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10 FE practice problems: beams and column buckling, with solutions

By Dr. Mohamed Elansary, PhD Environmental EngineeringUpdated

These ten original problems practice beams and column buckling, a topic from the FE Civil exam specification, using the Mechanics of Materials, Simply Supported Beams part of the FE Reference Handbook 10.6. Each problem gives the situation and the values with units. Work it with the handbook PDF open and commit to an answer before you open the solution. Every solution shows the handbook page, the equation, the substitution with units, a size check, and the mistake behind each wrong option, so a wrong pick tells you exactly what to fix.

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, Beams and column buckling

A simply supported beam spans 42 ft. A concentrated load of 16 kips acts 6 ft from the left support. Neglecting the beam's weight, the maximum bending moment is most nearly:

  • A 84.0 kip·ft
  • B 96.0 kip·ft
  • C 494 kip·ft
  • D 82.3 kip·ft

Handbook: Mechanics of Materials, Simply Supported Beams, FE Reference Handbook 10.6

Show the worked solution
Answer
D (82.3 kip·ft)
Given
L = 42 ft, P = 16 kips, a = 6 ft
Find
maximum bending moment (kip·ft)
Handbook
Mechanics of Materials, Simply Supported Beam Slopes and Deflections (maximum moment column), page 140
Equation
RA = Pb/L; Mmax = Pab/L under the load
Substitute
  1. Reaction at the left support: RA = P·b/L = 16(36)/42 = 13.71 kips
  2. Mmax (under the load) = RA·a = P·a·b/L = 16(6)(36)/42 = 82.3 kip·ft
Result
82.3 kip·ft, 3 significant figures
Check
Mmax is below PL/4 = 168 kip·ft, the largest value any position of the load can produce.
Why the others are wrong
  • A: used the uniform-load coefficient L/8
  • B: multiplied the load by its distance from the support
  • C: multiplied the reaction at A by the wrong segment length

Problem 2 · FE Civil, Beams and column buckling

A simply supported beam spans 12 ft. A concentrated load of 9 kips acts 4 ft from the left support. Neglecting the beam's weight, the maximum bending moment is most nearly:

  • A 48.0 kip·ft
  • B 27.0 kip·ft
  • C 36.0 kip·ft
  • D 24.0 kip·ft

Handbook: Mechanics of Materials, Simply Supported Beams, FE Reference Handbook 10.6

Show the worked solution
Answer
D (24.0 kip·ft)
Given
L = 12 ft, P = 9 kips, a = 4 ft
Find
maximum bending moment (kip·ft)
Handbook
Mechanics of Materials, Simply Supported Beam Slopes and Deflections (maximum moment column), page 140
Equation
RA = Pb/L; Mmax = Pab/L under the load
Substitute
  1. Reaction at the left support: RA = P·b/L = 9(8)/12 = 6.000 kips
  2. Mmax (under the load) = RA·a = P·a·b/L = 9(4)(8)/12 = 24.0 kip·ft
Result
24.0 kip·ft, 3 significant figures
Check
Mmax is below PL/4 = 27.0 kip·ft, the largest value any position of the load can produce.
Why the others are wrong
  • A: multiplied the reaction at A by the wrong segment length
  • B: used PL/4, which applies only when the load is at midspan
  • C: multiplied the load by its distance from the support

Problem 3 · FE Civil, Beams and column buckling

A simply supported beam spans 15.3 m. A concentrated load of 240 kN acts 6.6 m from the left support. Neglecting the beam's weight, the maximum bending moment is most nearly:

  • A 450 kN·m
  • B 1,580 kN·m
  • C 1,190 kN·m
  • D 901 kN·m

Handbook: Mechanics of Materials, Simply Supported Beams, FE Reference Handbook 10.6

Show the worked solution
Answer
D (901 kN·m)
Given
L = 15.3 m, P = 240 kN, a = 6.6 m
Find
maximum bending moment (kN·m)
Handbook
Mechanics of Materials, Simply Supported Beam Slopes and Deflections (maximum moment column), page 140
Equation
RA = Pb/L; Mmax = Pab/L under the load
Substitute
  1. Reaction at the left support: RA = P·b/L = 240(8.7)/15.3 = 136.5 kN
  2. Mmax (under the load) = RA·a = P·a·b/L = 240(6.6)(8.7)/15.3 = 901 kN·m
Result
901 kN·m, 3 significant figures
Check
Mmax is below PL/4 = 918 kN·m, the largest value any position of the load can produce.
Why the others are wrong
  • A: divided by 2L instead of L
  • B: multiplied the load by its distance from the support
  • C: multiplied the reaction at A by the wrong segment length

Problem 4 · FE Civil, Beams and column buckling

A simply supported beam spans 44 ft. A concentrated load of 30 kips acts 40 ft from the left support. Neglecting the beam's weight, the maximum bending moment is most nearly:

  • A 54.5 kip·ft
  • B 165 kip·ft
  • C 109 kip·ft
  • D 330 kip·ft

Handbook: Mechanics of Materials, Simply Supported Beams, FE Reference Handbook 10.6

Show the worked solution
Answer
C (109 kip·ft)
Given
L = 44 ft, P = 30 kips, a = 40 ft
Find
maximum bending moment (kip·ft)
Handbook
Mechanics of Materials, Simply Supported Beam Slopes and Deflections (maximum moment column), page 140
Equation
RA = Pb/L; Mmax = Pab/L under the load
Substitute
  1. Reaction at the left support: RA = P·b/L = 30(4)/44 = 2.727 kips
  2. Mmax (under the load) = RA·a = P·a·b/L = 30(40)(4)/44 = 109 kip·ft
Result
109 kip·ft, 3 significant figures
Check
Mmax is below PL/4 = 330 kip·ft, the largest value any position of the load can produce.
Why the others are wrong
  • A: divided by 2L instead of L
  • B: used the uniform-load coefficient L/8
  • D: used PL/4, which applies only when the load is at midspan

Problem 5 · FE Civil, Beams and column buckling

A simply supported steel beam spans 4.0 m and carries a uniform load of 17 kN/m (including its own weight). E = 200 GPa and I = 100 × 10⁶ mm⁴. The maximum deflection is most nearly:

  • A 27.2 mm
  • B 0.567 mm
  • C 5.67 mm
  • D 2.83 mm

Handbook: Mechanics of Materials, Simply Supported Beam Slopes and Deflections, FE Reference Handbook 10.6

Show the worked solution
Answer
D (2.83 mm)
Given
w = 17 kN/m, L = 4.0 m, E = 200 GPa, I = 100 × 10⁶ mm⁴, si =
Find
maximum deflection (mm)
Handbook
Mechanics of Materials, Simply Supported Beam Slopes and Deflections (table), page 140
Equation
δmax = 5wL⁴/(384EI) (simply supported, uniform load)
Substitute
  1. δ = 5wL⁴/(384EI) = 5(17 N/mm)(4,000 mm)⁴/[384(200,000 N/mm²)(100 × 10⁶ mm⁴)] = 2.83 mm
Result
2.83 mm, 3 significant figures
Check
δ/L = 1/1,412; the deflection occurs at midspan, where the moment wL²/8 is largest.
Why the others are wrong
  • A: used the cantilever formula wL⁴/(8EI)
  • B: left out the 5 in 5wL⁴/(384EI)
  • C: used half the moment of inertia

Problem 6 · FE Civil, Beams and column buckling

A simply supported steel beam spans 4.3 m and carries a uniform load of 18 kN/m (including its own weight). E = 200 GPa and I = 560 × 10⁶ mm⁴. The maximum deflection is most nearly:

  • A 0.143 mm
  • B 6.87 mm
  • C 0.715 mm
  • D 1.14 mm

Handbook: Mechanics of Materials, Simply Supported Beam Slopes and Deflections, FE Reference Handbook 10.6

Show the worked solution
Answer
C (0.715 mm)
Given
w = 18 kN/m, L = 4.3 m, E = 200 GPa, I = 560 × 10⁶ mm⁴, si =
Find
maximum deflection (mm)
Handbook
Mechanics of Materials, Simply Supported Beam Slopes and Deflections (table), page 140
Equation
δmax = 5wL⁴/(384EI) (simply supported, uniform load)
Substitute
  1. δ = 5wL⁴/(384EI) = 5(18 N/mm)(4,300 mm)⁴/[384(200,000 N/mm²)(560 × 10⁶ mm⁴)] = 0.715 mm
Result
0.715 mm, 3 significant figures
Check
δ/L = 1/6,010; the deflection occurs at midspan, where the moment wL²/8 is largest.
Why the others are wrong
  • A: left out the 5 in 5wL⁴/(384EI)
  • B: used the cantilever formula wL⁴/(8EI)
  • D: treated the total load wL as a midspan point load, PL³/(48EI)

Problem 7 · FE Civil, Beams and column buckling

A simply supported beam spans 4.0 m. A concentrated load of 85 kN acts 0.5 m from the left support. Neglecting the beam's weight, the maximum bending moment is most nearly:

  • A 85.0 kN·m
  • B 37.2 kN·m
  • C 260 kN·m
  • D 18.6 kN·m

Handbook: Mechanics of Materials, Simply Supported Beams, FE Reference Handbook 10.6

Show the worked solution
Answer
B (37.2 kN·m)
Given
L = 4.0 m, P = 85 kN, a = 0.5 m
Find
maximum bending moment (kN·m)
Handbook
Mechanics of Materials, Simply Supported Beam Slopes and Deflections (maximum moment column), page 140
Equation
RA = Pb/L; Mmax = Pab/L under the load
Substitute
  1. Reaction at the left support: RA = P·b/L = 85(3.5)/4.0 = 74.38 kN
  2. Mmax (under the load) = RA·a = P·a·b/L = 85(0.5)(3.5)/4.0 = 37.2 kN·m
Result
37.2 kN·m, 3 significant figures
Check
Mmax is below PL/4 = 85.0 kN·m, the largest value any position of the load can produce.
Why the others are wrong
  • A: used PL/4, which applies only when the load is at midspan
  • C: multiplied the reaction at A by the wrong segment length
  • D: divided by 2L instead of L

Problem 8 · FE Civil, Beams and column buckling

A simply supported beam spans 13 ft. A concentrated load of 35.5 kips acts 9 ft from the left support. Neglecting the beam's weight, the maximum bending moment is most nearly:

  • A 320 kip·ft
  • B 98.3 kip·ft
  • C 57.7 kip·ft
  • D 49.2 kip·ft

Handbook: Mechanics of Materials, Simply Supported Beams, FE Reference Handbook 10.6

Show the worked solution
Answer
B (98.3 kip·ft)
Given
L = 13 ft, P = 35.5 kips, a = 9 ft
Find
maximum bending moment (kip·ft)
Handbook
Mechanics of Materials, Simply Supported Beam Slopes and Deflections (maximum moment column), page 140
Equation
RA = Pb/L; Mmax = Pab/L under the load
Substitute
  1. Reaction at the left support: RA = P·b/L = 35.5(4)/13 = 10.92 kips
  2. Mmax (under the load) = RA·a = P·a·b/L = 35.5(9)(4)/13 = 98.3 kip·ft
Result
98.3 kip·ft, 3 significant figures
Check
Mmax is below PL/4 = 115 kip·ft, the largest value any position of the load can produce.
Why the others are wrong
  • A: multiplied the load by its distance from the support
  • C: used the uniform-load coefficient L/8
  • D: divided by 2L instead of L

Problem 9 · FE Civil, Beams and column buckling

A simply supported beam spans 7.5 m. A concentrated load of 20 kN acts 1.9 m from the left support. Neglecting the beam's weight, the maximum bending moment is most nearly:

  • A 83.6 kN·m
  • B 18.8 kN·m
  • C 28.4 kN·m
  • D 37.5 kN·m

Handbook: Mechanics of Materials, Simply Supported Beams, FE Reference Handbook 10.6

Show the worked solution
Answer
C (28.4 kN·m)
Given
L = 7.5 m, P = 20 kN, a = 1.9 m
Find
maximum bending moment (kN·m)
Handbook
Mechanics of Materials, Simply Supported Beam Slopes and Deflections (maximum moment column), page 140
Equation
RA = Pb/L; Mmax = Pab/L under the load
Substitute
  1. Reaction at the left support: RA = P·b/L = 20(5.6)/7.5 = 14.93 kN
  2. Mmax (under the load) = RA·a = P·a·b/L = 20(1.9)(5.6)/7.5 = 28.4 kN·m
Result
28.4 kN·m, 3 significant figures
Check
Mmax is below PL/4 = 37.5 kN·m, the largest value any position of the load can produce.
Why the others are wrong
  • A: multiplied the reaction at A by the wrong segment length
  • B: used the uniform-load coefficient L/8
  • D: used PL/4, which applies only when the load is at midspan

Problem 10 · FE Civil, Beams and column buckling

A simply supported beam spans 12.8 m. A concentrated load of 250 kN acts 1.3 m from the left support. Neglecting the beam's weight, the maximum bending moment is most nearly:

  • A 400 kN·m
  • B 2,580 kN·m
  • C 292 kN·m
  • D 146 kN·m

Handbook: Mechanics of Materials, Simply Supported Beams, FE Reference Handbook 10.6

Show the worked solution
Answer
C (292 kN·m)
Given
L = 12.8 m, P = 250 kN, a = 1.3 m
Find
maximum bending moment (kN·m)
Handbook
Mechanics of Materials, Simply Supported Beam Slopes and Deflections (maximum moment column), page 140
Equation
RA = Pb/L; Mmax = Pab/L under the load
Substitute
  1. Reaction at the left support: RA = P·b/L = 250(11.5)/12.8 = 224.6 kN
  2. Mmax (under the load) = RA·a = P·a·b/L = 250(1.3)(11.5)/12.8 = 292 kN·m
Result
292 kN·m, 3 significant figures
Check
Mmax is below PL/4 = 800 kN·m, the largest value any position of the load can produce.
Why the others are wrong
  • A: used the uniform-load coefficient L/8
  • B: multiplied the reaction at A by the wrong segment length
  • D: divided by 2L instead of L

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

Frequently asked questions

Where is beams and column buckling in the FE Reference Handbook?

Look in the Mechanics of Materials, Simply Supported Beams part of FE Reference Handbook 10.6. Each solution gives the exact page, so you can practice finding it in the PDF the way you will on exam day.

Are these real FE exam questions?

No. They are original problems generated from the handbook formulas and checked by code. Real exam questions are confidential, and sharing them breaks the NCEES agreement every examinee accepts.

How do I check my answer before opening the solution?

Check the units of your result and whether its size makes sense for the situation. Each solution ends with the same kind of check, so you can compare your habit with ours.

Sources

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