العربية

FE

Saudi Council of Engineers FE Exam: Civil Engineering Part: free practice test

Start with 10 free questions to see what needs more practice. Finish to see your score, weakest units and every option explained.

Back to Saudi Council of Engineers FE Exam: Civil Engineering Part

Practice test questions with answers and explanations

Try the questions in the test above first, then open each one here to see the correct answer and why every other option is wrong.

01A solid circular steel shaft of diameter 40 mm and length 2 m with shear modulus G=80 GPa carries a torque of 1 kN·m. What is the total angle of twist between its ends?

Practice question Unit: Structural Analysis

  1. A5.70° ✓ Correct answer

    Correct: J=\pi d^4/32=2.513\times10^5 mm⁴ and \varphi=TL/(GJ)=0.0995 rad =5.70^\circ.

  2. B11.4°

    This results from using J=\pi d^4/64 (the bending second moment) instead of the polar moment, which doubles the angle.

  3. C2.85°

    This results from using J=\pi d^4/16, twice the correct polar moment.

  4. D0.0995°

    This is the value in radians read as if it were degrees.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Civil Engineering Standards (2014), Topic T1: Structural Analysis, indicator CE-T1-01; standard formula: \varphi=TL/(GJ)

02A T-section has a top flange of 200 mm × 40 mm and a web of 40 mm × 160 mm below it, giving a total depth of 200 mm. A bending moment of 10 kN·m causes tension in the bottom fibres. What is the tensile stress at the bottom of the section?

Practice question Unit: Structural Analysis

  1. A27.0 MPa ✓ Correct answer

    Correct: \bar y=64.4 mm from the top, I=50.3\times10^6 mm⁴, and \sigma=10^7\times135.6/(50.3\times10^6)=27.0 MPa.

  2. B12.8 MPa

    This results from using the distance to the top fibre (64.4 mm) instead of the bottom fibre.

  3. C19.9 MPa

    This results from using h/2=100 mm as if the section were symmetric.

  4. D92.1 MPa

    This results from ignoring the parallel-axis terms when computing I.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Civil Engineering Standards (2014), Topic T1: Structural Analysis, indicator CE-T1-02; standard formula: \sigma=Mc/I with the parallel-axis theorem

03A 150 mm concrete slab (concrete unit weight 24 kN/m³) carries a 1.5 kPa floor finish and a 3 kPa live load. What is the factored line load on a supporting beam with a tributary width of 3 m? Neglect the beam self-weight and use 1.2D + 1.6L.

Practice question Unit: Structural Design

  1. A21.42 kN/m

    This is 1.4D alone over 3 m, which does not govern because 1.2D + 1.6L is larger.

  2. B24.30 kN/m

    This is the unfactored service load (D + L): (5.1 + 3) × 3 = 24.30 kN/m.

  3. C32.76 kN/m ✓ Correct answer

    Correct: D = 0.15 × 24 + 1.5 = 5.1 kPa, wu = 1.2 × 5.1 + 1.6 × 3 = 10.92 kPa, and over 3 m this is 32.76 kN/m.

  4. D35.28 kN/m

    This swaps the factors (1.6D + 1.2L) and then multiplies by 3 m.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Civil Engineering Standards (2014), Topic T2: Structural Design, indicator CE-T2-01

04In a sieve analysis of a sand, the cumulative percentages retained on the 4.75, 2.36, 1.18, 0.60, 0.30 and 0.15 mm sieves are 2, 12, 30, 55, 80 and 95%, respectively. What is the fineness modulus of the sand?

Practice question Unit: Materials

  1. A0.95

    This value comes from adding the individual percentages retained on each sieve (95%) instead of the cumulative ones, a common mistake.

  2. B3.26

    This value comes from adding the cumulative percentages passing (326) instead of those retained, which reverses the meaning of the fineness modulus.

  3. C4.57

    This value comes from dividing the sum (274) by the number of sieves (6) instead of by 100.

  4. D2.74 ✓ Correct answer

    Fineness modulus = sum of cumulative percentages retained ÷ 100 = (2 + 12 + 30 + 55 + 80 + 95) ÷ 100 = 2.74.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Civil Engineering Standards (2014), Topic T3: Materials, indicator CE-T3-01; standard formula: FM = Σ(cumulative % retained) / 100

05A fully saturated soil has void ratio e = 0.65 and specific gravity of solids G_s = 2.70. What is the saturated unit weight \gamma_{sat}? (Take \gamma_w = 9.81 kN/m³)

Practice question Unit: Geotechnical Engineering

  1. A16.1 kN/m³

    This is the dry unit weight G_s\gamma_w/(1+e), not the saturated one.

  2. B10.1 kN/m³

    This is the submerged (buoyant) unit weight (G_s-1)\gamma_w/(1+e).

  3. C19.9 kN/m³ ✓ Correct answer

    \gamma_{sat}=(G_s+e)\gamma_w/(1+e)=3.35\times 9.81/1.65\approx 19.9 kN/m³, so it is correct.

  4. D26.5 kN/m³

    This is the unit weight of the solid grains alone, G_s\gamma_w, without voids.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Civil Engineering Standards (2014), Topic T4: Geotechnical Engineering, indicator CE-T4-03

06A flat sign panel of frontal area 2 m² faces an air stream of density 1.2 kg/m³ at 20 m/s, with drag coefficient C_D = 1.2. What is the drag force?

Practice question Unit: Water Resources Engineering

  1. A1152 N

    The factor \tfrac{1}{2} of the dynamic pressure was omitted, doubling the force.

  2. B28.8 N

    The velocity was not squared; drag is proportional to the square of velocity.

  3. C576 N ✓ Correct answer

    F_D = \tfrac{1}{2} C_D \rho V^2 A = 0.5 \times 1.2 \times 1.2 \times 400 \times 2 = 576 N.

  4. D480 N

    The drag coefficient C_D was left out; it enters the formula directly.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Civil Engineering Standards (2014), Topic T5: Water Resources Engineering, indicator CE-T5-01; standard formula: F_D = \tfrac{1}{2} C_D \rho V^2 A

07A contaminant moves with groundwater in a sand aquifer with hydraulic conductivity K = 1 \times 10^{-4} m/s, porosity n = 0.25 and hydraulic gradient i = 0.005. About how many days does it take to travel 60 m to a well?

Practice question Unit: Environmental Engineering

  1. A347 days ✓ Correct answer

    Seepage velocity v = Ki/n = 2\times10^{-6} m/s = 0.1728 m/day, so t = 60/0.1728 \approx 347 days.

  2. B1389 days

    This results from using the Darcy velocity without dividing by porosity.

  3. C5556 days

    This results from multiplying the velocity by porosity instead of dividing.

  4. D87 days

    This results from dividing the velocity by porosity squared.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Civil Engineering Standards (2014), Topic T6: Environmental Engineering, indicator CE-T6-02

08A car descends a 4% downgrade at 90 km/h and the braking friction coefficient is 0.30. What is the braking distance? Use d_b = V^2/[254(f + G)], where G is the road grade (negative downhill).

Practice question Unit: Transportation Engineering

  1. A94 m

    It comes from adding the grade to friction as if the road were uphill. Here the road descends, so the distance gets longer.

  2. B106 m

    This is the braking distance on a level road. It ignores the effect of the downgrade.

  3. C61 m

    It results from halving the correct distance through a formula slip (using 508 instead of 254).

  4. D123 m ✓ Correct answer

    The denominator is 254 × (0.30 − 0.04) = 66.0 and the numerator is 8100, so the distance is about 123 m.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Civil Engineering Standards (2014), Topic T7: Transportation Engineering, indicator CE-T7-01

09What is the present worth of equal annual payments of 10,000 SAR made at the end of each year for 10 years, at 6% interest per year?

Practice question Unit: Construction Management

  1. A73,601 SAR ✓ Correct answer

    The present worth is P = A\,[(1+i)^n-1]/[i(1+i)^n] = 10{,}000\times7.3601 \approx 73{,}601 SAR.

  2. B94,340 SAR

    This equals 100{,}000/1.06, which discounts the total of the payments by only one year.

  3. C100,000 SAR

    This is the plain sum of the payments with no discounting, ignoring the time value of money.

  4. D131,808 SAR

    This is the future worth of the series, A\,[(1+i)^n-1]/i, not its present worth.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Civil Engineering Standards (2014), Topic T8: Construction Management, indicator CE-T8-01

10An aerial camera takes square photographs of 230 mm × 230 mm at a scale of 1 : 10,000 with 60% forward overlap. What is the ground distance between two successive exposure stations (air base)?

Practice question Unit: Surveying

  1. A920 m ✓ Correct answer

    The photo covers 2300 m on the ground and 40% of it is new coverage, so the air base = 2300 × 0.40 = 920 m.

  2. B1380 m

    This uses the 60% overlap instead of the 40% of new coverage.

  3. C460 m

    This is half the correct air base, from dividing by 2 without reason.

  4. D2300 m

    This is the ground coverage of one photo, not the spacing between exposures; the overlap is ignored.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Civil Engineering Standards (2014), Topic T9: Surveying, indicator CE-T9-02

Keep preparing for Saudi Council of Engineers FE Exam: Civil Engineering Part

See also: Engineering Exams