Trusses live inside Statics, which NCEES weights at 8 to 12 questions on the 110-question FE Civil exam, roughly 7 to 11 percent. 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 (7–11%) 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
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.)
A) 4.5 kN (C)
B) 6.0 kN (T)
C) 7.5 kN (T)
D) 7.5 kN (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 2
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 AB? (Report T/C and magnitude.)
A) 6 kN (T)
B) 8.5 kN (C)
C) 8.5 kN (T)
D) 12 kN (C)
Answer: B) 8.5 kN (C)
Step 1 – Find the support reactions:
The pin at A provides \( A_x \) and \( R_{Ay} \); the roller at C provides only \( R_{Cy} \). The only external load is \( 12\,\mathrm{kN} \) downward at D \( (4,\,0) \). With no horizontal loads, \( \sum F_x = 0 \) gives \( A_x = 0 \).
Take moments about A so that both components at A drop out — the load acts at \( x = 4\,\mathrm{m} \) and \( R_{Cy} \) at \( x = 8\,\mathrm{m} \) (each moment = force multiplied by its perpendicular arm):
$$ \sum M_A = 0: \quad R_{Cy}(8) - 12(4) = 0 \quad \Rightarrow \quad R_{Cy} = \frac{48}{8} = 6 \,\mathrm{kN} \;(\uparrow) $$
$$ \sum F_y = 0: \quad R_{Ay} + R_{Cy} - 12 = 0 \quad \Rightarrow \quad R_{Ay} = 12 - 6 = 6 \,\mathrm{kN} \;(\uparrow) $$
As expected by symmetry (the load acts at midspan).
Step 2 – Identify the geometry at Joint A:
Joint A is at \( (0,\,0) \). Two members connect here:
- AD → horizontal, along x-axis to \( (4,\,0) \)
- AB → diagonal to \( (4,\,4) \), angle \( \theta = \arctan(4/4) = 45^\circ \)
Step 3 – Apply equilibrium at Joint A (\( \sum F_y = 0 \)):
The vertical forces at Joint A are:
- \( R_{Ay} = 6 \,\mathrm{kN} \) (upward)
- \( F_{AB} \sin 45^\circ \) (vertical component of member AB)
- \( F_{AD} \) has no vertical component (horizontal member)
$$ \sum F_y = 0: \quad 6 + F_{AB} \sin 45^\circ = 0 $$
Step 4 – Solve for \( F_{AB} \):
$$ F_{AB} = \frac{-6}{\sin 45^\circ} = \frac{-6}{0.7071} = -8.49 \,\mathrm{kN} $$
Step 5 – Interpret the sign:
The negative sign means our assumed direction was wrong → member AB is in compression.
$$ |F_{AB}| = 8.49 \approx 8.5 \,\mathrm{kN} \;(\text{C}) $$
Correct answer: (B)
Problem 3
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. One member listed below is a zero-force member. Which one?
A) BE
B) EF
C) AE
D) CD
Answer: A) BE
Step 1 – Identify given information: Same Pratt truss. At joint B \((3,0)\): members AB and BC are horizontal (collinear), member BE is vertical. No external load at B.
Step 2 – Zero-force member rule (FE Handbook): At a joint where two members are collinear and a third is non-collinear with no external load, the non-collinear member carries zero force.
Step 3 – Apply at joint B:
AB and BC are collinear (both horizontal). BE is vertical (non-collinear). No external load at B.
\(\sum F_y = 0: \quad F_{BE} = 0\)
Step 4 – Answer: BE is the zero-force member. Option (B) EF carries compression (top chord); (C) AE carries compression (diagonal); (D) CD carries force (bottom chord). Zero-force members maintain structural stability but carry no force under this particular loading.
Correct answer: (A)
Problem 4
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 vertical member BD?
A) 0 kN
B) 4.5 kN (tension)
C) 6.0 kN (compression)
D) 7.5 kN (tension)
Answer: A) 0 kN
Step 1 – Identify given information: Warren-type truss with \(12\,\mathrm{kN}\) load at B. At joint D: members AD and DC are horizontal (collinear), BD is vertical. No external load at D.
Step 2 – Zero-force member identification (FE Handbook): At a joint where two members are collinear and a third member is non-collinear, if there is no external load at that joint, the non-collinear member is a zero-force member.
Step 3 – Apply at joint D:
AD and DC are both horizontal (collinear). BD is vertical (non-collinear). No external load acts at D.
\(\sum F_y = 0: \quad F_{BD} = 0\)
Step 4 – Answer: \(F_{BD} = 0\,\mathrm{kN}\). BD is a zero-force member. Options (B), (C), and (D) all assign nonzero forces, which would violate vertical equilibrium at joint D. Zero-force members provide stability but carry no load under this loading.
Correct answer: (A)
Problem 5
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?
A) 7.5 kN (compression)
B) 7.5 kN (tension)
C) 10.0 kN (compression)
D) 6.0 kN (compression)
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 6
Planar truss with joints \( A(0,\,0) \), \( B(3,\,0) \), \( C(6,\,0) \), \( D(9,\,0) \), \( E(3,\,3) \), \( F(6,\,3) \). Supports: A is a pin, D is a roller. Members: AB, BC, CD, BE, CF, EF, AE, EC, FD. Loads: \( 20\,\mathrm{kN} \) downward at E and \( 10\,\mathrm{kN} \) downward at F. What is the force in member EF?
A) 13.33 kN (tension)
B) 13.33 kN (compression)
C) 10.00 kN (tension)
D) 18.86 kN (compression)
Answer: B) 13.33 kN (compression)
Step 1 – Identify given information: Pratt-type truss: pin at A \((0,0)\), roller at D \((9,0)\), top chord at \(y = 3\). Loads: \(20\,\mathrm{kN}\) down at E \((3,3)\) and \(10\,\mathrm{kN}\) down at F \((6,3)\). Find force in member EF (top chord).
Step 2 – Support reactions (whole-truss equilibrium): The pin at A provides \(A_x\) and \(A_y\); the roller at D provides only \(D_y\). With no horizontal loads, \(\sum F_x = 0\) gives \(A_x = 0\). Take moments about A (each term is force multiplied by its perpendicular arm):
\(\sum M_A = 0: \quad D_y(9) - 20(3) - 10(6) = 0 \quad \Rightarrow \quad D_y = \frac{120}{9} = 13.333\,\mathrm{kN}\;(\uparrow)\)
\(\sum F_y = 0: \quad A_y = 20 + 10 - 13.333 = 16.667\,\mathrm{kN}\;(\uparrow)\)
Step 3 – Method of sections or joints: Use the method of joints at joint F. Member EF is horizontal along the top chord.
Step 4 – Joint F equilibrium:
First get the diagonal FD from joint D, where the only members are CD (horizontal) and FD:
\(\sum F_y = 0 \text{ at D}: \quad D_y + F_{FD}\sin 45^{\circ} = 0 \quad \Rightarrow \quad F_{FD} = -\frac{13.333}{0.7071} = -18.86\,\mathrm{kN}\) (C)
At joint F, \(\sum F_y = 0\) gives \(F_{CF} = 3.33\,\mathrm{kN}\) (T). Then the horizontal equilibrium at F:
\(\sum F_x = 0: \quad -F_{EF} + F_{FD}\cos 45^{\circ} = 0 \quad \Rightarrow \quad F_{EF} = (-18.86)(0.7071) = -13.333\,\mathrm{kN}\)
The negative sign indicates compression.
Step 5 – Answer: \(F_{EF} = 13.33\,\mathrm{kN}\) (compression). Top chord members in a loaded truss typically carry compression. Option (A) gives tension; (C) \(10\,\mathrm{kN}\) is the load at F; (D) \(18.86\,\mathrm{kN}\) is incorrect.
Correct answer: (B)
Problem 7
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.)
A) 10 kN (T)
B) 10 kN (C)
C) 15 kN (T)
D) 15 kN (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; (D) \(15\,\mathrm{kN}\) (C) is incorrect magnitude.
Correct answer: (B)
Problem 8
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 member BE?
A) 9.0 kN (compression)
B) 9.0 kN (tension)
C) 4.5 kN (compression)
D) 0 kN
Answer: A) 9.0 kN (compression)
Step 1 – Identify given information: Parallel-chord truss: A pin \((0,0)\), C roller \((6,0)\), E at \((3,3)\). Load: \(9\,\mathrm{kN}\) down at E. BE is a vertical member from B \((3,0)\) to E \((3,3)\).
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 \(9\,\mathrm{kN}\) load acts at \(x = 3\,\mathrm{m}\), \(C_y\) at \(x = 6\,\mathrm{m}\)):
\(\sum M_A = 0: \quad C_y(6) - 9(3) = 0 \quad \Rightarrow \quad C_y = \frac{27}{6} = 4.5\,\mathrm{kN}\;(\uparrow)\)
\(\sum F_y = 0: \quad A_y = 9 - 4.5 = 4.5\,\mathrm{kN}\;(\uparrow)\) — as expected by symmetry (the load acts at midspan).
Step 3 – Analyze joint E:
At joint E, the \(9\,\mathrm{kN}\) load acts downward. Members DE and EF are horizontal (no vertical component). The only member with a vertical component at E is BE (vertical).
\(\sum F_y = 0: \quad F_{BE} - 9 = 0 \quad \Rightarrow \quad F_{BE} = 9\,\mathrm{kN}\)
Since BE is compressed (load pushes down, member resists by compression): \(9\,\mathrm{kN}\) (C).
Step 4 – Answer: \(F_{BE} = 9\,\mathrm{kN}\) (compression). The vertical member directly transfers the full applied load. Option (B) gives tension; (C) \(4.5\,\mathrm{kN}\) is the reaction; (D) \(0\,\mathrm{kN}\) would require the load to be balanced by other members.
Correct answer: (A)
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.