joints we need to analyze would be the number of joints in the truss less one.

The joint that was not analyzed is a great way to double-check our answers. The values we've computed must comply with the equilibrium principle.

Joint D

To illustrate, let's analyze the final joint, joint D, and see if our answers are correct. Applying the equilibrium principle:

\(\sum{F_v}=0]\space{\uparrow_+}\)

\(38.88\sin60-38.88\sin60=0\)

\(0=0\)

\(\sum{F_h}=0]\space{\leftarrow_+}\)

\(27.31+11.55-38.88\cos60-38.88\cos60=0\)

\(0=0\)

Summarise Your Results

We have finally solved all 11 bar forces! After analysis, it is always a good idea to summarise your results using a table. When summarizing, it's always great to categorize your results for easy reference (top chords, bottom chords, and web members).

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We can solve this truss example using the method of sections also.