PE_GENIUS_EXAM_C_GUMROAD

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

Practice Exam C – 1 st Edition www.pegenius.com 5 303. 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 0.025 for both pipes, determine the total input flow Q. Assume minor losses are negligible. A. 1.3 cfs B. 2.6 cfs C. 2.0 cfs D. 3.5 cfs 304. Given the following time intervals and vehicle count passing a freeway segment, determine the peak hour factor. Time Number of Vehicles 7:00 – 7:15 759 7:15 – 7:30 805 7:30 – 7:45 825 7:45 – 8:00 847 8:00 – 8:15 855 8:15 – 8:30 819 8:30 – 8:45 798 8:45 – 9:00 770 A. 0.65 B. 0.98 C. 0.45 D. 0.80 Q PIPE A, DIAMETER = 5”, LENGTH = 100 FT PIPE B, DIAMETER = 3”, LENGTH = 75 FT

Practice Exam C – 1 st Edition www.pegenius.com 8 309. Given a horizontal curve with the following parameters, determine the length of the curve: Radius: , = 700 +/ Tangent: 0 = 300 +/ A. 433 ft B. 475 ft C. 567 ft D. 643 ft 310. Pictured below is a continuous cantilever retaining wall with clayey gravel soil. Given the parameters shown determine the resultant Rankine Active Earth Force Resultant P a acting transverse to each foot of wall. A. 0.5 k/ft B. 1.4 k/ft C. 1.9 k/ft D. 2.3 k/ft 6’-0” 1’-6” 3’-0” P a 8’-0” 2’-6” 3’-0” ! = #$% '() * = $+°

Practice Exam C – 1 st Edition www.pegenius.com 11 315. Three excavation cross sections taken 70 feet apart are provided in the figure below. Use the prismoidal formula to estimate the total volume of soil excavated in cubic yards. A. 1894 CY B. 2983 CY C. 3867 CY D. 5049 CY 316. The following figure shows a simply supported beam with a uniform load. Two options are being considered for the structural steel section to be used for the beam: • Hollow circular section with 1/2" thick wall. • Box section with 1/2" thick wall. Based on the parameters shown, determine which section will result in a lower deflection. A. Option 1 B. Option 2 C. Both options will have the same shear stress. D. Impossible to determine based on information.

Practice Exam C – 1 st Edition www.pegenius.com 14 321. Given the following parameters, determine the void ratio for the saturated soil sample: Specific Gravity: 34 = 2.7 Water Content: 6 = 0.25 A. 0.53 B. 0.68 C. 0.71 D. 0.84 322. Given the following figures, determine which option approximately represents the proper location of the shear center for a structural steel channel section: A. Option 1 B. Option 2 C. Option 3 D. Option 4

Practice Exam C – 1 st Edition www.pegenius.com 17 327. Your town's Public Works Department has plans to upsize an old rainfall runoff basin culvert in 5 years. In order to help the department, decide if the project can wait 5 years or needs to be advanced, determine the probability of flooding twice in the next 5 years. Assume the culvert was originally designed for a storm with a return period of 25 years. A. 5.2% B. 3.3% C. 2.5% D. 1.4% 328. You are working on a project staffed with 65 full-time employees. Unfortunately, one of the tradesmen suffers a minor arm injury following 105 days without an incident. Determine the Safety Incidence Rate. A. 2.33 B. 3.67 C. 4.25 D. 5.50

Practice Exam C – 1 st Edition www.pegenius.com 20 333. Three concrete sample cylinders are taken for testing during a pour for foundation designed to have a compressive strength f’ c of 6000 psi. Based on the table shown, determine the minimum value X that would be acceptable to validate the design compressive strength. A. 4800 psi B. 5400 psi C. 6000 psi D. 6200 psi Sample f' c (psi) 1 6700 2 6500 3 X 334. Given the following parameters for a soil sample, determine the total unit weight: Void Ratio G = 0.66 Water Content 6 = 20% Weight of Solid in Sample: H $ = 10 IJ Volume of Solid in Sample: F $ = 260 %: % Unit Weight of Water: D & = 62.4 &*+ A. 105 pcf B. 115 pcf C. 125 pcf D. 140 pcf

Practice Exam C – 1 st Edition www.pegenius.com 21 335. A 75-foot diameter tank filled is being drained through a 10-inch diameter short pipe orifice. Given that the height of water reaches 20 feet above the orifice centerline, determine the time it will take to drain the tank to the orifice level. A. 53 min B. 69 min C. 88 min D. 96 min 336. The following figure shows a plan for a grid of 12 square spread foundation to be poured. The dimensions of each foundation are 9 feet by 9 feet with a bottom of foundation depth that reaches 10 feet below grade. Given the following information, determine the cost of hauling all the excavation soil away from site (assume no soil excavated for the foundations will remain on site). Volume per Truck Haul: 15 ;L % /ℎNOI Cost of Truck Hauling: $400/ℎNOI Swell Factor: 3 & = 10% A. $10,800 B. $11,100 C. $12,500 D. $14,200 75’-0” 20’-0”

Practice Exam C – 1 st Edition www.pegenius.com 22 337. Given the following fixed-end beam supporting point load P, determine the location of the maximum moment. A. Point A B. Point B C. Point C D. Point D 338. 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 ultimate tensile strength? A. Point A B. Point B C. Point C D. Point D

Practice Exam C – 1 st Edition www.pegenius.com 23 339. Determine the internal force in member b based on the provided truss geometry and loads. A. 106 kip B. 75 kip C. 181 kip D. 0 kip 340. The following figure shows a simply supported beam with a uniform load. Two options are being considered for the structural steel section to be used for the beam: • Wide flange section composed of 1" steel plate. • Box section composed of 1" steel plate. Based on the parameters shown, determine which section will result in the lower shear stress. A. Option 1 B. Option 2 C. Both options will have the same shear stress. D. Impossible to determine based on information.

Practice Exam C – 1 st Edition www.pegenius.com 24 Solution Key Problem Solution Problem Solution 301 C 321 B 302 B 322 A 303 A 323 D 304 B 324 A 305 B 325 B 306 C 326 D 307 A 327 D 308 B 328 B 309 C 329 D 310 D 330 B 311 B 331 C 312 B 332 A 313 A 333 B 314 A 334 C 315 C 335 C 316 A 336 A 317 D 337 D 318 A 338 C 319 C 339 A 320 C 340 B Please visit www.pegenius.com/errata for any updates to these solutions.

Practice Exam C – 1 st Edition www.pegenius.com 25 Solutions 301. Answer: C) 25 days Relevant PE Civil Handbook Section: 2.4.1 Critical path is that which has the longest duration to complete required activities. Upon reviewing the schedule as shown, one can see that the longest duration path is D-E-A-F-G which results in a duration of 27. If activity A is 1 day in duration, then path D-E-A-F-G has a duration of 25 days. Accordingly, path D-B-C-G now becomes the critical path because it has a longer duration of 25 days. 302. Answer: B) $ 34,000 Relevant PE Civil Handbook Section: 2.2 Calculate the time it will take 3 bulldozers to move all aggregate: 25,000 ;L % 2 ∗ 90 ;L % /ℎ8 = 139 ℎ8 Calculate the bulldozer rental cost: 2 ∗ 139 ℎ8 ∗ $30 ℎ8 = $ 8,340 Calculate crew cost: 139 ℎ8 ∗ $180 ℎ8 = $ 25,020 Calculate total cost: 0T/NI = $8,340 + $ 25,020 = $ 33,360

Practice Exam C – 1 st Edition www.pegenius.com 26 303. Answer: A) 1.3 cfs Relevant PE Civil Handbook Section: 6.2.3 Head loss for a parallel system is equal for each pipe: ℎ '()*)#+ = ℎ ', = ℎ '- Head loss per the Darcy-Weisbach equation: ℎ ' = + ∗ V W ∗ X . 2Y Calculate velocity in Pipe A: 5 +/ = 0.025 ∗ 100 +/ 5 12 +/ ∗ X , . 2 Z32.2 +/ ' [ X , = 7.32 +/ ' Calculate velocity in Pipe B: 5 +/ = 0.025 ∗ 75 +/ 3 12 +/ ∗ X - . 2 Z32.2 +/ ' [ X - = 6.55 +/ ' Total input flow is equal to flow through Pipes A and B: \ )*)#+ = \ , + \ - Calculate total input flow: \ )*)#+ = ] , X , + ] - X - \ )*)#+ = ^ ∗ (5/12 +/) . 4 ∗ Z7.32 +/ ' [ + ^ ∗ a 3 12 +/b . 4 ∗ Z6.55 +/ ' [ ≈ 1.3 *+'

Practice Exam C – 1 st Edition www.pegenius.com 27 304. Answer: B) 0.98 Relevant PE Civil Handbook Section: 5.1.3 Time Number of Vehicles 7:00 – 7:15 759 7:15 – 7:30 805 7:30 – 7:45 825 7:45 – 8:00 847 8:00 – 8:15 855 8:15 – 8:30 819 8:30 – 8:45 798 8:45 – 9:00 770 The peak hour factor (PHF) is based on the highest number of vehicles within an hour. The 7:30-8:30 hour has the largest total volume of vehicles: F = 825 + 847 + 855 + 819 = 3346 Maximum Volume in a 15 Minute Period (8:00-8:15): F_15 = 855 Peak Hour Factor: Eef = F/(4 ∗ F_15 ) = 3346/(4 ∗ 855) − 0.98 305. Answer: B) 21 ft Relevant PE Civil Handbook Section: 3.10 The maximum allowable slope for Type A soil with a vertically sided lower portion and a maximum depth of 12 feet is 1:1 (Horizontal: Vertical) Accordingly: h /01 = 2 ∗ Z8 +/ ∗ 1 1 [ + 5 +/ = 21 +/

Practice Exam C – 1 st Edition www.pegenius.com 28 306. Answer: C) 2.25 ft Relevant PE Civil Handbook Section: 6.4.7 Equation to determine the super critical flow depth based on subcritical velocity and depth: ; 2 = − 1 2 ; . + i 2 ∗ X . . ∗ ; . Y + ; . . 4 ; 2 = − 1 2 (3 +/) + j 2 ∗ (8 *+') . ∗ (3 +/) Z32.2 +/ ' . [ + (3 +/) . 4 ; 2 ≈ 2.25 +/ 307. Answer: A) 2.5 cfs Relevant PE Civil Handbook Section: 6.2.5.5 Basic flow equation for a V-notch weir: \ = k ∗ e 3 . Discharge coefficient (given): k = 2.5 Head above weir bottom: e = 1 +/ Flow: \ = 2.5 ∗ 1 3 . = 2.5 *+'

Practice Exam C – 1 st Edition www.pegenius.com 29 308. Answer: B) 1100 psf Relevant PE Civil Handbook Section: 3.4 Concrete Weight: 6 ! = 150 &*+ Column Load: E = 10 $ Weight of Pier: H 405" = 150 &*+ ∗ 2 +/ ∗ 2 +/ ∗ 3 +/ = 1.8 $ Weight of Footing: H ')6 = 150 &*+ ∗ 5 +/ ∗ 5 +/ ∗ 1.5 +/ = 5.625 $ Total Load on Soil: E )*)#+ = 20 $ + 1.8 $ + 5.625 $ = 27.425 $ Gross Bearing Pressure: l = 17.425 $ 5 +/ ∗ 5 +/ ≈ 1100 &'+ 309. Answer: C) 567 ft Relevant PE Civil Handbook Section: 5.3.1 Consider the relationships between tangent, radius, intersection angle, and length: 0 = , tan p 2 V = ^ , p 180° Solve for intersection angle: p = 2 tan (2 0 , p = 2 tan (2 (300 +/) 700 +/ = 46.4° Solve for length based on intersection angle and radius: L = ^ × 700 +/ × (46.4°) 180° = 567 +/

Practice Exam C – 1 st Edition www.pegenius.com 30 310. Answer: D) 2.3 k/ft Relevant PE Civil Handbook Section: 3.1.2 Total Height of Wall: e = 8 +/ + 2.5 +/ = 10.5 +/ Effective Friction Angle: p = 29° Unit Weight of Soil: D = 120 &*+ Rankine Active Earth Coefficient: t # = tan . Z45 − p 2 [ t # = tan . Z45° − 29° 2 [ = 0.35 Active Earth Pressure Resultant: E # = t # D e . 2 E # = (0.35) (120 &*+)(10.5 +/) . 2 = 2315.25 IJ/+/ E # = 2.3 $/+/ 311. Answer: B) 1.3 kip Relevant PE Civil Handbook Section: 4.1 Calculate moment at base: u 7 = E ∗ I = 320 IJ ∗ 4 +/ = 1280 IJ − +/ u 7 = 1280 lb − ft Calculate force at anchor based on moment: E #1!8*" = u 7 2 ∗ L where moment arm d is: L = 6 %: = 0.5 +/ E #1!8*" = 1280 IJ − +/ 2 ∗ 0.5 +/ = 1280 IJ ≈ 1.3 $%&

Practice Exam C – 1 st Edition www.pegenius.com 31 312. Answer: B) 2.0 k Relevant PE Civil Handbook Section: 1.5 Given weight of the crane: H !"#15 = 5 $ Take moment about front wheel: H 971:+5 × 25 +/ − H !"#15 × 10 +/ = 0 H 971:+5 = 5 $ × 10 +/ 25 +/ = 2 $ 313. Answer: A) 4.6 ksi Relevant PE Civil Handbook Section: 1.6.7 Location of maximum bending stress corresponds to the location of maximum moment which occurs at the support point of a cantilever beam. Determine the maximum moment: u /#; = 6I . 2 u /#; = 2 $I+ × (4 +/) . 2 = 16 $ ⋅ +/ Maximum flexural stress is computed using the formula below: + 9_/#; = u /#; ; z where: y is the distance to the extreme fiber from the centroid of the section. In our case, this is the outer radius. I is the moment of inertia. The moment of inertia for a pipe section is calculated as follows:

Practice Exam C – 1 st Edition www.pegenius.com 32 z = ^(, * = − , 0 = ) 4 Inner Radius: , 0 = 5 %: Outer Radius: , * = 5 %: + 0.5 %: = 5.5 %: Moment of Inertia: z = ^[(5.5 %:) = − (5%:) = ] 4 = 227.8 %: = Maximum Bending Stress: + 9_/#; = 16 $ ∙ +/ × 12 × 5.5 %: 227.8 %: = + 9_/#; = 4.6 $'% 314. Answer: A) 420 psi Relevant PE Civil Handbook Section: 1.6.8 E !" = ^ . ~z (tV) . where: L = length (given) K = effective-length factor E = modulus of elasticity (given) I = Moment of Inertia Moment of Inertia: z = Jℎ % 12 = (6 %:) = 12 = 108 %: = For a column with fixed-free end conditions, the effective-length factor is: t = 2.0 Calculate the critical buckling load: E !" = ^ . × 1600 $'% × 108 %: = (2.0 × 14 +/ × 12) . E !" = 15.1 $

Practice Exam C – 1 st Edition www.pegenius.com 33 We can use the critical buckling load to determine the critical buckling stress as follows: 2 !" = E !" ] 2 !" = 15.1 $ (6 %:) . = 0.42 $'% 2 !" ≈ 420 &'% 315. Answer: C) 3867 CY Relevant PE Civil Handbook Section: 2.1.2 Prismoidal Formula: F = V Z ] 2 + 4] / + ] . 6 [ Distance between cross sections: V = 70 +/ Volume Estimate: F = 70 +/ Z 1200 + 4 × 1600 + 1350 6 [ = 104416.7+/ % F = 104416.7 +/ % = 3867 k

Practice Exam C – 1 st Edition www.pegenius.com 34 316. Answer: A) 4.6 ksi Relevant PE Civil Handbook Section: 1.5.7 & 1.6.7 Deflection for beam under uniform load: ∆= 56I = 384~z A larger moment of inertia will result in a lower deflection. The moment of inertia for a hollow circular section can be calculated based on the difference of moment of inertia of two circular sections: Inner Radius: , 0 = 6 %: Outer Radius: , * = 6 %: + 6.5 %: = 6.5 %: Outer Section Moment of Inertia: z * = ^, * = 4 = ^ × (6.5 %:) = 4 = 1401.98 %: = Inner Section Moment of Inertia: z 0 = ^, 0 = 4 = ^ × (6 %:) = 4 = 1017.88 %: = Moment of Inertia of Hollow Section: z !0"!7+#" = z * − z 0 = 384.10 %: = The moment of inertia for a box section can be calculated based on the difference of moment of inertia of two rectangular sections: Outer Section Moment of Inertia: z * = J * ℎ * % 12 = (12.5 %:) = 12 = 2034.51 %: = Inner Section Moment of Inertia: z 0 = J * ℎ * % 12 = (12 %:) = 12 = 1728 %: = Moment of Inertia of Tube Section: z 9*; = z * − z 0 = 306.51 %: = z !0"!7+#" > z 9*; Therefore, circular section will minimize deflection.

Practice Exam C – 1 st Edition www.pegenius.com 35 317. Answer: D) 2634 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: 2 ? = (1 + 0.3) × 105 &*+ × 20 +/ + 120 &*+ × 20 +/ = 5130 &'+ Pore Water Pressure: O = 62.4 &*+ × 40 +/ = 2496 &'+ Effective Stress: 2 ? @ = 2 ? − O 2 ? @ = 5130 &'+ − 2496 &'+ = 2634 &'+ 318. Answer: A) 15 mph Relevant PE Civil Handbook Section: 5.1.1 Space mean speed can be determined based on free flow speed, density, and jam density. O $ = O ' Ç1 − $ $ A É Free flow speed (miles per hour): O ' = 50 Ñ&ℎ Density (vehicles per mile): $ = 350 X&Ñ Jam Density (vehicles per mile): $ A = 500 X&Ñ Calculate space mean speed: O $ = 50 Ñ&ℎ ∗ Z1 − 350 X&Ñ 500 X&Ñ [ = 15 Ñ&ℎ

Practice Exam C – 1 st Edition www.pegenius.com 36 319. Answer: C) 102.8 ft Relevant PE Civil Handbook Section: 5.3.1 Elevation at PVC: BCD = 98 +/ Length of Curve: V = 750 +/ Grade of Back Tangent: Y 2 = 3.0% Grade of Forward Tangent: Y . = −4.0% Horizontal Distance to Max Elevation: Ö / = Y 2 V Y 2 − Y . Ö / = 3% ∗ 750 +/ 3% − (−4%) = 321.43 +/ Curve Max Elevation: ~V /#; = BCD + Y 2 Ö / + Ö / . a Y . − Y 2 2V b ~V /#; = 98 +/ + 3.0% × 321.43 +/ + (321.43 +/) . Z −4% − 3% 2 ∗ 750 +/ [ ~V /#; = 102.82 +/ 320. Answer: C) 114.8 ft Relevant PE Civil Handbook Section: 2.1.3 Determine Height of Instrument (HI) between TP 1 and TP 2: ez . = 0E . + f3 . ez . = 120.4 +/ + 1.8 +/ = 122.2 +/ Determine elevation at TP 1: 0E 2 = ez . − Ü3 . 0E 2 = 122.2 +/ − 5.1 +/ 0E 2 = 117.1 +/ Determine elevation at Height of Instrument (HI) between TP 1 and BM: ez 2 = 0E 2 + f3 2 ez 2 = 117.1 +/ + 2.3 +/ = 119.4+/ Determine elevation at BM: Üu = ez 2 − Ü3 2 Üu = 119.4 +/ − 4.6 +/ Üu = 114.8 +/

Practice Exam C – 1 st Edition www.pegenius.com 37 321. Answer: B) 0.68 Relevant PE Civil Handbook Section: 3.8.3 The relationship between water, degree of saturation, void ratio, and specific gravity is as follows: 6 = 3 × G 34 For saturated soil samples, the degree of saturation S is 1.0. 0.25 = 1.0 × G 2.7 G = 0.68 322. Answer: A) Option 1 Relevant PE Civil Handbook Section: N/A Shear center of a channel section is located outside of the web, opposite the flanges.

Practice Exam C – 1 st Edition www.pegenius.com 38 323. Answer: D) 1,400,000 Relevant PE Civil Handbook Section: 6.2.2 Reynolds Number is calculated as follows: ,G = X, á Flow Velocity: X = 30 +//' Hydraulic Radius , 8 = ]/E , 8 = ^8 . 2^8 = 8 2 , 8 = 6 %: Kinematic Viscosity of Water: á = 1.059 × 10 (3 +/ . /'G* Reynolds Number: ,G = 30 +/ ' × 6 %: 1.059 × 10 (3 +/ . /'G* ,G = 1416431 ≈ 1,400,000 324. Answer: A) 5.1 cfs Relevant PE Civil Handbook Section: 6.4.5 Manning’s Roughness Coefficient for a Cast Iron Pipe: : = 0.013 Depth of Flow: ; = 8 %: Radius: 8 = 6 %: Unfilled Segment Height: ℎ = 2 ∗ 6 %: − 8 %: = 4 %: à = 2 ∗ cos (2 Z 8 − ℎ ℎ [ à = 2 ∗ cos (2 Z 6 %: − 4 %: 4 %: [ à = 2.09 8NL Calculate Area of Flow: ] = ^8 . − 8 . (à − sin à) 2 ] = ^(6 %:) . − (6 %:) . (2.09 − sin 2.09) 2 ] = 0.63 +/ . Calculate Wetted Perimeter: E = 8à E = 0.5 +/ ∗ 2.09 8NL = 1.05 +/

Practice Exam C – 1 st Edition www.pegenius.com 39 Calculate Hydraulic Radius: , = ] E , = 0.63 +/ . 1.05 +/ = 0.6 +/ Calculate Flow Based on Manning’s Equation: \ = 1.49 : ], ./% 3 2/. \ = 1.49 0.013 (0.63 +/ . )(0.6 +/) ./% (0.01) 2/. \ = 5.1 *+' 325. Answer: B) 1.8 Relevant PE Civil Handbook Section: 1.5.8 Total Height of Wall: e = 8 +/ + 2 +/ = 10 +/ Unit Weight of Soil: D = 115 &*+ Active Earth Pressure Resultant: E # = 1.8 $/+/ Weight of Soil Above Wall Footing: H $*0+ = 5 +/ × 8 +/ × 115 &*+ = 4600 $/+/ Weight of Wall: H &#++ = [2 +/ × (10 +/) + 2 +/ × 5 +/] × 150 &*+ H &#++ = 4.5 $/+/ ç = 0.35 Horizontal Shear Demand: F :5/#1: = E # = 1.8 $/+/ Friction Force Resisting Sliding: F "5$0$) = (H &#++ + H $*0+ ) × ç F "5$0$) = 3.185 $/+/ Safety Factor: 3f = F "5$0$) F :5/#1: 3f = 3.185 $/+/ 1.8 $/+/ ≈ 1.8

Practice Exam C – 1 st Edition www.pegenius.com 40 326. Answer: C) 0.75 cfs Relevant PE Civil Handbook Section: 6.5.2 Peak runoff flow can be calculated based on the rational formula: \ = k ∗ z ∗ ] Runoff flow prior to extending the cemetery: \ = (0.4 ∗ 5 N* + 0.1 ∗ 2 N*) ∗ 12 %: ℎ8 = 26.4 *+' 327. Answer: D) 1.4% Relevant PE Civil Handbook Section: 6.5.1.5 Probability of occurrence in one year: = 1 8G/O8: &G8%TL = 1 f = 1 25 = 4% Probability of occurrence for 2 times in the next 5 years: é! :! (é − :) ! E 1 (1 − E) F(1 where é = 5 ;GN8' , : = 2 , E = 4% P (2,5) = 3! .!(3(.)! 0.04 . (1 − 0.04) 3(. = 1.4 %

Practice Exam C – 1 st Edition www.pegenius.com 41 328. Answer: B) 3.67 Relevant PE Civil Handbook Section: 2.6.1 Safety Incidence Rate is calculated as follows: z, = é × 200,000/0 Number of Incidents: é = 1 Total Number Worked by All Employees: 0 = 65 ∗ 8 ℎ8 LN; ∗ 105 LN; = 54600 ℎ8 Incidence Rate: z, = 1 ∗ 200,000/54600 = 3.67 329. Answer: D) 3 in Relevant PE Civil Handbook Section: 2.5.2.3 Per the relevant section, concrete cast against and permanently in contact with ground requires a cover of 3 inches for reinforcement. The same value can be found in the requirements of Building Code Requirements for Structural Concrete (ACI 318).

Practice Exam C – 1 st Edition www.pegenius.com 42 330. Answer: B) 520 psf Relevant PE Civil Handbook Section: 3.5 Load on Soil: E '1: = 150 $ Width of Foundation: Ü = 6 +/ Length of Foundation: V = 8 +/ Depth of Soil: ê = 10 +/ Pressure Based on 2:1 Vertical Distribution: & = E '1: (Ü + ê)(V + ê) & = 150 $ (6 +/ + 10 +/)(8 +/ + 10 +/) ≈ 520 &'+ 331. Answer: C) 5000 LB Relevant PE Civil Handbook Section: N/A Per the OSHA Code of Federal Regulations 1926.502 - Fall protection systems and practices: “ Anchorages used for attachment of personal fall arrest equipment shall be independent of any anchorage being used to support or suspend platforms and capable of supporting at least 5,000 pounds (22.2 kN) per employee attached ”

Practice Exam C – 1 st Edition www.pegenius.com 43 332. Answer: A) 1.9 ksf Relevant PE Civil Handbook Section: 3.4.2 Width of Footing: Ü = 7 +/ Length of Footing: V = 7 +/ Vertical Load: E = 100 $ Lateral Load: F = 15 $ Distance from Base of Footing: ℎ = 5 +/ Moment at Base of Footing: u = F × ℎ u = 15 $ ∗ 5 +/ = 75 $ ⋅ +/ Eccentricity: G = u E = 75 $ ⋅ +/ 100 $ G = 0.75 +/ Check that eccentricity is less than B/6: G < Ü 6 = 7 +/ 6 = 1.12 +/ Since eccentricity is less than B/6, minimum bearing pressure can be calculated as follows: l /#; = E ÜV Z1 + 6G Ü [ l /#; = 100 $ 7 +/ × 7 +/ Z1 + 6 × 1.12 +/ 7 +/ [ l /#; = 1.9 $'+

Practice Exam C – 1 st Edition www.pegenius.com 44 333. Answer: B) 5400 psi Relevant PE Civil Handbook Section: 2.5.1 Arithmetic average of test results needs to be higher than +’* : + ! @ = 6000 &'% 6000 = 6700 + 6500 + h 3 h = 4800 For a target of +’* greater than 5000 psi, the minimum value of test result cannot fall below f’c by 10%: h /01 = 0.90 ∗ 6000 = 5400 &'% → h = 5400 &'% 334. Answer: C) 125 pcf Relevant PE Civil Handbook Section: 3.8.3 We can calculate specific gravity based on weight and volume of solids: 4 $ = H $ F $ D & = 10 IJ 100 %: % × 1 (12 %:) % × 62.4 &*+ = 2.77 We can calculate total unit weight based on water content, specific gravity, and void ratio: D ) = (1 + 6)4 $ D & 1 + G D ) = (1 + 0.2) × 2.77 × 62.4 &*+ 1 + 0.66 ≈ 125 &*+

Practice Exam C – 1 st Edition www.pegenius.com 45 335. Answer: C) 88 min Relevant PE Civil Handbook Section: 6.2.6 Coefficient of Discharge (given): k : = 0.8 Area of Orifice: ] J = ^ ∗ L J . 4 ] J = ^ ∗ (10 %:) . 4 = 0.55 +/ . Height of Fluid Above Orifice: ℎ = 20 +/ Gravity: Y = 32.2 +//' . Discharge Flow from Orifice: \ = k : ] J î2Yℎ \ = 0.85 × 78.5 × √2 × 32.2 × 20 \ = 16.6 *+' Volume of Water Above Orifice: F = ^ ∗ L )#1K . 4 ∗ ℎ F = ^ ∗ (75 +/) . 4 ∗ 20 +/ = 88,357.3 +/ % Discharge Time: / = F \ = 88,357.3 +/ % 16.6 *+' = 5322 ' ≈ 89 Ñ%:

Practice Exam C – 1 st Edition www.pegenius.com 46 336. Answer: A) $10,800 Relevant PE Civil Handbook Section: 2.1.1 Volume of a Single Foundation: F = (9 +/) . × 10 +/ = 810 +/ % Total Volume of Undisturbed (banked) Soil: F - = 12 × 810 +/ % = 9720 +/ % Total Volume of Loose Soil: F - = 12 × 810 +/ % = 9720 +/ % F L = Z1 + 3 & 100 [ F - F L = Z1 + 10 100 [ × 9720 +/ % F L = 10962 +/ % = 396 ;L % Haul’s Required: # eNOI' = 396 ;L % × 1 ℎNOI 15 ;L % = 26.4 ℎNOI' → 27 8GlO%8GL Cost of Hauls: kT'/ = 27 ℎNOI' × $ 400 ℎNOI = $ 10,800

Practice Exam C – 1 st Edition www.pegenius.com 47 337. Answer: D) Point D Relevant PE Civil Handbook Section: 4.1 Since maximum moment for fixed-end beams occurs at the supports, we can eliminate Points B and C as possible options. Since load P occurs closer to the right restraint, it will cause a larger flexural demand there rather than the left restraint. Therefore, Point D is the answer. 338. Answer: C) Point C 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).

Practice Exam C – 1 st Edition www.pegenius.com 48 339. Answer: A) 106 kip Relevant PE Civil Handbook Section: 1.5 Vertical Summation of forces at Point 3 = 0: f 9 '%:p − 75 $%& = 0 f 9 '%:p = 75 $%& where: p = 45 LGY8GG Solve for f 9 : f 9 = 75 $%& '%: 45 = 106 $%& 340. Answer: B) Option 2 Relevant PE Civil Handbook Section: 1.6.7 For reference, beam shear stress is calculated as follows: ó = F\ zJ F = 'ℎGN8 +T8*G \ = +%8'/ ÑTÑG:/ T+ N8GN z = ÑTÑG:/ T+ %:G8/%N J = 6%L/ℎ T+ 'G*/%T: For a wide flange beam, the maximum shear stress can be approximated as: ó /#; = ] &59 F ] &59 = *8T'' − 'G*/%T:NI N8GN T+ JGNÑ 6GJ Given that the box section has approximately twice the web area of the wide flange section, it can be deduced that Option 2 will have a lower shear stress.

MMM$F;=;D?KI$9EC F; =;D?KI ;N7C FH;F