How to Use the Method of Sections?

Sample Solution

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

Bar Forces BD, BC, and AC

Strategically Place the Cutting Plane

We start by placing a cutting plane through a maximum of three members. We do this because we can only apply three equilibrium equations for 2D trusses. It ensures we only have a maximum of three unknowns.

As a demonstration, let's solve internal bar forces \(F_{BD}\), \(F_{BC}\), and \(F_{AC}\). We place a cutting plane that will cut these members. It will divide the truss into two free-body diagrams (FBD), as shown. The cut members will have their internal forces exposed.

All Parts of the Structure Must Be In Equilibrium

Since we have divided the truss into two parts, we can choose either one or the other part to solve for bar forces \(F_{BD}\), \(F_{BC}\), and \(F_{AC}\). In this example, we'll choose the left-side part.

Let's start by solving \(F_{BD}\). We do that by applying the moment equilibrium equation at \(C\) because it is at this point where bar forces \(F_{BC}\) and \(F_{AC}\) will intersect. A moment at \(C\) will cancel these bar forces. Usually, this is the strategy for solving bar forces – find the point where two or more forces will meet and apply \(\sum M=0\) at the intersection point to find the unknown bar force.

Assuming all bar forces are in tension, we compute the bar forces:

\(\sum{M}_C=0]\circlearrowright_+\)

\(6.33(3)+20(2.6)+F_{BD}(2.6)=0\)

\(F_{BD}=-27.3kN\)

We can continue using other equilibrium equations to solve for the other bar forces.

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

\(6.33-F_{BC}\sin60=0\)

\(F_{BC}=7.31kN\)

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

\(7.31\cos60+20+(-27.3)+F_{AC}-20=0\)

\(F_{AC}=23.65kN\)

If our answer is negative, we had the wrong assumed direction for said bar force. In this case, the direction of \(F_{BD}\) should be the other way around.

Part of the analysis is also to know whether these axial forces are either tension \(T\) or compression \(C\). In this case, the two forces \(F_{AC}\) and \(F_{BC}\) cause their respective members to elongate; So, the internal bar forces in these members are in tension. If these two end forces cause the member to compress, such as \(F_{BD}\), it is a compressive force.

When we finally solve for the unknowns, we can proceed to find another section and repeat the process.

Full Solution

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

Bar Force AB

Let's place a cut section through members \(AB\) and \(AC\) and choose the left side. We only have one unknown: \(F_{AB}\). Assuming it is a tensile force, let's solve for it:

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

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

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

The computed force \(F_{AB}\) should be acting toward the joint, meaning it is a compressive force.

Bar Force DF, EF, and EG

Let's place a cut section through members \(DF\), \(EF\), and \(EG\) and choose the right side. We have three unknowns: \(F_{DF}\), \(F_{EF}\), and \(F_{EG}\). Assuming all are tensile forces, let's solve these:

\(\sum{M}_E=0]\circlearrowleft_+\)

\(F_{DF}(2.6)-10(3)=0\)

\(F_{DF}=11.54kN (T)\)

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

\(10+F_{EF}\sin60=0\)

\(F_{EF}=-11.55kN (C)\)

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

\(-(-11.55)\cos60-11.54-F_{EG}=0\)

\(F_{EG}=-5.77kN (C)\)

The computed forces \(F_{EF}\) and \(F_{EG}\) should be acting toward the joint, meaning these are compressive forces.

Bar Force FG

Let's place a cut section through members \(FG\) and \(EG\) and choose the right side. We only have one unknown: \(F_{FG}\). Assuming it is a tensile force, let's solve for it:

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

\(F_{FG}\sin60-10=0\)

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

Bar Force DE and CE

Let's place a cut section through members \(DF\), \(DE\), and \(CE\) and choose the right side. We have two unknowns: \(F_{DE}\) and \(F_{CE}\). Assuming all are tensile forces, let's solve these:

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

\(43.67+F_{DE}\sin60-10=0\)

\(F_{DE}=-38.88kN (C)\)

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

\(-(-38.88\cos60)-11.54-F_{CE}=0\)

\(F_{CE}=7.9kN (T)\)

The computed force \(F_{DE}\) should be acting toward the joint, meaning it is a compressive force.

Bar Force CD

Let's place a cut section through members \(BD\), \(CD\), and \(CE\) and choose the right side. We have only one unknown: \(F_{CD}\). Assuming it is a tensile force, let's solve it:

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

\(43.67-10-F_{CD}\sin60=0\)

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

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 joints also.