10 FE practice problems: mechanics of materials, with solutions
These ten original problems practice mechanics of materials, a topic from the FE Civil exam specification, using the Mechanics of Materials, Uniaxial Loading and Deformation 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, Mechanics of materials
A steel rod with a diameter of 1.25 in. and a length of 7 ft carries an axial tension of 10 kips. E = 29,000 ksi. The elongation of the rod is most nearly:
Handbook: Mechanics of Materials, Uniaxial Loading and Deformation, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (0.0236 in.)
- Given
- d = 1.25 in., L = 7 ft, P = 10 kips, E = 29,000 ksi, si =
- Find
- elongation δ (in.)
- Handbook
- Mechanics of Materials, Uniaxial Loading and Deformation, page 131
- Equation
δ = PL/(AE)- Substitute
A = πd²/4 = π(1.25 in.)²/4 = 1.227 in.²δ = PL/(AE) = (10 kips)(7 ft × 12)/[(1.227 in.²)(29,000 ksi)] = 0.0236 in.
- Result
- 0.0236 in., 3 significant figures
- Check
- the strain δ/L = 0.000281 is small, as expected in the elastic range.
- Why the others are wrong
- A: used d² for the area, leaving out π/4
- C: left the length in ft
- D: used πd² for the area (the diameter as if it were the radius)
Problem 2 · FE Civil, Mechanics of materials
A rectangular beam 225 mm wide and 700 mm deep carries a maximum bending moment of 265 kN·m. The maximum bending stress is most nearly:
Handbook: Mechanics of Materials, Stresses in Beams, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (14.4 MPa)
- Given
- b = 225 mm, h = 700 mm, M = 265 kN·m, si =
- Find
- maximum bending stress σmax (MPa)
- Handbook
- Mechanics of Materials, Stresses in Beams (normal stress due to bending), page 135
- Equation
σ = Mc/I, with I = bh³/12 and c = h/2- Substitute
I = bh³/12 = (225)(700)³/12 = 6,431,000,000 mm⁴; c = h/2 = 350 mmσmax = Mc/I = (265 × 10⁶ N·mm)(350 mm)/(6,431,000,000 mm⁴) = 14.4 MPa
- Result
- 14.4 MPa, 3 significant figures
- Check
- the same result comes from σ = 6M/(bh²), the section-modulus form S = bh²/6.
- Why the others are wrong
- A: used I = bh³/3 (about the base) instead of bh³/12 (about the centroid)
- C: used the full depth h for c instead of h/2
- D: used half the moment
Problem 3 · FE Civil, Mechanics of materials
A bridge girder 40.0 m long, free to expand, warms by 39°C. Its coefficient of thermal expansion is 11.7 × 10⁻⁶ /°C. The change in length is most nearly:
Handbook: Mechanics of Materials, Thermal Deformations, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (18.3 mm)
- Given
- alpha = 11.7 × 10⁻⁶, L = 40.0 m, dT = 39°C, si =
- Find
- change in length δt (mm)
- Handbook
- Mechanics of Materials, Thermal Deformations, page 131
- Equation
δt = αL(T - To)- Substitute
δt = αL(T - To) = (11.7 × 10⁻⁶ /°C)(40.0 m × 1,000)(39°C) = 18.3 mm
- Result
- 18.3 mm, 3 significant figures
- Check
- the strain αΔT = 0.000456 is tiny, so δ is a small fraction of the length.
- Why the others are wrong
- A: doubled the free expansion as if both ends moved by δ
- C: used half the temperature change
- D: converted the temperature change to °F while using α per °C
Problem 4 · FE Civil, Mechanics of materials
A steel rod with a diameter of 0.625 in. and a length of 19 ft carries an axial tension of 22 kips. E = 29,000 ksi. The elongation of the rod is most nearly:
Handbook: Mechanics of Materials, Uniaxial Loading and Deformation, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (0.564 in.)
- Given
- d = 0.625 in., L = 19 ft, P = 22 kips, E = 29,000 ksi, si =
- Find
- elongation δ (in.)
- Handbook
- Mechanics of Materials, Uniaxial Loading and Deformation, page 131
- Equation
δ = PL/(AE)- Substitute
A = πd²/4 = π(0.625 in.)²/4 = 0.3068 in.²δ = PL/(AE) = (22 kips)(19 ft × 12)/[(0.3068 in.²)(29,000 ksi)] = 0.564 in.
- Result
- 0.564 in., 3 significant figures
- Check
- the strain δ/L = 0.00247 is small, as expected in the elastic range.
- Why the others are wrong
- B: used half the cross-sectional area
- C: used d² for the area, leaving out π/4
- D: left the length in ft
Problem 5 · FE Civil, Mechanics of materials
A solid circular steel shaft with a diameter of 2.5 in. transmits a torque of 6 kip·in. The maximum shear stress in the shaft is most nearly:
Handbook: Mechanics of Materials, Torsion, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (1.96 ksi)
- Given
- d = 2.5 in., T = 6 kip·in., si =
- Find
- maximum shear stress τmax (ksi)
- Handbook
- Mechanics of Materials, Torsion, page 134
- Equation
τ = Tr/J, with J = πd⁴/32 for a solid circular shaft- Substitute
J = πd⁴/32 = π(2.5 in.)⁴/32 = 3.835 in.⁴; c = d/2 = 1.25 in.τmax = Tc/J = (6 kip·in.)(1.25 in.)/(3.835 in.⁴) = 1.96 ksi
- Result
- 1.96 ksi, 3 significant figures
- Check
- the result equals 16T/(πd³), the closed form for a solid shaft.
- Why the others are wrong
- A: used πd⁴/16 for J (twice the polar moment), which halves the stress
- B: used J = πr⁴/2 with the diameter in place of r
- C: left π out of J = πd⁴/32
Problem 6 · FE Civil, Mechanics of materials
A steel rod with a diameter of 1.0 in. and a length of 14 ft carries an axial tension of 10 kips. E = 29,000 ksi. The elongation of the rod is most nearly:
Handbook: Mechanics of Materials, Uniaxial Loading and Deformation, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (0.0738 in.)
- Given
- d = 1.0 in., L = 14 ft, P = 10 kips, E = 29,000 ksi, si =
- Find
- elongation δ (in.)
- Handbook
- Mechanics of Materials, Uniaxial Loading and Deformation, page 131
- Equation
δ = PL/(AE)- Substitute
A = πd²/4 = π(1.0 in.)²/4 = 0.7854 in.²δ = PL/(AE) = (10 kips)(14 ft × 12)/[(0.7854 in.²)(29,000 ksi)] = 0.0738 in.
- Result
- 0.0738 in., 3 significant figures
- Check
- the strain δ/L = 0.000439 is small, as expected in the elastic range.
- Why the others are wrong
- A: used πd² for the area (the diameter as if it were the radius)
- C: used half the cross-sectional area
- D: left the length in ft
Problem 7 · FE Civil, Mechanics of materials
A bridge girder 47.0 m long, free to expand, warms by 49°C. Its coefficient of thermal expansion is 11.7 × 10⁻⁶ /°C. The change in length is most nearly:
Handbook: Mechanics of Materials, Thermal Deformations, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (26.9 mm)
- Given
- alpha = 11.7 × 10⁻⁶, L = 47.0 m, dT = 49°C, si =
- Find
- change in length δt (mm)
- Handbook
- Mechanics of Materials, Thermal Deformations, page 131
- Equation
δt = αL(T - To)- Substitute
δt = αL(T - To) = (11.7 × 10⁻⁶ /°C)(47.0 m × 1,000)(49°C) = 26.9 mm
- Result
- 26.9 mm, 3 significant figures
- Check
- the strain αΔT = 0.000573 is tiny, so δ is a small fraction of the length.
- Why the others are wrong
- A: converted the temperature change to °F while using α per °C
- B: used half the temperature change
- D: doubled the free expansion as if both ends moved by δ
Problem 8 · FE Civil, Mechanics of materials
A steel rod with a diameter of 28 mm and a length of 3.9 m carries an axial tension of 130 kN. E = 200 GPa. The elongation of the rod is most nearly:
Handbook: Mechanics of Materials, Uniaxial Loading and Deformation, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (4.12 mm)
- Given
- d = 28 mm, L = 3.9 m, P = 130 kN, E = 200 GPa, si =
- Find
- elongation δ (mm)
- Handbook
- Mechanics of Materials, Uniaxial Loading and Deformation, page 131
- Equation
δ = PL/(AE)- Substitute
A = πd²/4 = π(28 mm)²/4 = 615.8 mm²δ = PL/(AE) = (130 kN × 1,000)(3.9 m × 1,000)/[(615.8 mm²)(200 GPa × 1,000)] = 4.12 mm
- Result
- 4.12 mm, 3 significant figures
- Check
- the strain δ/L = 0.00106 is small, as expected in the elastic range.
- Why the others are wrong
- B: used d² for the area, leaving out π/4
- C: used πd² for the area (the diameter as if it were the radius)
- D: used half the cross-sectional area
Problem 9 · FE Civil, Mechanics of materials
A steel rod with a diameter of 1.0 in. and a length of 20 ft carries an axial tension of 20 kips. E = 29,000 ksi. The elongation of the rod is most nearly:
Handbook: Mechanics of Materials, Uniaxial Loading and Deformation, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (0.211 in.)
- Given
- d = 1.0 in., L = 20 ft, P = 20 kips, E = 29,000 ksi, si =
- Find
- elongation δ (in.)
- Handbook
- Mechanics of Materials, Uniaxial Loading and Deformation, page 131
- Equation
δ = PL/(AE)- Substitute
A = πd²/4 = π(1.0 in.)²/4 = 0.7854 in.²δ = PL/(AE) = (20 kips)(20 ft × 12)/[(0.7854 in.²)(29,000 ksi)] = 0.211 in.
- Result
- 0.211 in., 3 significant figures
- Check
- the strain δ/L = 0.000878 is small, as expected in the elastic range.
- Why the others are wrong
- B: used half the cross-sectional area
- C: used πd² for the area (the diameter as if it were the radius)
- D: left the length in ft
Problem 10 · FE Civil, Mechanics of materials
A steel rod with a diameter of 1.0 in. and a length of 4 ft carries an axial tension of 19 kips. E = 29,000 ksi. The elongation of the rod is most nearly:
Handbook: Mechanics of Materials, Uniaxial Loading and Deformation, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (0.0400 in.)
- Given
- d = 1.0 in., L = 4 ft, P = 19 kips, E = 29,000 ksi, si =
- Find
- elongation δ (in.)
- Handbook
- Mechanics of Materials, Uniaxial Loading and Deformation, page 131
- Equation
δ = PL/(AE)- Substitute
A = πd²/4 = π(1.0 in.)²/4 = 0.7854 in.²δ = PL/(AE) = (19 kips)(4 ft × 12)/[(0.7854 in.²)(29,000 ksi)] = 0.0400 in.
- Result
- 0.0400 in., 3 significant figures
- Check
- the strain δ/L = 0.000834 is small, as expected in the elastic range.
- Why the others are wrong
- B: put the radius into πd²/4, so the area is 4 times too small
- C: used d² for the area, leaving out π/4
- D: left the length in ft
A new problem posts every day inside the lab, with the full worked solution the same evening.
Frequently asked questions
Where is mechanics of materials in the FE Reference Handbook?
Look in the Mechanics of Materials, Uniaxial Loading and Deformation 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
- NCEES FE Civil CBT exam specifications (PDF). Retrieved October 3, 2026.
- NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.