Civil Engineering & Architecture Final Examination. Parts A, B & C ANSWER KEY

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1 Civil Engineering & Architecture Final Examination Parts A, B & C ANSWER KEY Spring 2009 PRACTICE EXAM For Teacher Use ONLY Copyright Page- 1 -

2 Part A Multiple Choice Questions Question Answer CEA Assessment Concepts 1 C Unit 1 Overview of Civil Engineering and Architecture 2 B Unit 1 Overview of Civil Engineering and Architecture 3 A Unit 1 Overview of Civil Engineering and Architecture 4 C Unit 1 Overview of Civil Engineering and Architecture 5 C Unit 1 Overview of Civil Engineering and Architecture 6 B Unit 2 Introduction to Projects 7 D Unit 2 Introduction to Projects 8 A Unit 2 Introduction to Projects 9 D Unit 2 Introduction to Projects 10 D Unit 2 Introduction to Projects 11 C Unit 3 Project Planning 12 C Unit 3 Project Planning 13 A Unit 3 Project Planning 14 B Unit 3 Project Planning 15 B Unit 4 Site Planning 16 A Unit 4 Site Planning 17 C Unit 4 Site Planning 18 A Unit 4 Site Planning 19 C Unit 4 Site Planning 20 D Unit 4 Site Planning 21 B Unit 4 Site Planning 22 C Unit 4 Site Planning 23 B Unit 5 Architecture 24 A Unit 5 Architecture 25 B Unit 5 Architecture 26 B Unit 5 Architecture 27 A Unit 5 Architecture 28 B Unit 5 Architecture 29 D Unit 5 Architecture 30 C Unit 5 Architecture 31 D Unit 6 Structures - 32 C Unit 6 Structures 33 D Unit 6 Structures 34 D Unit 6 Structures 35 C Unit 6 Structures 36 A Unit 6 Structures 37 D Unit 6 Structures 38 C Unit 7 Presentations and Reviews 39 B Unit 7 Presentations and Reviews 40 D Unit 7 Presentations and Reviews Answer Breakdown: A-8; B-10; C-11; D-11 Copyright Page- 2 -

3 Part A Scoring Conversion Chart Raw Score Conversion Raw Score Conversion Raw Score Conversion Raw Score Conversion Part B High School Performance Exam 1) Tasks (6 Points: 1 point each) 1.1 Conducts a percolation test to determine the size of the absorption field: C 1.2 Develops the legal documents for an easement to allow the construction of a city bike path: N 1.3 Knows building codes for commercial and residential buildings: B 1.4 Creates an artistic rendering of a retirement home for a presentation: A 1.5 Creates complete working drawing package including site work for a 10,000 square foot commercial building: B 1.6 Selects door style for an entrance to a Victorian home: A 2) Community Center building Site Considerations (4 Points: 1 point apiece any 4) Public Ingress and Egress traffic entrances and exits Room for growth of facilities- expansion capabilities Zoning- of property and adjacent properties Topography- flat or hilly Soil Characteristics bearing capacity / drainage Location of site- near transportation routes and target population Parking lot size number of spaces required Copyright Page- 3 -

4 3) Matching (6 Points: 1 point each) 3.1 Regulations E 3.2 Covenants C 3.3 Deed restrictions D 3.4 Zoning A 3.5 Ownership F 3.6 Development B 4) Review the topographic map below and answer the following questions. (4 Points: 1 point each). 4.1 What is the highest elevation shown on the topographic map? 2277 feet 4.2 Circle an area of steep slopes. Several possible answers.( see below) 4.3 What is the contour interval in feet? 20 feet 4.4 Contour lines along streams and swales form V s that point: (upstream or downstream) Circle either upstream or downstream to complete the sentence. High Point Elev. 2277ft Copyright Page- 4 -

5 5) For each symbol indicate if it is used on a Electrical plan, Plumbing plan, or Heating, Ventilation, and Air-Conditioning (HVAC) plan by placing a E, P, or HVAC in the blank next to each symbol. (5 Points: 1 point each) P HVAC E E P 6) (8 Points Total: 1 point for each correct HI and 1 point for correct elevation, 2 points for correct closure error). Point BS HI FS ELEV BM 5.26 ft ft ft TP ft ft 1.76 ft ft TP ft ft ft ft BM 8.04 ft ft Closure Error: ft ft = 0.01 ft Copyright Page- 5 -

6 7) Heat Loss (9 Points: 1 point for heat loss thru wall not including door and windows; 1 point for heat loss thru windows; 1 point for heat loss thru door; 1 point for total heat loss thru wall including door, and windows; 1 point for area calculations; 1 point for U value of wall; 1 point for U value of door; 1 point for U value of windows; and 1 point for correct temperature used in formula). (Answer Precision =0.000) Heat Loss Formula: Wall without door and windows heat loss Q = A U ΔT Q = 241 ft² X.079 BTU/H X 47.8 F Q = BTU/H ft² F Window heat loss Q = A U ΔT Q = 20 ft² X BTU/H X 47.8 F Q = BTU/H ft² F Door heat loss Q = A U ΔT Q = 39 ft² X.599 BTU/H X 47.8 F Q = BTU/H ft² F Total heat loss for wall including doors and windows: Q = BTU/H ft² F Area Calculations: Area of wall: 10 X 30 = 300 ft² Area of window: 2(60 X 24 ) = 2880 in² 2880 in² / 144 in²/ft² = 20 ft² Area of door: 72 X 78 = 5616 in² 5616 in² / 144 in²/ft² = 39 ft² Total area of the wall minus door and windows: 300 ft² - (20 ft² + 39 ft²) = 241 ft² U = 1/R-value U value of wall without door and windows R = sum of the materials R-values in the given wall 3 5/8 BRICK: X.02 = AIR SPACE =.68 3 RIGID INSULATION: 3 X 3.6 = CONCRETE BLOCK (EMPTY CORE) = 1.11 U = 1 / ( ) U = 1 / U =.079 Copyright Page- 6 -

7 U value of windows Two 60 X 24 single glazed sliding windows =.88 U = 1/.88 U = U value of door 72 X 78 double flush door = 1.67 U = 1/1.67 U =.599 ΔT = The difference between the desired inside temperature and the outside temperature of the area. ΔT = 65 F 17.2 F ΔT = 47.8 F Copyright Page- 7 -

8 8) Calculate the Shear (End Beam Reactions) and the Maximum Moment. You may choose to complete your work using Shear and Moment Diagrams (below) or using traditional calculations. (8 Points: 1 point for R A and 1 point for R B ; 2 points for shear diagram or shear calculations, 1 point for work shown, and 2 points for Moment diagram or Moment calculations, 1 point for work shown.) (Answer Precision =0.00) Span = 24 feet W 1 = 100 lb/ft (girder weight) P 1 = 4,000 lbs. at 8 feet from left support P 2 = 4,000 lbs. at 16 feet from left support R A =5, 200 lbs R B = 5, 200 lbs M max = 39, 200 ft- lbs Copyright Page- 8 -

9 PART C College Credit Performance Exam 1) Structural Analysis (9 Points Total: see breakdown continued on following page) a) DL+LL = 55 psf + 50 psf = 105 psf Length of Beam = 25 ft, Tributary Width = 6-6 W= (105 psf )( 6.5 ft) = lb/ft b) Reaction 8,531.3 lb (2 points) Maximum Moment 53,320.3 ft-lb (2 points) Reactions R 1 =R 2 Beam is symmetrically loaded R 1 =R 2 = wl/2 = (682.5 lb/ft) (25) / 2 = 8,531.3 lb (1 point for formula, 1 point for answer) Moment M= wl 2 / 8 = (682.5lb/ft) (25) 2 / 8 = 53,320.3 ft-lb (1 point for formula, 1 point for answer) Shear and Moment Diagrams not required for Answers but may help obtain answer. Shear diagram, not required. Moment diagram, not required Copyright Page- 9 -

10 c) Deflection Deflection limited to L/360 ( or other ratio ) with L equal to length of beam in inches compare to actual deflection (1 point: for any item listed below) Ceiling cracks in plaster Roof ponding (flat roofs) Visual or psychological reasons Copyright Page- 10 -

11 2) Soil Testing (9 Points: 1 point for each correct entry in the Presentation Table) Item No 4 sieve No 40 sieve Bottom Pan Mason Jar Weight, grams Weight, grams Soil Sample weight, grams Item No 4 sieve and retained soil GRAVEL 20.0 No 40 sieve and retained soil Med. and coarse SAND Bottom pan and soil Fine SAND, SILT and CLAY Weight, Item grams Mason jar and soil Mason jar and fine SAND Soil in mason jar 75.9 Fine SAND in mason jar SILT and CLAY in mason jar Fine SAND in pan SILT and CLAY in pan Fraction in mason jar Soil Sample weight, grams PRESENTATION OF RESULTS Item Soil Sample weight, grams Percent in soil sample Copyright Page- 11 -

12 GRAVEL Med. and coarse SAND Fine SAND SILT and CLAY Total SAND % 77.6 (1 point per entry in the table above) 3) Footing Size (6 Points: see breakdown) The Soil is capable of resisting a total bearing pressure of force of 2500 lb/ft 2 Using the following formula: Pressure = Load /Area q= P/A q= 2,500 lb/ft 2 allowable bearing capacity of the soil Deduct the weight of the footing; the footing thickness is 12 inches (1.00 ft). Weight of Footing =1.00 ft thick x 150 lb/ft 2 Weight of Footing = 150 lb/ft 2 Soil Capacity Available = 2,500 lb/ft 2-150lb/ft 2 Soil Capacity Available = 2,350 lb/ft 2 = qnet Total Load of Footing = 50,000lb + 65,000lb = 105,000lb Pressure = Load /Area q= P/A Rearranging the formula so that we can get the required Area of the footing Area= P/ q net Area Required = 105,000 lb / 2,350 lb/ft2 -= ft2 (1 Footing Size = 6.68 ft x 6.68 ft go to next 6 increment USE 7-0 x 7-0 Square footing point) 4) Architectural Options (4 Points: 2 responses for 2 points for each) Some possible answers: Building style/detailing should be compatible with adjacent properties Proper building scale relative to adjacent structures and for property itself Possible town/village architectural requirements (architectural review boards) Using materials that are native to the area (both in color and texture) Copyright Page- 12 -

13 5) Survey (8 Points Total: 1 point for each correct HI and 1 point for correct elevation, 2 points for correct closure error). Point BS HI FS ELEV BM ft ft ft TP ft ft ft ft TP ft ft 1.33 ft ft BM 4.04 ft ft Closure Error: ft ft = 0.02 ft Copyright Page- 13 -

14 6) Water Pressure (14 Points: see breakdown) a) Static water head Static water head = WSEL at water tower Elev of water supply connection at street Static water head = 641 feet feet Static water head = 142 feet b) Static water pressure Static water pressure = static water head * (psi/2.31 ft) Static water pressure = 142 ft * (psi/2.31 ft) (1 point for substitution) Static water pressure = 61.5 psi c) Yes 61.5 psi > 40 psi (1 point for Yes, 1 point for comparing to 40 psi) d) All head losses are represented as equivalent lengths and include straight pipe and fittings. The Hazen-Williams formula for calculating head loss due to friction in straight pipes is: h f = * L * Q 1.85 (1 point for formula) C 1.85 * d Where: h f is head loss due to friction, feet L is the length of pipe = 500 feet Q is the flowrate of water through the pipe = 50 gpm C is the Hazen-Williams coefficient = 130 d is the diameter of the pipe = 2 inches h L = * 340 feet * (50 gpm) 1.85 (130) 1.85 * (2 inch) h L = 30.6 feet (1 point for substitution) e) First Floor Static water head = WSEL at water tower elevation of FIRST floor Static water head = 641 feet feet Static water head at FIRST floor= 121 feet TDH (Total Dynamic Head) = Static Water Pressure head loss TDH (Total Dynamic Head) = 121 feet 30.6 feet TDH (Total Dynamic Head) = 90.4 feet * psi/2.31 ft TDH (Total Dynamic Head) = 39.1 psi Copyright Page- 14 -

15 f) NO 39.1 psi < 40 psi (1 point for No 1 point for comparing to 40 psi) Copyright Page- 15 -