Trusses live inside Statics, which NCEES weights at 8 to 12 questions on the 110-question FE Civil exam. That makes truss work worth more points than almost any single topic you can learn in an afternoon. The eight problems below are the ones we would hand a candidate who has two weeks left before the appointment.
Trusses are among the most reliably testable statics topics because the method is mechanical once you know it: solve the global reactions, pick a joint or a cut, write equilibrium, read the sign. Every problem here comes with two solution figures, one showing the support reactions and one showing the free-body diagram at the cut or joint, so you see the geometry rather than reconstructing it from a paragraph of text.
Where this sits on the exam: trusses live inside Statics, which NCEES weights at 8–12 questions of the 110-question FE Civil exam. Every problem below shows the truss, and every solution shows the reactions and the free-body diagram — not just the arithmetic.
What NCEES Tests in Truss Analysis
Most truss items ask for the force in one named member and give you four choices that differ in magnitude and sense. The usual routes are the method of joints, which is fastest when the member you want touches a joint with two unknowns, and the method of sections, which is fastest when the member sits in the middle of a long span. Support reactions come first either way.
Zero-force members show up often, usually as a standalone question worth thirty seconds if you recognize the two-member unloaded joint or the three-member joint with two collinear members. Sign convention matters as much as arithmetic: assume tension, and a negative answer means compression. The exam expects the answer as a magnitude plus a sense, so 12.5 kN compression and 12.5 kN tension are different options.
Method of Joints or Method of Sections?
Both give the same answer. Picking the wrong one is what costs you time on exam day.
| Method of joints | Method of sections | |
|---|---|---|
| How it works | Isolate one joint at a time and apply ΣFx = 0 and ΣFy = 0. | Cut the truss into two parts and apply equilibrium to one whole side. |
| Unknown limit | Two unknown members per joint (two equations). | Three unknown members per cut (three equations). |
| Use it when | You need forces in several members, or the member sits next to a support. | You need one specific member deep inside the truss. |
| Watch out for | Chaining errors — one bad joint poisons every joint after it. | Cutting through more than three unknown members, which leaves the cut unsolvable. |
Whichever you choose, solve the global support reactions first. Almost every truss problem on the exam is quicker once the reactions are on the diagram.
8 Free Truss Practice Problems
Work each one on paper before revealing the solution. Each worked solution opens with the support reactions drawn on the truss, then the free-body diagram of the joint or cut that produces the answer — the same two sketches you should be drawing yourself under exam conditions.
Problem 1 — D. Frames and trusses
Truss geometry: A pin at \( (0,\,0) \), C roller at \( (8,\,0) \), joint D at \( (4,\,0) \), joint B at \( (4,\,4) \). Members: AD, DC, AB, BC, BD. A \( 12\,\mathrm{kN} \) downward load is applied at joint D. What is the axial force in member BD? (Report T/C.)
Answer: C) 12 kN (T)
Step 1 – Identify given information: Truss: A pin \((0,0)\), C roller \((8,0)\), D \((4,0)\), B \((4,4)\). Load: \(12\,\mathrm{kN}\) downward at D. BD is vertical.
Step 2 – Equilibrium at joint D (FE Handbook): Exam tip: the support reactions are not needed for this member — joint D can be solved directly. (For completeness: \(\sum M_A = 0: \; C_y(8) - 12(4) = 0 \Rightarrow C_y = 6\,\mathrm{kN}\), and \(\sum F_y = 0\) gives \(A_y = 6\,\mathrm{kN}\).) At joint D, members AD and DC are both horizontal (along x-axis) and contribute no vertical force. Only BD has a vertical component.
Step 3 – Vertical equilibrium at D:
$$\sum F_y = 0: \quad F_{BD} - 12 = 0 \quad \Rightarrow \quad F_{BD} = 12\,\mathrm{kN}$$
BD pulls upward on D to resist the downward load, so BD is in tension.
Step 4 – Answer: \(F_{BD} = 12\,\mathrm{kN}\) (tension). The vertical member carries the entire applied load because it is the only member at D with a vertical component. Option (A) \(0\,\mathrm{kN}\) would apply only if there were no external load; (B) \(6\,\mathrm{kN}\) incorrectly halves the load; (D) \(12\,\mathrm{kN}\) (C) gives the wrong sign.
Correct answer: (C)
Problem 2 — D. Frames and trusses
Three-member triangular truss: A pin \( (0,\,0) \), B roller \( (4,\,0) \), C \( (4,\,3) \). Members: AB, AC, BC. A \( 6\,\mathrm{kN} \) horizontal load acts to the right at joint C. What is the axial force in member AC? (T/C.)
Answer: C) 7.5 kN (T)
Step 1 – Identify given information: Triangular truss: A pin \((0,0)\), B roller \((4,0)\), C \((4,3)\). Load: \(6\,\mathrm{kN}\) horizontal (right) at C. AC goes from \((0,0)\) to \((4,3)\): a 4-3-5 triangle.
Step 2 – Support reactions (whole-truss equilibrium): The pin at A provides \(A_x\) and \(A_y\); the roller at B provides only \(B_y\).
\(\sum F_x = 0: \quad A_x + 6 = 0 \quad \Rightarrow \quad A_x = -6\,\mathrm{kN}\) (it acts \(6\,\mathrm{kN}\) to the left)
Take moments about A: the \(6\,\mathrm{kN}\) horizontal load acts at height \(y = 3\,\mathrm{m}\) (the perpendicular arm of a horizontal force is its height), and \(B_y\) acts at \(x = 4\,\mathrm{m}\):
$$\sum M_A = 0: \quad B_y(4) - 6(3) = 0 \quad \Rightarrow \quad B_y = \frac{18}{4} = 4.5\,\mathrm{kN}\;(\uparrow)$$
\(\sum F_y = 0: \quad A_y + B_y = 0 \quad \Rightarrow \quad A_y = -4.5\,\mathrm{kN}\) (it acts downward)
Step 3 – Geometry of member AC:
\(\cos\theta = 4/5 = 0.8\), \(\sin\theta = 3/5 = 0.6\)
Step 4 – Joint C equilibrium:
At joint C, member BC is vertical. The horizontal equilibrium gives:
$$\sum F_x = 0: \quad 6 - F_{AC}\cos\theta = 0$$
\(F_{AC} = \frac{6}{4/5} = 6 \times \frac{5}{4} = 7.5\,\mathrm{kN}\) (tension)
Step 5 – Answer: \(F_{AC} = 7.5\,\mathrm{kN}\) (tension). The inclined member AC resists the horizontal load. Option (A) \(4.5\,\mathrm{kN}\) is the force in the vertical member BC (equal in magnitude to the reaction \(B_y\)), not AC; (B) \(6.0\,\mathrm{kN}\) ignores the angle amplification; (D) gives compression.
Correct answer: (C)
Problem 3 — D. Frames and trusses
A simply supported triangular truss has joints A (pin) and C (roller) \( 6\,\mathrm{m} \) apart. Joint B is located \( 4\,\mathrm{m} \) above the midpoint of AC. A \( 12\,\mathrm{kN} \) downward load is applied at joint B. What is the force in member AB?
Answer: A) 7.5 kN (compression)
Step 1 – Identify given information: Triangular truss: A (pin) and C (roller) are \(6\,\mathrm{m}\) apart. B is \(4\,\mathrm{m}\) above the midpoint. Load: \(12\,\mathrm{kN}\) downward at B.
Step 2 – Support reactions (whole-truss equilibrium): The pin at A provides \(A_x\) and \(A_y\); the roller at C provides only \(C_y\). With no horizontal loads:
$$\sum F_x = 0: \quad A_x = 0$$
Take moments about A so that \(A_x\) and \(A_y\) drop out — the \(12\,\mathrm{kN}\) load acts at \(x = 3\,\mathrm{m}\) and \(C_y\) acts at \(x = 6\,\mathrm{m}\) (each moment = force multiplied by its perpendicular arm, counterclockwise positive):
$$\sum M_A = 0: \quad C_y(6) - 12(3) = 0 \quad \Rightarrow \quad C_y = \frac{36}{6} = 6\,\mathrm{kN}\;(\uparrow)$$
$$\sum F_y = 0: \quad A_y + C_y - 12 = 0 \quad \Rightarrow \quad A_y = 12 - 6 = 6\,\mathrm{kN}\;(\uparrow)$$
As expected by symmetry (the load acts at midspan). Geometry from A to B: horizontal \(= 3\,\mathrm{m}\), vertical \(= 4\,\mathrm{m}\), hypotenuse \(= 5\,\mathrm{m}\) (3-4-5 triangle). So \(\sin\theta = 4/5\), \(\cos\theta = 3/5\).
Step 3 – Joint B equilibrium (method of joints):
By symmetry, \(F_{AB} = F_{CB}\). Vertical equilibrium at B:
$$\sum F_y = 0: \quad 2F_{AB}\sin\theta = 12$$
$$2F_{AB}\left(\frac{4}{5}\right) = 12 \quad \Rightarrow \quad F_{AB} = \frac{12 \times 5}{2 \times 4} = 7.5\,\mathrm{kN}$$
Since the members push inward at B, they are in compression.
Step 4 – Answer: \(F_{AB} = 7.5\,\mathrm{kN}\) (compression). Option (B) gives tension (wrong sign); (C) \(10\,\mathrm{kN}\) uses the wrong trigonometric ratio; (D) \(6\,\mathrm{kN}\) equals the reaction, not the member force.
Correct answer: (A)
Problem 4 — D. Frames and trusses
Warren-type truss with joints \( A(0,\,0) \) (pin), \( D(4,\,0) \), \( C(8,\,0) \) (roller), and \( B(4,\,3) \). Members: AB, BC, AD, DC, BD. A \( 12\,\mathrm{kN} \) downward load is applied at joint B. What is the force in member AD?
Answer: A) 8.0 kN (tension)
Step 1 – Identify given information: Warren-type truss: A pin at \((0,0)\), D at \((4,0)\), C roller at \((8,0)\), B at \((4,3)\). Load: \(12\,\mathrm{kN}\) down at B. At joint D the chords AD and DC are collinear and no load acts there, so \(F_{BD} = 0\) (zero-force member).
Step 2 – Support reactions (whole-truss equilibrium): The pin at A provides \(A_x\) and \(A_y\); the roller at C provides only \(C_y\). With no horizontal loads, \(\sum F_x = 0\) gives \(A_x = 0\). Take moments about A (the \(12\,\mathrm{kN}\) load acts at \(x = 4\,\mathrm{m}\), \(C_y\) at \(x = 8\,\mathrm{m}\)):
$$\sum M_A = 0: \quad C_y(8) - 12(4) = 0 \quad \Rightarrow \quad C_y = \frac{48}{8} = 6\,\mathrm{kN}\;(\uparrow)$$
\(\sum F_y = 0: \quad A_y = 12 - 6 = 6\,\mathrm{kN}\;(\uparrow)\) — as expected by symmetry. Member AB forms a 4-3-5 triangle: \(\cos\theta = 4/5\), \(\sin\theta = 3/5\).
Step 3 – Joint A equilibrium:
\(\sum F_y = 0: \quad 6 + F_{AB}\sin\theta = 0 \quad \Rightarrow \quad F_{AB} = \frac{-6}{3/5} = -10\,\mathrm{kN}\) (compression)
\(\sum F_x = 0: \quad F_{AD} + F_{AB}\cos\theta = 0 \quad \Rightarrow \quad F_{AD} = -(-10)(4/5) = 8\,\mathrm{kN}\) (tension)
Step 4 – Answer: \(F_{AD} = 8\,\mathrm{kN}\) (tension). The bottom chord carries tension while the inclined members carry compression, which is typical for top-loaded trusses. Option (B) gives compression; (C) and (D) have wrong magnitudes.
Correct answer: (A)
Problem 5 — D. Frames and trusses
Parallel-chord truss with joints \( A(0,\,0) \) (pin), \( B(3,\,0) \), \( C(6,\,0) \) (roller at C), \( D(0,\,3) \), \( E(3,\,3) \), \( F(6,\,3) \). Members: AB, BC, DE, EF, AD, BE, CF, DB, BF. A \( 9\,\mathrm{kN} \) downward load is applied at joint E. What is the force in diagonal member DB?
Answer: A) 6.36 kN (tension)
Step 1 – Identify given information: Parallel-chord truss: A pin \((0,0)\), C roller \((6,0)\), \(9\,\mathrm{kN}\) down at E \((3,3)\). Member DB goes from D \((0,3)\) to B \((3,0)\), making a \(45^{\circ}\) angle.
Step 2 – Support reactions: With no horizontal loads, \(\sum F_x = 0\) gives \(A_x = 0\). Take moments about A (the \(9\,\mathrm{kN}\) load acts at \(x = 3\,\mathrm{m}\), \(C_y\) at \(x = 6\,\mathrm{m}\)):
$$\sum M_A = 0: \quad C_y(6) - 9(3) = 0 \quad \Rightarrow \quad C_y = \frac{27}{6} = 4.5\,\mathrm{kN}\;(\uparrow)$$
$$\sum F_y = 0: \quad A_y = 9 - 4.5 = 4.5\,\mathrm{kN}\;(\uparrow)$$
Step 3 – Method of joints at D: At joint D, member AD is vertical, member DE is horizontal, and DB is the only diagonal. First find AD from joint A: \(\sum F_y = 0: \quad 4.5 + F_{AD} = 0 \Rightarrow F_{AD} = -4.5\,\mathrm{kN}\), i.e. \(4.5\,\mathrm{kN}\) (C), so AD pushes joint D upward.
Step 4 – Solve:
Vertical equilibrium at joint D (DB pulls the joint toward B, downward at \(45^{\circ}\)):
$$\sum F_y = 0: \quad 4.5 - F_{DB}\sin 45^{\circ} = 0$$
\(F_{DB} = \frac{4.5}{\sin 45^{\circ}} = 4.5\sqrt{2} = 6.364\,\mathrm{kN}\) (tension)
Step 5 – Answer: \(F_{DB} = 6.36\,\mathrm{kN}\) (tension). Diagonal web members in trusses typically carry tension when the top chord is compressed. Option (B) gives compression; (C) \(9\,\mathrm{kN}\) is the applied load; (D) \(4.5\,\mathrm{kN}\) is the reaction.
Correct answer: (A)
Problem 6 — D. Frames and trusses
Pratt-type truss: A pin \( (0,\,0) \), B \( (3,\,0) \), C \( (6,\,0) \), D roller \( (9,\,0) \), E \( (3,\,3) \), F \( (6,\,3) \). Members: AB, BC, CD, BE, CF, EF, AE, EC, FD. Loads: \( 10\,\mathrm{kN} \) downward at E and \( 10\,\mathrm{kN} \) downward at F. What is the axial force in member EF? (T/C.)
Answer: B) 10 kN (C)
Step 1 – Identify given information: Pratt truss: A pin \((0,0)\), D roller \((9,0)\). Loads: \(10\,\mathrm{kN}\) down at E \((3,3)\) and \(10\,\mathrm{kN}\) down at F \((6,3)\).
Step 2 – Support reactions (whole-truss equilibrium): The pin at A provides \(A_x\) and \(R_{Ay}\); the roller at D provides only \(R_{Dy}\). With no horizontal loads, \(\sum F_x = 0\) gives \(A_x = 0\). Take moments about A (the loads act at \(x = 3\,\mathrm{m}\) and \(x = 6\,\mathrm{m}\); \(R_{Dy}\) at \(x = 9\,\mathrm{m}\)):
$$\sum M_A = 0: \quad R_{Dy}(9) - 10(3) - 10(6) = 0 \quad \Rightarrow \quad R_{Dy} = \frac{90}{9} = 10\,\mathrm{kN}\;(\uparrow)$$
$$\sum F_y = 0: \quad R_{Ay} = 10 + 10 - 10 = 10\,\mathrm{kN}\;(\uparrow)$$
Step 3 – Method of joints: At joint A (\(45^{\circ}\) geometry for AE): \(\sum F_y = 0: \; 10 + F_{AE}\sin 45^{\circ} = 0 \Rightarrow F_{AE} = -14.14\,\mathrm{kN}\) (C). At joint B the chords AB and BC are collinear with no load, so \(F_{BE} = 0\). At joint E (\(x\) right, \(y\) up, tension positive):
$$\sum F_y = 0: \quad -F_{AE}\sin 45^{\circ} - F_{EC}\sin 45^{\circ} - 10 = 0 \Rightarrow F_{EC} = 0$$
$$\sum F_x = 0: \quad F_{EF} - F_{AE}\cos 45^{\circ} = 0 \Rightarrow F_{EF} = (-14.14)(0.707) = -10\,\mathrm{kN}$$
\(F_{EF} = 10\,\mathrm{kN}\) (compression)
Step 4 – Answer: \(F_{EF} = 10\,\mathrm{kN}\) (compression). The top chord carries compression, as expected in a top-loaded Pratt truss. Option (A) gives tension; (C) \(14.1\,\mathrm{kN}\) is the diagonal force; (D) \(0\,\mathrm{kN}\) is wrong.
Correct answer: (B)
Problem 7 — D. Frames and trusses
Warren-type truss: A pin \( (0,\,0) \), B \( (4,\,0) \), C roller \( (8,\,0) \), D \( (2,\,3) \), E \( (6,\,3) \). Members: AB, BC, AD, DB, DE, BE, EC. A \( 15\,\mathrm{kN} \) downward load is applied at joint B. What is the axial force in member DE? (T/C.)
Answer: B) 10 kN (C)
Step 1 – Identify given information: Warren truss: A pin \((0,0)\), C roller \((8,0)\), D \((2,3)\), E \((6,3)\). Load: \(15\,\mathrm{kN}\) down at B \((4,0)\).
Step 2 – Support reactions (whole-truss equilibrium): The pin at A provides \(A_x\) and \(R_{Ay}\); the roller at C provides only \(R_{Cy}\). With no horizontal loads, \(\sum F_x = 0\) gives \(A_x = 0\). Take moments about A (the \(15\,\mathrm{kN}\) load acts at \(x = 4\,\mathrm{m}\), \(R_{Cy}\) at \(x = 8\,\mathrm{m}\)):
$$\sum M_A = 0: \quad R_{Cy}(8) - 15(4) = 0 \quad \Rightarrow \quad R_{Cy} = \frac{60}{8} = 7.5\,\mathrm{kN}\;(\uparrow)$$
\(\sum F_y = 0: \quad R_{Ay} = 15 - 7.5 = 7.5\,\mathrm{kN}\;(\uparrow)\) — as expected by symmetry (the load acts at midspan).
Step 3 – Method of joints/sections for DE: Using the method of sections with a vertical cut through DE, DB, and AB, and taking moments about B:
Member DE is horizontal (top chord); DB and AB both pass through B, so they drop out of the moment equation:
\(\sum M_B = 0: \quad R_{Ay}(4) + F_{DE}(3) = 0 \Rightarrow F_{DE} = -\frac{7.5 \times 4}{3} = -10\,\mathrm{kN}\) (compression)
Step 4 – Answer: \(F_{DE} = 10\,\mathrm{kN}\) (compression). The top chord of a Warren truss under downward loading is in compression. Option (A) gives tension; (C) \(15\,\mathrm{kN}\) equals the applied load; (C) \(15\,\mathrm{kN}\) (C) is incorrect magnitude.
Correct answer: (B)
Problem 8 — D. Frames and trusses
Pin-connected braced frame (analyze as a truss): A pin \( (0,\,0) \), B roller \( (4,\,0) \), C \( (0,\,3) \), D \( (4,\,3) \). Members: AB, AC, BD, CD, and diagonal AD. Lateral loads: \( 5\,\mathrm{kN} \) to the right at C and \( 5\,\mathrm{kN} \) to the right at D. What is the axial force in diagonal AD? (T/C.)
Answer: C) 12.5 kN (T)
Step 1 – Identify given information: Braced frame: A pin \((0,0)\), B roller \((4,0)\), C \((0,3)\), D \((4,3)\). Diagonal AD goes from \((0,0)\) to \((4,3)\): a 3-4-5 triangle. Loads: \(5\,\mathrm{kN}\) right at C and \(5\,\mathrm{kN}\) right at D.
Step 2 – Support reactions (whole-frame equilibrium): The pin at A provides \(A_x\) and \(A_y\); the roller at B provides only \(B_y\).
\(\sum F_x = 0: \quad A_x + 5 + 5 = 0 \quad \Rightarrow \quad A_x = -10\,\mathrm{kN}\) (it acts \(10\,\mathrm{kN}\) to the left)
Take moments about A. The perpendicular arm of a horizontal force is its height: both \(5\,\mathrm{kN}\) loads act at \(y = 3\,\mathrm{m}\); \(B_y\) acts at \(x = 4\,\mathrm{m}\) (counterclockwise positive):
$$\sum M_A = 0: \quad B_y(4) - 5(3) - 5(3) = 0 \quad \Rightarrow \quad B_y = \frac{30}{4} = 7.5\,\mathrm{kN}\;(\uparrow)$$
\(\sum F_y = 0: \quad A_y + B_y = 0 \quad \Rightarrow \quad A_y = -7.5\,\mathrm{kN}\) (it acts downward: the lateral loads try to overturn the frame and the pin holds it down)
Step 3 – Find \(F_{AD}\) using joint equilibrium: AD has \(\cos\theta = 4/5\), \(\sin\theta = 3/5\). At joint B: \(\sum F_y\) gives \(F_{BD} = 7.5\,\mathrm{kN}\) (C) and \(\sum F_x\) gives \(F_{AB} = 0\). Horizontal equilibrium at joint A:
$$\sum F_x = 0: \quad A_x + F_{AD}\cos\theta = 0 \Rightarrow -10 + F_{AD}\left(\frac{4}{5}\right) = 0$$
\(F_{AD} = 12.5\,\mathrm{kN}\) (tension)
Step 4 – Answer: \(F_{AD} = 12.5\,\mathrm{kN}\) (tension). The diagonal brace resists lateral loads through tension. Option (A) \(7.5\,\mathrm{kN}\) is the reaction; (B) \(10\,\mathrm{kN}\) is the total horizontal load; (D) gives compression instead of tension.
Correct answer: (C)
Using the FE Reference Handbook for Trusses
The Statics section prints the equilibrium equations and short statements of the method of joints and the method of sections, nothing more. There is no worked example, no zero-force member rule, and no sign convention spelled out. Search it to confirm a formula, not to learn a procedure. If you are reading that page for the first time on exam day, you have already lost the question.
Four Mistakes That Cost Points
- Cutting or picking a joint before solving the reactions Almost every truss problem needs the support reactions before anything else, because your section or joint equation will contain one of them. Sum moments about a support, get the other reaction, then sum vertical forces. Skipping this and starting at an interior joint leaves you with three unknowns and no way forward.
- Reading tension and compression from intuition instead of the sign Assume every unknown member is in tension, draw the force pulling away from the joint, and let the algebra decide. A negative result is compression. If you flip arrows mid-problem to match what the truss looks like it is doing, you will get the magnitude right and the answer choice wrong.
- Missing zero-force members before you start calculating Scan the joints first. A two-member joint with no load and no reaction, or a three-member joint with two collinear members and no load, gives you zero-force members for free. Finding them can collapse a truss that looked like ten unknowns into three, and sometimes the zero-force member is the one being asked about.
- Cutting through more than three unknown members The method of sections gives you three equilibrium equations in the plane, so a cut can expose at most three unknown member forces and still be solvable. If your cut crosses four, move it, or eliminate one unknown first with a zero-force check or a single joint solved by the method of joints.
Frequently Asked Questions
Method of joints or method of sections: which should I use on the FE?
Use the method of sections when the question names one member somewhere in the middle of the truss; one cut and one moment equation can get it. Use the method of joints when the member frames into a joint with only two unknowns, usually near a support. On a timed exam, sections is often two minutes faster.
How many truss questions are on the FE Civil exam?
NCEES does not publish a count by subtopic. The specification puts Statics at 8 to 12 questions out of 110, and trusses are one of the standard Statics subtopics alongside friction, centroids, and moments of inertia. Plan on seeing at least one or two truss items, and treat them as points you should not be giving away.
What are the zero-force member rules?
Two rules cover almost every exam case. If two non-collinear members meet at a joint with no external load or support reaction there, both carry zero force. If three members meet at a joint, two of them collinear, and there is no external load, the odd member out carries zero force. A load or reaction at the joint voids both rules.
Does the FE Reference Handbook give truss formulas?
It gives the equilibrium equations and a brief description of both methods in the Statics section, and that is the extent of it. There is no zero-force member rule, no sign convention, and no worked example. The handbook is searchable on screen during the exam, but searching for a procedure you never practiced costs more time than it saves.
Keep Going
Trusses are one slice of Statics — work the rest with our FE Civil statics practice problems, browse free problems across all 14 knowledge areas, or see how Statics fits the wider exam in the FE Civil topic breakdown.