Mechanics of materials carries 8 to 12 questions on the FE Civil exam and feeds directly into the structural and geotechnical items, so points here pay twice. The area rewards a small set of well-drilled moves: draw the shear and moment diagram, get the section property, apply the right stress formula, check the sign.
Every solution below shows the diagram or the stress element, not only the arithmetic, because the setup is where these problems are won or lost. Where a handbook table replaces a derivation, such as the standard beam deflection cases, the solution names the case to look up instead of integrating.
Exam weight: NCEES lists Mechanics of Materials at 8-12 questions (7-11%) of the 110-question FE Civil exam. Work each problem below on paper first, then reveal the worked solution — reading a solution you have not attempted builds recognition, not recall.
What NCEES Tests in Mechanics of Materials
NCEES specifies 8 to 12 mechanics of materials questions out of 110, about 7 to 11 percent, tying it with statics as the largest mechanics area. The core is stress and strain: axial members under load and temperature change, Hooke's law and Poisson's ratio, thin-walled pressure vessels, torsion of circular shafts, and bending stress from M c over I with transverse shear from V Q over I b.
Beyond that you get shear and moment diagrams, beam deflection from the handbook's standard cases, combined loading where axial and bending stresses superpose on the same fiber, principal stresses and Mohr's circle, and Euler buckling of columns with the effective length factor that depends on end conditions. Statically indeterminate axial members appear as compatibility problems.
5 Free Mechanics of Materials Practice Problems
Each problem below comes from the PECivilClick FE Civil question bank and matches the style, difficulty and format of the real exam. Attempt each one under a three-minute limit — roughly the pace the exam demands.
Problem 1
A steel rod carries axial load \( P = 200\,\mathrm{kN} \) with cross-sectional area \( A = 500\,\mathrm{mm^2} \). Given \( E = 200\,\mathrm{GPa} \) and \( \nu = 0.30 \), what is the lateral strain \( \varepsilon_{\mathrm{lat}} \)?
A) \( -6.0 imes 10^{-4} \)
B) \( -2.0 imes 10^{-3} \)
C) \( 6.0 imes 10^{-4} \)
D) \( 2.0 imes 10^{-3} \)
Answer: A) \( -6.0 imes 10^{-4} \)
Step 1 – Identify given information:
| Parameter | Value |
|---|---|
| \( P \) | \( 200 \;\text{kN} \) |
| \( A \) | \( 500 \;\text{mm}^2 \) |
| \( E \) | \( 200 \;\text{GPa} \) |
| \( \nu \) | \( 0.30 \) |
Step 2 – Poisson's ratio relates lateral and axial strain (FE Handbook):
\( \varepsilon_{\text{lat}} = -\nu \varepsilon_{\text{axial}} \), where \( \varepsilon_{\text{axial}} = \frac{\sigma}{E} = \frac{P}{AE} \)
Step 3 – Solve:
Axial stress: \( \sigma = \frac{P}{A} = \frac{200{,}000}{500 \times 10^{-6}} = 400 \;\text{MPa} \)
Axial strain: \( \varepsilon_x = \frac{400 \times 10^6}{200 \times 10^9} = 0.002 \)
Lateral strain: \( \varepsilon_{\text{lat}} = -0.30 \times 0.002 = -6.0 \times 10^{-4} \)
Step 4 – Answer: The lateral strain is \( -6.0 \times 10^{-4} \). The negative sign indicates the rod contracts laterally when stretched axially (Poisson effect). Option (B) forgets to apply \( \nu \); options (C) and (D) have incorrect signs.
Correct answer: (A)
Problem 2
A plane stress state has \( \sigma_x = 80\,\mathrm{MPa} \), \( \sigma_y = -20\,\mathrm{MPa} \), and \( \tau_{xy} = 30\,\mathrm{MPa} \). What is the maximum in-plane shear stress \( \tau_{\mathrm{max}} \) (MPa)?
A) 40
B) 50
C) 58.3
D) 70
Answer: C) 58.3
Step 1 – Identify the given stress state:
| Parameter | Value |
|---|---|
| \( \sigma_x \) | \( 80 \;\text{MPa} \) (tension) |
| \( \sigma_y \) | \( -20 \;\text{MPa} \) (compression) |
| \( \tau_{xy} \) | \( 30 \;\text{MPa} \) |
Step 2 – Recall the maximum in-plane shear stress formula from the FE Handbook. This equals the radius of Mohr's circle:
\( \tau_{\max} = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} \)
Step 3 – Compute \( \frac{\sigma_x - \sigma_y}{2} \):
\( \frac{\sigma_x - \sigma_y}{2} = \frac{80 - (-20)}{2} = \frac{100}{2} = 50 \;\text{MPa} \)
Note: Since \( \sigma_y \) is negative (compression), subtracting it adds to the difference: \( 80 - (-20) = 80 + 20 = 100 \).
Step 4 – Substitute into the formula:
\( \tau_{\max} = \sqrt{(50)^2 + (30)^2} = \sqrt{2{,}500 + 900} = \sqrt{3{,}400} = 58.3 \;\text{MPa} \)
Note: The center of Mohr's circle is at \( \sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2} = \frac{80 + (-20)}{2} = 30 \;\text{MPa} \), and the radius (\( \tau_{\max} \)) = 58.3 MPa.
Correct answer: (C)
Problem 3
For \( \sigma_x = 80\,\mathrm{MPa} \), \( \sigma_y = -20\,\mathrm{MPa} \), and \( \tau_{xy} = 30\,\mathrm{MPa} \), what are the orientations of the principal stress planes \( \theta_p \) (degrees) relative to the +x-axis?
A) (−74.5°; 15.5°)
B) (−60.0°; 30.0°)
C) (−45.0°; 45.0°)
D) (−15.5°; 74.5°)
Answer: A) (−74.5°; 15.5°)
Step 1 – Identify given information:
| Parameter | Value |
|---|---|
| \( \sigma_x \) | \( 80 \;\text{MPa} \) |
| \( \sigma_y \) | \( -20 \;\text{MPa} \) |
| \( \tau_{xy} \) | \( 30 \;\text{MPa} \) |
Step 2 – Principal angle formula (FE Handbook):
\( \tan(2\theta_p) = \frac{2\tau_{xy}}{\sigma_x - \sigma_y} \)
This gives two angles \( 90^\circ \) apart — the orientations of the two principal planes.
Step 3 – Solve:
\( \tan(2\theta_p) = \frac{2(30)}{80 - (-20)} = \frac{60}{100} = 0.6 \)
\( 2\theta_p = \arctan(0.6) = 30.96^\circ \)
\( \theta_p = 15.5^\circ \)
The second principal plane is \( 90^\circ \) away:
\( \theta_{p2} = 15.5^\circ - 90^\circ = -74.5^\circ \)
Step 4 – Answer: The principal planes are at \( (-74.5^\circ,\; 15.5^\circ) \). On Mohr's circle, the two principal directions are always \( 90^\circ \) apart in physical space (\( 180^\circ \) on the circle). Options (B) and (C) correspond to common errors of using incorrect tangent arguments.
Correct answer: (A)
Problem 4
For \( \sigma_x = 80\,\mathrm{MPa} \), \( \sigma_y = -20\,\mathrm{MPa} \), and \( \tau_{xy} = 30\,\mathrm{MPa} \), what are the principal stresses \( (\sigma_1,\, \sigma_2) \) in MPa?
A) (70; −10)
B) (88.3; −28.3)
C) (58.3; 30.0)
D) (110; −50)
Answer: B) (88.3; −28.3)
Step 1 – Identify given information:
| Parameter | Value |
|---|---|
| \( \sigma_x \) | \( 80 \;\text{MPa} \) |
| \( \sigma_y \) | \( -20 \;\text{MPa} \) |
| \( \tau_{xy} \) | \( 30 \;\text{MPa} \) |
Step 2 – Principal stress formulas (FE Handbook):
The principal stresses are the center of Mohr's circle plus/minus the radius:
\( \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} \)
Step 3 – Solve:
Average stress (center of Mohr's circle):
\( \sigma_{\text{avg}} = \frac{80 + (-20)}{2} = \frac{60}{2} = 30 \;\text{MPa} \)
Radius (from MOM001):
\( R = \sqrt{\left(\frac{80-(-20)}{2}\right)^2 + 30^2} = \sqrt{50^2 + 30^2} = \sqrt{3400} = 58.3 \;\text{MPa} \)
Therefore:
\( \sigma_1 = 30 + 58.3 = 88.3 \;\text{MPa} \)
\( \sigma_2 = 30 - 58.3 = -28.3 \;\text{MPa} \)
Step 4 – Answer: The principal stresses are \( (88.3,\; -28.3) \) MPa. Option (A) incorrectly uses only half the difference; option (D) doubles the actual values. The negative sign on \( \sigma_2 \) confirms one principal plane is in compression.
Correct answer: (B)
Problem 5
An aluminum bar (\( E = 70\,\mathrm{GPa} \)) has length \( L = 2.0\,\mathrm{m} \) and area \( A = 800\,\mathrm{mm^2} \). If it is subjected to an axial tensile load \( P = 50\,\mathrm{kN} \) (elastic), what is the elongation \( \delta \) (mm)?
A) 0.9
B) 1.79
C) 2.7
D) 3.6
Answer: B) 1.79
Step 1 – Identify given information:
| Parameter | Value |
|---|---|
| \( E \) | \( 70 \;\text{GPa} \) |
| \( L \) | \( 2.0 \;\text{m} \) |
| \( A \) | \( 800 \;\text{mm}^2 \) |
| \( P \) | \( 50 \;\text{kN} \) |
Step 2 – Axial deformation formula (FE Handbook):
\( \delta = \frac{PL}{AE} \)
Step 3 – Solve:
\( \delta = \frac{50{,}000 \times 2.0}{800 \times 10^{-6} \times 70 \times 10^9} = \frac{100{,}000}{56{,}000{,}000} = 1.786 \times 10^{-3} \;\text{m} = 1.79 \;\text{mm} \)
Step 4 – Answer: The elongation is \( 1.79 \;\text{mm} \). A common mistake is to forget unit conversion for area (\( 800 \;\text{mm}^2 = 800 \times 10^{-6} \;\text{m}^2 \)), which would yield incorrect results.
Correct answer: (B)
Using the FE Reference Handbook for Mechanics of Materials
Two handbook resources decide your speed here: the table of simply supported and cantilever beam deflection cases, and the Mohr's circle and principal stress relations. Find both before exam day and note the sign convention the handbook uses for shear stress on the circle, because it differs from some textbooks and flips which principal plane you report.
Four Mistakes That Cost Points
- Using the wrong I for the axis that matters Bending about the strong axis uses I about that axis; buckling uses the smallest I of the section unless bracing prevents it. Reading the wrong column of the section table produces an answer that looks reasonable and is one of the offered choices.
- Putting c at the wrong distance In M c over I, c is measured from the neutral axis to the fiber you care about, which is only half the depth for a symmetric section. For a tee or an unsymmetric built-up shape the top and bottom values differ, and the larger one governs.
- Ignoring the effective length factor in buckling Euler's formula uses K L, not L. A fixed-pinned column is not a pinned-pinned column, and forgetting K changes the critical load by a factor of two or four. Identify the end conditions from the sketch before you touch the equation.
- Superposing stresses of different types Axial and bending stresses add algebraically because both act normal to the section. Shear and normal stresses do not add; they combine through Mohr's circle or the principal stress equations. Adding a torsional shear to a bending normal stress is a common and expensive slip.
Frequently Asked Questions
How many mechanics of materials questions are on the FE Civil exam?
NCEES specifies 8 to 12 questions out of 110, roughly 7 to 11 percent. It shares top billing with statics among the mechanics areas, and the skills carry over into the structural questions, where beam behavior, section properties and column capacity reappear in a design context.
Do I have to draw shear and moment diagrams on the exam?
Usually you only need one value, not the full diagram, but sketching it is still the fastest reliable route. Draw shear from the reactions, remember that moment is the area under the shear diagram, and maximum moment occurs where shear crosses zero. That single rule answers a surprising share of these questions.
Is Mohr's circle tested on the FE Civil exam?
Yes, principal stresses and maximum in-plane shear stress are part of the specification. You are rarely asked to draw the circle; you are asked for the numbers. The handbook gives the principal stress and maximum shear equations directly, so plugging in sigma x, sigma y and tau xy is often faster than sketching.
Do I need to memorize beam deflection formulas?
No. The NCEES FE Reference Handbook includes standard deflection and slope cases for simply supported and cantilever beams under point loads, uniform loads and end moments. Your job is recognizing which case the problem matches, and knowing that superposition lets you add two cases when the loading is a combination of them.
Keep Going
Work through the other FE Civil knowledge areas with more free practice problems, review the full FE Civil exam topic breakdown, or plan your preparation with our FE exam study guide and study timeline.