Module 4 · Lesson 4.2
Method of joints and method of sections
Two ways to get member forces, and how to choose between them.
Why this matters
If you need every member force, work joint by joint. If you need one member in the middle of a big truss, cutting straight through it gets you there in one step. Choosing well is the difference between five minutes and fifty.
By the end of this lesson you should be able to
- Solve a joint by resolving in two directions
- Cut a section and take moments to isolate one member
- Interpret the sign as tension or compression
In the method of joints, you isolate one joint at a time and resolve horizontally and vertically. Because a joint is a point, there is no moment equation — just two equations, so you can only solve a joint with two unknown member forces. Always assume every unknown member is in tension, pulling away from the joint. If the answer comes out negative, the member is in compression. Sticking to that habit removes almost all sign confusion.
Try it
Work joint by joint
Pick a joint, isolate it, and resolve. Start where only two member forces are unknown.
Isolate joint
Joint A has a support reaction and two members, so it is a sensible starting point.
Worked example
Member forces in a triangular truss
Given
- Span A to B = 6.0 m, apex C is 4.0 m above the chord, 3.0 m from each support
- Pin at A, roller at B
- Vertical load of 24 kN downwards at C
Find
The forces in members AB and AC, stating tension or compression.
Assumptions
- Pinned joints
- Load applied at a joint
- Member self-weight ignored
The method of sections cuts an imaginary line right through the truss, including the member you want. You then treat everything on one side as a free body. Because you now have a whole body rather than a point, you get the moment equation back — and taking moments about the point where the other two cut members meet leaves your target member as the only unknown.
Practice
In the truss above, if the apex load is increased to 40 kN, what is the force in the bottom chord AB? Give the magnitude in kN (it stays in tension).
Practice
In the triangular truss above (span 6.0 m, apex 4.0 m high), if the apex load is 18 kN, what is the force in the sloping member AC? Give the magnitude in kN.
Practice
For that same truss under the 18 kN apex load, what is the force in the bottom chord AB? Give the magnitude in kN (it is in tension).
Worked example
One cut, one member force: the method of sections
Given
- Parallel-chord truss, span 24.0 m in six panels of 4.00 m, depth 3.00 m
- Simply supported, with 60.0 kN hanging at each of the five interior bottom joints
- Wanted: the force in the bottom chord of the third panel, between 8.00 and 12.0 m
Find
The bottom chord force, without working joint by joint across the truss.
Practice
A parallel-chord truss spans 24.0 m in 4.00 m panels with a depth of 3.00 m, carrying 60.0 kN at each of the five interior bottom joints. What is the support reaction, in kN?
Practice
For that truss, take a section through the third panel and moments about the top-chord joint at 8.00 m. What is the force in the bottom chord, in kN?
Practice
If the same truss were built 4.00 m deep instead of 3.00 m, what would that bottom chord force become, in kN?
Summary
- Method of joints: two equations per joint, so start where only two forces are unknown
- Always assume tension; a negative answer means compression
- Method of sections: cut through the target member and take moments about where the other cut members meet
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint