King Saud University
College of Engineering
Civil Engineering Department
FINAL EXAM
CE 417: Construction Methods and Equipment – 1ed Semester 1432/1433 h
Tuesday, 16 SAFAR, 1433 H – 7th of Jan. 2012
Time allowed: 3 hrs
Student name
Student number
Section
Student No. in class
Total number of Questions: 6
Attempt all questions
Questions
Maximum Marks
Part 1
20
Part 2, Q # 1
16
Q#2
16
Q#3
16
Q#4
16
Q#5
16
Marks obtained
Total marks
Total marks obtained (in words):____________________________
_______
100
PART I: Answer the following questions (Open Book). (20%)
1) State 3 out 7 factors that reduce the construction cost during construction phase?
2) Efficient management of the earthmoving process requires 3 considerations, state them.
3) State the 2 major factors controlling the shovel production.
4) Which condition the retarder chart is used, in positive or negative condition?
5) What is the different between the compaction and consolidation of soil in term of time requirement?
6) There are 2 methods for producing prestress in concrete members, what they are?
7) State the 3 general categories of construction delay:
8) State 2 out of 7 major elements of safety program?
PART I: Answer the following questions (Open Book). (20%)
1) State 3 out 7 factors that reduce the construction cost during construction phase?
 Good work planning.
 Careful selection and training of workers and
managers.
 Efficient scheduling of labor, materials, and
equipment.
 Proper organization of work.
 Use of laborsaving techniques such as
prefabrication and preassembly.
 Minimizing rework through timely quality
control.
 Preventing accidents through good safety
procedures.
2) Efficient management of the earthmoving process requires 3 considerations, state them.
 accurate estimating of work quantities and job conditions,
 proper selection of equipment, and
 competent job management?
3) State the 2 major factors controlling the shovel production.
 the swing angle and
 Lost time during the production cycle.
4) Which condition the retarder chart is used, in positive or negative condition?
Retarder chart is used in negative resistance factor condition, while performance chart is used in
positive resistance factor condition.
5) What is the different between the compaction and consolidation of soil in term of time requirement?
Consolidation may require months or years to complete, whereas compaction is accomplished in a
matter of hours.
6) There are 2 methods for producing prestress in concrete members, what they are?
pretensioning and posttensioning.
7) State the 3 general categories of construction delay:
 those beyond the control of either the contractor or owner ("acts of God"),
 those under the control of the owner, and
 those under the control of the contractor.
8) State 2 out of 7 major elements of safety program?
1. A formal safety training program for all new employees.
2. Periodic refresher training for each worker.
3. A formal supervisory safety training program for all
supervisors.
4. A program of regular site visits by safety personnel to
review and control job hazards.
5. Provision of adequate personal protective equipment,
first-aid equipment, and trained emergency personnel.
6. An established procedure for the emergency evacuation of
injured workers.
7. Provisions for maintaining safety records and reporting
accidents in compliance with OSHA requirements.
PART II: Solve the rest of the questions with taking the middle value of any 2 range values in any
table.
Question 1 (16%)
A triangular pile with 6 m width and 180 m length had been excavated from a trench in a street in
Riyadh city by a hydraulic excavator that has the following specifications and conditions:
 Heaped bucket capacity is 1 m3.
 Material is dry common earth.
 Maximum depth of cut is 8 m.
 Bucket fill factor is 0.95.
o
 Average swing is 90 .
 Job and Management condition is good.
Accordingly, answer the following questions:
a) Determine the average depth of this trench that has a 1 m width and 200 m length.
b) Estimate the job production in LCM.
c) How many working hours had been taken to finish this job?
Solution
a)
B = 6 m, L = 180 m, and R = 32o (Table 2-6).
4𝑉
𝐵 = √𝐿×tan 𝑅  V =
Load factor = 0.8
𝐵2 ×𝐿×tan 𝑅
4
= 1012.29 LCM
(Table 2-5)
Pile volume = 1012.29 × 0.8 = 809.83 BCM
Trench volume = 1 m× 200 m× Average Depth
809.83 BCM
Pile volume = Trench volume  Average Depth = 200 m × 1m = 4 m
b)
Cycle output (C) = 160 cycle /60 minutes
(Table 3-3)
Maximum Depth
4𝑚
Depth of Cut = Average Depth × 100 = 8 𝑚× 100 = 50%
Swing-depth factor (S) = 1.10
(Table 3-4)
Bucket Volume (V) = 1 LCM
Bucket fill factor (B) = 0.95
Job Efficiency (E) = 0.75
(Table 2-1)
Adjustment factor = 0.925
(Table 3-5)
Production = C × S × V × B × E × Adjustment factor
= 160 × 1.1 × 1 LCM × 0.95 × 0.75 × 0.925 = 116 LCM/hr
c)
Pile volume
No. of working hours = Job Production =
1012.29 LCM
116 LCM/hr
= 8.7 ≈ 9 working hours
Question 2 (16%)
You are given the following data for a scraper job: a. number of scrapers are seven single engines
overhauling; b. tandem pusher will be used; c. the scraper will carry 19.6 BCM (full load=34,020); d.
same route will be used for haul and return; e. chain loading method (pusher cycle time is 0.9 min); f.
scraper fixed cycle time = 1.3 min; g. efficiency factor is 0.85 and job conditions are average; h. use
figure 4-4 and 4-5 for estimating travel time. Sections of the haul route from the cut to the fill area as
following. Determine the production for the seven scrapers?
Section
Height Horizontal
increased increased
(m)
(m)
Distance
(m)
Grade
(%)
Rolling
Effective Travel time
Resistance
Grade (%)
(min.)
factor (kg/ton)
1
-1.424
+47.434
50
2
-6.364
+636.396
70
3
+2.121
+212.132
90
4
0
+210
100
Total travel time
Question 3 (16%)
Design the Stringer of a concrete slab formwork (including checking of shore strength) having the following
information:
 Design Load = 6.22 kPa
 Assume all members are continuous over three or
more span.
 All lumber will be Southern Pine.
 50×200 mm (2×8) lumber at 600 mm will be used for  Limit deflection to 1/360 of span length.
Joist.
 Commercial shores of 20 kN capacity will be used.
 100×200 mm (4×8) lumber at 2 m will be used for
 Assume dry condition & < 7 day load
stringer.
Solution
Stringer Design
w = (2 m2/m)  (6.22 KN/m2) = 12.44 KN/m
Allowable stress of lumber (@Southern Pine). (Table 13-8)
Fb = 9653 kPa, Fv = 1241 kPa, E = 11×106 kPa
Properties of lumber (@4×8). (Table 13-7)
d = 7.25 mm, A = 16.37×103 mm2, S = 5.024×105 mm3, I = 46.26106 mm4
For continues over three or more span
a) Bending
1
1
100  9653  5.024  105  2
100  Fb S  2

 = 1,974.45 mm.
l=

 =
1000 
12.44
1000  w 

b) Shear
1.11 Fv A
1.11 1241 16.37  103
 2d =
 2  7.25 = 1,827.19 mm.
l=
1000 w
1000
12.44
c) Deflection
13
73.8  11 106  46.26  106 
73.8  EI 


l=
  =
1000 
12.44
1000  w 

d) Check for shore strength
20
 1.6 m
l=
12.44

13
= 2,542.82 mm
 Shore strength governs in this design and the maximum allowable span is 1.6m. Select a 1.5 m)
shore spacing for design.
Question 4 (16%)
A crawler tractor cost SR1,000,000 has an estimated salvage value of SR200,000 and has a 5 year life.
a. Find the annual depreciation and book value at the end of each year using sum-of-the-years’ –
digits method of depreciation.
b. Calculate the average hourly owning cost for the first year of life of the tractor if the tractor is
operated 2000 hours during the year. The rate for interest, taxes, and insurance is 12% and the
rate for storage and miscellaneous cost is 4%.
Solution
a.
Sum of the year digit of 7 years = 1 + 2 + 3 + 4 + 5 = 15
5
D1 = 15 (1000,000 – 200,000) = 266,666.67
4
D2 = 15 (1000,000 – 200,000) = 213,333.33
3
D3 = 15 (1000,000 – 200,000) = 160,000
2
D4 = 15 (1000,000 – 200,000) = 106,666.67
1
D5 = 15 (1000,000 – 200,000) = 53,333.33
Year
Depreciation
0
1
2
3
4
5
0
266,666.67
213,333.33
160,000
106,666.67
53,333.33
Book Value
(End of Year)
SR 1000,000
733,333.33
520,000.00
360,000.00
253,333.33
200,000
b.
Cost
Calculation
SR/h
Owning cost
1. Depreciation cost
D1 = 266,666.67/2000
2. Investment, tax,
insurance, and
storage
Cost Rate =12+ 4 = 16%
Average investment = (1000,000+200,000)/2 = SR 600,000
Investment, tax, insurance, and storage = SR 600,000/2000
Total Owning Cost
133.333
133.33 + 48
48
181.333
Question 5 (16%)
A wheel-tractor-towed tamping foot compactor works under the following conditions: average speed =
8.0 km/h, compacted lift thickness = 15.2 cm, effective roller width = 3 m, job efficiency = 0.75, and
number of passes = 8. Thus, answer the following:
a) Estimate the production in compacted cubic meters per hour?
b) In compacting a highway project, How far of linear length can this compactor compact a 15 m
width one way highway within 1 week? Assume: 7 working hour/day & 6 working day/week.
Solution
a)
S = Average speed = 8 km/hr
(Table 5-3) taking the lower value
L= Compacted lift thickness = 15.2 cm,
W = Effective roller width = 3 m,
E = Job efficiency = 0.75
P = Number of passes = 8
Production (CCM/hr) =
10×W×S×L×E
P
=
10 × 3 × 8 × 15.2 × 0.75
8
= 342 CCM/hr
b)
15.2 cm
Compacted Volume = 100 cm/m × 15 m × length
 Number of hours =
Compacted Volume
Production
 Compacted Volume = 7 hr/day × 5 day/week × Production = Production
342 CCM/hr
 Length = 35 ℎ𝑟/𝑤𝑒𝑒𝑘 × 0.152CM ×15 CM = 5250 m/week = 5.25 km/week
OR
a)
S = Average speed = (8+16)/2 = 12 km/hr
L= Compacted lift thickness = 15.2 cm,
W = Effective roller width = 3 m,
E = Job efficiency = 0.75
P = Number of passes = 8
Production (CCM/hr) =
10×W×S×L×E
P
=
(Table 5-3) taking the average value
10 × 3 × 12 × 15.2 × 0.75
8
= 506.25 CCM/hr
b)
15.2 cm
Compacted Volume = 100 cm/m × 15 m × length
 Number of hours =
Compacted Volume
Production
 Compacted Volume = 7 hr/day × 5 day/week × Production = Production
506.25 CCM/hr
 Length = 35 ℎ𝑟/𝑤𝑒𝑒𝑘 × 0.152CM ×15 CM = 7770 m/week = 7.77 km/week
Figure 4-4, 4-5,
Table 13-5A, 13-7, 13-8