Frames and Trusses

NCEES FE Civil spec: Statics, D. Frames and trusses. FE Reference Handbook: Statics chapter, statically determinate trusses.

Trusses are assemblies of straight members joined at pins, loaded only at the joints, so every member is a two-force member in pure tension or compression. Frames contain at least one multi-force member, so you take them apart and analyze each piece. Both appear on almost every FE Civil exam.

What the NCEES spec covers

Statics section D, "Frames and trusses", covers member forces by the method of joints and the method of sections, zero-force members, and pin forces in simple frames. Structural Engineering later revisits trusses for deflection and determinacy, but the force analysis is tested here.

Key concepts

Method of joints. Draw an FBD of a joint and apply $\sum F_{x} = 0$ and $\sum F_{y} = 0$ . Start at a joint with no more than two unknowns.
Method of sections. Cut through no more than three members with unknown forces and apply all three equilibrium equations to one side. Best when you need one member in the middle of a truss.
Zero-force members. At an unloaded joint with two non-collinear members, both are zero. At an unloaded joint with three members, two of them collinear, the third is zero.
Sign convention. Assume every unknown member force is tension (pulling away from the joint). A negative result means compression.
Frames. Pins between members carry equal and opposite forces on the two connected parts.

Core formulas

Joint: \sum F_{x} = 0, \sum F_{y} = 0 Section: \sum F_{x} = 0, \sum F_{y} = 0, \sum M = 0

Determinacy check for a plane truss (m members, r reactions, j joints):

m + r = 2 j (determinate) m + r > 2 j (indeterminate) m + r < 2 j (unstable)

Member force components follow the member slope: $F_{x} = F \frac{Δ x}{L}$ , $F_{y} = F \frac{Δ y}{L}$ .

Where it is in the FE Reference Handbook

The Statics chapter has a short section on statically determinate trusses that describes the method of joints and the method of sections. Determinacy and stability rules are also discussed in the structural portion of the Civil Engineering chapter.

Worked example

A truss has joints A (0, 0), D (4, 0), C (8, 0) and B (4, 3), in meters, with members AB, BC, AD, DC and BD. A is a pin, C is a roller, and a 12 kN load acts down at D. Find the forces in AB, AD and BD.

Determinacy: m = 5, r = 3, j = 4, so m + r = 8 = 2j. Determinate.

Reactions: by symmetry $A_{y} = C_{y} = 6 kN$ , $A_{x} = 0$ .

Joint A. AB has length 5 m, so its direction components are 4/5 and 3/5.

\sum F_{y} = 6 + F_{A B} (\frac{3}{5}) = 0 \Rightarrow F_{A B} = - 10 kN (compression)

\sum F_{x} = F_{A D} + F_{A B} (\frac{4}{5}) = 0 \Rightarrow F_{A D} = 8 kN (tension)

Joint D. AD and DC are collinear and horizontal, so vertical equilibrium involves only BD and the load:

\sum F_{y} = F_{B D} - 12 = 0 \Rightarrow F_{B D} = 12 kN (tension)

Check at joint B: the vertical components of AB and BC (each 10 kN compression, 6 kN vertical) push up on B with 12 kN, which balances the 12 kN tension in BD pulling down. ✓

Common mistakes

Starting at a joint with three unknowns. Find the reactions first, then pick a joint with two.
Mixing up tension and compression. Assume tension and let the sign tell you.
Cutting through four unknown members with a section, which cannot be solved with three equations.
Treating a frame member that carries a mid-span load as a two-force member.
Missing zero-force members, which wastes time on joints that need no calculation.

Check your answers with the truss calculator, and read the method of joints example.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

At an unloaded pin joint in an ideal truss, only two non-collinear members meet and there is no support reaction. What are the member forces? See the worked solution or practice the Trusses deck
A symmetric pin joint is supported by two ideal two-force members, each at 45° to the horizontal, and carries a 20 kN downward load. What is the axial force magnitude in each member? See the worked solution or practice the Trusses deck
A simply supported Pratt truss spans 30 m in five 6 m panels with a height of 4 m. Vertical loads of 40 kN act at the two top-chord joints located 6 m from each support. Using the method of sections, the force in the bottom-chord panel at midspan is most nearly: See the worked solution or practice the Trusses deck
A triangular truss has joints A (pin support, left), C (roller support, right) and B (apex). The span from A to C is 8 m, B is directly above midspan, and the apex height is 3 m. A 10 kN vertical load acts downward at B. Using the method of joints, the force in member AB is most nearly: See the worked solution or practice the Trusses deck
A planar truss has 12 members and 7 joints. It is supported by a pin at one joint and a roller at another. Neglecting the arrangement of members, the truss is classified as: See the worked solution or practice the Trusses deck
At an ideal truss joint B, two members extend horizontally to the left and right. A third member BD extends vertically upward to D. A 12 kN external load acts downward at B, and no support reaction acts at B. All member forces are axial. Find the force in BD and state whether it is tension or compression. See the worked solution or practice the Trusses deck

Related Calculators and Examples

Frequently Asked Questions

When should I use the method of sections instead of the method of joints?

Use the method of sections when you need the force in one or two members in the middle of a truss. Cut through no more than three unknown members and take moments about the point where two of them intersect.

How do I spot zero-force members?

At an unloaded joint with only two non-collinear members, both are zero-force members. At an unloaded joint with three members where two are collinear, the third member carries zero force.

How do I know if a truss is statically determinate?

For a plane truss, compare m + r with 2j, where m is members, r is reactions and j is joints. Equal means determinate (if stable), greater means indeterminate, less means unstable.

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