Free FE Civil practice problems

Ten real questions from our FE Civil bank — one from each knowledge area, with the same exam-realistic figures and fully worked solutions our members get. No signup, no email. Try each one first, then reveal the solution.

10 with figures10 knowledge areasFull worked solutions

1
medium
A. Earned value & CPM scheduling
In the activity-on-node network shown, activity A (6 d) runs alone from Start to End, while activities B (5 d) and C (4 d) run in series on the parallel branch. Both branches begin at the Start milestone (day 0). The total float of activity A is most nearly:
1. 3 d
2. 4 d
3. 0 d
4. -3 d
Show the answer & worked solution
Answer: A.3 d
Project duration $= \max(d_A,\, d_B + d_C) = \max(6,\, 9) = 9$ d (the longest path is critical). For A: $ES_A = 0$ , $EF_A = 6$ ; $LF_A = 9$ , $LS_A = 3$ . $TF_A = LF_A - EF_A = 9 - 6 = 3$ d. Distractors: $0$ (treating A as critical when it is not the longest path); only C's 4 d taken as the slack (ignoring B); and $d_A-(d_B+d_C) = -3$ d (LS−LF order reversed — an ES/LS sign slip).
2
medium
A. Fluid properties, statics & energy/momentum
Water flows steadily through a pipe that contracts from a diameter $D_1 = 200$ mm to $D_2 = 100$ mm. The upstream velocity is $V_1 = 2$ m/s. The velocity in the smaller section is most nearly:
1. 32 m/s
2. 4 m/s
3. 8 m/s
4. 1 m/s
Show the answer & worked solution
Answer: C.8 m/s
Continuity $A_1V_1=A_2V_2$ with $A\propto D^2$ gives $V_2=V_1(D_1/D_2)^2=2\,(200/100)^2=8$ m/s. Distractors: the diameter ratio raised to the FOURTH power ( $V_1(D_1/D_2)^4$ , the area-ratio squaring applied a second time); the DIAMETER ratio used un-squared ( $V_1 D_1/D_2$ , forgetting $A\propto D^2$ ); and the ratio both inverted and un-squared ( $V_1 D_2/D_1$ ).
3
medium
A. Index properties, phase relations & effective stress
At the Memphis levee borrow area the water table sits at depth 2 m. Above it is moist soil with $\gamma = 17.5$ kN/m³; below it is saturated soil with $\gamma_{sat} = 19.5$ kN/m³. The effective vertical stress at point A, 3 m below the water table, is most nearly:
1. 93.5 kPa
2. 64.1 kPa
3. 34.6 kPa
4. 44.4 kPa
Show the answer & worked solution
Answer: B.64.1 kPa
$\sigma = 17.5(2) + 19.5(3) = 93.5$ kPa; $u = \gamma_w h_w = 9.81(3) = 29.4$ kPa; $\sigma' = \sigma - u = 64.1$ kPa. Distractors: report total stress $\sigma$ (pore pressure $u$ forgotten); apply $u$ over the FULL depth $h_1+h_2$ (water table mis-placed at the surface); and double-count buoyancy by using $\gamma'$ below AND subtracting $u$ .
4
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:
1. 60 MPa
2. 120 MPa
3. 10 MPa
4. 15 MPa
Show the answer & worked solution
Answer: A.60 MPa
$I=\dfrac{bh^3}{12}=4166667\,\text{mm}^4$ , $c=h/2=50\,\text{mm}$ ; $\sigma=\dfrac{Mc}{I}=\dfrac{6M}{bh^2}=\dfrac{6(5\times10^6)}{(50)(100)^2}=60$ MPa. Distractors: used $I=bh^3/3$ instead of $bh^3/12$ (¼ of the true stress); used $c=h$ instead of $h/2$ (twice the stress); and dropped the factor 6 (used $\sigma=M/bh^2$ ).
5
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:
1. 16 kN
2. 18 kN
3. 32 kN
4. 20 kN
Show the answer & worked solution
Answer: D.20 kN
ΣM about B: $R_A=\dfrac{P(L-a)}{L}+\dfrac{wL}{2}=\dfrac{12(4)}{6}+\dfrac{4\cdot6}{2}=20$ kN. Distractors: the entire distributed load assigned to A instead of half ( $+wL$ , giving 32 kN), the point-load moment arm taken as $a$ instead of $(L-a)$ — moments summed about the wrong support — yielding R_B (16 kN), and the point load split evenly regardless of its position ( $P/2$ , giving 18 kN).
6
medium
A. Determinate analysis, internal forces, deflection & buckling
A simply supported beam of span $L=6\,\text{m}$ carries a uniformly distributed load $w=12\,\text{kN/m}$ over its full length (see the shear and moment diagrams). The maximum bending moment is most nearly:
1. 54 kN·m
2. 216 kN·m
3. 36 kN·m
4. 9 kN·m
Show the answer & worked solution
Answer: A.54 kN·m
$M_{max}=\dfrac{wL^2}{8}=\dfrac{12(6)^2}{8}=54$ kN·m at midspan (where $V=0$ ); the support shear is $V_{max}=wL/2=36$ kN. Distractors: used $wL^2/2=216$ kN·m (the cantilever fixed-end value, $4\times$ too big); dropped an $L$ to get $wL/8=9$ ; and used $wL^2/12=36$ kN·m (the fixed–fixed end moment, not the simple-span midspan value).
7
medium
A. Angles, distances, traverse & areas
A four-sided closed traverse has corner coordinates (easting, northing in m): A (0, 0), B (120, 10), C (140, 90), D (20, 110). By the coordinate (shoelace) method, the enclosed area is most nearly:
1. 15400 m²
2. 23000 m²
3. 11500 m²
4. 5750 m²
Show the answer & worked solution
Answer: C.11500 m²
Shoelace: $A=\tfrac12\left|\sum (x_i y_{i+1}-x_{i+1}y_i)\right|=\tfrac12|23000|=11500$ m². Distractors: the ÷2 forgotten (doubled cross-product); the cross-product divided by the vertex count $n=4$ instead of 2 (an "averaging" slip); and a bounding-box estimate (max E × max N), which over-counts a non-rectangular figure.
8
medium
A. Horizontal & vertical geometric design
A simple horizontal circular curve has radius $R=300\,\text{m}$ and deflection angle $\Delta=40^\circ$ (see figure). The length of curve $L$ (PC to PT along the arc) is most nearly:
1. 209 m
2. 252 m
3. 105 m
4. 205 m
Show the answer & worked solution
Answer: A.209 m
$L=R\,\Delta\,\dfrac{\pi}{180}=300\cdot40\cdot\dfrac{\pi}{180}=209.44$ m. Distractors: returned the long chord $C=2R\sin(\Delta/2)$ (the straight PC–PT distance, slightly shorter than the arc); used $R\tan\Delta$ (a tangent, not an arc); and used $\Delta/2$ instead of the full central angle (half the true length). A frequent slip is $L=R\Delta$ with $\Delta$ in degrees, which forgets the $\pi/180$ and runs about $57\times$ too large.
9
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:
1. $13,333
2. $12,000
3. $8,593
4. $20,593
Show the answer & worked solution
Answer: D.$20,593
$(A/P,0.06,15)=\dfrac{i(1+i)^n}{(1+i)^n-1}=0.10296$ , so $A=$ $200,000 $\times\ 0.10296=$ $20,593. Distractors: $P/n$ ignoring interest, the interest-only payment $P\cdot i$ , and the sinking-fund factor (A/F) used in place of A/P.
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:
1. 6.68 m³/s
2. 5.84 m³/s
3. 0.211 m³/s
4. 9.89 m³/s
Show the answer & worked solution
Answer: A.6.68 m³/s
$A = by = 3.6$ m², $P = b + 2y = 5.4$ m, $R = A/P = 0.667$ m. $Q = \dfrac{1}{n}A R^{2/3}\sqrt{S} = 6.683$ m³/s. Distractors: the flow depth $y$ used in place of the hydraulic radius $R$ ; $R$ to the first power instead of $R^{2/3}$ ; and the slope used as $S$ rather than $\sqrt{S}$ .

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