Materials is worth 4 to 6 questions on the 110-question FE Civil exam, and the problems are unusually mechanical: an absolute volume mix design, a water-cement ratio, a sieve analysis, a point read off a stress-strain curve. Almost every one is solvable in under three minutes if you know which handbook table it comes from.
These free practice problems are built the way the real items are: numeric answers with distractors that come from realistic mistakes, such as using bulk specific gravity where SSD belongs, or reading ultimate strength when the question asks for yield. Each one has a full worked solution and the handbook section you should have opened.
Exam weight: NCEES lists Materials at 4-6 questions (4-6%) 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 Materials
NCEES lists Materials as 4 to 6 of the 110 questions and breaks it into three areas: mix design for concrete and asphalt; test methods and specifications for metals, concrete, aggregates, asphalt, and wood; and the physical and mechanical properties of those same materials. In practice that means absolute volume batching, water-cement ratio, aggregate gradation and fineness modulus, and Superpave or Marshall mix parameters.
The properties half leans on the stress-strain diagram: proportional limit, yield by the 0.2 percent offset, ultimate strength, modulus of elasticity, ductility measured as percent elongation, and toughness as area under the curve. Expect questions on concrete compressive strength from cylinder tests, modulus of rupture, steel grades and their yield stresses, thermal expansion of a restrained member, and the difference between hardness, strength, and stiffness.
Materials appears in both halves of the exam session, so do not treat it as an afterthought you review the night before.
5 Free 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 concrete mix uses \( 360\,\mathrm{kg} \) of cement per m\(^{3}\) with a water-cement ratio (w/c) of \( 0.50 \). What mass of mixing water is required (kg per m\(^{3}\))?
A) 160
B) 180
C) 200
D) 220
Answer: B) 180
Step 1 – Identify given information:
| Cement content | \( C = 360 \;\text{kg/m}^3 \) |
| Water-cement ratio | \( w/c = 0.50 \) |
| Find | Mass of mixing water \( W \) (kg/m\(^{3}\)) |
Step 2 – Water-cement ratio definition:
From the FE Handbook, the water-cement ratio is defined as:
\( w/c = \dfrac{W}{C} \)
Rearranging to solve for water: \( W = (w/c) \times C \)
Step 3 – Substitute and solve:
\( W = 0.50 \times 360 = 180 \;\text{kg/m}^3 \)
Since \( \rho_{water} = 1 \;\text{kg/L} \), this also equals approximately 180 L of water.
Step 4 – Answer: The required water mass is \( 180 \;\text{kg/m}^3 \). Option (A) 160 kg would correspond to \( w/c = 0.44 \); option (C) 200 kg to \( w/c = 0.56 \); option (D) 220 kg to \( w/c = 0.61 \). Only 180 kg satisfies the specified ratio.
Correct answer: (B)
Problem 2
A mix uses \( 560\,\mathrm{lb} \) of cement per yd\(^{3}\) and a w/c of \( 0.45 \). Approximately how many gallons of water are required per yd\(^{3}\)?
A) 24
B) 30.2
C) 35
D) 40
Answer: B) 30.2
Step 1 – Water weight:
\(W = 0.45 \times 560 = 252 \text{ lb}\)
Step 2 – Convert to gallons:
\(\text{gallons} = \frac{252}{8.34} = 30.2 \text{ gal}\)
Correct answer: (B)
Problem 3
A batch contains cement = \( 300\,\mathrm{kg} \) (SG = 3.15), water = \( 180\,\mathrm{kg} \), coarse aggregate = \( 1050\,\mathrm{kg} \) (SG = 2.65), fine aggregate = \( 700\,\mathrm{kg} \) (SG = 2.60), and air = \( 2\% \). What is the approximate yield of this batch (m\(^{3}\))?
A) 0.86
B) 0.92
C) 0.96
D) 1.02
Answer: C) 0.96
Step 1 – Identify given information:
| Cement | \( 300 \;\text{kg},\; SG = 3.15 \) |
| Water | \( 180 \;\text{kg} \) |
| Coarse aggregate | \( 1050 \;\text{kg},\; SG = 2.65 \) |
| Fine aggregate | \( 700 \;\text{kg},\; SG = 2.60 \) |
| Air | \( 2\% \) |
Step 2 – Yield by absolute volume method:
Yield is the total volume of all components. Each solid component volume is \( V_i = m_i / (SG_i \times 1000) \).
Step 3 – Compute each volume and sum:
\( V_c = \dfrac{300}{3150} = 0.095 \;\text{m}^3 \)
\( V_w = \dfrac{180}{1000} = 0.180 \;\text{m}^3 \)
\( V_{ca} = \dfrac{1050}{2650} = 0.396 \;\text{m}^3 \)
\( V_{fa} = \dfrac{700}{2600} = 0.269 \;\text{m}^3 \)
\( V_{air} = 0.020 \;\text{m}^3 \)
\( Y = 0.095 + 0.180 + 0.396 + 0.269 + 0.020 = 0.960 \;\text{m}^3 \)
Step 4 – Answer: The batch yield is approximately \( 0.96 \;\text{m}^3 \), meaning this batch produces slightly less than 1.00 m\(^{3}\). If the design target is 1.00 m\(^{3}\), all ingredient masses would need to be scaled up by \( 1/0.96 \approx 1.04 \).
Correct answer: (C)
Problem 4
Using the absolute volume method for \( 1.00\,\mathrm{m^3} \) of concrete: cement = \( 350\,\mathrm{kg} \) (SG = 3.15), water = \( 175\,\mathrm{kg} \), entrapped air = \( 2\% \), coarse aggregate = \( 1050\,\mathrm{kg} \) (SG = 2.65), and fine aggregate has SG = 2.65. What is the required fine aggregate mass (kg)?
A) 620
B) 710
C) 790
D) 880
Answer: C) 790
Step 1 – Identify given information:
| Cement | \( 350 \;\text{kg},\; SG = 3.15 \) |
| Water | \( 175 \;\text{kg} \) |
| Entrapped air | \( 2\% \) |
| Coarse aggregate | \( 1050 \;\text{kg},\; SG = 2.65 \) |
| Fine aggregate SG | \( 2.65 \) |
| Total volume | \( 1.00 \;\text{m}^3 \) |
Step 2 – Absolute volume method:
Each component volume equals \( V_i = m_i / (SG_i \times 1000) \). The sum of all volumes must equal 1.00 m\(^{3}\):
\( V_w + V_c + V_{air} + V_{ca} + V_{fa} = 1.00 \)
Step 3 – Compute each volume and solve:
\( V_w = \dfrac{175}{1000} = 0.175 \;\text{m}^3 \)
\( V_c = \dfrac{350}{3150} = 0.111 \;\text{m}^3 \)
\( V_{air} = 0.02 \;\text{m}^3 \)
\( V_{ca} = \dfrac{1050}{2650} = 0.396 \;\text{m}^3 \)
\( V_{fa} = 1.00 - (0.175 + 0.111 + 0.02 + 0.396) = 1.00 - 0.702 = 0.298 \;\text{m}^3 \)
\( m_{fa} = V_{fa} \times SG_{fa} \times 1000 = 0.298 \times 2650 = 789.7 \approx 790 \;\text{kg} \)
Step 4 – Answer: The required fine aggregate mass is approximately \( 790 \;\text{kg} \). This problem tests the absolute volume method, a fundamental mix design calculation. Each wrong option results from arithmetic errors or using incorrect SG values.
Correct answer: (C)
Problem 5
A mix requires \( 185\,\mathrm{kg} \) of water per m\(^{3}\) and a w/c of \( 0.42 \). What cement content is required (kg per m\(^{3}\))?
A) 370
B) 405
C) 440
D) 485
Answer: C) 440
Step 1 – Identify given information:
| Water content | \( W = 185 \;\text{kg/m}^3 \) |
| Water-cement ratio | \( w/c = 0.42 \) |
| Find | Cement content \( C \) (kg/m\(^{3}\)) |
Step 2 – Rearrange the w/c ratio formula:
From \( w/c = W/C \), solving for cement:
\( C = \dfrac{W}{w/c} \)
Step 3 – Substitute and solve:
\( C = \dfrac{185}{0.42} = 440.5 \approx 440 \;\text{kg/m}^3 \)
Step 4 – Answer: The required cement content is approximately \( 440 \;\text{kg/m}^3 \). A lower w/c ratio demands more cement for the same water content, producing stronger concrete. Options (A) and (B) underestimate cement; option (D) overestimates it.
Correct answer: (C)
Using the FE Reference Handbook for Materials
Materials answers live in two places. Search absolute volume for concrete batching and the specific gravity relationships, and go to the Civil Engineering Materials pages for asphalt mix properties, aggregate gradation, and steel designations. The stress-strain definitions sit in Mechanics of Materials, not Materials. Learn both locations by name before exam day so you are not searching a generic word like strength under time pressure.
Four Mistakes That Cost Points
- Ignoring the aggregate moisture correction Mix design proportions assume saturated surface-dry aggregate. When the problem gives a moisture content above or below absorption, you must adjust both the aggregate batch weight and the mixing water. Candidates compute the design batch correctly and then report it without the correction.
- Reading ultimate strength instead of yield On a stress-strain curve the peak is ultimate strength, but design questions usually want yield, found at the proportional limit or by the 0.2 percent offset line. If the curve has no sharp knee, the offset construction is required, not the highest point on the plot.
- Confusing the specific gravity terms in asphalt volumetrics Asphalt problems throw Gmb, Gmm, Gsb, and Gse at you in one sentence. Air voids use bulk and maximum theoretical specific gravity; VMA uses bulk aggregate specific gravity. Picking the wrong G gives a clean-looking number that often matches a distractor.
- Treating hardness, strength, and stiffness as one property Modulus of elasticity is stiffness, yield is strength, Brinell or Rockwell numbers are hardness, and area under the curve is toughness. Questions often describe a material qualitatively and ask which property changed; heat treating steel can raise hardness while cutting ductility with no change in modulus.
Frequently Asked Questions
How many Materials questions are on the FE Civil exam?
NCEES allocates 4 to 6 questions to Materials out of 110 on the FE Civil exam. That is a small block, but the questions are among the fastest on the test because most reduce to one handbook equation or a single table lookup. Getting all of them is realistic and buys time for longer structural or water resources problems.
Do I need to memorize concrete mix design procedures?
No. The absolute volume method, specific gravity relationships, and water-cement ratio are all in the FE Reference Handbook. What you need is the sequence: convert each ingredient weight to volume using its specific gravity and the unit weight of water, sum them, and let the aggregate fill the remainder of the cubic yard or cubic meter.
Is asphalt mix design tested as heavily as concrete?
Concrete appears more often, but asphalt shows up regularly through volumetrics: air voids, voids in mineral aggregate, voids filled with asphalt, and effective binder content. The arithmetic is short once you have the right specific gravities identified. Practice a few volumetric problems so the symbols stop looking interchangeable under exam pressure.
What is the fastest way to study Materials for the FE Civil exam?
Work problems rather than reading. Spend one session on concrete batching and moisture corrections, one on asphalt volumetrics and gradation, and one on stress-strain interpretation and material property comparisons. Three focused sessions cover the entire NCEES outline for this area, which is why Materials is usually the best return per hour on the whole exam.
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.