Method of Sections

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When analyzing a truss, it is important to understand what kind of truss you are dealing with, as well as when to employ the correct method of analysis[1]. The method of sections is an alternative method to the Method of Joints for determining internal forces within a truss system.

This section will explain how to identify when the method of sections will be needed, and also how to apply the method of sections to a truss system.

Important Terms

Terms Definitions
Truss System A structural system (made with wooden,metal etc.) which consist of similar members with similar dimensional properties (ie. thicknesses) that are joined linked together at a node, also known as a joint. When external forces are applied, these members are put under tensional and compressional forces. [2]
Internal Force Forces within a member(s) that are transferred from external loads acting the structure. These forces are also known as resisting forces[3]; when a structure is in equilibrium with external loads acting on it, these are the forces (along with support reactions) that keep it that way.
Support reaction External supports (eg. rollers, pins etc.) that produce forces which are applied to the truss system in order to keep the structure in equilibrium.
Joint The intersection where 2 or more members meet in a truss system.[4]
Free body diagram Diagram demonstrating all known and unknown forces acting on the system. This also includes the direction and magnitude of the known forces.

Method of Analysis

Generally, when trying to solve for all forces in a truss, we use the Method of Joints, which goes through the truss one joint at a time solving for the member forces.The purpose of using the 'method of sections' is to be able to efficiently determine forces for specific members in a truss system. This is done by cutting through the members of interest and using Equilibrium and Compatibility equations, solve for the internal member forces.[5]


  1. The system must be stable and internally/externally determinate (check by using determinacy equations found in Truss Stability)
  2. You can only solve for 3 unknown members at a time


  1. Check for Determinacy and Stability
  2. Find the Support Reactions
  3. Use an imaginary line to cut through the sections that you want to find the forces
  4. Solve using Equilibrium and Compatibility Equations
  • hint: Taking a moment at a joint with multiple members running through it might make it easier to solve the system, since any force running through the joint cannot cause a moment on it.[6]

Special Cases

In some instances, a truss will not lend itself readily to the analysis by method of sections, when this occurs, there is a special trick that may be employed in order to enable a truss' internal forces to be easily analyzed using the method of sections.

This special case, will occur if there is one more force unknown than the number of equilibrium equations. If this is the case, once again a cut-section must be made through the truss.[7]


If we are confronted with a K-truss, shown in the image above, abide by the following procedure:

  1. Make a cut-section, labeled 1-1, vertically through the k-truss, going around the joint in the middle of the truss where the two diagonal members meet. [7]
    • This first cut is done to reduce the number of unknowns in your equilibrium equations.
  2. Calculate the moment about either point G or point B, to solve for either F-BC or F-GF, depending on how you calculated the moment.[7]
    • Calculating the moment about point G will solve for F-BC, while calculating the moment about point B will solve for F-GF
  3. Now that you have solved for one of the internal forces, create a new cut-section labeled 2-2, as seen in the image on the right. [7]
  4. Create a sum of forces in the horizontal and vertical directions, and solve the truss as found in the regular procedure. [7] Notice how you now have 3 unknowns? This makes it possible to follow the regular procedure described earlier.

Sample Problem

Find the forces in members ED,EB and AB (assume that structure is determinate and stable both externally and internally).

Sample Problem.jpg

1) Find Support Reactions

Support reactions.jpg

$ \sum M_{A} = 0\hspace{2 mm}(anticlockwise\hspace{1 mm}as\hspace{1 mm}positive) $

$ (C_{y}\hspace{1 mm})(20m)-(2 KN)(6 m)-(1 KN)(5m)= 0 $

$ (20m \hspace{1 mm}C_{y}\hspace{1 mm})-(12\hspace{1 mm}KN.m)-(5\hspace{1 mm}KN.m)=0 $

$ C_{y}= 0.85\hspace{1 mm}KN $

$ +\uparrow \sum F_{y}=0 $

$ (A_{y}\hspace{1 mm})+(C_{y}\hspace{1 mm})-(1\hspace{1 mm}KN)=0 $

$ (A_{y}\hspace{1 mm})+(0.85\hspace{1 mm}KN)-(1\hspace{1 mm}KN)=0 $

$ A_{y}= 0.15\hspace{1 mm}KN $

$ +\rightarrow \sum F_{x}=0 $

$ (A_{x}\hspace{1 mm}) + (2\hspace{1 mm}KN)=0 $

$ A_{x}=-2\hspace{1 mm}KN=2\hspace{1 mm}KN\hspace{1 mm}\leftarrow $

2) Cut the desired members in order to find unknown internal forces

Cut sections1.jpg

$ \sum M_{B} = 0\hspace{2 mm}(anticlockwise\hspace{1 mm}as\hspace{1 mm}positive) $

$ (-F_{ED}\hspace{1 mm})(6\hspace{1 mm}m)+(1\hspace{1 mm}KN)(5\hspace{1 mm}m)-(0.15\hspace{1 mm}KN)(10\hspace{1 mm}m)=0 $

$ (-6m\hspace{1 mm}F_{ED}\hspace{1 mm})+(5\hspace{1 mm}KN.m)-(1.5\hspace{1 mm}KN.m)=0 $

$ F_{ED}=0.583\hspace{1 mm}KN\hspace{1 mm}(Tension) $

$ \tan \alpha =6\hspace{1 mm}/\hspace{1 mm}5 $

$ \alpha=50.2^\circ $

$ +\uparrow \sum F_{y}=0 $

$ (-F_{EB}\hspace{1 mm}\hspace{1 mm}\sin \alpha )-(1\hspace{1 mm}KN)+(0.15\hspace{1 mm}KN)=0 $

$ (-F_{EB}\hspace{1 mm}\sin(0.583)\hspace{1 mm})-(1\hspace{1 mm}KN)+(0.15\hspace{1 mm}KN)=0 $

$ F_{EB}=-83.5\hspace{1 mm}KN $

$ F_{EB}=83.5\hspace{1 mm}KN\hspace{1 mm}(Compression) $

$ +\rightarrow \sum F_{x}=0 $

$ (-2\hspace{1 mm}KN)+(F_{AB})+(F_{EB}\hspace{1 mm}\cos \alpha)+\hspace{1 mm}(F_{ED})=0 $

$ (-2\hspace{1 mm}KN)+(F_{AB})+(-83.5\hspace{1 mm}\cos(50.2)\hspace{1 mm}KN)+(0.583\hspace{1 mm}KN)=0 $

$ F_{AB}=54.9\hspace{1 mm}KN\hspace{1 mm}(Tension) $


  1. Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.). Stamford, CT: Cengage Learning.
  7. 7.0 7.1 7.2 7.3 7.4