notes for mar 22nd 23rd and 24th classes part 2

2010

School of PE

Professional Engineer

by George Stankiewicz, P.E., LEED ® A. P.

C I V I L E N G I N E E R ahmed youssef ([email protected])

This copy is given to the following student as part of School of PE course. Not allowed to distribute to others.

School of PE

Civil Engineer Refresher Course - CONSTRUCTION Workshop Questions and Solutions 2

TABLE OF CONTENTS

Workshop Questions and Solutions ............................................................................................3

1. Surveying .........................................................................................................................3

2. Construction Management .............................................................................................12

3. Materials ........................................................................................................................24

This symbol represents topics within the Refresher Course that are part of

the subject matter which will further help your understanding.

The information is intented for self-study and may not be

reviewed during the refresher course.

ahmed youssef ([email protected])


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WORKSHOP QUESTIONS AND SOLUTIONS

1. SURVEYING

1. - Question 130,000-yd3 of banked soil from a borrow pit is

stockpiled before being trucked to the jobsite. The soil has 18% swell and

shrinkage of 8%. The final volume of the compacted soil is most nearly:

a. 119,600-yd3

b. 124,600-yd3

c. 125,400-yd3

d. 135,400-yd3

Solution: Shrinkage is measured with respect to the bank condition.

V compacted = (100% - % shrinkage) V bank 100% Vc = (100% -8%) (130,000-yds3) = 119,600-yd3 (answer) 100%

2. - Question A 30-ft wide trapezoidal shaped earthen stream diversion channel is cut along a 2-mile stretch of rolling level terrain. The depth of the channel at station 52+25 is 8’-6” deep and at station 53+75 is 12’-6” deep. The bottom of the channel is a constant 12-ft wide and parallel with the surface cut. The volume of excavated material between the referenced stations is most nearly:

a. 1225-yd3 b. 1225-ft3 c. 1415-yd3 d. 32225-ft3

Solution: Distance between stations: 53+75 – 52+25 = 150-ft End Area at stations: Sta 52+25 Area = A = ((a +b)h) ÷ 2 = ((30-ft + 12-ft)8.5-ft) ÷ 2 = 178.5-ft2 Sta 53+75 Area = A = ((a +b)h) ÷ 2 = ((30-ft + 12-ft)12.5-ft) ÷ 2 =262.5-ft2 Calculate Volume [(178.5-ft2 + 262.5-ft2) ÷ 2 x 150-ft] ÷ (1-yd3/27-ft3) = 1225-yd3 (answer)



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3. - Question Geotechnical project specifications require that the soil’s optimum moisture content of 24% be maintained during the roadbed’s construction. The field moisture test finds that five-pounds of soil has a water content of 11%. The amount of water that must be added during the day’s planned production of 1,250-tons to achieve the optimum moisture content is most nearly:

a. 4.5-lbs b. 4,7000-gal c. 35,000-gal d. 292,500-gal

Solution: The total mass of the moist soil is equal to the dry soil and water content. The mass of water is equal to 11% of the dry soil. This relationship is represented in the following equation: Mtotal = Msoil + Mwater Mtotal = Msoil + 0.11 Msoil Mtotal = 1.11 Msoil Solve for the mass of the dry soil and water: Msoil = Mtotal = 5-lb = 4.50-lb 1.11 1.11 Mwater = 0.11Msoil = (0.11) (4.50-lb) = .50-lb To raise the water content from 11% to 24%, the earthwork contractor must add 13% by mass of water. ΔMwater = (ΔWrequired) (Msoil) = (0.13) (4.50-lb) = 0.585-lbm of water per 5-lbs of soil OR = 0.117-lb of water / lb of soil Convert the results of the required additional water and apply it to the day’s planned production: 1,250-ton x 2,000-lb/ton = 2,500,000-lb of soil 2,500,000-lb of soil x 0.117-lbm of water/lb of soil = 292,500-lbm of water 292,500-lb of water ÷ 62.4-lb of water/ft3 = 4,688-ft3 4,688-ft3 x 7.48gal/ft3 = 35,062.5-gallons



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4. - Question The Table below provides the end area between stations 51+00 and 57+00. Use the Table to answer the following:

1. The net excavation from Station 51+00 to

57+00 is most nearly:

a. 1000-yd3 borrow

b. 1200-yd3 waste

c. 1400-yd3 borrow

d. 1600-yd3 waste

2. An earthwork contractor will use his fleet of 20-yd3 dump trucks to

move the waste or borrow soil. Using a swell of 10% the number of

dump truck loads needed are most nearly:

a. 50-truck loads

b. 62-truck loads

c. 82-truck loads

d. 92-truck loads

End Area Station Cut Fill

(ft2) (ft2) 51 + 00 0 250 52 + 00 0 300 53 + 00 0 435 54 + 00 0 550 54 + 30 150 0 55 + 00 650 0 56 + 00 850 0 57 + 00 380 0



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Solution

End Area (ft2) Station Distance Cut Fill cut vol fill vol (ft) (sf) (sf) (cy) (cy)

51+00 250 100 1019

52+00 300 100 1361

53+00 435 100 1824

54+00 0 550 30 83 306

54+30 150 0 70 1037

55+00 650 100 2778

56+00 850 100 2278

57+00 380 TOTAL 6176 4510

1. Cut volume – Fill volume = Net excavation

6176-yd3 – 4510-yd3 = 1666-yd3 of soil to be trucked off-site (Answer)

2. Number of truck loads = (waste x swell) / yd3 per truck load

Number of truck loads = (1666-yd3 x 1.1) / 20-yd3/truck load = 92-truck loads (answer)



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5. - Question On a 5-acre level terrain building site, an earthwork

contractor has instructed her crew to strip and grub the topsoil of a 60,000-

ft2 proposed building pad to a minus 2-ft sub-grade. The soil has a swell of

40% and an angle of repose at 30°. The diameter of the stockpile is most

nearly:

a. 120-ft

b. 130-ft

c. 140-ft

d. 150-ft

Solution:

Determine the cubic volume of the cut and the swell of the soil:

60,000-ft2 x 2-ft x 1.40 (40% swell) = 168,000-ft3 or 6,222-yd3

Evaluate the question using the equation for the volume of a cone and the maximum incline of

the sides of the cone are at the natural angle of repose equal to the angle of internal friction.

Check the maximum height based on the natural angle of repose.

r = h ÷ tan α = h ÷ tan 30° = 1.73h

Using the equation to find the Volume of the cone, solve for h, the Height:

V = π r2 h 3

168,000-ft3 = (π (1.73h)2 x h) ÷ 3 = π h3 h= (168,000-ft3 ÷ π)1/3 = 37.55-ft Solve for r: 168,000-ft3 = (π r2 h) ÷ 3 r = 65.33 x 2 = 130.76-ft diameter (answer = b)



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6. - Question An earthwork contract was awarded to excavate and

backfill the foundation of a proposed 50-ft by 50-ft office building. The

existing grade elevation is 437.5-ft while the sub-base for the below grade

basement is 432.5-ft. The concrete contractor requires a 3-ft perimeter

walkway in order to place the concrete formwork. Soil conditions are

classified as OSHA Type B soils which require 1H:1V side slopes. The

bank volume (yd3) to be stockpiled and used for backfill is most nearly:

a. 110 b. 278 c. 463 d. 690

Solution: The foundation excavation can be described as an inverted truncated

pyramid. Compute the earthwork volume using the buildings dimensions and add a 3-ft

perimeter walkway around the building for the workers erecting the concrete formwork.

Equation for the volume of a truncated pyramid:

Volume = V1 = h/3 (A1 + A2 + √(A1 x A2))

Compute depth of foundation = h = 437.5-ft – 432.5-ft = 5.0-ft

A1 = Area of the base of truncated pyramid = (50 + 3 + 3) (50 + 3 + 3) = 3,136-ft2

A2 = Area of the top of truncated pyramid = (50 + 3 + 3 + 5 + 5) (50 + 3 + 3 + 5 + 5) =

4,356-ft2

V1 = 5/3 (3,136 + 4,356 + (√(3,136 x 4,356)

V1 = 18,647-ft3 = 690-yd3

Compute the volume of the basement and subtract this from the total excavation to

determine the volume of backfill material.

Building Volume = 50’ x 50’ x 5’ = 12,500-ft3 = 463yd3

Backfill stockpile required = 690-yd3 - 463-yd3 = 278-yd3



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7. - Question The information given in Figure 1 is translated from the

surveyor’s field book data for a proposed local borrow pit. The volume of

excavation in yd3 is most nearly?

a. 2,050-yd3

b. 7,275 yd3

c. 9,275 yd3

d. 10,250-yd3

Solution:

V = ((33.4 x 1) + (34.2 x 2) + (32.9 x 1) + (30.6 x 2) + (31.8 x 1) + (32.8 x 1) + (35.7 x 3) + (33.0 x 1)) (2500 ÷ (4 x 27))

V = (2500 ÷ (4 x 27))

V = 9,273-yd3 (answer)

Figure 1 [not to scale]



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8. - Question Based on the information provided in Figure 1, the

elevation of BM2 is most nearly:

a. 109.53-ft

b. 117.85-ft

c. 124.47-ft

d. 24.47-ft

[not to scale]

FIGURE 1

Solution:

BM + BS = HI

HI – FS = TP Elevation

Remember to always check the summation of the back sight and foresight with the elevation

change.

Point BS HI FS Elevation BM1 12.64 112.64 100.00 TP1 10.88 120.41 3.11 109.53 TP2 9.72 127.57 2.56 117.85 BM2 3.10 124.47

+33.24 -8.77 +24.47

BS 9.72

BS 12.64

FS 3.10

FS 3.11

BM1 Elev. 100.00

TP1

TP2

BM2

BS 10.88

FS 2.56



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9. - Question The centerline separation distance between the sanitary manhole located at sta 52+00 and sta 62+00 is 20 stations. The invert elevation of the 24-in diameter RCP at station 52+00 is 345.5-ft. The pipe’s downward slope is 3% to sta 54+50 where the downward slope changes to 5/8” per foot through to sta 62+00. The invert elevation of the pipe entering the manhole at station 62+00 is most nearly:

a. 296.5-ft b. 298.5-ft c. 299.0-ft d. 302.5-ft

Solution: Distance between sta 52+00 and 54+50 = 250-ft 250-ft x 3% slope = 7.5-ft Distance between sta 54+50 and 62+00 = 750-ft 750-ft x (.625-in/ft) = 468.75-in Calculate elevation using the results: Elev. 345.5-ft – 7.5-ft – 39.0625-ft = 298.94-ft (answer)



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2. CONSTRUCTION MANAGEMENT

10. - Question The third floor of a 5-story 450,000-ft2 (gross area)

office building includes a 45-ft x 65-ft open-air atrium through to the first

floor. The project is being prepared for the placement of concrete on the

composite steel deck on the third floor. The specification requires a curing

agent to be applied no later than 45-minutes after machine trowel. Job

records show that (14) 55-gallon drums were used to apply the curing

agent on the 1st and 2nd floors. Using the recorded application rate, the

number of drums needed and gallons remaining for the project are most

nearly:

a) 20-drums; 4-gallons remain

b) 21-drums; 36-gallons remain

c) 22-drums; 49-gallons remain

d) 22-drums; 50-gallons remain

Solution:

Floor Gross ft2 Net ft2

Curing Agent Used

ft2

Curing Agent

Required ft2

5th 90,000 90,000 90,000 4th 90,000 90,000 90,000 3rd 90,000 87,075 87,075 2nd 90,000 87,075 87,075 1st 90,000 90,000 90,000

Column Totals 450,000 177,075 267,075

Curing agent application rate from job records:

177,075-ft2 ÷ 14-drums = 12,648.21-ft2/drum ÷ 55-gal/drum = 230-ft2/gal

Curing agent needed to complete the project:

267,075-ft2 ÷ 230-ft2/gal = 1,161.20-gallons ÷ 55-gal/drum = 21.11 drums

Purchase 22 drums; 49-gallons remain (answer)



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11. - Question An electrical contractor is preparing to place the duct bank concrete encasement for the high voltage feeders between manhole # 11 and # 12, total distance is 323-ft. The duct bank is 2-ft x 3-ft and holds nine 4.5-in Schedule 80 PVC conduits. The amount of concrete to be ordered is most nearly:

a. 60-yd3 b. 63-yd3 c. 72-yd3 d. 75-yd3

Solution:

Nominal 4-in Schedule 80 PVC pipe is 4.5-in O.D.

Total Volume = 2’ x 3’ x 323’ = 1938-ft3 / 27-ft3/CY = 71.77-yd3

Total Deduct = 9 x ( π ((4.5”/12”)/ft)2) x 323’ = 321-ft3 / 27-ft3/CY = 11.89-yd3

4

Total Concrete = 71.77-yd3 – 11.89-yd3 = 59.88-yds3 (answer)



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12. - Question The overall foundation plan for a new residence is shown in the drawing below. The foundation consists of an 838-LF perimeter spread footing measuring 18-in x 12-in and three column footings measuring 4-ft square by 12-in. During excavation, it was discovered that the soil conditions required that the footing width be increased by 70% in order to meet the required bearing with no change to depth of the foundation wall. The amount of additional concrete needed to meet the new requirement is most nearly:

a. 31.4-yd3 b. 35.9-yd3 c. 48.3-yd3 d. 82.2-yd3

Crawl Space

Crawl Space

Column footings

4’ x 4’ (typical)

Foundation Plan (Footing Layout)

Not to scale



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Solution:

Calculate the cubic volume of concrete using the given parameters:

Footing:

838-ft x 1.5-ft x 1-ft = 1257-ft3 = 46.56-yd3

Column Footing:

3-ea x 4-ft x 4-ft x 1-ft = 48-ft3 = 1.78-yd3

Total Concrete:

46.56-yd3 + 1.78-yd3 = 48.34-yd3

Calculate a 70% increase:

Footing:

838-ft x (1.5-ft x 1.7) x 1-ft = 2,137-ft3 = 79.14-yd3

Column Footing:

3-ea x (4-ft x 1.7 x 4-ft x 1.7) x 1-ft = 139-ft3 = 5.14-yd3

Total Concrete:

79.14-yd3 + 5.14-yd3 = 84.28-yd3

Calculate Net Increase:

84.28-yd3 - 48.34-yd3 = 35.94-yd3

(Answer = b)



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13. - Question A city engineer is deciding which of two alternatives to

select. If the interest rate is 10% per year, the Benefit-Cost analysis ratio of

the alternative to be selected is most nearly:

a. 1.125

b. 1.179

c. 1.307

d. 1.387

Alternative Cost Annual Benefit Salvage Useful Life

A $48,000 $13,000 $0 6 yrs

B $40,000 $12,000 $0 6 yrs

Solution:

Alternative A:

PW of Benefits = $13,000 (P/A, 10%, 6)

= $13,000 (4.355)

= $56,615

PW of Costs = $48,000

B/C = $56,615/$48,000 = 1.179

Alternative B:

PW of Benefits = $12,000 (P/A, 10%, 6)

= $12,000 (4.355)

= $52,260

PW of Costs = $40,000

B/C = $52,260/$40,000 = 1.3065

Since B/C of Alternative B is more than Alternative A, select Alternative B.

(answer = c)



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14. - Question With interest at 10%, the benefit-cost ratio for this

municipal government project is most nearly:

Initial cost $200,000

Additional costs at end of years 1 & 2 $30,000/yr

Benefits at end of years 1 & 2 $0

Annual benefits from year’s 3 to 10 $90,000/yr

a. 1.05

b. 1.25

c. 1.35

d. 1.57

Solution:

PW Cost = $200,000 + $30,000 (P/A, 10%, 2)

= $200,000 + $30,000 (1.736)

= $252,080

PW Benefits = $90,000 (P/A,10%,8) (P/F, 10%, 2)

= $90,000 (5.335) (0.8264)

= $396,800

B/C = $396,800/$252,080 = 1.574 (answer = d)

n-10

6 5 4 3 2 1

0

10

30k

i = 10%

1 2 8 7 6 5 4 3 2 1

7 8 9

30k



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15. - Question An excavating contractor determined through a test

strip, that a smooth steel-wheel vibrating roller compactor operating at 3-

mph can compact a 6-inch layer of borrow material to a proper density in

four passes. The width of the smooth steel drum is 8.0-ft and operates 50-

min per hour. The number of compacting rollers required to maintain a

material delivery rate of 540-bank cubic yards per hour is most nearly (1

bank cubic yard = 0.83 compacted cubic yard):

a) 1

b) 2

c) 3

d) 4

Solution: Calculate roller production.

Step 1: Compute a conversion factor to standardize the various units

of measure:

(5280 ft/mile) x (1 cubic yard / 27 cubic feet) x (1 foot / 12 inches) =

5280/27/12 = 16.3

Step 2: Compute compacted cubic yards (CCY) per hour:

Compacted cubic yards per hour = (16.3 x 8-ft x 3-mph x 6-inch x 50-

min/60-min) ÷ 4 passes

= 489 CCY/hr

Step 3: Compute compacted cubic yards (CCY) per hour to bank

cubic yards (BCY)

489-CCY/hr ÷ 0.83-CCY/BCY = 589.2 bank cubic yards per hour

Step 4: Compute the number of roller compactors needed.

Thus, 540 BCY/hr ÷ 589.2 BCU/hr = 0.92 < 1-roller compactor

Therefore, only one roller is required to keep up with the delivery of borrow

material.



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16. - Question Common ways to present the project schedule include

all except:

a. Gantt charts

b. Milestone charts

c. PERT charts

d. Project Network Diagrams with dates added

Solution: C - A PERT chart represents task dependencies but does not

represent activity durations or dates.

17. - Question The method of shortening a schedule that involves

breaking or changing logical relationships is known as:

a. GERT

b. Fast tracking

c. Crashing

d. Precedence mapping

Solution: B. Beginning construction before design is completed is an

example of fast tracking.

18. - Question Schedule control is concerned with all of the following

except:

a. Influencing the factors which create schedule changes

b. Determining that the schedule has changed

c. Resource requirement updates

d. Managing changes when and as they occur.

Solution: C - Resource planning to meet new or additional requirements is

a c

ost management process



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19. - Question A finish-to-finish activity is best described as:

a. The “from” activity must finish before the “to” activity can finish

b. The “from” activity must finish before the “to” activity can start

c. The “from” activity must start before the “to” activity can start

d. The “from” activity must start before the “to” activity can finish

Solution: A. The "from" activity appears first and the "to" activity

appears second in the sequence.

20. - Question Sources of historical information for activity duration

include all except:

a. Project files

b. Commercial estimating databases

c. Expert judgment

d. Project team knowledge

Solution: C Expert judgment is the synthesis of information from multiple

sources.

21. - Question Sheet rock cannot be installed in the new house you

are building until a building inspector approves the plumbing, electrical,

HVAC, and insulation. This is an example of:

a. A constraint

b. An assumption

c. A start-to-finish dependency

d. A finish-to-start dependency

Solution: D. The inspections must be finished before the sheet-rocking

can start.



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22. - Question A project requires that activities A and B can be

started independently and activities C and D are independent but can be

started after A and B have been completed. Activity E can be started only

when all activities A, B, C, and D have been completed. The correct

network for this project is:

a. b. c. d. Solution: a=answer

2

5

A

3

4

B

E

C

D

1

2

5

A

3

4

B

E

C

D

1

1

4

A

2

3

B

E C

D 0

5

3 B

4

5

C

E

D

2 1 A



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23. - Question Given the network shown in the figure below:

1. Determine the total duration to complete the project

a) 20

b) 23

c) 25

d) 26

2. Identify the critical path:

a) A-C-D-E-G-H

b) A-C-D-G-H

c) A-B-D-E-G-H

d) A-C-D-F-G-H

G A

B E

C

D F 2 6

7

3 5

2 2

9

H

5

Activity node

Activity duration

G

2



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Solution:

1. Total duration: 26

2. Critical Path: A-C-D-E-G-H

G A

B E

C

D F

17 24 8

2

26

17

H

5



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3. MATERIALS

24. - Question A concrete batch plant provided the Design Strength

for the engineer’s mix design to match the project’s specification

requirement for a 4000 psi Tremie pump mix. The w/c ratio is most nearly:

a) 05

b) .34

c) .38

d) .45

Solution:

From the given information: Cement + Flyash =

658-lbs

Water = 30.0-gal x 8.345-lbs/gal = 250.35-lbs

w/c = 250.35 / 658 = .38

Design Strength 4000 psi Tremie Mix

Material Units Check Quantity

Cement Lbs 559

Flyash Lbs 99

Sand Lbs 1245

Stone Lbs 1750

Water Gal 30.0

ADMIX 1 Ozs 3.3

ADMIX 2 Ozs 26.3

Slump Inch 5 - 7

Air % 6 ± 1.5



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25. - Question A jobsite delivery ticket is shown below for a grout

mix. The actual w/c ratio is most nearly:

a) .30

b) .32

c) .34

d) .36

Batch Plt. 37 Load 210 Ticket 1487-TCC Volume 10-CY Mix Description SOG 4000-psi Grout Truck 41 w/c Date/Time 12/12/11 10:48 Material Target Actual Status Moisture Material Target Actual Status Sand 18,580-lb 18,640-lb Done 4.4 % Cl 0-oz 0-oz WR 0-oz 0-oz Retarder 0-oz 0-oz Air Entrain. 0-oz 0-oz MR 0-oz 0-oz HRWR 0-oz 0-oz Calcium 0-oz 0-oz NC Accel 2564-oz 2592-oz Done Type I 9,060-lb 9,100-lb Done Water 2,420-lb 2,394-lb Done Flyash 1,600-lb 1,595-lb Done

Solution:

Note, although there is a significant amount of Non Chlorinated Accelerator (NC ACC) in the grout mix, the product is a water reducing agent and is not added to the w/c ratio. [( 2,592-oz ÷ 128-oz/gal * 8.345-gal/lb = 167-lb] Calculate the total weight of the water: Water = (2,394-lb) + (Sand 18,640-lb x 4.4%) = 3,214-lb Calculate the total Weight of Cement: Type1 9,100-lb + FlyAsh 1,595-lb = 10,695-lb Calculate the w/c ratio. Since the units are the same, the ratio can be directly calculated: w/c = 3,214-lb ÷ 10,695-lb = .300 (answer = a)



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26. - Question A concrete wall form tie with an ultimate strength

capacity of 2,400-lbs is designed for use in a concrete wall form with an 18-

in center to center spacing. During formwork inspection it was discovered

that the actual spacing is 24-in center to center. Using the actual spacing,

the percent change in load on the wall form tie is most nearly:

a) 25%

b) 33%

c) 56.25%

d) 78%

Solution:

Determine design tributary area: 1.5-ft x 1.5-ft = 2.25-ft2

Determine actual tributary area: 2-ft x 2-ft = 4-ft2

Determine design ft2 load: 2400-lbs / 2.25-ft2 = 1066.67-lbs/ft2

Determine actual load on 4-ft2: 1066.67- lbs/ft2 x 4-ft2 = 4266.68-lbs

Increase in load is: 4266.68-lbs – 2400-lbs = 1866.68-lbs

Calculate % change = 1 - (4266.68-lbs / 2400-lbs) = 78%

The load increase is 78% on the wall form tie.



notes for mar 22nd 23rd and 24th classes part 2

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