Post on 16-Apr-2017
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 (ahmed_usief@yahoo.com)
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 (ahmed_usief@yahoo.com)
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 3
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)
ahmed youssef (ahmed_usief@yahoo.com)
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 4
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
ahmed youssef (ahmed_usief@yahoo.com)
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 5
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
ahmed youssef (ahmed_usief@yahoo.com)
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 6
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)
ahmed youssef (ahmed_usief@yahoo.com)
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 7
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)
ahmed youssef (ahmed_usief@yahoo.com)
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 8
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
ahmed youssef (ahmed_usief@yahoo.com)
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 9
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]
ahmed youssef (ahmed_usief@yahoo.com)
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 10
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
ahmed youssef (ahmed_usief@yahoo.com)
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 11
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)
ahmed youssef (ahmed_usief@yahoo.com)
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 12
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)
ahmed youssef (ahmed_usief@yahoo.com)
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 13
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)
ahmed youssef (ahmed_usief@yahoo.com)
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 14
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
ahmed youssef (ahmed_usief@yahoo.com)
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 15
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)
ahmed youssef (ahmed_usief@yahoo.com)
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 16
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)
ahmed youssef (ahmed_usief@yahoo.com)
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 17
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
ahmed youssef (ahmed_usief@yahoo.com)
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 18
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.
ahmed youssef (ahmed_usief@yahoo.com)
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 19
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
ahmed youssef (ahmed_usief@yahoo.com)
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 20
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.
ahmed youssef (ahmed_usief@yahoo.com)
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 21
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
ahmed youssef (ahmed_usief@yahoo.com)
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 22
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
ahmed youssef (ahmed_usief@yahoo.com)
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 23
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
ahmed youssef (ahmed_usief@yahoo.com)
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 24
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
ahmed youssef (ahmed_usief@yahoo.com)
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 25
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)
ahmed youssef (ahmed_usief@yahoo.com)
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 26
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.
ahmed youssef (ahmed_usief@yahoo.com)
This copy is given to the following student as part of School of PE course. Not allowed to distribute to others.