Only Civil Engineering & Architecture Practice Engineering & Architecture. Final Examination . Parts...

Civil Engineering & Architecture Final Examination

Parts A, B & C ANSWER KEY

Fall/Winter 2005-06

For Teacher Use ONLY

Project Lead The Way® Copyright 2005 - Page- 1 -

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Part A – Multiple Choice Questions

Question Answer CEA Assessment Concepts 1 B Unit 1 – Overview of Civil Engineering and Architecture 2 D Unit 1– Overview of Civil Engineering and Architecture 3 B Unit 1 – Overview of Civil Engineering and Architecture 4 C Unit 1 – Overview of Civil Engineering and Architecture 5 A Unit 1– Overview of Civil Engineering and Architecture 6 B Unit 2 – Introduction to Projects 7 A Unit 2 – Introduction to Projects 8 C Unit 2 – Introduction to Projects 9 A Unit 2 – Introduction to Projects

10 C Unit 2 – Introduction to Projects 11 B Unit 3 – Project Planning 12 C Unit 3 – Project Planning 13 D Unit 3 – Project Planning 14 D Unit 3 – Project Planning 15 B Unit 4 – Site Planning 16 C Unit 4 – Site Planning 17 C Unit 4 – Site Planning 18 A Unit 4 – Site Planning 19 B Unit 4 – Site Planning 20 D Unit 4 – Site Planning 21 D Unit 4 – Site Planning 22 A Unit 4 – Site Planning 23 D Unit 5 – Architecture 24 D Unit 5 – Architecture 25 B Unit 5 – Architecture 26 A Unit 5 – Architecture 27 A Unit 5 – Architecture 28 B Unit 5 – Architecture 29 D Unit 5 – Architecture 30 B Unit 5 – Architecture 31 A Unit 6– Structures 32 C Unit 6 – Structures 33 A Unit 6 – Structures 34 A Unit 6 – Structures 35 B Unit 6 – Structures 36 A Unit 6 – Structures 37 C Unit 6 – Structures 38 C Unit 7 – Presentations and Reviews 39 D Unit 7 – Presentations and Reviews 40 C Unit 7 – Presentations and Reviews

Answer Breakdown: A-11; B-10; C-10; D-9


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Part A Scoring Conversion Chart

Raw Score Conversion Raw Score Conversion Raw Score Conversion Raw Score Conversion 40 50 30 38 20 25 10 13 39 49 29 36 19 24 9 11 38 48 28 35 18 23 8 10 37 46 27 34 17 21 7 9 36 45 26 33 16 20 6 8 35 44 25 31 15 19 5 6 34 43 24 30 14 18 4 5 33 41 23 29 13 16 3 4 32 40 22 28 12 15 2 3 31 39 21 26 11 14 1 1

Part B – High School Performance Exam 1) Civil Engineer/ Architect

(5 points - 1 point for each correct answer)

• Determining whether or not to change the zoning designation of a parcel of land from residential to commercial: ____N___

• Selecting the siding to be placed on a Community Center: ____A____ • Working with contractors throughout the construction of a project: ___B___ • Laying out and locating septic systems in a development: ___C_____ • Discussing the use of a commercial building with a client: _____A___

2) Determine if the following information would be gathered or tasks performed during

Site Visit (SV) activities or during Site Discovery (SD) activities by placing an SV or SD next to the task. (6 points: 1 point for each correct answer). Orientations: __SV_____

Covenants: _____SD___

Perc test: ___SV____

Site history: __SD_____

Liens: ___SD____

Metes and Bounds: ____SD

3) Determine the elevation of the boulder using the following information collected during a differential leveling activity. (4 points: 1point for each correct HI and 2 points for correct elevation).

Point BS HI FS ELEV

BM 5.26 ft 239.82 ft 234.56 ft

boulder 4.11 ft 242.17 1.76 ft 238.06 ft

curb 1.70 ft


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4) Create a bubble diagram from the floor plan (8 points: ½ point for each bubble properly placed, ½ point for each proper link) Day Care 1st Floor Plan

Bath

Hallway

Foyer/Office Kitchen

Recreation Infant Room

Class 2

Class 1

5) How would the following factors affect your bubble diagram above?

(2points: 1 point for each correct answer).

• Solar orientation: _Structures and sides of structures can be oriented to the

direction of the sun in order to maximize sunlight or, conversely, minimize

sunlight. Minimize / Maximize Heating and Cooling effects.__

• Wind orientation:_ Building location will consider wind orientation depending on

the function of the building and on energy efficiency


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6) Soil Sample (9 points: 1 point for each correct entry in the Presentation Table – next page) (These are shown only for information needed to obtain final results)

Item Weight, grams

No 4 sieve 230.5

No 40 sieve 252.6

Bottom Pan 300.1

Mason Jar 152.0

Item Weight, grams

Soil Sample

wgt, grams

No 4 sieve and retained soil 241.6 GRAVEL 11.1 No 40 sieve and retained soil 365.2 Med. and coarse SAND 112.6 Bottom pan and soil 412.0 Fine SAND, SILT and CLAY 111.9

Item Weight, grams

Fraction in mason jar

Soil Sample wgt, grams

Mason jar and soil 263.9

Mason jar and fine SAND 232.4

Soil in mason jar 111.9

Fine SAND in mason jar 80.4 0.72

SILT and CLAY in mason jar 31.5 0.28

Fine SAND in pan

80.4

SILT and CLAY in pan

31.5


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(1 point per entry in the Presentation of Results table - see below.)

PRESENTATION OF RESULTS

Item Soil Sample

weight, grams Percent in soil sample GRAVEL 11.1 4.7

Med. and coarse SAND 112.6 47.8

Fine SAND 80.4 34.1

Total SAND, %

81.9 SILT and CLAY 31.5 13.4


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7) Heat loss 8-points total – see distribution below Heat Loss Formula: Q = A U ∆T (1point for correct formula) Q = 82 x .037 x 47 (1point for correct values inserted in formula) Q = 142.589 BTU per hour ft² °F (1point correct answer)

(4 points for work for the calculated values- see below) Calculations generated for use in the formula above A = Area of the wall minus windows and doors Area of the wall: 10 ‘ x 9’ = 90 ft2 Area of the window: 24” x 48” = 1152 in2 (1 point) Area of the window: 1152 in2 / (144 in2 /ft2) = 8 ft2

A = Total area of the wall minus the window: 90 ft2 - 8 ft2 = 82 ft2 Area = 82 ft2 (1 point)

U = 1/R value R = sum of the materials R values in the given wall Brick 90lb/cu.ft. = .20 x 4 = .80 Air film space = .68 R Value of Wall= 0.8+0.68 +25.2 = 26.68 (1 point) U = 1/R U = 1/26.68 or .037 (1 point)

∆T = the differential temperature = the needed temperature in the room minus the average temperature for the area.

∆ T = 72 degrees – 25 degrees

∆T = 47 degrees (1 point)


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8) Shear & Moment

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 points each for RA and RB reaction; 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.)

Span = 25 feet W1= 200 lb/ft (girder weight)

P1= 1000 lbs. at 12.5 feet from left support

SHEAR (lb)

MOMENT (lb-ft) RA = _____3,000 lb__________ RB = __3,000lb__________ Calculations on next page


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ftlbM

ftftlb

M

ftftlb

ftlb

M

wlM

lbRlbftftR

ftRlbftlbft

ftRftftlbftlb

ftRftftftlbdP

RRRR

A

A

A

A

Aportleftfromcedislb

AB

BA

−=

=

+=

=

==

−+=

−+=

−+=

−==

218758

)25)(280(

8

)25))(5.12

1000()200((

8

300075000)25(

)25(62500125000

)25()5.12)(25(200)5.12(10000

)25()5.12)(25(200)))(1((0

0

2

2

2

sup___tan

Moment

Shear


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PART C – College Credit Performance Exam 1) Structural Analysis (10 points)

a) Calculate the beam loading, using the beam that carries the greatest load. (4 points) DL+LL = 55 psf + 80 psf = 135psf Length of Beam = 25 ft Tributary Width = 6’-0” (1 point) Beam loading = Tributary Width x (DL + LL) (1 point) = 6 feet x (55 lb/ft2 + 80 lb/ft2) (1 point) = 810 lb/ft (1 point)

b) Calculate End Reactions and Moment for the beam selected in Part a)

(4 points)

Reaction ________________ 10,125lb________ (2points) Maximum Moment_________38,880 ft-lb _______ (2 points) (Therefore, 1 point formula, 1 point answer) Reactions

R1=R2 Beam is symmetrically loaded R1=R2 = wL/2 = (810lb/ft) (25) / 2 = 10,125lb

Moment M= wL2 / 8 = (810 lb/ft) (25) 2 / 8 = 63,281.25 ft-lb Shear and Moment Diagrams not required for Answers but may help obtain answer. Work must be shown to earn credit.

c) Identify which Girder carries the greatest load. Explain how you determined this. (2 points) Girder on Column Line 5 between Columns F and Column G. This girder carries the greatest load because the Tributary Width = ½(25ft) + ½ (25ft) = 25 feet of floor While the Girders on Column Line 6 carry less of a load, with a Tributary Width = ½(25 ft) +1/2(20ft)=22.5ft. The span of girders between Column E and F are the same as Columns F and G. Therefore, select Girder that carries larger Tributary Area.


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2) Identify 4 structural components of the building.

A

B

E C D Structural Components- Label with arrows (Total 4-POINTS) (1 point for each member identified AND labeled)

A - Roof Joists (Open Web Steel Joists)

B - Columns

C - Lateral Bracing (for the structural framework)

D - Beams/Girders

E - Concrete Deck NOTE: Only 4 structural components need to be identified


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3) Determine footing Size (8 points)

The soil is capable of resisting a total bearing pressure of force of 3000 lb/ft2 Using the following formula: Pressure = Load /Area q= P/A (1 point) q= 3000 lb/ft2 q= allowable bearing capacity of the soil Deduct the weight of the footing; the footing thickness is 16 inches (1.33 ft). Weight of Footing= (1.33 ft thick) x 150 lb/ft2 = footing weight in lb/ft2 Weight of Footing = 200 lb/ft2 (1 point) qnet =Soil Capacity Available = 3000 lb/ft2 - 200lb/ft2

Soil Capacity Available = 2800 lb/ft2 = qnet (1 point) Total Load of Footing = DL + LL + Column Load Total Column Load = 64,000 lb + 75,000 lb + 10ft(53 lb/ft) = 139,530 lb (1 point) Pressure = Load /Area q= P/A Rearranging the formula so that we can get the required Area of the footing Area = P/ q net (1 point)

Area Required = 139,530 lb / 2800 lb/ft2 = 49.83 ft2 (1 point)

Footing Size =7.06 ft x7.06 ft go to next 6” increment (1 point) Footing Size 7’-6” X 7’-6” (1 point)

USE 7’-6” x 7’-6” Square footing

4) Architectural Options (4 points)

- Neighborhood characteristics

- Location (geographical)/ Climate

- Covenants in the community (restrictions for materials)

- Low maintenance materials(as it a vacation home)


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5) Calculate the closure error for the differential survey notes below.

(8 points: 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 239.82 ft 234.56 ft

TP1 4.11 ft 242.17 ft 1.76 ft 238.06 ft

TP2 12.24 ft 232.59 ft 11.82 ft 230.35 ft

BM 8.04 ft 234.55 ft

Closure Error: 234.56 ft – 234.55 ft = 0.01 ft


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6) Head Loss - (8 Points Total)

hf = 10.44 * L * Q1.85 C1.85 * d4.8655

Total head loss = head loss from 2” pipe + head loss from ¾” pipe and fittings Head loss from 2” pipe: L = 100 ft hf = 10.44 * 150ft * (15 gpm)1.85 (2 points) 1401.85 * (2 inch)4.8655 hf = 0.86 feet (1 point) Head loss from 3/4” pipe: L = 75 ft + (4 feet/fitting * 5 fittings) = 95 feet (1 point) hf = 10.44 * 95ft * (15 gpm)1.85 (2 points) 1401.85 * (3/4 inch)4.8655 hf = 64.53 feet (1 point) Total Head Loss: Total head loss = 0.86 ft + 64.53 ft Total head loss = 65.39 ft (1 point)

7) Predevelopment peak discharge - (8 Points Total)

Answers may vary depending on coefficients used from the Rational method Runoff

Q=CiA C = 0.7 asphalt surface 0.7 to 0.95 (1 point) i = 3.5 in/hr, (1 point) A = 10 acres, (1 point) Q = 0.7 * 3.5 in/hr * 10 acres = 24.5 cfs, (1 point) Post development peak discharge (redevelopment) Q=CiA C = 0.4 asphalt surface 0.3 to 0.5 (1 point) i = 3.5 in/hr, (1 point) A = 10 acres, (1 point) Q = 0.4 * 3.5 in/hr * 10 acres = 14.0 cfs, (1 point)

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