Last modified on 15 August 2013, at 21:35

# Statics/Method of Sections

The method of sections is another method to determine forces in members of a truss structure. In order to find unknown forces in using the method of sections, sections of the truss structure must be isolated. The net force on the entire isolated section must be zero since the isolated section does not move (if it did move it wouldn't be a statics problem). This method is often faster because instead of moving from joint to joint with the method of joints, the members of interest can be immediately isolated.

## Example 1Edit

### QuestionEdit

The truss pictured to the right is secured with pin mounts to the brown concrete block. A 100kg mass is hanging from the rightmost joint. You can ignore the weight of the truss structure. Calculate the tension or compression in member A-B.

### AnswerEdit

This problem could be solved using the method of joints, but you would have to start at the joint above the weight then solve many other joints between the first joint and member A-B. The fifth joint solved would give member A-B's loading. It is easier using the method of sections.

The method of joints isolates a joint to find unknown forces. The method of sections is the same except an entire section is isolated. It should be obvious at this point that there cannot be any net force or moment on the section, if there was the section would move.

In order to find the stress in member A-B it is convenient to isolate the section of the truss as indicated by the image to the left. In problems where the method of joints was used, a joint was isolated by cutting surrounding members. Cutting is also used to isolate a section, three of the members of the truss were cut to isolate the section in this example. Because all cut members were part of a truss, all members supported an axial force of tension or compression. After being cut, all members must have the same axial force applied to support the section.

Equilibrium equations can now be written.

$\sum F_x = 0 = -T_1 - T_2\cos(60) - T_3$

$\sum F_y = 0 = T_2\sin(60) - 981N$

The equation summing forces in the Y direction only has one unknown because all cut members except A-B are horizontal.

$\ 981 N = T_2\sin(60)$

$\ T_2 = 1130 N$

Because $T_2$ is positive, member A-B is in 1130N of tension.

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