PE_GENIUS_EXAM_A_GUMROAD

Practice Exam A – 1 st Edition www.pegenius.com 2

Practice Exam A – 1 st Edition www.pegenius.com 5 103. The parallel pipe system shown has a head loss due to friction of 5 feet. Given the pipe parameters shown, and a Darcy-Weisbach friction factor of 0.018 for both pipes, determine the total input flow Q in cubic feet per second. A. 2 cfs B. 4 cfs C. 6 cfs D. 8 cfs 104. The activity-on-node network pictured represents a construction schedule. If the schedule's critical path is dependent on Activities AEG (shown in RED), determine the maximum length of Activity D to the nearest day. A. 5 days B. 7 days C. 9 days D. 11 days

Practice Exam A – 1 st Edition www.pegenius.com 8 109. You are designing a cantilever retaining wall for your new home in Beverly Hills, California. The hillside in your neighborhood is known for saturated soft clay that is prone to collapse and has already damaged the third guest house of many residents. Given the parameters in the following figure, determine the factor of safety against overturning for the retaining wall. Ignore any lateral resistance provided by passive pressure but consider the weight of soil above the toe in your calculation. A. 1.9 B. 3.8 C. 1.2 D. 2.4 110. The pictured beam overhanging a support is to be designed for uniform dead load w D and uniform live load w L based on the following strength load combination: 1.2DL + 1.6LL Determine the beam shear at location B based on the factored load demand. A. 5.5 kip B. 6.5 kip C. 7.5 kip D. 8.5 kip

Practice Exam A – 1 st Edition www.pegenius.com 11 115. Two beams with equal section and material properties are both loaded at midpoint with load P. Given that Beam 2 deflects 2.5 inches, determine the deflection associated with Beam 1. A. 0.09 in B. 0.18 in C. 0.83 in D. 1.25 in

Practice Exam A – 1 st Edition www.pegenius.com 14 119. Given the following soil profile, determine the effective stress at bottom of Soil Layer 2: A. 2404 psf B. 2954 psf C. 3652 psf D. 4202 psf 120. A 16-foot long square 18” x 18” column supports axial load P. The column is fixed at the base and unrestrained against any sort of movement at the top. Given the brace points pictured in each direction, determine the critical slenderness ratio to be considered for design of this column. A. 30 B. 35 C. 40 D. 45

Practice Exam A – 1 st Edition www.pegenius.com 17 125. The pictured two-lane roadway is congested in the direction of Lane A every evening during rush hour. A study was conducted, and Lane A was found to have a peak hour factor of 0.75 and a peak rate of flow of 150 vehicles per 15 minutes. Given that Lane B has an hourly volume during rush hour equal to one-third the volume of Lane A, determine the average vehicle headway anticipated for Lane B during rush hour in seconds per vehicle. A. 16 seconds per vehicle B. 24 seconds per vehicle C. 32 seconds per vehicle D. 48 seconds per vehicle 126. A new 14-inch diameter welded steel pipe (C = 130), 300 feet in length, is used as a firewater line in an industrial facility. Given a required outlet pressure of 20 psi and a flow of 1.67 cubic feet per second, calculate the necessary inlet pressure. Assume changes in head loss are only due to friction. A. 22.5 psi B. 38.6 psi C. 33.1 psi D. 27.1 psi

Practice Exam A – 1 st Edition www.pegenius.com 20 131. A mobile crane is being used to organize structural steel beams in a laydown yard. The crane utilizes rigging cables produced by Very Good Rigging Company (VGRC) Inc. VGRC has specified the maximum tension load allowed for each cable is 5 kip. If length of the beam L equals 20 feet, what is the maximum beam weight allowed in pounds per linear foot? Assume each cable supports the same amount of load. A. 256 plf B. 325 plf C. 354 plf D. 389 plf 132. Consider a saturated soil sample with a void ratio e of 0.80. Determine the sample’s water content if the saturated unit weight is found to equal twice as much as the unit weight of water: A. 0.20 B. 0.26 C. 0.29 D. 0.34

Practice Exam A – 1 st Edition www.pegenius.com 21 133. Shaft ABCDE is subjected to clockwise torques 590 lb-ft, 880 lb-ft and 750 lb-ft at points B, C and D respectively. Supports A and E are both unyielding. The diameter of the shaft varies as shown in the figure. Given modulus of rigidity G = 11.5 x 10 6 psi , determine the reaction at support E . A. 308.2 lb-ft B. 1018.8 lb-ft C. 1110.2 lb-ft D. 1911.8 lb-ft 134. Runoff from a storm is measured over a period of 25 hours using a catchment with an area of 25 acres. Given the following direct runoff hydrograph and a peak flow rate of 40 cubic feet per second, calculate the rainfall excess that resulted from the storm: A. 0.5 in B. 1.2 in C. 4.5 in D. 2.6 in

Practice Exam A – 1 st Edition www.pegenius.com 22 135. Given the soil profile pictured below, determine the effective stress at the bottom of soil layer 2. Assume groundwater table at the surface. A. 1478 psf B. 1872 psf C. 3705 psf D. 1833 psf 136. The pipe pictured below has an inner radius r of 18 inches and a flow depth h of 9 inches. Given a slope S of 1% and Manning’s constant n of 0.025, determine the flow rate Q in cubic feet per second. A. 7.4 cfs B. 15.0 cfs C. 4.8 cfs D. 23.2 cfs

Practice Exam A – 1 st Edition www.pegenius.com 23 137. The cantilever retaining wall pictured below is 25 feet in length. Determine the horizontal component of the total active earth pressure resultant force acting on the wall. A. 155 kip B. 236 kip C. 179 kip D. 486 kip 138. Municipal records show a flood devastated a rural town 40 years ago. Risk assessors employed by the town have determined that the probability a flood of the same magnitude will not occur over the next 20 years is 80%. In nearly how many years would you anticipate a similar flood event to occur from present? A. 50 years B. 75 years C. 90 years D. 125 years

Practice Exam A – 1 st Edition www.pegenius.com 24 139. Given a required water-cement ratio of 0.5, you are preparing a concrete mix to place the 1-foot-thick square spread footing pictured below. If 4 sacks of cement are used in the mixture, what volume of water is required? Assume each bag of cement weighs 94 pounds. Assume a specific weight of water of 62.4 pounds per cubic foot. A. 1.9 cubic feet B. 2.5 cubic feet C. 3.8 cubic feet D. 4.6 cubic feet 140. Pictured below is a stress-strain curve for a carbon steel rod subject to a tension test via a universal testing machine. At what point did the tensile stress on the rod reach the steel’s yield strength? A. Point A B. Point B C. Point C D. Point D

Practice Exam A – 1 st Edition www.pegenius.com 25 Solution Key Problem Solution Problem Solution 101 C 121 B 102 B 122 A 103 A 123 C 104 C 124 B 105 B 125 B 106 D 126 C 107 A 127 B 108 D 128 A 109 B 129 C 110 A 130 C 111 B 131 C 112 A 132 C 113 C 133 D 114 B 134 D 115 A 135 D 116 B 136 C 117 C 137 D 118 D 138 A 119 B 139 C 120 C 140 B Please visit www.pegenius.com/errata for any updates to these solutions.

Practice Exam A – 1 st Edition www.pegenius.com 26 Solutions 101. Answer: C) 4.7 kip Relevant PE Civil Handbook Section: 4.1 Calculate moment at wall: ) # = * # ∗ , $ 2 = 250 &,! ∗ (5 !0) $ 2 ) # = 3125 lb − ft Calculate force at anchor based on moment: 9 %&!'() = ) # : where moment arm d is: : = 8 (< = 0.67 !0 9 %&!'() = 3125,@ − !0 0.67!0 9 %&!'() ≈ 4665 lb ≈ 4.7 kip 102. Answer: B) 3150 vph Relevant PE Civil Handbook Section: 5.1 The rate of flow in vehicles per hour can be found by multiplying speed by density: E = FG Speed can be determined based on free flow speed, density, and jam density. F = HHF(1 − G G * ) Free flow speed (miles per hour): HHF = 60 I&ℎ Density (vehicles per mile): G = 175 E&I Jam Density (vehicles per mile): G * = 250 E&I Calculate speed: F = 60 I&ℎ ∗ K1 − 175 E&I 250 E&I L = 18 I&ℎ Calculate rate of flow: E = (18 I&ℎ)(175 E&I) = 3150 E&ℎ

Practice Exam A – 1 st Edition www.pegenius.com 27 103. Answer: A) 2 cfs Relevant PE Civil Handbook Section: 6.2.3 Head loss for a parallel system is equal for each pipe: ℎ +,-(-%. = ℎ +/ = ℎ +0 Head loss per the Darcy-Weisbach equation: ℎ + = ! ∗ M G ∗ E $ 2N Calculate velocity in Pipe A: 5 !0 = 0.018 ∗ 100 !0 4 12 !0 ∗ E / $ 2 K32.2 !0 ' L E / = 7.72 !0 ' Calculate velocity in Pipe B: 5 !0 = 0.018 ∗ 200 !0 6 12 !0 ∗ E 0 $ 2 K32.2 !0 ' L E 0 = 6.69 !0 ' Total input flow is equal to flow through Pipes A and B: P -(-%. = P / + P 0 Calculate total input flow: P -(-%. = R / E / + R 0 E 0 P -(-%. = S ∗ T 4 12 !0U $ 4 ∗ K7.72 !0 ' L + S ∗ T 6 12 !0U $ 4

Practice Exam A – 1 st Edition www.pegenius.com 28 104. Answer: C) 9 days Relevant PE Civil Handbook Section: 2.4 A project’s critical path is the minimum length it takes to complete the schedule. If the critical path is based on total length of activities AEG, then the total maximum length of activities ADF will be one less than the length of AEG. Total Length of AEG: RVW = 15 :XY' + 7 :XY' + 12 :XY' = 34 :XY' Maximum Length of ADF (to keep AEG as critical path): RGH = RVW − 1 = 33 :XY' Solving for Maximum Length of Activity D: 33 :XY' = R + G + H 33 :XY' = 15 :XY' + G + 9 :XY' G = 9 :XY' 105. Answer: B) 6% Relevant PE Civil Handbook Section: 3.8.3 Assume a value of 1.0 for the total volume: Z - = 1.0 !0 1 Z - = Z 2 + Z 3 1.0 = Z 2 + Z 3 Determine volume of solids and volume of voids based on void ratio, e: [ = Z 2 Z 3 → [ = 0.3 1.0 = Z 2 + Z 3 Z 2 = 0.3Z 3 1.0 = 1.3Z 3 Z 3 = 0.77 Z 2 = 0.23 Determine the volume of water based on degree of saturation, S: F = 0.45 F = Z 4 Z 2 Z 4 = 0.45Z 2 → Z 4 = 0.10 Determine the water content: I 4 = ] 4 ∗ Z 4 = 6.5 lb I 3 = & 6)7 ∗ Z - = 108 lb * = I 4 I 3

Practice Exam A – 1 st Edition www.pegenius.com 29 * = 6.5 ,@ 108 ,@ ≈ 6% 106. Answer: D) 500 kip Relevant PE Civil Handbook Section: 4.1 A truss functions to resist load by developing axial forces in its members. Given the truss shown in the problem statement and associated loading, we can deduce that the top chord is in compression and the bottom chord is in tension as shown in the following image. We can think of truss action analogous to that of a deep beam where the top flange is in compression and the bottom flange is in tension. Calculate the maximum moment on the truss based on uniform load over the length: ) # = * # ∗ M $ 8 = K 10 _(& !0 L (40 !0) $ 8 ) # = 2000 _(& − !0 Calculate the tension based on depth of the truss: : = 4 !0 ` = ) # : = 2000 _(& − !0 4 !0 = 500 _(& COMPRESSION TENSION

Practice Exam A – 1 st Edition www.pegenius.com 30 107. Answer: A) 15.4 cfs Basic flow equation for a rectangular weir: P = 2/3 b@√2N d^(3/2) Discharge coefficient (given): b = 0.6 Head above weir bottom: d = 1 !0 Effective width based on weir width: @ = @ 489) − 0.1 ∗ f ∗ d where: f = <gI@[h i! ji<0hXj0(i<' = 2 @ 489) = *[(h ,[<N0ℎ = 5 !0 @ = 5 !0 − 0.1 ∗ 2 ∗ 1 !0 = 4.8 !0 Gravity constant: N = 32.2 !0 ' $ Flow: P = 15.4 !0 1 ' 108. Answer: D) $257,000 Relevant PE Civil Handbook Section: 2.1.1 Bank volume: Z : = (150 !0) $ ∗ 12 !0 ∗ 1 Y: 1 27 !0 1 = 10000 Y: 1 Swell Factor: '*[,, = 0.2 Loose Volume: Z . = 1 + '*[,, 1 ∗ Z : = 12000 Y: 1 Number of Hauls Required: < '%#. = Z . 14 Y: 1 = 857.14 → 858 ℎXg,' h[kg(h[: Total Hauling Cost: ji'0 '%#. = < '%#. ∗ $300 ℎXg, ≈ $257000

Practice Exam A – 1 st Edition www.pegenius.com 31 109. Answer: B) 1.3 Relevant PE Civil Handbook Section: 3.1 The factor of safety against overturning is calculated by taking the ratio of resisting moment to overturning moment. Resisting moment is caused by the self-weight of concrete and soil above the wall footing resisting overturning of the wall. Overturning moment is caused by active pressure of soil retained by the wall. Both overturning moment and resisting moment are calculated based on rotation about the toe. Moments are calculated in units of pound-foot per foot of wall. Calculate the resisting moment based on weight and distance to center of each component from toe: m 4%.. = 1.5 !0 ∗ 12 !0 ∗ 150 &j! = 2700 &,! n 4%.. = 4.75 !0 m_!0N = 2.5 !0 ∗ (4.0 !0 + 1.5 !0 + 8.0 !0) ∗ 150 &j! = 5062.5 &,! n +-; = 6.75 !0 m -(83(9. = 4 !0 ∗ 3 !0 ∗ 120 &j! = 1440 &,! n -(83(9. = 2 !0 m '88.3(9. = 12 !0 ∗ 8 !0 ∗ 120 &j! = 11520 &,! n '88.3(9. = 9.5 !0 ) )8393-9&; = m 4%.. ∗ n 4%.. + m +-; ∗ n +-; + m -(83(9. ∗ n -(83(9. + m '88.3(9. ∗ n '88.3(9. ) )8393-9&; = 159316.9 ,@ ∗ !0 !0 Calculate the overturning moment based on the active earth pressure. The equation for active earth pressure: & % = & 2 ∗ _ % − 2 ∗ j ∗ p_ % where: & 2 = E[h0(jX, &h[''gh[ _ % = qX<_(<[ Xj0(E[ [Xh0ℎ &h[''gh[ ji[!!(j([<0 j = jiℎ['(i< (N(E[<) Rankine active earth pressure coefficient, since ϕ = 0 (saturated clay): _ % = 1 Vertical pressure: & 2 = r ∗ d = 120 &j! ∗ (12 !0 + 2.5 !0) = 1740 &'! Active earth pressure: & % = 1740 &'! ∗ 1 − 2 ∗ 275 &'! ∗ √1 = 1190 &'! Resultant force due to active earth pressure: q % = 1 2 ∗ & % ∗ d = 1 2 ∗ 1190 &'! ∗ (12 !0 + 2.5 !0) = 8627.5 &,!

Practice Exam A – 1 st Edition www.pegenius.com 32 Since the resultant force acts at H/3 above the base (triangular distribution), we can calculate the overturning moment as follows: ) (28)-#)&9&; = q % ∗ d 3 = 8627.5 &,! ∗ (12 !0 + 2.5 !0)/3 = 125098.8 ,@ ∗ !0 !0 Determine Factor of Safety: HF = ) )8393-9&; ) (28)-#)&9&; ≈ 1.3 110. Answer: A) 5.5 kip Relevant PE Civil Handbook Section: 4.1 (see shears, moments, and deflections table) Consider the factored load demand based on the given load combination 1.2DL + 1.6LL: * # = 1.2 ∗ 2 _,! + 1.6 ∗ 4 _,! = 8.8 _,! Calculate the reaction at support C by taking the moment about support A (consider counterclockwise rotation as positive): 8.8 _,! ∗ 5 !0 ∗ 5 !0 2 − 8.8 _,! ∗ 20 !0 ∗ 20 !0 2 + q < ∗ 20 !0 = 0 q < = 82.5 _(& Calculate the shear at 10 feet away from support C: Z 0 = 8.8 _,! ∗ 10 !0 − 82.5 _(& Z 0 = 5.5 _(& Refer to the following diagram for reactions and beam shear:

Practice Exam A – 1 st Edition www.pegenius.com 33 111. Answer: B) 188 ft Relevant PE Civil Handbook Section: 5.3 Length of Curve: M = 500 !0 Grade of Back Tangent: N = = 3.0% Grade of Forward Tangent: N $ = −5.0% Distance to Max Elevation: n > = N = M N = − N $ = 3.0% × 500 !0 3.0% − (−5.0%) ≈ 188 !0

Practice Exam A – 1 st Edition www.pegenius.com 34 112. Answer: A) 10 min Relevant PE Civil Handbook Section: 6.2 Calculate the rate of discharge of the 6 inch diameter pipe: P = R ∗ E R = S ∗ : ?9?8 $ 4 where: P = :('jℎXhN[ hX0[ (< jg@(j ![[0 &[h '[ji<: E = :('jℎXhN[ E[,ij(0Y (< ![[0 &[h '[ji<: : ?9?8 = :(XI[0[h i! &(&[ Solving for Q : R = S ∗ (0.5 !0) $ 4 = 0.2 !0 $ P = 0.2 !0 $ ∗ 14 !0 ' = 2.8 !0 1 ' Time can be found by dividing volume to be filled by the rate of discharge: 0 = Z P where: Z = Ei,gI[ (< ![[0 P = :('jℎXhN[ hX0[ (< jg@(j ![[0 &[h '[ji<: 0 = 0(I[ (< I(<g0[' Solving for V : Z = S ∗ : -%&@ $ 4 ∗ ℎ Z = S ∗ (15 !0) $ 4 ∗ (9 !0) = 1590.4 !0 1 Solving for t : 0 = Z P = 1590.4 !0 1 2.8 !0 1 ' ∗ 1 I(< 60 ' = 9.5 I(< 0 ≈ 10 I(<

Practice Exam A – 1 st Edition www.pegenius.com 35 113. Answer: C) 8 ft Relevant PE Civil Handbook Section: 3.10 OSHA 1926 Subpart P Table B-1 identified the maximum allowable slope for Type C soil as 1 ½:1 (H:V). Accordingly: t = 12 !0 ∗ 1 1.5 = 8 !0 Note that the total excavation depth is: 8 !0 + 5 !0 = 13 !0 < 20 !0 If total depth was greater than 20 feet, an engineered shoring system would be required. 114. Answer: B) 385 k Relevant PE Civil Handbook Section: 2.3.4 Cross-Sectional Area of Pile: R = S 4 ∗ (12 (<) $ = 452.4 (< $ Max Pile Driving Stress for Conventionally Reinforced Pile: 0.85! ! " = 4250 &'( Max Pile Driving Force: 9 A = 4250 &'( × 452.4 (< $ 9 A ≈ 385 _

Practice Exam A – 1 st Edition www.pegenius.com 36 115. Answer: A) 0.09 in Relevant PE Civil Handbook Section: 4.1.7 Maximum beam deflection for load P at midpoint: v = 9 ∗ M 1 48 ∗ V ∗ w Deflection for Beam 1: Δ = = 9 ∗ M = 1 48 ∗ V = ∗ w = Where: M = = 8 !0 Deflection for Beam 2: 2.5 (< = 9 ∗ M $ 1 48 ∗ V $ ∗ w $ Where: M $ = 24 !0 Since section and material properties are equal: V = = V $ w = = w $ Dividing deflection for Beam 1 by deflection for Beam 2: Δ = 2.5 (< = (8 !0) 1 (24 !0) 1 Δ = = 0.1 (< 116. Answer: B) 972.1 ft Relevant PE Civil Handbook Section: 5.3.1 Entry Grade: W = = − 0.03 Exit Grade: W $ = 0.015 Length of Curve: M = 8250 !0 − 8000 !0 = 250 !0 Distance to jogger: n = 8200 !0 − 8000 !0 = 200 !0 Rate of change of vertical curve grade: q = W $ − W = M = 0.015 + 0.03 2 50 !0 = 1.8 ∗ 10 ,B Elevation at jogger: Y C = q ∗ n $ 2 + W = ∗ n + Y 0D< Y C = (1.8 ∗ 10 ,B ) ∗ (200 !0) $ 2 + (−0.03) ∗ (200 !0) + 968.5 !0 Y C = 966.1 !0 Elevation at top of jogger’s head: Y -(? = Y C + 6 !0 = 972.1 !0

Practice Exam A – 1 st Edition www.pegenius.com 37 117. Answer: C) 0.75 cfs Relevant PE Civil Handbook Section: 4.5.2 Peak runoff flow can be calculated based on the rational formula: P = b ∗ w ∗ R Runoff flow prior to extending the cemetery: P (.6 = (0.15 ∗ 0.75 Xj + 0.25 ∗ 2.5 Xj + 0.1 ∗ 2 Xj) ∗ 10 (< ℎh = 9.38 j!' Runoff flow after extending the cemetery: P &84 = (0.25 ∗ (2.5 Xj + 0.75 Xj) + 0.1 ∗ 2 Xj) ∗ 10 (< ℎh = 10.13 j!' Change in flow: yP = P &84 − P (.6 = 10.13 j!' − 9.38 j!' = 0.75 j!' 118. Answer: D) Type V, Exposure Categories F & S Relevant PE Civil Handbook Section: 2.5.2.2 Concrete placed in high sulfur environments can be prone to degradation and sulfate attacks. ACI 318 specifies that concrete cement type V is to be used for sulfate resistance. Additionally, exposure categories F (for freeze/thaw) and S (sulfates) are both applicable in this scenario.

Practice Exam A – 1 st Edition www.pegenius.com 38 119. Answer: B) 2954 psf Relevant PE Civil Handbook Section: 3.3 Effective stress is calculated based on the portion of stress that may be resisted by soil contact. To calculate effective stress, first calculate total stress at the bottom of layer 2, then subtract pore water pressure. Note that the layer of water above soil layers is ignored in this calculation. Total Stress: z 2 = 110 &j! ∗ 20 !0 ∗ (1 + 0.25) + 135 &j! ∗ 20 !0 = 5450 &'! Pore Water Pressure: g = 62.4 &j! ∗ 40 !0 = 2496 &'! Effective Stress: z′ 2 = z 2 − g = 5450 &'! − 2496 &'! = 2954 &'!

Practice Exam A – 1 st Edition www.pegenius.com 39 120. Answer: C) 40 Relevant PE Civil Handbook Section: 1.6.8 The slenderness ratio is computed is using following formula: |M/h where: | = [!![j0(E[ ,[<N0ℎ ji[!!(j([<0 M = g<@hXj[: ,[<N0ℎ h = hX:(g' i! NYhX0(i< The critical slenderness ratio shall be the largest of the ratio in either direction: Fq !) = IXn } | C ∗ M C h C , | 7 ∗ M 7 h 7 The radius of gyration is based on the section’s area (A) and moment of inertia (I). Since the column is square, radii of gyration in each direction are equal: h C = h 7 = Ä w R = Ä 1 12 ∗ (18 (<) ∗ (18 (<) 1 (18 (<) ∗ (18 (<) = 5.2 (< The effective length coefficient value, K, is considered based on a fixed base and free headed support: | = 2 Given that: | C = | 7 h C = h 7 The critical slenderness ratio will be controlled by the maximum unbraced length. In this case, that is in the y-direction: M 7 = 8 !0 Fq !) = 2 ∗ 8 !0 ∗ 12 (< 1 !0 5.2 (< ≈ 40

Practice Exam A – 1 st Edition www.pegenius.com 40 121. Answer: B) STA 21+47 Relevant PE Civil Handbook Section: 5.2.1 The location of the point of tangent (PT) can be related to the point of intersection, curve length, and tangent length through the following equation: F`R EF = F`R EG − ` + M where: F`R EF = F`R $H + 00 ` = 400 !0 M = 2 ∗ S ∗ q ∗ Å 360° Solve for length M by finding the intersection angle Å : Å = 2 ∗ tan ,= K ` q L = 2 ∗ tan ,= K 400 !0 900 !0 L = 47.9° M = 2 ∗ S ∗ 900!0 ∗ 47.9° 360° = 752.8 !0 Solve for STA PI: F`R $HIJJ = F`R EG − 400!0 + 752.8 !0 F`R EG = F`R $= + 47 122. Answer: A) 41 cubic yards Relevant PE Civil Handbook Section: 2.1 (no swell factor required for this problem) Depth of Footing: : + = 5 !0 Depth of Excavation: : 8C = : + + 3 !0 = 8 !0 Width of Footing @ + = 8 !0 Width of Excavation: @ 8C = @ + + 2 ∗ 3 !0 = 14 !0 Area of Excavation: R 8C = (@ 8C ) $ = 196 !0 $ Volume of Excavation: Z 8C = R 8C ∗ : 8C = 1568 !0 1 Area of Footing: R + = Ö@ + Ü $ = 64 !0 $ Thickness of Footing: 0 + = 2 !0 Width of Pier: @ ? = 2 !0 Area of Pier: R ? = (2 !0) $ = 4 !0 $ Height of Pier Below Grade: ℎ ? = 3 !0 Volume of Foundation: Z + = R + ∗ 0 + + R ? ∗ ℎ ? = 140 !0 1 Volume of Fill per Foundation:

Practice Exam A – 1 st Edition www.pegenius.com 41 Z +9.. = Z 8C − Z + = 1568 !0 1 − 140 !0 1 = 1428 !0 1 Total Net Excavation (cubic yards): Z &8- = 8 ∗ (Z 8C − Z +9.. ) ∗ 1 Y: 1 27 !0 1 ≈ 41 123. Answer: C) 135 pcf Relevant PE Civil Handbook Section: 3.9.2 Dry density achieved by proctor compaction test: ] 6 !"# = 118 &j! Relative compaction: qb = 0.95 Relationship between dry density found by test and actual dry density: qb = ] 6 ] 6 !"# → 0.95 = ] 6 (118 &j!) Solve for actual dry density: ] 6 = 0.95 ∗ 118 &j! = 112.1 &j! Solve for unit weight/unit density based on water content: ] = ] 6 ∗ (1 + *) ] = 112.1 &j! ∗ (1 + 0.2) = 134.5 &j!

Practice Exam A – 1 st Edition www.pegenius.com 42 124. Answer: B) 149 ksi Relevant PE Civil Handbook Section: 1.6.7.2 The location of the maximum moment load in a cantilever beam is at the support. Solve for the moment at support: ) = 20 _(& ∗ 60 (< + 1 2 ∗ } 80 _(& !0 ∗ (1 !0) (12 (<) ∗ (80 (< ) }160 (< + (80 (<) 3 ) = 50,977.8 _(& − (< Maximum flexural stress is computed using the formula below: ! : = )Y w where: w = IiI[<0 i! (<[h0(X Y = :('0X<j[ i! 0ℎ[ [n0h[I[ !(@[h !hiI 0ℎ[ j[<0hi(: i! 0ℎ[ '[j0(i< Solve for f b : ! : = 50,977.8 _(& − (< ∗ T 16 (< 2 U 1 12 ∗ (8 (<) ∗ (16 (<) 1

Practice Exam A – 1 st Edition www.pegenius.com 43 ! : ≈ 149 _'( 125. Answer: B) 24 seconds per vehicle Relevant PE Civil Handbook Section: 5.1.3 Peak hour factor for Lane A: 9dH / = 0.75 Peak hourly flow for Lane A: E ?8%@/ = 150 E[ℎ 15 I(< ∗ 4 15 I(< 1 ℎh = 600 E[ℎ/ℎh Hourly volume for Lane A: 9dH / = Z '(#).7/ E ?8%@/ 0.75 = Z '(#).7/ 600 E[ℎ ℎh Z '(#).7/ = 450 E[ℎ ℎh Hourly volume for Lane B: Z '(#).70 = Z '(#).7/ 3 = 150 E[ℎ ℎh Hourly volume to headway relationship: Z '(#).70 = 3600 '[j ℎh ℎ[X:*XY : Average headway at Lane B: ℎ[X:*XY : = 24 '[j E[ℎ(j,[

Practice Exam A – 1 st Edition www.pegenius.com 44 126. Answer: C) 33.1 psi Relevant PE Civil Handbook Section: 6.3.1.2 Pressure drop in a pipe can be determined by its relation to head loss. y& = & 9& − & (#- = ℎ + ∗ r Where: ℎ + = ℎ[X: ,i'' :g[ 0i !h(j0(i< r = '&[j(!(j *[(Nℎ0 i! !,g(: K62.4 ,@ !0 1 !ih *X0[hL Using the Hazen-Williams equation, we can determine the approximate head loss: ℎ + = 4 .73 ∗ M ∗ P =.LH$ b =.LH$ ∗ G B.LM where: M = ,[<N0ℎ i! &(&[ (![[0) = 300 !0 P = !,i* (j g@(j ![[0 &[h '[ji<: ) = 1150 j !' b = higNℎ<['' ji[!!(j([<0 (g<(0,['') = 130 G = :(XI[0[h i! &(&[ (! [[0 ) = 14/1 2 (< & 9& − 20 &'( = 4 .73 ∗ (300 !0) ∗ (1.67 j !' ) =.LH$ (130) =.LH$ ∗ (14/1 2 (<) B.LM ∗ 62.4 ,@ !0 1 & 9& = 33.1 &'( 127. Answer: B) 3472 cubic yards Relevant PE Civil Handbook Section: 2.1.2 Volume of soil can be estimated based on the average end area method: Z 3(9. = R 3-%)- + R 8&6 2 ∗ M where: M = 200 !0 R 3-%)- = 215 !0 + 30 !0 ∗ 25 !0 = 562.5 !0 $ R 8&6 = 21 ∗ 30 !0 ∗ 25 !0 = 375 !0 $ Z 3(9. = 2562.5 !0 $ + 375 !0 $ ∗ 200 !0 ∗ 1 Y: 1 27 !0 1 = 3472 Y: 1

Practice Exam A – 1 st Edition www.pegenius.com 45 128. Answer: A) 0.64 in Relevant PE Civil Handbook Section: 3.2 Calculate the effective stress at the mid-depth of the soil layer: z 2 " = (125 &j! − 62.4 &j!) ∗ 15 !0 2 = 469.5 &'! Calculate past pressure based on overconsolidation ratio of existing soil: ábq = z ? " z 2 " z ? " = ábq ∗ z 2 " = 3 ∗ 469.5 &'! = 1408.5 &'! Using 2V:1H stress distribution, calculate the stress increase at mid-depth of the layer: yz = 300 _(& ∗ 1000 ,@ _(& (6 !0 + 7.5 !0) ∗ (40 !0 + 7.5 !0) = 467.84 &'! z 2 " + yz = 469.5 &'! + 467.84 &'! = 937.34 &'! < z ? " (1408.5 &'!) Calculate OCR after the footing is applied and loaded: ábq = z ? " z 2 " + yz = 1.5 > 1 ∴ Soil is overconsolidated Calculate the primary consolidation based on an overconsolidated clay: ' = b ) 1 + [ J d,iN =J z 2 " + yz z 2 " ' = 0.02 1 + 0.7 ∗ 15 !0 ∗ 12 (< !0 ∗ log =J 469.5 &'! + 467.84 &'! 469.5 &'! = 0.64 (<

Practice Exam A – 1 st Edition www.pegenius.com 46 129. Answer: C) 3.33 Relevant PE Civil Handbook Section: 3.4 Calculate the eccentricity, e: [ = ) N 9 = 700 _(& ∗ !0 (300 _(&) = 2.33 !0 > å 6 = 2 !0 w! [ < å 6 0ℎ[<: k >%C = 9 (å ∗ M) + (6 ∗ )) (å ∗ M $ ) w! [ > å 6 0ℎ[<: k >%C = 2 ∗ 9 3 ∗ å ∗ T M 2 − [U Calculate maximum bearing pressure: k >%C = (2 ∗ 300 _(&) K3 ∗ 12 !0 ∗ K 12 !0 2 − 2.33!0L L = 4.5 _'! Calculate the Factory of Safety: HF = k &8- k >%C = 15 _'! 4.5 _'! = 3.33

Practice Exam A – 1 st Edition www.pegenius.com 47 130. Answer: C) 19 ft Relevant PE Civil Handbook Section: 6.2 Detention time: 0 = 4 ℎh Flow into the tank: P = 500 N&I Radius of the tank: h = 20 !0 Volume of tank, where h represents the height required for water detention: Z = S ∗ h $ ∗ ℎ Volume based on flow over time: Z = P ∗ 0 Setting the two volumes equal: P ∗ 0 = S ∗ h $ ∗ ℎ 500 N&I ∗ 4 ℎh ∗ 60 I(< 1 ℎh ∗ 1 !0 1 7.48 NX, = S ∗ (20 !0) $ ∗ ℎ Solving for h: ℎ = 12.8 !0 To obtain the total minimum tank height, we add height of freeboard and siltation to h: ℎ - = 12.8 !0 + 2 !0 + 4 !0 ≈ 19 !0 131. Answer: C) 354 plf Relevant PE Civil Handbook Section: 4.1 Consider weight of the steel beam, W. m = * ∗ M where: * = *[(Nℎ0 (< &ig<:' &[h ,(<[Xh !ii0 M = ,[<N0ℎ i! @[XI Weight supported by each cable: m !%:.8 = * ∗ M 2 Tension in each cable based on able: ` = m !%:.8 cos (45°) Solve for weight per foot based on tension capacity: 5 _(& ∗ 1000 ,@ _(& = * ∗ 20 !0 2 cos (45°) * = 354 &,!

Practice Exam A – 1 st Edition www.pegenius.com 48 132. Answer: C) 0.29 Relevant PE Civil Handbook Section: 3.8.3 Consider the relationship between saturated unit weight and water: r 3%- = r 4 (FW + [) 1 + [ Given r 3%- = 2r 4 : 2r 4 = r 4 (FW + [) 1 + [ Solving for the specific gravity, SG : 2 = W + 0.80 1 + 0.8 ; W = 2.8 FW = 2.8 Calculate water content based on its relationship with specific gravity SG and degree of saturation, S : * = F ∗ [ FW *ℎ[h[ F = 1.0 !ih 'X0ghX0[: 'XI&,[' * = 1.0 ∗ 0.8 2.8 = 0.29

Practice Exam A – 1 st Edition www.pegenius.com 49 133. Answer: D) 1911.8 lb-ft Relevant PE Civil Handbook Section: 1.6.5 Let ` % and ` 8 be the support reactions at points A and E respectively. In equilibrium: ` % + ` 8 = 590 + 880 + 750 ` % + ` 8 = 2220 Since the support is unyielding, the angle of twist at point A with respect to point E shall be zero. Shown below is the shafts torque diagram. Equilibrium equation for twist: ë / O = ë / 0 + ë 0 < + ë < A + ë A O = 0 To compute for the angle of twist, use: ë = FP QR where: ` = 0ih'(i< M = ,[<N0ℎ W = Ii:g,g' i! h(N(:(0Y í = &i,Xh IiI[<0 i! (<[h0(X The polar moment of inertia of a solid circular section is:32 πD 4 Accordingly: ë / O = ` % (60 (<) (11.5 ∗ 10 S &'() ì S(5 (<) B 32 î + (` % − 590 ,@ ⋅ !0)(40 (<) (11.5 ∗ 10 S &'() ì S(5 (<) B 32 î + (` % − 1470 ,@ ⋅ !0)(75 (<) (11.5 ∗ 10 S &'() ì S(1 0 (<) B 32 î + (` % − 2220 ,@ ⋅ !0)(75 (<) (11.5 ∗ 10 S &'() ì S(15 (<) B 32 î = 0 Simplifying: ë / O = ` % (60 (<) (5 (<) B + (` % − 590 ,@ ⋅ !0)(40 (<) (5 (<) B + (` % − 1470 ,@ ⋅ !0)(75 (<) (10 (<) B + (` % − 2220 ,@ ⋅ !0)(75 (<) (15 (<) B = 0 ` % = 318.16 ,@ ⋅ !0

Practice Exam A – 1 st Edition www.pegenius.com 50 With this value of T a and using the equilibrium equation: ` 8 = 1911.84 ,@ ⋅ !0 134. Answer: D) 4.0 in Relevant PE Civil Handbook Section: 6.5 Peak Flow Rate: P = 40 !0 1 /' Storm Peak Duration: ` ?8%@ = 20 ℎh − 5 ℎh = 15 ℎh Storm Complete Duration: ` 3-()> = 25 ℎh Catchment area: R = 200 Xjh[ ∗ 43560 !0 $ 1 Xjh[ = 8.7 ∗ 10 M !0 $ 1) Calculate volume of runoff based on the hydrograph: Z )#&(++ = P ∗ ` 3-()> + ` ?8%@ 2 = 40 !0 1 ' ∗ 25 ℎh + 15 ℎh 2 ∗ 3600 ' 1 ℎh Z )#&(++ = 2.88 ∗ 106!0 1 2) Calculate the excess rainfall based on the catchment area: Vnj['' = Z )#&(++ R = 4.0 (<

Practice Exam A – 1 st Edition www.pegenius.com 51 135. Answer: D) 1833 psf Relevant PE Civil Handbook Section: 3.3 Effective stress is calculated based on the portion of stress that may be resisted by soil contact. To calculate effective stress, first calculate total stress at the bottom of layer 2, then subtract pore water pressure. Total Stress: z 2 = 115 &j! ∗ 15 !0 + (1 + 0.2) ∗ 110 &j! ∗ 15 !0 = 3705 &'! Pore Water Pressure: g = 62.4 &j! ∗ 30 !0 = 1872 &'! Effective Stress: z 2 " = z 2 − g = 3705 &'! − 1872 &'! = 1833 &'! 136. Answer: C) 4.8 cfs Relevant PE Civil Handbook Section: 6.4.5 Flow Depth: ℎ = 9(< Radius: h = 18(< Calculate the angle theta for a partially filled pipe: ë = 2 ∗ cos ,= h − ℎ ℎ = 2.09 hX:(X<' Calculate area of flow: R = h $ (ë − '(<ë) 2 = 1.38 !0 $ Calculate wetted perimeter: 9 = hë = 37.7(< Calculate hydraulic radius: q = R 9 = 0.440 Calculate flow based on Manning’s equation:

Practice Exam A – 1 st Edition www.pegenius.com 52 P = 1.49 < Rq $/1 F =/$ = 4.76 !0 1 ' 137. Answer: D) 486 kip Relevant PE Civil Handbook Section: 3.1.3 (different trigonometric equation used which yields the same results) Calculate the Rankine coefficient of active earth pressure: | % = ji'ñ ∗ ji'ñ − pcos $ ñ − cos $ Å ji'ñ + pcos $ ñ − cos $ Å | % = ji' 15° ∗ ji' 15° − √cos $ 15° − cos $ 30° ji' 15° + √cos $ 15° − cos $ 30° = 0.37 Calculate the horizontal component of the coefficient: | %' = | %' ∗ ji' ñ = 0.37 ∗ ji'15° = 0.36 Calculate the horizontal component of the resultant force: q %' = 1 2 r d $ | %' M = 1 2 ∗ 120 &j! ∗ (30 !0) $ ∗ 0.36 ∗ (25 !0) = 486,000 ,@

Practice Exam A – 1 st Edition www.pegenius.com 53 138. Answer: A) 50 years Relevant PE Civil Handbook Section: 6.5 The probability that an event will not occur in N years is defined by the following equation: q = (1 − &) U *ℎ[h[ & = X<<gX, &hi@X@(,(0Y Given the findings of the risk assessors, we are able to solve for the yearly probability of the event: 0.80 = (1 − &) $J & = 0.011 Calculate the return period based on the annual probability: ` = 1 & = 90 Yh Calculate the number of years from today when the flood can be expected occur: < = 90 Yh − 40 Yh = 50 Yh 139. Answer: C) 3.8 cubic feet Relevant PE Civil Handbook Section: 2.5.2 Volume of Footing: Z +((-9&; = 1 !0 ∗ 4 !0 ∗ 4 !0 = 16 !0 1 Weight of Cement: m !8>8&- = 5 ∗ 94 ,@ = 470 ,@ Water-Cement Ratio: * !) = 0.5 Weight of Water required: * !) = m 4%-8) m !8>8&- m 4%-8) = 0.5 ∗ 470 ,@ = 235 ,@ Volume of Water Required: Z 4%-8) = m 4%-8) r 4%-8) = 3.8 !0 1

Practice Exam A – 1 st Edition www.pegenius.com 54 140. Answer: B) Point B Relevant PE Civil Handbook Section: 1.6 The steel yield strength is reached after the steel has reached its elastic limit (Point A) and begins to behave inelastically (Point B). Once loaded past its yield strength, the steel experiences inelastic deformation and is able to be stressed until it reaches its ultimate tensile strength (Point C). The steel is then loaded until fracture (Point D).

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