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10 FE practice problems: truss member forces (method of joints), with solutions

By Dr. Mohamed Elansary, PhD Environmental EngineeringUpdated

These ten problems ask for the force in a truss member and whether it is in tension or compression, a staple of the Statics area in the FE Civil, Mechanical, and Other Disciplines specifications. Solve each one with a free-body diagram at the right joint before you open the solution, and check your sign convention: a sign error flips tension into compression and turns a right number into a wrong answer.

How to use these problems

Give each problem an honest attempt before you open the solution: write the given values with units, find the equation in the FE Reference Handbook PDF, and commit to an answer. Then compare line by line. If you picked a wrong option, read the note for that option; each one names the mistake that produces it.

The problems

Problem 1 · FE Civil, Truss member force

A three-member plane truss has joints A (0, 0), B (28, 23) and C (54, 0), with coordinates in ft. The members are AB, BC and AC. A is a pin support and C is a roller on a horizontal surface. Joint B carries a downward load of 48.0 kips and a horizontal load of 18.5 kips in the +x direction. The force in member AC is most nearly:

  • A 24.0 kips (T)
  • B 31.0 kips (T)
  • C 37.0 kips (C)
  • D 37.0 kips (T)

Handbook: Statics, Plane Truss: Method of Joints, FE Reference Handbook 10.6

Show the worked solution
Answer
D (37.0 kips in tension)
Given
L = 54 ft, a = 28 ft, h = 23 ft, P = 48.0 kips, H = 18.5 kips
Find
force in member AC, magnitude and tension (T) or compression (C)
Handbook
Statics, Plane Truss: Method of Joints, page 98
Equation
Method of joints: sum Fx = 0 and sum Fy = 0 at each joint; reactions from sum of moments = 0; tension positive
Substitute
  1. Reactions, sum of moments about A = 0: Cy = (P·a + H·h)/L = ((48.0 kips)(28 ft) + (18.5 kips)(23 ft))/(54 ft) = 32.77 kips
  2. Sum Fy = 0: Ay = P - Cy = 15.23 kips; sum Fx = 0: Ax = -H = -18.5 kips
  3. Length AB = √(a² + h²) = √(28² + 23²) ft = 36.24 ft
  4. Joint A, sum Fy = 0: Ay + F_AB(h/AB) = 0, so F_AB = -Ay·AB/h = -(15.23 kips)(36.24 ft)/(23 ft) = -24.00 kips
  5. Joint A, sum Fx = 0: Ax + F_AB(a/AB) + F_AC = 0, so F_AC = H + Ay·a/h = 18.5 kips + (15.23 kips)(28 ft)/(23 ft) = 37.04 kips
  6. F_AC = 37.04 kips, so AC carries 37.0 kips in tension (T)
Result
37.0 kips (T), 3 significant figures
Check
with all three member forces, joint B also balances (sum Fx = 0 and sum Fy = 0 including the load); a positive member force is tension, a negative one compression.
Why the others are wrong
  • A: took the force in the inclined member AB as the force in AC
  • B: used h/a instead of a/h when taking the horizontal part of AB at joint A
  • C: has the right size but the wrong sense; equilibrium of joint A puts AC in tension

Problem 2 · FE Civil, Truss member force

A three-member plane truss has joints A (0, 0), B (8.5, 7.5) and C (15.5, 0), with coordinates in m. The members are AB, BC and AC. A is a pin support and C is a roller on a horizontal surface. Joint B carries a downward load of 155 kN. The force in member AC is most nearly:

  • A 61.8 kN (T)
  • B 79.3 kN (T)
  • C 176 kN (T)
  • D 87.8 kN (T)

Handbook: Statics, Plane Truss: Method of Joints, FE Reference Handbook 10.6

Show the worked solution
Answer
B (79.3 kN in tension)
Given
L = 15.5 m, a = 8.5 m, h = 7.5 m, P = 155 kN
Find
force in member AC, magnitude and tension (T) or compression (C)
Handbook
Statics, Plane Truss: Method of Joints, page 98
Equation
Method of joints: sum Fx = 0 and sum Fy = 0 at each joint; reactions from sum of moments = 0; tension positive
Substitute
  1. Reactions, sum of moments about A = 0: Cy = P·a/L = (155 kN)(8.5 m)/(15.5 m) = 85.00 kN
  2. Sum Fy = 0: Ay = P - Cy = 70.00 kN; sum Fx = 0: Ax = 0
  3. Length AB = √(a² + h²) = √(8.5² + 7.5²) m = 11.34 m
  4. Joint A, sum Fy = 0: Ay + F_AB(h/AB) = 0, so F_AB = -Ay·AB/h = -(70.00 kN)(11.34 m)/(7.5 m) = -105.8 kN
  5. Joint A, sum Fx = 0: F_AB(a/AB) + F_AC = 0, so F_AC = Ay·a/h = (70.00 kN)(8.5 m)/(7.5 m) = 79.33 kN
  6. F_AC = 79.33 kN, so AC carries 79.3 kN in tension (T)
Result
79.3 kN (T), 3 significant figures
Check
with all three member forces, joint B also balances (sum Fx = 0 and sum Fy = 0 including the load); a positive member force is tension, a negative one compression.
Why the others are wrong
  • A: used h/a instead of a/h when taking the horizontal part of AB at joint A
  • C: used the whole load P as the reaction at A
  • D: assumed each support carries P/2, but B is not at midspan

Problem 3 · FE Civil, Truss member force

A three-member plane truss has joints A (0, 0), B (6.0, 4.0) and C (8.0, 0), with coordinates in m. The members are AB, BC and AC. A is a pin support and C is a roller on a horizontal surface. Joint B carries a downward load of 100 kN. The force in member AB is most nearly:

  • A 135 kN (C)
  • B 25.0 kN (C)
  • C 180 kN (C)
  • D 45.1 kN (C)

Handbook: Statics, Plane Truss: Method of Joints, FE Reference Handbook 10.6

Show the worked solution
Answer
D (45.1 kN in compression)
Given
L = 8.0 m, a = 6.0 m, h = 4.0 m, P = 100 kN
Find
force in member AB, magnitude and tension (T) or compression (C)
Handbook
Statics, Plane Truss: Method of Joints, page 98
Equation
Method of joints: sum Fx = 0 and sum Fy = 0 at each joint; reactions from sum of moments = 0; tension positive
Substitute
  1. Reactions, sum of moments about A = 0: Cy = P·a/L = (100 kN)(6.0 m)/(8.0 m) = 75.00 kN
  2. Sum Fy = 0: Ay = P - Cy = 25.00 kN; sum Fx = 0: Ax = 0
  3. Length AB = √(a² + h²) = √(6.0² + 4.0²) m = 7.211 m
  4. Joint A, sum Fy = 0: Ay + F_AB(h/AB) = 0, so F_AB = -Ay·AB/h = -(25.00 kN)(7.211 m)/(4.0 m) = -45.07 kN
  5. F_AB = -45.07 kN, so AB carries 45.1 kN in compression (C)
Result
45.1 kN (C), 3 significant figures
Check
with all three member forces, joint B also balances (sum Fx = 0 and sum Fy = 0 including the load); a positive member force is tension, a negative one compression.
Why the others are wrong
  • A: used the reaction at C in the equilibrium of joint A
  • B: set the member force equal to the reaction without resolving it along the member
  • C: used the whole load P as the reaction at A

Problem 4 · FE Civil, Truss member force

A three-member plane truss has joints A (0, 0), B (26, 17) and C (37, 0), with coordinates in ft. The members are AB, BC and AC. A is a pin support and C is a roller on a horizontal surface. Joint B carries a downward load of 15.5 kips and a horizontal load of 4.0 kips in the +x direction. The force in member AC is most nearly:

  • A 5.81 kips (T)
  • B 8.24 kips (T)
  • C 8.24 kips (C)
  • D 5.06 kips (T)

Handbook: Statics, Plane Truss: Method of Joints, FE Reference Handbook 10.6

Show the worked solution
Answer
B (8.24 kips in tension)
Given
L = 37 ft, a = 26 ft, h = 17 ft, P = 15.5 kips, H = 4.0 kips
Find
force in member AC, magnitude and tension (T) or compression (C)
Handbook
Statics, Plane Truss: Method of Joints, page 98
Equation
Method of joints: sum Fx = 0 and sum Fy = 0 at each joint; reactions from sum of moments = 0; tension positive
Substitute
  1. Reactions, sum of moments about A = 0: Cy = (P·a + H·h)/L = ((15.5 kips)(26 ft) + (4.0 kips)(17 ft))/(37 ft) = 12.73 kips
  2. Sum Fy = 0: Ay = P - Cy = 2.770 kips; sum Fx = 0: Ax = -H = -4.0 kips
  3. Length AB = √(a² + h²) = √(26² + 17²) ft = 31.06 ft
  4. Joint A, sum Fy = 0: Ay + F_AB(h/AB) = 0, so F_AB = -Ay·AB/h = -(2.770 kips)(31.06 ft)/(17 ft) = -5.062 kips
  5. Joint A, sum Fx = 0: Ax + F_AB(a/AB) + F_AC = 0, so F_AC = H + Ay·a/h = 4.0 kips + (2.770 kips)(26 ft)/(17 ft) = 8.237 kips
  6. F_AC = 8.237 kips, so AC carries 8.24 kips in tension (T)
Result
8.24 kips (T), 3 significant figures
Check
with all three member forces, joint B also balances (sum Fx = 0 and sum Fy = 0 including the load); a positive member force is tension, a negative one compression.
Why the others are wrong
  • A: used h/a instead of a/h when taking the horizontal part of AB at joint A
  • C: has the right size but the wrong sense; equilibrium of joint A puts AC in tension
  • D: took the force in the inclined member AB as the force in AC

Problem 5 · FE Civil, Truss member force

A three-member plane truss has joints A (0, 0), B (5.0, 8.0) and C (15.5, 0), with coordinates in m. The members are AB, BC and AC. A is a pin support and C is a roller on a horizontal surface. Joint B carries a downward load of 170 kN. The force in member AC is most nearly:

  • A 72.0 kN (C)
  • B 72.0 kN (T)
  • C 106 kN (T)
  • D 136 kN (T)

Handbook: Statics, Plane Truss: Method of Joints, FE Reference Handbook 10.6

Show the worked solution
Answer
B (72.0 kN in tension)
Given
L = 15.5 m, a = 5.0 m, h = 8.0 m, P = 170 kN
Find
force in member AC, magnitude and tension (T) or compression (C)
Handbook
Statics, Plane Truss: Method of Joints, page 98
Equation
Method of joints: sum Fx = 0 and sum Fy = 0 at each joint; reactions from sum of moments = 0; tension positive
Substitute
  1. Reactions, sum of moments about A = 0: Cy = P·a/L = (170 kN)(5.0 m)/(15.5 m) = 54.84 kN
  2. Sum Fy = 0: Ay = P - Cy = 115.2 kN; sum Fx = 0: Ax = 0
  3. Length AB = √(a² + h²) = √(5.0² + 8.0²) m = 9.434 m
  4. Joint A, sum Fy = 0: Ay + F_AB(h/AB) = 0, so F_AB = -Ay·AB/h = -(115.2 kN)(9.434 m)/(8.0 m) = -135.8 kN
  5. Joint A, sum Fx = 0: F_AB(a/AB) + F_AC = 0, so F_AC = Ay·a/h = (115.2 kN)(5.0 m)/(8.0 m) = 71.98 kN
  6. F_AC = 71.98 kN, so AC carries 72.0 kN in tension (T)
Result
72.0 kN (T), 3 significant figures
Check
with all three member forces, joint B also balances (sum Fx = 0 and sum Fy = 0 including the load); a positive member force is tension, a negative one compression.
Why the others are wrong
  • A: has the right size but the wrong sense; equilibrium of joint A puts AC in tension
  • C: used the whole load P as the reaction at A
  • D: took the force in the inclined member AB as the force in AC

Problem 6 · FE Civil, Truss member force

A three-member plane truss has joints A (0, 0), B (2.0, 3.0) and C (7.5, 0), with coordinates in m. The members are AB, BC and AC. A is a pin support and C is a roller on a horizontal surface. Joint B carries a downward load of 160 kN and a horizontal load of 55 kN in the +x direction. The force in member AB is most nearly:

  • A 115 kN (C)
  • B 77.7 kN (C)
  • C 172 kN (C)
  • D 95.3 kN (C)

Handbook: Statics, Plane Truss: Method of Joints, FE Reference Handbook 10.6

Show the worked solution
Answer
A (115 kN in compression)
Given
L = 7.5 m, a = 2.0 m, h = 3.0 m, P = 160 kN, H = 55 kN
Find
force in member AB, magnitude and tension (T) or compression (C)
Handbook
Statics, Plane Truss: Method of Joints, page 98
Equation
Method of joints: sum Fx = 0 and sum Fy = 0 at each joint; reactions from sum of moments = 0; tension positive
Substitute
  1. Reactions, sum of moments about A = 0: Cy = (P·a + H·h)/L = ((160 kN)(2.0 m) + (55 kN)(3.0 m))/(7.5 m) = 64.67 kN
  2. Sum Fy = 0: Ay = P - Cy = 95.33 kN; sum Fx = 0: Ax = -H = -55 kN
  3. Length AB = √(a² + h²) = √(2.0² + 3.0²) m = 3.606 m
  4. Joint A, sum Fy = 0: Ay + F_AB(h/AB) = 0, so F_AB = -Ay·AB/h = -(95.33 kN)(3.606 m)/(3.0 m) = -114.6 kN
  5. F_AB = -114.6 kN, so AB carries 115 kN in compression (C)
Result
115 kN (C), 3 significant figures
Check
with all three member forces, joint B also balances (sum Fx = 0 and sum Fy = 0 including the load); a positive member force is tension, a negative one compression.
Why the others are wrong
  • B: used the reaction at C in the equilibrium of joint A
  • C: resolved AB with the horizontal run a instead of the rise h at joint A (cosine for sine)
  • D: set the member force equal to the reaction without resolving it along the member

Problem 7 · FE Civil, Truss member force

A three-member plane truss has joints A (0, 0), B (22, 18) and C (44, 0), with coordinates in ft. The members are AB, BC and AC. A is a pin support and C is a roller on a horizontal surface. Joint B carries a downward load of 21.0 kips and a horizontal load of 1.5 kips in the +x direction. The force in member AB is most nearly:

  • A 9.89 kips (C)
  • B 15.6 kips (C)
  • C 15.6 kips (T)
  • D 33.2 kips (C)

Handbook: Statics, Plane Truss: Method of Joints, FE Reference Handbook 10.6

Show the worked solution
Answer
B (15.6 kips in compression)
Given
L = 44 ft, a = 22 ft, h = 18 ft, P = 21.0 kips, H = 1.5 kips
Find
force in member AB, magnitude and tension (T) or compression (C)
Handbook
Statics, Plane Truss: Method of Joints, page 98
Equation
Method of joints: sum Fx = 0 and sum Fy = 0 at each joint; reactions from sum of moments = 0; tension positive
Substitute
  1. Reactions, sum of moments about A = 0: Cy = (P·a + H·h)/L = ((21.0 kips)(22 ft) + (1.5 kips)(18 ft))/(44 ft) = 11.11 kips
  2. Sum Fy = 0: Ay = P - Cy = 9.886 kips; sum Fx = 0: Ax = -H = -1.5 kips
  3. Length AB = √(a² + h²) = √(22² + 18²) ft = 28.43 ft
  4. Joint A, sum Fy = 0: Ay + F_AB(h/AB) = 0, so F_AB = -Ay·AB/h = -(9.886 kips)(28.43 ft)/(18 ft) = -15.61 kips
  5. F_AB = -15.61 kips, so AB carries 15.6 kips in compression (C)
Result
15.6 kips (C), 3 significant figures
Check
with all three member forces, joint B also balances (sum Fx = 0 and sum Fy = 0 including the load); a positive member force is tension, a negative one compression.
Why the others are wrong
  • A: set the member force equal to the reaction without resolving it along the member
  • C: has the right size but the wrong sense; equilibrium of joint A puts AB in compression
  • D: used the whole load P as the reaction at A

Problem 8 · FE Civil, Truss member force

A three-member plane truss has joints A (0, 0), B (4.0, 4.0) and C (11.0, 0), with coordinates in m. The members are AB, BC and AC. A is a pin support and C is a roller on a horizontal surface. Joint B carries a downward load of 70 kN. The force in member AB is most nearly:

  • A 63.0 kN (T)
  • B 49.5 kN (C)
  • C 63.0 kN (C)
  • D 99.0 kN (C)

Handbook: Statics, Plane Truss: Method of Joints, FE Reference Handbook 10.6

Show the worked solution
Answer
C (63.0 kN in compression)
Given
L = 11.0 m, a = 4.0 m, h = 4.0 m, P = 70 kN
Find
force in member AB, magnitude and tension (T) or compression (C)
Handbook
Statics, Plane Truss: Method of Joints, page 98
Equation
Method of joints: sum Fx = 0 and sum Fy = 0 at each joint; reactions from sum of moments = 0; tension positive
Substitute
  1. Reactions, sum of moments about A = 0: Cy = P·a/L = (70 kN)(4.0 m)/(11.0 m) = 25.45 kN
  2. Sum Fy = 0: Ay = P - Cy = 44.55 kN; sum Fx = 0: Ax = 0
  3. Length AB = √(a² + h²) = √(4.0² + 4.0²) m = 5.657 m
  4. Joint A, sum Fy = 0: Ay + F_AB(h/AB) = 0, so F_AB = -Ay·AB/h = -(44.55 kN)(5.657 m)/(4.0 m) = -63.00 kN
  5. F_AB = -63.00 kN, so AB carries 63.0 kN in compression (C)
Result
63.0 kN (C), 3 significant figures
Check
with all three member forces, joint B also balances (sum Fx = 0 and sum Fy = 0 including the load); a positive member force is tension, a negative one compression.
Why the others are wrong
  • A: has the right size but the wrong sense; equilibrium of joint A puts AB in compression
  • B: assumed each support carries P/2, but B is not at midspan
  • D: used the whole load P as the reaction at A

Problem 9 · FE Civil, Truss member force

A three-member plane truss has joints A (0, 0), B (9.5, 6.5) and C (14.0, 0), with coordinates in m. The members are AB, BC and AC. A is a pin support and C is a roller on a horizontal surface. Joint B carries a downward load of 115 kN and a horizontal load of 10 kN in the +x direction. The force in member BC is most nearly:

  • A 101 kN (T)
  • B 69.9 kN (C)
  • C 101 kN (C)
  • D 94.9 kN (C)

Handbook: Statics, Plane Truss: Method of Joints, FE Reference Handbook 10.6

Show the worked solution
Answer
C (101 kN in compression)
Given
L = 14.0 m, a = 9.5 m, h = 6.5 m, P = 115 kN, H = 10 kN
Find
force in member BC, magnitude and tension (T) or compression (C)
Handbook
Statics, Plane Truss: Method of Joints, page 98
Equation
Method of joints: sum Fx = 0 and sum Fy = 0 at each joint; reactions from sum of moments = 0; tension positive
Substitute
  1. Reactions, sum of moments about A = 0: Cy = (P·a + H·h)/L = ((115 kN)(9.5 m) + (10 kN)(6.5 m))/(14.0 m) = 82.68 kN
  2. Sum Fy = 0: Ay = P - Cy = 32.32 kN; sum Fx = 0: Ax = -H = -10 kN
  3. Length BC = √((L - a)² + h²) = √(4.5² + 6.5²) m = 7.906 m
  4. Joint C, sum Fy = 0: Cy + F_BC(h/BC) = 0, so F_BC = -Cy·BC/h = -(82.68 kN)(7.906 m)/(6.5 m) = -100.6 kN
  5. F_BC = -100.6 kN, so BC carries 101 kN in compression (C)
Result
101 kN (C), 3 significant figures
Check
with all three member forces, joint B also balances (sum Fx = 0 and sum Fy = 0 including the load); a positive member force is tension, a negative one compression.
Why the others are wrong
  • A: has the right size but the wrong sense; equilibrium of joint C puts BC in compression
  • B: assumed each support carries P/2, but B is not at midspan
  • D: left the horizontal load out of the moment equation for the reactions

Problem 10 · FE Civil, Truss member force

A three-member plane truss has joints A (0, 0), B (3.0, 7.0) and C (10.0, 0), with coordinates in m. The members are AB, BC and AC. A is a pin support and C is a roller on a horizontal surface. Joint B carries a downward load of 75 kN. The force in member BC is most nearly:

  • A 31.8 kN (T)
  • B 74.2 kN (C)
  • C 31.8 kN (C)
  • D 106 kN (C)

Handbook: Statics, Plane Truss: Method of Joints, FE Reference Handbook 10.6

Show the worked solution
Answer
C (31.8 kN in compression)
Given
L = 10.0 m, a = 3.0 m, h = 7.0 m, P = 75 kN
Find
force in member BC, magnitude and tension (T) or compression (C)
Handbook
Statics, Plane Truss: Method of Joints, page 98
Equation
Method of joints: sum Fx = 0 and sum Fy = 0 at each joint; reactions from sum of moments = 0; tension positive
Substitute
  1. Reactions, sum of moments about A = 0: Cy = P·a/L = (75 kN)(3.0 m)/(10.0 m) = 22.50 kN
  2. Sum Fy = 0: Ay = P - Cy = 52.50 kN; sum Fx = 0: Ax = 0
  3. Length BC = √((L - a)² + h²) = √(7.0² + 7.0²) m = 9.899 m
  4. Joint C, sum Fy = 0: Cy + F_BC(h/BC) = 0, so F_BC = -Cy·BC/h = -(22.50 kN)(9.899 m)/(7.0 m) = -31.82 kN
  5. F_BC = -31.82 kN, so BC carries 31.8 kN in compression (C)
Result
31.8 kN (C), 3 significant figures
Check
with all three member forces, joint B also balances (sum Fx = 0 and sum Fy = 0 including the load); a positive member force is tension, a negative one compression.
Why the others are wrong
  • A: has the right size but the wrong sense; equilibrium of joint C puts BC in compression
  • B: used the reaction at A in the equilibrium of joint C
  • D: used the whole load P as the reaction at C

A new problem posts every day inside the lab, with the full worked solution the same evening.

Frequently asked questions

What method should I use for truss problems on the FE?

The method of joints is the most direct when the member you need meets a joint with two unknowns. The method of sections is faster for a member in the middle of a long truss. Both are covered in the Statics chapter of the handbook.

How do I tell tension from compression?

Assume every unknown member force pulls away from the joint (tension). A positive result is tension and a negative result is compression.

Are these the same problems that post in the lab?

No. The site uses its own seeds, so these problems never repeat the daily problems posted inside the community.

Sources

  1. NCEES FE Civil CBT exam specifications (PDF). Retrieved October 3, 2026.
  2. NCEES FE Mechanical CBT exam specifications (PDF). Retrieved October 3, 2026.
  3. NCEES FE Other Disciplines CBT exam specifications (PDF). Retrieved October 3, 2026.
  4. NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.