Free FE Civil practice exam

Acceleration is the second derivative of position. First derivative (velocity): $v(t) = 3t^2 - 12t$. Second derivative (acceleration): $a(t) = 6t - 12$. At $t = 3$: $a(3) = 18 - 12 = 6$ m/s squared. The value -9 is the velocity at $t = 3$ ($v(3) = 27 - 36 = -9$), the classic mistake of stopping at the first derivative. The value 0 is the velocity at the turning point, and 18 is 6t without the constant.

1. Velocity is the first derivative:
   $v(t) = 3t^2 - 12t$
2. Acceleration is the second derivative:
   $a(t) = 6t - 12$
3. Evaluate at $t = 3$:
   $a(3) = 6(3) - 12 = 6\,\text{m/s}^2$

FE Reference Handbook 10.6, p. 102

Compute the expected value of the risky bid: $E(A) = 40{,}000(0.6) + (-10{,}000)(0.4) = 24{,}000 - 4{,}000 = 20{,}000$ dollars. Bid B is a certain 25,000 dollars. Since $25{,}000 > 20{,}000$, Bid B is preferred by 5,000 dollars. "Bid A, by 5,000 dollars" reverses the comparison, picking the risky bid. "Bid B, by 15,000 dollars" uses 15,000 dollars, which comes from treating the loss as a positive payoff so that $E(A) = 24{,}000 + 4{,}000 = 28{,}000$ and then taking the wrong difference. "They are equal" ignores the probability weighting entirely.

1. Expected value of Bid A (keep the loss negative):
   $E(A) = 40{,}000(0.6) + (-10{,}000)(0.4)$
2. Evaluate each term:
   $24{,}000 + (-4{,}000) = 20{,}000 \text{ dollars}$
3. Bid B is certain:
   $E(B) = 25{,}000 \text{ dollars}$
4. Compare and take the difference:
   $E(B) - E(A) = 25{,}000 - 20{,}000 = 5{,}000 \text{ dollars (Bid B preferred)}$

FE Reference Handbook 10.6, p. 67

A master's degree sets the qualifying-experience requirement at three years. The one year the engineer worked before earning the master's does not count toward that three-year post-degree requirement, so the full three years still must be completed after the master's. The 2 years option uses the doctorate reduction by mistake. The 1 year option wrongly credits the pre-master's year against the requirement. The 4 years option uses the bachelor's requirement and ignores the master's reduction entirely. The answer is three years.

1. The highest qualifying degree is the master's, which sets the experience requirement at three years.
   $T_{req} = 3\ \text{years (master's)}$
2. The one year worked before the master's does not count toward the post-degree requirement, so it contributes zero.
   $T_{credit,\,pre} = 0$
3. The remaining required experience equals the full three-year post-master's requirement.
   $T_{remaining} = 3 - 0 = 3\ \text{years}$

FE Reference Handbook 10.6, p. 8-9, Model Law §130.10 B.2.a

Disbenefits are negative effects on the public, so they are subtracted from the benefits in the numerator: $B/C = (B - D)/C = (8{,}000{,}000 - 1{,}200{,}000)/5{,}500{,}000 = 6{,}800{,}000/5{,}500{,}000 = 1.24$. The 1.45 option ignores the disbenefits entirely ($8{,}000{,}000/5{,}500{,}000$). The 1.19 option incorrectly adds the disbenefits to the cost denominator ($8{,}000{,}000/6{,}700{,}000$). The 1.86 option incorrectly subtracts the disbenefits from the cost denominator ($8{,}000{,}000/4{,}300{,}000$) instead of from the benefits.

1. Subtract disbenefits from benefits in the numerator:
   $B - D = 8{,}000{,}000 - 1{,}200{,}000 = 6{,}800{,}000$
2. Divide by the present worth of agency costs:
   $B/C = \frac{6{,}800{,}000}{5{,}500{,}000} = 1.24$
3. Since B/C = 1.24 > 1, the project is economically justified.

FE Reference Handbook 10.6, p. 237, Benefit-Cost Analysis

Sum moments about B to eliminate $B_y$ and find $A_y$ directly, or sum moments about A to find $B_y$ first. Taking moments about A: $B_y(8) - 10(10) = 0$, so $B_y = 12.5$ kN. Then $\sum F_y = 0$: $A_y + 12.5 - 10 = 0$, giving $A_y = -2.5$ kN, i.e. 2.5 kN downward. The overhang load actually pulls the pin down (it acts as a tie-down). The 12.5 kN upward choice reports $B_y$ instead of $A_y$. The 5.0 kN choice splits the load evenly, valid only for a midspan load between the supports. The 2.5 kN upward choice has the right magnitude but the wrong direction, and the sign matters here because an overhang can reverse the reaction.

1. Sum moments about A (CCW positive), load is 10 m from A, $B_y$ is 8 m from A:
   $B_y(8) - 10(10) = 0 \quad \Rightarrow \quad B_y = 12.5\,\text{kN}$
2. Apply vertical equilibrium:
   $A_y + B_y - 10 = 0 \quad \Rightarrow \quad A_y = 10 - 12.5 = -2.5\,\text{kN}$
3. The negative sign means $A_y$ acts downward, so the pin holds the beam down against the overhang load:
   $A_y = 2.5\,\text{kN (downward)}$

FE Reference Handbook 10.6, p. 95, Equilibrium Requirements

Treat the notch as negative area. The full rectangle has $A_1 = 100 \times 120 = 12{,}000$ mm$^2$ with centroid at $y_1 = 60$ mm. The removed notch has $A_2 = 40 \times 40 = 1{,}600$ mm$^2$ with its centroid at $y_2 = 120 - 20 = 100$ mm. Using subtraction: $\bar{y} = (12{,}000(60) - 1{,}600(100))/(12{,}000 - 1{,}600) = 560{,}000/10{,}400 = 53.8$ mm. Removing material from the top pulls the centroid down below mid-height. The 60 mm option ignores the notch. The 50.0 mm option is the geometric mid-height. The 46.2 mm option over-corrects, as if material were added at the top instead of removed.

1. Full rectangle:
   $A_1 = 100 \times 120 = 12{,}000\,\text{mm}^2, \quad y_1 = 60\,\text{mm}$
2. Removed notch (centroid 20 mm below the top):
   $A_2 = 40 \times 40 = 1{,}600\,\text{mm}^2, \quad y_2 = 100\,\text{mm}$
3. Composite centroid (subtract the hole):
   $\bar{y} = \frac{12{,}000(60) - 1{,}600(100)}{12{,}000 - 1{,}600} = \frac{720{,}000 - 160{,}000}{10{,}400} = \frac{560{,}000}{10{,}400} = 53.8\,\text{mm}$

FE Reference Handbook 10.6, p. 96, Centroid of Area

Change in kinetic energy is $\tfrac{1}{2}m(v_2^2 - v_1^2)$. Square each speed separately first: $25^2 - 10^2 = 625 - 100 = 525$. Then $\tfrac{1}{2}(1200)(525) = 315{,}000\,\text{J}$. The $135\,\text{kJ}$ choice squares the velocity difference $(25-10)^2 = 225$ instead of the difference of squares, giving $\tfrac{1}{2}(1200)(225) = 135{,}000\,\text{J}$. The $162\,\text{kJ}$ choice comes from a partial-squaring slip. The $270\,\text{kJ}$ choice both drops the one-half and uses the squared difference: $(1200)(225) = 270{,}000\,\text{J}$.

1. Change in kinetic energy:
   $\Delta T = \tfrac{1}{2}m(v_2^2 - v_1^2)$
2. Difference of squares (not the square of the difference):
   $v_2^2 - v_1^2 = 25^2 - 10^2 = 625 - 100 = 525\,\text{m}^2/\text{s}^2$
3. Evaluate:
   $\Delta T = \tfrac{1}{2}(1{,}200)(525) = 315{,}000\,\text{J} = 315\,\text{kJ}$

FE Reference Handbook 10.6, p. 107

The total load is $w \times (L/2) = 8 \times 3 = 24$ kN, and its resultant acts at the centroid of the loaded region, 1.5 m from the left support. Take moments about the right support (B) to find the left reaction $R_A$: the resultant is $6 - 1.5 = 4.5$ m from B, so $R_A \times 6 = 24 \times 4.5 = 108$, giving $R_A = 18$ kN. The 12 kN choice splits the total load evenly (24/2), ignoring that the load is closer to A. The 24 kN choice reports the entire load as the reaction. The 6 kN choice is the right reaction $R_B = 24 - 18 = 6$ kN, the other support.

1. Resultant of the partial load:
   $W = w \cdot \frac{L}{2} = 8 \times 3 = 24\,\text{kN at } 1.5\,\text{m from A}$
2. Sum moments about B (resultant is 4.5 m from B):
   $R_A \times 6 = 24 \times 4.5 = 108$
3. Solve for the left reaction:
   $R_A = \frac{108}{6} = 18\,\text{kN}$

FE Reference Handbook 10.6, p. 135

The modulus of rupture estimates concrete tensile (flexural) strength: $f_r = 7.5\sqrt{f'_c} = 7.5\sqrt{4{,}000} = 7.5 \times 63.2 = 474 \text{ psi}$. The 30,000 psi choice forgets the square root and computes $7.5 \times 4{,}000$. The 170 psi choice takes the square root of the whole product $\sqrt{7.5 \times 4{,}000}$ instead of multiplying by the root of $f'_c$. The 400 psi choice uses the rough 10% rule of thumb ($0.10 \times 4{,}000$) rather than the modulus-of-rupture formula the problem specifies.

1. Modulus of rupture relationship:
   $f_r = 7.5\sqrt{f'_c}$
2. Take the square root of the compressive strength:
   $\sqrt{4{,}000} = 63.2$
3. Multiply:
   $f_r = 7.5 \times 63.2 = 474 \text{ psi}$

FE Reference Handbook 10.6, p. 126

Continuity at a junction: inflow equals total outflow. $Q_{\text{in}} = A_{\text{in}} v = \frac{\pi}{4}(0.3)^2(2) = 0.07069 \times 2 = 0.1414$ m³/s. The combined outlet area is $A_{\text{out}} = \frac{\pi}{4}(0.2^2 + 0.15^2) = \frac{\pi}{4}(0.0625) = 0.04909$ m². With equal velocities, $v_{\text{out}} = Q_{\text{in}}/A_{\text{out}} = 0.1414/0.04909 = 2.88$ m/s. The 2.0 m/s choice assumes the velocity is unchanged. The 4.5 m/s choice uses only the 200 mm pipe area. The 8.0 m/s choice uses only the 150 mm pipe area.

1. Inlet flow rate:
   $Q_{\text{in}} = \frac{\pi}{4}(0.3)^2(2) = 0.1414\,\text{m}^3/\text{s}$
2. Combined outlet area:
   $A_{\text{out}} = \frac{\pi}{4}(0.2^2 + 0.15^2) = 0.04909\,\text{m}^2$
3. Outlet velocity (equal in both pipes):
   $v_{\text{out}} = \frac{Q_{\text{in}}}{A_{\text{out}}} = \frac{0.1414}{0.04909} = 2.88\,\text{m/s}$

FE Reference Handbook 10.6, p. 185

Shrinkage means compacted fill occupies less volume than the bank material it came from. Compacted volume = bank volume $\times (1 - \text{shrinkage}) = 5{,}400 \times (1 - 0.15) = 5{,}400 \times 0.85 = 4{,}590$ yd³. The 6,210 yd³ choice adds 15 percent instead of subtracting it. The 810 yd³ choice reports only the lost volume ($5{,}400 \times 0.15$). The 6,353 yd³ choice divides by 0.85 (that direction is for finding required bank volume from a known fill).

1. Apply the shrinkage factor to the bank volume:
   $V_{\text{fill}} = V_{\text{bank}} (1 - S) = 5{,}400 (1 - 0.15)$
2. Compute:
   $V_{\text{fill}} = 5{,}400 \times 0.85 = 4{,}590 \text{ yd}^3$

FE Reference Handbook 10.6, p. 315

When both conjugate depths are known, the energy loss has a closed form: $\Delta E = (y_2 - y_1)^3/(4 y_1 y_2)$. $\Delta E = (2.5 - 0.5)^3/(4 \times 0.5 \times 2.5) = (2.0)^3/5.0 = 8/5 = 1.60$ m. The 6.40 m option omits the factor of 4 in the denominator: $(2.0)^3/(0.5 \times 2.5) = 6.40$. The 2.00 m option just reports the depth difference $y_2 - y_1 = 2.0$. The 4.00 m option drops $y_2$ from the denominator: $(2.0)^3/(4 \times 0.5) = 4.00$.

1. Energy loss across a hydraulic jump:
   $\Delta E = \frac{(y_2 - y_1)^3}{4 y_1 y_2}$
2. Substitute:
   $\Delta E = \frac{(2.5 - 0.5)^3}{4(0.5)(2.5)}$
3. Compute:
   $\Delta E = \frac{(2.0)^3}{5.0} = \frac{8}{5} = 1.60\,\text{m}$

FE Reference Handbook 10.6, p. 303

BOD remaining is the organic matter not yet decomposed: $L_t = L_0 e^{-kt} = 320\,e^{-0.18 \times 4} = 320\,e^{-0.72} = 320(0.487) = 155.8$ mg/L. Note the BOD exerted ($L_0(1 - e^{-kt}) = 164.2$) plus the remaining (155.8) sum to the ultimate 320. The 164.2 mg/L choice gives the BOD exerted rather than remaining. The 57.6 mg/L choice applies $kt$ as a percentage ($0.18 \times 4 = 0.72$, then $320 \times 0.18 = 57.6$), a formula confusion. The 320 mg/L choice reports the ultimate BOD, as if no decay had occurred.

1. BOD remaining (decay form):
   $L_t = L_0 e^{-kt} = 320\,e^{-0.18 \times 4}$
2. Evaluate the exponent:
   $e^{-0.72} = 0.487$
3. BOD remaining at 4 days:
   $L_t = 320 \times 0.487 = 155.8\,\text{mg/L}$

FE Reference Handbook 10.6, p. 327

Combination 1 is $1.4D = 1.4(100) = 140$ kips. Combination 2 is $1.2D + 1.6L = 1.2(100) + 1.6(10) = 120 + 16 = 136$ kips. Because dead load dominates and live load is small, Combination 1 (140) produces the larger factored load and controls. This is the case the 1.4D combination exists for. The 136 kips option is Combination 2, which is smaller here. The 120 kips option keeps only the 1.2D term and drops live load. The 110 kips option is the unfactored service load $D + L$.

1. Combination 1:
   $1.4D = 1.4(100) = 140 \text{ kips}$
2. Combination 2:
   $1.2D + 1.6L = 1.2(100) + 1.6(10) = 136 \text{ kips}$
3. Controlling: Combination 1 at 140 kips (dead load dominates).

FE Reference Handbook 10.6, p. 278

With full bracing, $M_n = M_p = F_y Z_x = 50 \times 100 = 5{,}000$ kip-in $= 416.7$ kip-ft. For ASD, divide by $\Omega_b = 1.67$: $M_n/\Omega_b = 416.7/1.67 = 249.5 \approx 250$ kip-ft. The 417 kip-ft choice is the nominal $M_n$ without applying $\Omega_b$. The 375 kip-ft choice is the LRFD design strength $\phi_b M_n = 0.90 \times 416.7$, but the problem asks for ASD. The 278 kip-ft choice uses an incorrect $\Omega_b = 1.50$.

1. Plastic moment:
   $M_p = F_y Z_x = 50 \times 100 = 5{,}000 \text{ kip-in} = 416.7 \text{ kip-ft}$
2. ASD allowable strength:
   $\frac{M_n}{\Omega_b} = \frac{416.7}{1.67} = 250 \text{ kip-ft}$

FE Reference Handbook 10.6, p. 287

Total stress: $\sigma = 118(4) + 128(6) = 472 + 768 = 1{,}240$ psf. Pore pressure (6 ft below WT): $u = 62.4(6) = 374$ psf. Effective: $\sigma' = 1{,}240 - 374 = 866 \approx 865$ psf. Or use buoyant weight: $\sigma' = 118(4) + (128-62.4)(6) = 472 + 393.6 = 865.6$ psf. The 1,240 value is the total stress without subtracting pore pressure. The 394 value is only the buoyant contribution of the saturated zone, missing the moist sand overburden. The 766 value comes from an error in handling the pore pressure or zone thickness; it does not match either the correct method (865 psf) or the listed partial methods.

1. Total stress:
   $\sigma = 118(4) + 128(6) = 472 + 768 = 1{,}240 \text{ psf}$
2. Pore pressure (6 ft below WT):
   $u = 62.4 \times 6 = 374.4 \text{ psf}$
3. Effective stress:
   $\sigma' = 1{,}240 - 374 = 865 \text{ psf}$

FE Reference Handbook 10.6, p. 269

q = kH(Nf/Nd) = (1×10⁻⁴)(8)(5/10) = (1×10⁻⁴)(8)(0.5) = 4.0×10⁻⁴ m³/s per m. The 1.6×10⁻³ option inverts the ratio (uses Nd/Nf = 2). The 8.0×10⁻⁴ option drops the ratio (uses 1.0). The 2.0×10⁻⁴ option halves the head (uses H = 4).

1. Apply the flow-net formula:
   $q = kH\frac{N_f}{N_d} = (1\times10^{-4})(8)(0.5)$
2. Compute:
   $q = 4.0\times10^{-4}\text{ m}^3\text{/s per m}$

FE Reference Handbook 10.6, p. 298

$M = R[1 - \cos(I/2)] = 1{,}000[1 - \cos(20\degree)] = 1{,}000(1 - 0.9397) = 1{,}000 \times 0.0603 = 60.3$ ft. The 234 ft choice used $\cos(I) = \cos(40\degree) = 0.766$ instead of $\cos(I/2)$: $1{,}000(1 - 0.766) = 234$. The 64.2 ft choice is the external distance $E = R[1/\cos(I/2) - 1] = 1{,}000(1/0.9397 - 1) = 64.2$, not the middle ordinate. The 30.2 ft choice halved the correct answer.

1. Middle ordinate:
   $M = R\left[1 - \cos\!\left(\frac{I}{2}\right)\right] = 1{,}000\left[1 - \cos(20\degree)\right]$
2. Evaluate:
   $M = 1{,}000(1 - 0.9397) = 60.3 \text{ ft}$

FE Reference Handbook 10.6, p. 309

Compacted volume $= 0.90 \times$ bank volume, so bank volume $=$ compacted volume $/ 0.90 = 9{,}000/0.90 = 10{,}000$ yd^3. Soil shrinks when compacted, so you must dig MORE bank material than the finished fill volume. The 8,100 yd^3 option multiplied by 0.90 instead of dividing: $9{,}000 \times 0.90$. The 9,000 yd^3 option ignored shrinkage entirely. The 9,900 yd^3 option added 10% instead of dividing by 0.90.

1. Relationship:
   $V_{\text{compacted}} = 0.90 \times V_{\text{bank}}$
2. Solve for bank volume:
   $V_{\text{bank}} = \frac{9{,}000}{0.90} = 10{,}000 \text{ yd}^3$

FE Reference Handbook 10.6, p. 315

$CV = BCWP - ACWP = 1{,}050{,}000 - 980{,}000 = +70{,}000$. A positive cost variance means the work performed is worth more than what was spent, so the project is under budget. The $-150{,}000$ option computed the schedule variance ($BCWP - BCWS = 1{,}050{,}000 - 1{,}200{,}000$) instead of the cost variance. The $-70{,}000$ option flipped the sign by subtracting in the wrong order. The $+220{,}000$ option used $BCWS - ACWP$, which is not a standard variance.

1. Cost variance:
   $CV = BCWP - ACWP = 1{,}050{,}000 - 980{,}000 = +70{,}000$
2. Positive CV means earned value exceeds actual cost: under budget.

FE Reference Handbook 10.6, p. 316