#### Read Microsoft Word - Civil PE Sample Exam 2008 errata 8-26-09.doc text version

`ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009Breadth Module p. 31, Breadth Question 106: Change line 4 as follows: Plastic limit 25p. 34, Breadth Question 113: Replace the existing figure with the following:BRIDGE STRUCTURE STATION 73+00 ELEV. = 365.94 PVCPVTg1 =L = 1,500 ft-3.0 %g2=PVI STATION 76+00 ELEV. = 334.56+2.0%Page 1 of 17ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009Breadth Module (Continued) p. 34, Breadth Question 114: Beginning with the April 2008 exam, traffic analysis is not covered in the Transportation portion of the Breadth exam. The Transportation portion of the Breadth exam will only contain questions about geometric design. The following is another example of a geometric design question. The back tangent of a horizontal curve has a bearing of N 654800 E. The ahead (forward) tangent has a bearing of S 7124 E. The bearing from the PI to the center is most nearly: (A) (B) (C) (D) S 248 E S 4000 E S 6048 E S 6140 EBreadth Solution: Solving for ,90- 6548¢ 00¢¢ = 2412¢ 90- 7124¢00¢¢ = 1836¢ D = 2412¢ + 1836¢ = 4248¢ D 2 = 4248¢ 2 = 2124¢Now solving for the angle subtended by the forward tangent to the line between the PI and center of curve,90-D 2 = 90- 2124¢ = 6836¢ Bearing = 7124¢ - 6836¢ = S 248¢ ETHE CORRECT ANSWER IS: (A)Page 2 of 17ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009Breadth Module (Continued) p. 139, Breadth Solution 106: Add the following equation to the solution: PI = 55 ­ 25 = 30 Construction Module p. 47, Construction Question 508: Replace the existing figure with the following:LEGEND ACTIVITY ES EF DURATION (days) FF TF LS LF FF, LAG=3 D 2A 1 3C 2E 1G 2B 1 4F 4H 3Page 3 of 17ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009Construction Module (Continued) p. 48, Construction Question 509: Change the first sentence as follows: In the activity-on-arrow network below, based on end-of-day calculations for starts and finishes, the early start of Activity N is most nearly: p. 50, Construction Question 511: Change the first sentence as follows: A formal CPM analysis for a project shows the planned costs to date are \$85,000, and the accounting department reports charges to the job of \$95,000. p. 54, Construction Question 518: Change line 2 as follows: The OSHA incidence rate for recordable cases is most nearly: p. 149, Construction Solution 504: Change line 5 as follows:Costs  9, 000  A P 5 yr  1, 000  0.10 10%Change line 8 as follows:Costs  10, 000  A P 5 yr  1, 000  A F 5 yr10% 10%p. 155, Construction Solution 518: Change the second sentence as follows: The term incidence rate for recordable cases means the number of injuries and illnesses requiring treatment beyond first aid, those that result in lost workdays, and those that result in restricted or light duty, per 100 full-time workers. Page 4 of 17ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009Geotechnical Module p. 64, Geotechnical Question 507: Replace the existing figure with the following:ORIGINAL GROUND SURFACE ELEV. 0'-0&quot;30 ft MAT FOUNDATION  = 105 pcfELEV (-)5'-0&quot;  = 115 pcf ELEV (-)10'-0&quot; NORMALLY CONSOLIDATED CLAY SATURATED UNIT WEIGHT = 95 PCF COMPRESSION INDEX = 0.29 VOID RATIO = 1.0GROUNDWATER TABLE SANDELEV (-)20'-0&quot; SANDNOT TO SCALEp. 69, Geotechnical Question 513: Change the first three sentences as follows: The liquefaction potential of a site is to be evaluated. For Layer 3, the design earthquake-induced average shear stress is 450 psf, and the maximum allowable cyclic stress ratio is 0.29. The factor of safety against liquefaction in Layer 3 is most nearly: Page 5 of 17ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009Geotechnical Module (Continued) p. 72, Geotechnical Question 515: Replace the existing figure of slope dimensions with the following: POTENTIAL FAILURE SURFACE BEDROCK 20 ft CLAY  = 120 pcf c = 750 psf 20 ft  = 0p. 74, Geotechnical Question 516: Replace the existing question and options with the following: The figure below shows the foundation and geotechnical data for a strip footing. To achieve a safety factor of 3, the allowable bearing capacity (psf) using the Terzaghi equation is most nearly: (A) (B) (C) (D) 1,995 2,187 2,730 5,985p. 75, Geotechnical Question 517: Options (A) through (D) are as follows: (A) (B) (C) (D) 1.6 1.7 1.9 2.3p. 162, Geotechnical Solution 507: Change line 6 as follows:s¢ = 5(105) + 5(115 - 62.4) + 5(95 - 62.4) = 951 psf voPage 6 of 17ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009Geotechnical Module (Continued) p. 165, Geotechnical Solution 515: Replace the existing solution with the following:D = 40D/H = 2  = 30stability number = 0.172 = c r @ 413 FS =cr cr = H 120 ´ 20c 750 = = 1.82 c r 413p. 165, Geotechnical Solution 516: Replace the existing solution with the following: Square footing B = L = 5 ft Df = 2 ft  = 115 pcf c = 200 psf  = 20 Given: Bearing-capacity factors Nc = 14.8 Nq = 6.4 N = 5.4 q ult  0.5 BN   cN c  DN q  0.5 115 pcf  5 ft  5.4    200 psf 14.8   115 pcf  2 ft  6.4   1, 552.5  2, 960  1, 472  5, 984.5 psf q allowable  5, 984.5  1, 995 3Given:Page 7 of 17ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009Geotechnical Module (Continued) p. 166, Geotechnical Solution 517:Replace resisting moment equation with the following: Resisting moment = (2)(9)(155)(4.5) + (5)(18)(155)(4.5) + (2)(18)(105)(8) = 105,570 ft-lb Replace calculation of FS with the following:FS =105,570 = 2.28 46, 200THE CORRECT ANSWER IS: (D) p. 166, Geotechnical Solution 518:Replace line 6 with the following:Overburden pressure at el.  20 ft, Po  5(120) 30 (115)  2,325 psf 2Structural Module p. 82, Structural Question 503:Options (A) through (D) are as follows: (A) (B) (C) (D) 250 75 50 36Page 8 of 17ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009Structural Module (Continued) p. 86, Structural Question 506:Add an assumption: Neglect block shear and the capacity of the connection. Options (A) through (D) are as follows: (A) (B) (C) (D) 2 1/2 × 2 1/2 × 3/8 3 1/2 × 3 1/2 × 5/16 3 1/2 × 3 1/2 × 3/8 4 × 4 × 3/8p. 100, Structural Question 518:Replace lines 5 and 6 with the following: Assuming adequate basic allowable stress, the design stress range (ksi) for the fatigue limit state for the beam during 25 years of service is most nearly: Options (A) through (D) are as follows: (A) (B) (C) (D) 16 24 29 38p. 172, Structural Solution 503:Change lines 6, 7, 8, and 9 as follows:Total weight Net uplift = 70 kips + 36 kips = 106 kips  0.6  63.6  ­50 kips  15.9 kips  36.1 kips (ASCE 7 - 05, part 2.4.1) Weight per leg = 106 kips/4 = 26.5 kips  0.6  15.9Page 9 of 17ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009Structural Module (Continued) pp. 174­175, Structural Solution 506:Replace the existing solution with the following:ASD Provisions: P P£ n Check tensile yield:Pn = Fy A g with t = 1.67(Eq. B3-2) (D2-1)\ 85 kips £Fy A g t=(36 ksi)A g 1.67 select smallest area with A g ³ 3.94 in.A g min ³(85 kips)(1.67) = 3.94 in 2 36 ksi= 3.94 in 2 Try3 1/2 × 3 1/2 × 5/16minA g = 4.21 in 2 ³ A g OK(D2-2)Check tensile rupture:Pn = Fu Ae with t = 2.0085 kips &lt; Fu A e 2.00Ae = A n U An = [(4.21 in ) ­ 2(7/8 in. + 1/16 in. + 1/16 in.)(5/16 in.)] An = 3.585 in2 2(p. 16.1-28)Page 10 of 17ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009Structural Module (Continued) Structural Solution 506 (continued):All elements of angle not connected,  U  1  Shear lag Per code, permissible to use largest &quot;U&quot; Case 8  Table D3.1 U = 0.6 (3 fasteners/line) Case 2 (Table D3.1)x 0.979 in. U = 1- =1= 0.837 l 6 in.Largest U = 0.837  lightest Ae = (3.585 in )(0.837) = 3.00 in2 2(58 ksi)(3.00 in 2 ) 85 kips £ = 87.0 kips 2.0Use 3 1/2 × 3 1/2 × 5/16OKTHE CORRECT ANSWER IS: (B) LRFD Provisions:Pu £ f t PnCheck tensile yield: Pn = F y A g Pu =  t FyAg with  t = 0.9 130 kips  0.9 (36 ksi) Ag(B3-1)(D2-1)Ag min ³(130 kips) = 4.01in 2 (0.9)(36 ksi)A g = 4.21 in 2 &gt; A g min = 4.01 in 2Try3 1/2 × 3 1/2 × 5/16Page 11 of 17ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009Structural Module (Continued) Structural Solution 506 (continued):Check tensile rupture:Pu £ f t Fu A e130 kips  0.75 (58 ksi) Ae Ae = AnU= [(4.21 in ) ­ (7/8 in. + 1/16 in. + 1/16 in.)(5/16 in.)(2)] (0.837) U = 0.837 (from ASD) Ae = 3.00 in2 2Pu   t Fu A e 130 kips  0.75 (58 ksi)(3.00 in ) = 130.5 in Use 3 1/2 × 3 1/2 × 5/162 2 OKTHE CORRECT ANSWER IS: (B) p. 177, Structural Solution 510:The following information clarifies the existing solution. The critical section for shear occurs at the distance d from the face of the columns. In this case, the section must be checked on the far, or exterior, side of the columns. From the critical bending shear calculation, 8'-0&quot; is the edge of footing to center of column 1'-0&quot; is the center of column to outside face of column 3'-0&quot; is d, effective footing depth 10'-0&quot; is the width of the footing.p. 182, Structural Solution 518:Replace calculation of FSR with the following:FSRæ 250(108 ) ö0.333 ÷ ÷ ç =ç = (57,077)0.333 = 38.4 ksi &gt; FTH = 24 ÷ ç 48 ´365´25 ÷ ÷ ç è øPage 12 of 17ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009Transportation Module p. 192, Transportation Solution 511:Replace lines 2, 3, 7, and 8 with the following:  25  V12  (30) 2  / (30  0.55) S 2  V22  V12  / (30  0.71) (150  24)  V22  (36.23) 2  / (30  0.71)S1  V12  V f2 / 30 fWater Resources and Environmental Module p. 122, Water Resources and Environmental Question 501:Change sentence 2 as follows: The elevation of the hydraulic grade line is 495 ft at the beginning of the pipe and 365 ft at the end of the pipe.p. 127, Water Resources and Environmental Question 510:Replace the existing question with the following: Groundwater flows through a sandstone aquifer (effective porosity = 0.3) that is shown in the figure below. The distance from the recharge area to the discharge point is 12 miles, and the head difference is 200 ft. The hydraulic conductivity is 15 ft/day. The time (years) it will take for water to travel from the recharge area to the discharge area is most nearly: (A) (B) (C) (D) 0.5 20 1,100 3,700RECHARGE AREA PIEZOMETRIC SURFACE h = 200 ft12 milesPage 13 of 17ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009Water Resources and Environmental Module (Continued) p. 201, Water Resources and Environmental Solution 501:Replace the existing solution with the following: At end of pipeVE 2 Ph = Pressure head (ft) = Ph = DH - h L 2gwhere H is total drop in hydraulic grade line, hL is the head loss caused by friction in the pipe, and VE is the velocity of flow at the end of the pipe. Find hL using Hazen-Williams equation (implied by use of C in problem).V = 1.318 CR 0.63s0.54 h D s = L and R = for pipes flowing full L 4 solve for h L æ V hL = ç ç ç æ ç ç 1.318C ç D ç 4 ç ç ç è è V= ö0.63 ÷ ÷ ÷ ø ö1 0.54 ÷ L ÷ ÷ ÷ ÷ ÷ ÷ ø 4 æ 8 ö2 ÷ pç ÷ ç 12 ø ÷ ç èQ 500 CON 1 gpm ft 3 min = A CON 7.48 gal 60 secV = 3.1916 ft/sec é 3.1916 hL = ê ê æ ê (1.318)(140) ç 8 ç ê ç 12(4) è ë DH = 495 - 365 = 130 ft ö0.63 ÷ ÷ ÷ ø ù1 0.54 15, 000 = 66.21 ft ú ú ú ú ûVE 2 (3.1916) 2 = = 0.158 ft 2g 2(32.2) Ph = 130 - 66.21 - 0.158 = 63.6 ft Ppsi = 63.6 ft 62.4 lb ft 3 ft 2 144 in 2not significant= 27.6 psiTHE CORRECT ANSWER IS: (B)Page 14 of 17ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009p. 203, Water Resources and Environmental Solution 503:Change line 14 as follows:2 = 12.65 × éê 7/(3.14 × 0.52 )ùú /(2 × 32.2) ë ûp. 206, Water Resources and Environmental Solution 506:Replace lines 6, 7, 8, and 9 with the following:Q H Hb 3.33  LH3/2  4.62 1      0.494 ft  3.33 4 ft   100.69  100.00  0.69 ft2/3H b 0.69   0.53  0.6 Ha 1.31Elevation at B = 100.2 ft + H = 100.2 + 0.494 = 100.69 ftp. 207, Water Resources and Environmental Solution 508:The following information clarifies the existing solution. Rainfall 1 is from time = 0 hr to time = 1 hr at an intensity of 1.5 in./hr. Rainfall 2 is from time = 1 hr to time = 2 hr at an intensity of 0.7 in./hr. The discharge in the watershed during the second hour will consist of: The second hour of Rainfall 1 (1.2 cfs/in. × 1.5 in./hr × 1 hr) plus the first hour of Rainfall 2 (0.5 cfs/in. × 0.7 in./hr × 1 hr), or 2.15 cfs.Page 15 of 17ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009Water Resources and Environmental Module (Continued) p. 208, Water Resources and Environmental Solution 510:Replace the existing solution with the following: Applying Darcy's equation, the Darcy velocity can be found as: dh h V = -K » -K dL Læ -200 ft ö ÷ = -15 ft/day ç ÷ ç ç12 ´ 5, 280 ft ÷ ÷ è ø = 0.047 ft/day (this is the Darcy velocity) The seepage velocity or the linear velocity is therefore: Darcy velocity 0.047 ft/day = = 0.157 ft/day porosity 0.3 \ Travel time = 12 ´ 5, 280 ft = 403,566 days = 1,105 years 0.157 ft/dayTHE CORRECT ANSWER IS: (C)Page 16 of 17ERRATA for Civil PE Sample Questions and Solutions Copyright 2008 ISBN: 978-1-932613-31-5 Errata posted 08-26-2009Water Resources and Environmental Module (Continued) p. 213, Water Resources and Environmental Solution 517:Replace the existing solution with the following: Sediment weight entering reservoir = total flow rate  solids concentration  conversion factor 9 3 3 5 = 2.1  10 ft /day  172 mg/L  28.32 L/ft  lb/4.54  10 mg 7 = 2.253  10 lb/day Sediment volume allotted Weight of sediment in reservoir = reservoir volume  reserved % 10 3 9 3 = 1.41  10 ft  22% = 3.102  10 ft = sediment volume allotted  solids specific weight 9 3 3 = 3.102  10 ft  80 lb/ft 11 = 2.482  10 lb = sediment weight  sludge weight entering reservoir/day 11 7 = 2.482  10 lb  2.253  10 lb/day 4 = 1.102  10 days 4 = 1.102  10 days  365 days/year 2 = 0.302  10 years = 30.2 yearsReservoir service lifePage 17 of 17`

#### Information

##### Microsoft Word - Civil PE Sample Exam 2008 errata 8-26-09.doc

17 pages

Find more like this

#### Report File (DMCA)

Our content is added by our users. We aim to remove reported files within 1 working day. Please use this link to notify us:

Report this file as copyright or inappropriate

199322

### You might also be interested in

BETA
Microsoft Word - MFE Review Notes A Overview.doc