Mechanics of Materials is 9-14 questions on the FE Mechanical exam. It sits directly between statics, which supplies the internal loads, and mechanical design, which applies them to real components, so it is the hinge of the whole mechanics block.

Most of the difficulty is bookkeeping rather than theory. Sections, axes, units and sign conventions are where the points go, not the constitutive relationships.

Exam weight: NCEES lists Mechanics of Materials at 9-14 questions (8-13%) of the 110-question FE Mechanical 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

The specification covers stress and strain including thermal effects, shear and moment diagrams, stress transformation and Mohr's circle, torsion of circular shafts, beam deflection, combined loading, and column buckling. Expect an axial member with a temperature change, a shaft sized for allowable shear stress, and a principal-stress calculation from a plane-stress state.

Combined loading is where mechanical candidates are pushed hardest: a shaft carrying torsion and bending simultaneously, resolved into a principal or maximum shear stress. That single problem type connects to the failure theories in Mechanical Design and Analysis.

5 Free Mechanics of Materials Practice Problems

Each problem below comes from the PECivilClick FE Mechanical 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. Shear and moment diagrams

For a simply supported beam with a concentrated load P at the center, what is the maximum bending moment?

A. Shear and moment diagrams figure for FE Civil practice problem 1

A) PL/4

B) PL

C) PL/8

D) PL/2

Answer: A) PL/4

For a simply supported beam of length L with a concentrated load P at the center:

Reactions: \(R_A = R_B = P/2\)

Maximum moment occurs at the center:

\(M_{max} = R_A \times \frac{L}{2} = \frac{P}{2} \times \frac{L}{2}\)

\(\boxed{M_{max} = \frac{PL}{4}}\)

Shear force and bending moment diagrams for this beam

Problem 2 — A. Shear and moment diagrams

A cantilever beam of length 3 m carries a uniformly distributed load of 4 kN/m. What is the maximum bending moment?

A. Shear and moment diagrams figure for FE Civil practice problem 2

A) 12 kN·m

B) 36 kN·m

C) 18 kN·m

D) 6 kN·m

Answer: C) 18 kN·m

For a cantilever with UDL \(w\) over length \(L\):

Maximum moment occurs at the fixed end:

\(M_{max} = \frac{wL^2}{2}\)

\(M_{max} = \frac{4 \times 3^2}{2} = \frac{36}{2}\)

\(\boxed{M_{max} = 18 \text{ kN·m}}\)

The moment is negative (causing hogging at the fixed end).

Shear force and bending moment diagrams for this cantilever

Problem 3 — A. Shear and moment diagrams

At what location does the maximum bending moment occur in a simply supported beam with a triangular load (zero at left, w\(_{0}\) at right)?

A. Shear and moment diagrams figure for FE Civil practice problem 3

A) At 2L/3 from the left support

B) At L/3 from the left support

C) At L/2

D) At L/\(\sqrt{3}\) from the left support

Answer: D) At L/\(\sqrt{3}\) from the left support

For triangular load \(w(x) = w_0 x/L\), total load \(= w_0 L/2\), acting at \(2L/3\) from left.

Left reaction: \(R_A = \frac{w_0 L}{6}\)

Shear at distance \(x\) from left:

\(V(x) = R_A - \frac{w_0 x^2}{2L} = \frac{w_0 L}{6} - \frac{w_0 x^2}{2L}\)

Maximum moment occurs where \(V = 0\):

\(\frac{w_0 L}{6} = \frac{w_0 x^2}{2L}\)

\(x^2 = \frac{L^2}{3}\)

\(\boxed{x = \frac{L}{\sqrt{3}} \approx 0.577L}\)

Shear force and bending moment diagrams for this beam

Problem 4 — A. Shear and moment diagrams

A simply supported beam of length 6 m carries a UDL of 10 kN/m. What is the maximum shear force?

A. Shear and moment diagrams figure for FE Civil practice problem 4

A) 45 kN

B) 30 kN

C) 15 kN

D) 60 kN

Answer: B) 30 kN

Total load: \(W = wL = 10 \times 6 = 60\) kN

By symmetry, reactions: \(R_A = R_B = W/2 = 30\) kN

Maximum shear occurs at the supports:

\(\boxed{V_{max} = 30 \text{ kN}}\)

The shear diagram is linear, going from +30 kN at left support to -30 kN at right support, passing through zero at midspan.

Shear force and bending moment diagrams for this beam

Problem 5 — B. Stress transformations and Mohr's circle

The center of Mohr's circle is located at:

A) At the origin

B) τxy on the τ-axis

C) (σx - σy)/2 on the σ-axis

D) (σx + σy)/2 on the σ-axis

Answer: D) (σx + σy)/2 on the σ-axis

For Mohr's circle construction:

\(\boxed{\text{Center: } C = \frac{\sigma_x + \sigma_y}{2}}\)

on the normal stress axis.

\(\text{Radius: } R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}\)

The center represents the average normal stress, and the radius represents the maximum shear stress.

Using the FE Reference Handbook for Mechanics of Materials

Everything here is tabulated: section properties, beam deflection cases, torsion formulas and the Euler buckling relationship. The trap is the beam deflection table, which is organised by load and support case, so you must read the diagram at the top of each entry rather than matching on the formula shape. Practise finding the simply-supported-with-central-point-load case and the cantilever-with-end-load case by sight.

Four Mistakes That Cost Points

Frequently Asked Questions

How many mechanics of materials questions are on the FE Mechanical exam?

NCEES specifies 9-14 questions out of 110, roughly 8-13 percent. Combined with Statics at 9-14 and Mechanical Design and Analysis at 10-15, the solid mechanics block is the largest part of the exam.

Do I need to draw shear and moment diagrams by hand?

Often you only need one value, not the whole diagram. Read the question first: if it asks for maximum moment on a standard load case, the handbook beam tables give it directly. Draw the diagram when the loading is non-standard or when the question asks where the maximum occurs.

How much does Mohr's circle actually appear?

Stress transformation is a reliable part of this area, and it can be done with the handbook equations rather than by drawing the circle. Knowing both routes helps: the equations are faster for a single principal stress, the sketch is faster for reasoning about orientation.

Is column buckling tested on FE Mechanical?

Yes, as Euler buckling with an effective length factor. Know the four standard end conditions and their factors, and remember that buckling uses the minimum second moment of area of the cross-section.

Keep Going

These topics feed into each other on the exam:

Browse every knowledge area from the free FE Mechanical practice problem hub, see what the full bank covers on the FE Mechanical exam prep page, or plan your schedule with the FE study timeline.