Free FE Civil practice problems

Ten real questions from our FE Civil bank, drawn from 10 of its 14 study topics, with fully worked solutions and the same figures our members get. No signup, no email. Try each one first, then reveal the solution.

10 with figures10 topicsFull worked solutions

1
medium
A. Earned value & CPM scheduling
In the activity-on-node network shown for a storm-sewer replacement, the durations are A 6, B 9, C 14, D 4, E 2, F 7 and G 10 days. A, B and C start at day 0; D follows A and B; E and F follow D; G follows C; E, F and G precede End. The total float of activity D is most nearly:
Figure — drag or pinch to move around
1. 4 d
2. 7 d
3. 9 d
4. 12 d
Show the answer & worked solution
Answer: A.4 d
Float $= LS - ES = LF - EF$ , with $ES$ = latest $EF$ of the predecessors and $LF$ = earliest $LS$ of the successors (FE Reference Handbook, Civil Engineering — Construction, Nomenclature). Forward pass: $ES_D = \max(EF_A,\, EF_B) = \max(6,\, 9) = 9$ d, so $EF_D = 9 + 4 = 13$ d; then $EF_E = 13 + 2 = 15$ , $EF_F = 13 + 7 = 20$ and $EF_G = 14 + 10 = 24$ d. The project takes $24$ d, set by the longest path (C–G). Backward pass: $LS_E = 24 - 2 = 22$ and $LS_F = 24 - 7 = 17$ d, so $LF_D = \min(22,\, 17) = 17$ d (F controls) and $LS_D = 17 - 4 = 13$ d. Float of D $= LS_D - ES_D = 13 - 9 = 4$ d (equally $LF_D - EF_D = 17 - 13$ ). Distractors: took the EARLIER finish at D's merge ( $ES_D = EF_A = 6$ d) in the forward pass, giving 7 d; took the LATER late start at D's burst ( $LF_D = LS_E = 22$ d) in the backward pass, giving 9 d; and measured D against the SHORTEST path through it (A–D–E) instead of the longest, reversing both rules, giving 12 d.
2
medium
A. Time value of money & equivalence
An engineer borrows $200,000 to be repaid in 15 equal end-of-year payments at 6% per year. The annual payment is most nearly:
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1. $8,593
2. $13,333
3. $20,593
4. $25,333
Show the answer & worked solution
Answer: C.$20,593
The capital recovery factor (FE Reference Handbook 10.6 p. 235) converts a present amount into equal end-of-year payments: $(A/P,\ 6\%,\ 15)=\dfrac{i(1+i)^n}{(1+i)^n-1}=\dfrac{0.06\,(1.06)^{15}}{(1.06)^{15}-1}=0.10296$ , printed as 0.1030 in the handbook's 6% factor table. The annual payment is $200,000 × 0.10296 = $20,593. Distractors: $P/n$ , spreading the principal evenly and ignoring interest ($13,333); the sinking-fund factor $(A/F)$ read in place of $(A/P)$ ($8,593), which treats the loan as a future sum; and simple interest on the full principal for all 15 years, spread evenly, $P(1+in)/n$ ($25,333), which charges interest on money already repaid.
3
medium
A. Fluid properties, statics & energy/momentum
Water flows steadily through a pipe that contracts from a diameter $D_1 = 200$ mm to $D_2 = 150$ mm. The upstream velocity is $V_1 = 1.5$ m/s. The velocity in the smaller section is most nearly:
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1. 0.844 m/s
2. 2 m/s
3. 2.67 m/s
4. 4.74 m/s
Show the answer & worked solution
Answer: C.2.67 m/s
Continuity (FE Reference Handbook 10.6 p. 185): the same flow $Q=Av$ passes both sections, so $A_1V_1=A_2V_2$ . With $A=\pi D^2/4$ the $\pi/4$ cancels, leaving $V_2=V_1\left(\dfrac{D_1}{D_2}\right)^2=1.5\left(\dfrac{200}{150}\right)^2=2.67$ m/s, faster in the smaller section, as it must be. Distractors: the diameter ratio used un-squared, $V_1D_1/D_2$ (2 m/s), forgetting that area goes as $D^2$ ; the area ratio inverted, $V_1(D_2/D_1)^2$ (0.844 m/s), which would have the water slow down in the smaller section; and the ratio squared twice, $V_1(D_1/D_2)^4$ (4.74 m/s), the area-ratio squaring applied a second time.
4
medium
A. Index properties, phase relations & effective stress
At the Memphis levee borrow area the water table is 3 m below the ground surface. The soil above it is moist, with $\gamma = 16$ kN/m³; below it the soil is saturated, with $\gamma_{sat} = 19.5$ kN/m³. The vertical effective stress at point A, 3.5 m below the water table, is most nearly:
Figure — drag or pinch to move around
1. 52.5 kPa
2. 69.7 kPa
3. 81.9 kPa
4. 116 kPa
Show the answer & worked solution
Answer: C.81.9 kPa
Effective stress (FE Reference Handbook, Civil Engineering — Geotechnical): $\sigma' = \sigma - u$ with $u = h_u\gamma_w$ . Total stress at A: $\sigma = (16\ \text{kN/m}^3)(3\ \text{m}) + (19.5\ \text{kN/m}^3)(3.5\ \text{m}) = 116.25$ kPa. A is 3.5 m below the water table, so $u = (9.81\ \text{kN/m}^3)(3.5\ \text{m}) = 34.34$ kPa. $\sigma' = 116.25 - 34.34 = 81.9$ kPa. Distractors: reported the total stress $\sigma$ (pore pressure never subtracted), 116 kPa; measured the pressure head from the ground surface ( $u = 9.81 \times 6.5$ ), as if the water table were at the surface, 52.5 kPa; and used the moist $\gamma = 16$ kN/m³ for the full 6.5 m, ignoring the change at the water table, 69.7 kPa.
5
medium
A. Stress, strain, deformation & Mohr's circle
A beam of solid rectangular cross-section, width $b=50\,\text{mm}$ and depth $h=100\,\text{mm}$ , carries an internal bending moment $M=5\,\text{kN}\cdot\text{m}$ . The maximum bending (flexural) stress is most nearly:
Figure — drag or pinch to move around
1. 10 MPa
2. 15 MPa
3. 40 MPa
4. 60 MPa
Show the answer & worked solution
Answer: D.60 MPa
Maximum bending stress (Mechanics of Materials — Stresses in Beams): $\sigma=\dfrac{Mc}{I}$ , with the centroidal $I=\dfrac{bh^3}{12}=\dfrac{(50)(100)^3}{12}=4.167\times10^{6}\ \text{mm}^4$ and $c=h/2=50\ \text{mm}$ , which for a rectangle is $\sigma=\dfrac{6M}{bh^2}$ . With $M=5\ \text{kN·m}=5\times10^{6}\ \text{N·mm}$ : $\sigma=\dfrac{6(5\times10^{6}\ \text{N·mm})}{(50\ \text{mm})(100\ \text{mm})^2}=60$ MPa. Distractors: used $I=bh^3/3$ , the area table's moment of inertia about the BASE, instead of the centroidal $bh^3/12$ (¼ of the true stress); dropped the factor 6 (used $\sigma=M/bh^2$ ); and divided by the plastic section modulus $Z=bh^2/4$ instead of the elastic $S=bh^2/6$ (two-thirds of the true stress).
6
medium
A. Equilibrium, trusses, centroids & friction
A simply-supported beam of span $L=6$ m carries a uniform load $w=4$ kN/m over its entire length plus a point load $P=12$ kN located 2 m from the left support (A). The pin reaction at A is most nearly:
Figure — drag or pinch to move around
1. 10 kN
2. 16 kN
3. 18 kN
4. 20 kN
Show the answer & worked solution
Answer: D.20 kN
Replace the uniform load by its resultant, $wL=(4\ \text{kN/m})(6\ \text{m})=24$ kN acting at midspan, 3 m from each support. Then take moments about the roller B, which removes $R_B$ (Statics — Equilibrium Requirements, $\Sigma M=0$ ): $R_A(6\ \text{m})=(12\ \text{kN})(4\ \text{m})+(24\ \text{kN})(3\ \text{m})$ , so $R_A=\dfrac{48+72}{6}=20$ kN. Check with $\Sigma F_y=0$ : $R_B=12+24-20=16$ kN. Distractors: the point-load moment arm taken as $a$ instead of $(L-a)$ — moments summed about the wrong support — yielding $R_B$ (16 kN); the point load split evenly regardless of its position ( $P/2$ , giving 18 kN); and the load intensity $w=4$ kN/m used as if it were the distributed load's resultant, instead of $wL=24$ kN, so A gets only $w/2$ from it (10 kN).
7
medium
A. Determinate analysis, internal forces, deflection & buckling
A simply supported beam of span $L=6\,\text{m}$ carries a uniformly distributed load $w=10\,\text{kN/m}$ over its full length (self-weight included) and a concentrated load $P=36\,\text{kN}$ at midspan, as shown. The maximum bending moment in the beam is most nearly:
Figure — drag or pinch to move around
1. 45 kN·m
2. 57 kN·m
3. 72 kN·m
4. 99 kN·m
Show the answer & worked solution
Answer: D.99 kN·m
Both loads are symmetric, so each one's maximum moment is at midspan and they superpose there (FE Reference Handbook, Mechanics of Materials — Simply Supported Beam Slopes and Deflections: $M_{max} = wL^2/8$ for the full uniform load and $PL/4$ for the central point load). $M_{max} = \dfrac{(10\ \text{kN/m})(6\ \text{m})^2}{8} + \dfrac{(36\ \text{kN})(6\ \text{m})}{4} = 45 + 54 = 99$ kN·m. Check by statics: each reaction is $(wL + P)/2 = 48$ kN, and at midspan $M = 48(3) - 10(3)^2/2 = 99$ kN·m. Distractors: kept only the uniform load's $wL^2/8$ and forgot the point load, 45 kN·m; gave the point load the uniform load's coefficient, $wL^2/8 + PL/8$ , 72 kN·m; and used the fixed-end moments $wL^2/12 + PL/8$ , which belong to a beam fixed at both ends, 57 kN·m.
8
medium
A. Angles, distances, traverse & areas
The corners of a four-sided parcel, taken in order around its boundary, have site coordinates (easting, northing, in m) of A (100, 100), B (320, 110), C (350, 290), D (120, 350). The area of the parcel is most nearly:
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1. 49,000 m²
2. 60,500 m²
3. 98,000 m²
4. 119,200 m²
Show the answer & worked solution
Answer: A.49,000 m²
Area by coordinates (FE Reference Handbook, Civil Engineering — Area formulas): $\text{Area} = \tfrac12\left[X_A(Y_B - Y_D) + X_B(Y_C - Y_A) + X_C(Y_D - Y_B) + X_D(Y_A - Y_C)\right]$ . Substituting, in m: $\tfrac12\left[100(110 - 350) + 320(290 - 100) + 350(350 - 110) + 120(100 - 290)\right] = \tfrac12\left[(-24{,}000) + 60{,}800 + 84{,}000 + (-22{,}800)\right] = \tfrac12(98{,}000) = 49{,}000$ m². Distractors: the ½ forgotten, 98,000 m²; the closing course left out (the products $x_Dy_A - x_Ay_D$ never formed, so the loop never returns to A), 60,500 m²; and only the $x_iy_{i+1}$ products summed and halved, the second set $x_{i+1}y_i$ never subtracted, 119,200 m².
9
medium
A. Horizontal & vertical geometric design
A simple horizontal circular curve has radius $R=300\,\text{m}$ and intersection angle $\Delta=80^\circ$ (see figure). The length of curve $L$ , from PC to PT along the arc, is most nearly:
Figure — drag or pinch to move around
1. 252 m
2. 386 m
3. 419 m
4. 503 m
Show the answer & worked solution
Answer: C.419 m
Length of curve (FE Reference Handbook, Civil Engineering — Horizontal Curves): $L = R\,I\,\dfrac{\pi}{180}$ with $I = \Delta$ in degrees. $L = (300\ \text{m})(80^\circ)\dfrac{\pi}{180} = 418.9$ m. Distractors: returned the tangent distance $T = R\tan(\Delta/2)$ , PI to PC, 252 m; returned the long chord $LC = 2R\sin(\Delta/2)$ , the straight PC–PT distance, 386 m; and measured along both tangents, PC–PI–PT $= 2T$ , 503 m. Using $R\Delta$ with $\Delta$ still in degrees, without the $\pi/180$ , gives about $57\times$ the arc.
10
medium
A. Hydrology & open-channel hydraulics
A rectangular concrete channel 3 m wide carries water at a uniform (normal) depth of 1.2 m. Manning's roughness is $n = 0.013$ and the bed slope is $S = 0.001$ . The discharge is most nearly:
Figure — drag or pinch to move around
1. 1.86 m³/s
2. 6.68 m³/s
3. 7.15 m³/s
4. 9.89 m³/s
Show the answer & worked solution
Answer: B.6.68 m³/s
Manning's equation (FE Reference Handbook, Civil Engineering — Hydraulics): $Q = \dfrac{K}{n}A R_H^{2/3} S^{1/2}$ with $K = 1.0$ in SI and $R_H = A/P$ . Here $A = by = (3)(1.2) = 3.6$ m², $P = b + 2y = 3 + 2(1.2) = 5.4$ m and $R_H = 3.6/5.4 = 0.667$ m, so $Q = \dfrac{1.0}{0.013}(3.6)(0.667)^{2/3}(0.001)^{1/2} = 6.68$ m³/s. Distractors: the velocity $V = \tfrac{1}{n}R_H^{2/3}S^{1/2} = 1.856$ m/s reported as the discharge (the $\times A$ forgotten), 1.86 m³/s; $R_H^{1/2}$ in place of $R_H^{2/3}$ (the square root taken of $R_H$ as well as of $S$ ), 7.15 m³/s; and the depth $y = 1.2$ m used as the hydraulic radius, 9.89 m³/s.

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