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The Group Assignment of the Course: CONSTRUCTION PROCESSES (BLDG492-6831 – UU) G ROUP F IRST N AME F AMILY N AME S TUDENT ID MARIO CESAR CASTRO LA BARRERA 40270431 TASHWINI MANJUNATH 40264245 SEYED ILIA MOUSAVI 40204305 JENIL RANA 40263428 SHIVA DARIHAKI 40221999 MD FAZLE RABBI 40239821 H ADI HATAMI RASTGOO BAJGIRAN 40277000 MARIA SHENODA 40270571 SAI KARTHIK VAKKANTHULA 40260589 Instructor: SHAHIN KARIMIDORABATI Concordia University Department of Building, Civil, & Environmental Engineering Fall 2023 Comments and Feedback (FOR TUTORS/MARKERS ONLY) Mark:

1) EARTHMOVING MATERIALS AND OPERATIONS A. C ALCULATIVE Q UESTION An earthwork excavation includes filling of a basement of dimensions 25m x 50m x 3m. The material to be excavated is common earth. On a sample of 120 grams taken from the borrow pit laboratory tests were conducted and it has been determined that the weight of the sample when oven dried is 100 grams. Independently, a modified proctor test has also been conducted and it has been determined that the required density is obtained at a moisture content of 22%. Given this, which of the following is correct? A. 6250 LCY and 260 kg/ m 3 of water to be added. B. 6945 BCY and 260 kg/ m 3 of water to be added. C. 6250 LCY and 260,254 kg of water to be added. D. 6945 BCY and 260,254 kg of water to be added. Sol. The Correct answer is D. Volume of fill = 25m x 50m x 5m = 6250 m 3 = 6250 CCY To convert bank volume to compacted volume, we use shrinkage factor. From Table 2 –5, shrinkage factor is 0.90 Volume of earthwork required = 6250 CCY / 0.75 = 6944.44 BCY say 6945 BCY The % moisture content = sample weight – dry weight / dry weight = 120 –100/ 100 = 20% As the moisture content in bank measure is less than that optimum moisture i.e., 20% < 22%, additional water is required to be added to excavated material. Weight of water in bank measure per cubic metre of soil = 1839 kg/ m 3 x 0.2 = 367.80 kg Weight of water in 8334 = 367.80 kg x 6945 m 3 = 2,554,371 kg Weight of water in compacted measure per cubic metre = 2047 kg/ m 3 x 0.22 = 450.34 kg Weight of water in 6250 = 450.34 kg x 6250 m 3 = 2,814,625 kg Difference = 2,814,625 – 2,554,371 = 260,254 kg (or) 260 m 3 B. N ON -C ALCULATIVE Q UESTION 2

Total volume of the trench= 2000 (m)* 2 (m)* 2.5 (m)= 10000 m 3 Productivity for a small sized excavator: V= 0.76 m 3 for best case scenario (according to table) C= 200 c/h (according to table) Depth of cut (% of maximum) = (2.5 m / 6 m) *100 = 41.67 % Angle of swing = 45 degree, for best case scenario Interpolation: S= 1.33 + ((1.28-1.33)/ (50-30)) *(41.67-30) = 1.30 (according to table) B= 1.10 for best case scenario (according to table) E= 0.63 (according to table) Adjustment factor for trench= 0.95 for best case scenario (according to table) Productivity = V*C*S*B*E*adjustment factor for trench = 0.76 (m 3 ) * 200 (c/h) * 1.3* 1.1* 0.63* 0.95 Productivity= 130.09 (m 3 /h) for small sized excavator Total amount of time needed to excavate the trench= 10000 (m 3 )/ 130.09 (m3/h) = 76.87 h Total to be paid for the small excavator= 76.87 (h)* 120 ($) = 9,224.4 $ Productivity for a medium sized excavator: V= 1.72 m 3 for best case scenario (according to table) C= 160 c/h (according to table) Depth of cut (% of maximum) = (2.5 m / 8.3 m) *100 = 30.12 % Angle of swing = 45 degree, for best case scenario S≈ 1.33 (according to table) (according to table) B= 1.10 for best case scenario (according to table) E= 0.63 (according to table) Adjustment factor for trench= 0.95 for best case scenario Productivity = V*C*S*B*E*adjustment factor for trench = 1.72 (m 3 ) * 160 (c/h) * 1.33* 1.1* 0.63* 0.95 Productivity= 240.97 (m 3 /h) for medium sized excavator 4

Total amount of time needed to excavate the trench= 10000 (m 3 )/ 240.97 (m 3 /h) = 41.50 h Total to be paid for the small excavator= 41.50 (h)* 210 ($) = 8,715 $ $ 8,715 < $ 9,224.4, So the medium sized excavator is better for these circumstances. D. N ON -C ALCULATIVE Q UESTION Which of the following statements is NOT correct regarding earthwork excavating equipment? A. The three criteria to select the proper shovel: Maximum depth needed, working radius, and dumping height. B. Heaped volume is the maximum volume that can be placed in the bucket without spillage based on the material’s angle of repose. C. Maximum productivity of a dragline is obtained with a minimum swing angle. D. Shovels are good for above track level conditions and cannot operate and dig below track level. Solution: The correct answer is A. The three selection criteria indicated in the option A are applicable to a hydraulic excavator but not to a shovel. 3) LIFTING EQUIPMENT E. C ALCULATIVE Q UESTION The maximum lifting capacity of a tower crane rigged with two-part line is 27,600 lb and the maximum hoist speed is 100 fpm, within the structural capacity constraints of the tower and jib configuration. For the same tower crane and jib configuration, what is the maximum lifting capacity and speed for a four-part line. A. 55,200 lb and 50 fpm B. 55,200 lb and 100 fpm C. 27,600 lb and 100 fpm. D. Unknown, more information required. 5

Solution: The correct answer is D. When the operating radius of a tower crane is less than 90 feet, a four-part line has more lifting capacity than a two-part line. When the operating radius is greater than 90 feet, the converse is true. When the operating radius is greater than 90 feet, the structural capacity of the tower crane controls the load-lifting ability, and when operating radius is less than 90 feet the hoisting system controls the load-lifting ability. 4) LOADING AND HAULING – PART 1 G. C ALCULATIVE Q UESTION Calculate the Max. Usable Pull for a vehicle with a weight of 12 tons, moving on a road with a grade of 5%, a penetration of 3 cm, and at an altitude of 1000 meters. Assume a coefficient of traction of 0.85. A. 11.5 tons B. 10.2 tons C. 6.5 tons D. 17.9 tons Solution: The correct answer is B. 1. Calculate Rolling Resistance Factor: Rolling resistance factor (Kg/t)=20+(6×penetration) Rolling resistance factor (Kg/t)=20+(6×3)=38 Kg/t 2. Calculate Grade Resistance Factor: Grade Resistance factor (kg/t)=10×grade (%) Grade Resistance factor (kg/t)=10×5=50 kg/t 3. Calculate Grade Resistance: Grade resistance (Kg)=Vehicle weight (t)×Grade Resistance factor (Kg/t) Grade resistance (Kg)=12×50=600 Kg 4. Calculate Effective Grade: Effective Grade(%)=Grade (%)+Rolling Resistance factor (Kg/t)10 Effective Grade (%)=5+3810=8.8% 5. Calculate Derating Factor: 7

Derating Factor (%)=Altitude (m) - 915×10−2 Derating Factor (%)=(1000−915)×10−2=8.5% 6. Calculate Max. Usable Pull: Max. Usable Pull=Coefficient of traction×Weight on driving wheels Max. Usable Pull=0.85×12=10.2 tons Therefore, the Max. Usable Pull for the given conditions is 10.2 tons. H. N ON -C ALCULATIVE Q UESTION What is the significance of the "Coefficient of Traction" in determining a vehicle's maximum usable pull? A. It represents the vehicle's total weight B. It indicates the force exerted at the rim of each driving wheel C. It influences the vehicle's pulling capability on different surfaces D. It measures the vehicle's horsepower Sol. The correct answer is C. The coefficient of traction is crucial because it influences how well a vehicle can grip different surfaces. A higher coefficient means better traction, allowing the vehicle to exert more force and pull heavier loads, especially on surfaces with lower friction. This directly impacts the vehicle's pulling capability on various terrains. 8

J. N ON -C ALCULATIVE Q UESTION A construction operation involves spreading of clean granular gravel for creation of subgrade of 6 metres width for a road. The subgrade has a camber of 1-in-50 (camber is the cross-slope of the road for effective drainage of rainwater). The following dozer configurations are available for selection and as a project manager what will be your recommendation? A. A dozer with a cushion blade with tilting adjustment B. A dozer with a universal blade with pitching adjustment C. A dozer with an angular blade with angling adjustment D. A dozer with a straight blade with tilting adjustment Sol. The correct answer is D A dozer can be used to both excavate and spread soils. Tilting is the adjustment operation for a dozer blade to achieve cross-slope when spreading or cutting soils. Of all the blades that are available for a dozer, cushion blade cannot be tilted. Hence, the only choice for the operation is a dozer with a straight blade with tilting adjustment. 6) LOADING AND HAULING – PART 3 K. C ALCULATIVE Q UESTION Estimate the total turnaround distance for a single-engine two-axle tractor based on the following information: Haul Route (single Pusher) Down a 5% grade, 500 m Up a 5% grade, 500 m Rolling resistance factor= 50 Kg/t Cycle time per hour= 4.41 min A. 250 m B. 300 m C. 350 m D.400 m 10

Sol: The correct answer is B 1. Calculate the fixed time from the table for single pusher: Spot time= 0.3 min Loading time= 0.6 min Maneuvering & Dump time= 0.7 min Fixed total time= 1.6 min 2. Total cycle time= variable time + fixed time 4.41 min= variable time + 1.6 Therefore, variable time= 2.81 min 3. Calculate the effective grade. Haul: -5 + 50/10 = 0% Return: 5+ 50/10= 10% Turnover: 0+ 50/10= 5% 4. Calculate the variable travel time for haul and return. Haul: 0.8 min (from the loaded graph) Return: 1.40 min (from the empty graph) Since 4.41= 0.8 + 1.40 + turnaround time + 1.6 Therefore, turnaround time = 4.41- (0.8+1.4+1.6) = 0.61 min 5. From the empty graph find the turnaround distance using 0.61 min and 5% effective grade. Therefore, turnaround distance = 300 m 11

tractor or another scraper”. [ S.W. Nunnally, “Loading & Hauling: Scrappers Operation and Employment” in Construction Methods & Management, Prentice-Hall Pearson, 8 th Edition, 2011, P. 64] 7) COMPACTING, GRADING, FINISHING, RIPPING M.C ALCULATIVE Q UESTION A contractor is faced with the decision of selecting between 2 compaction machinery (wheel tractor towed with tamping foot and a heavy pneumatic). It is known that the sandy soil weighs 1931.8 kg/BCM. The compacted lift has a thickness of 5 in. One machine has an effective roller width of 8m, while the other is equipped with a foot roller drum width of 4.5m. The cost per unit for the equipment with foot roller is $490/h, while the other is $460/h. What is the least cost-effective equipment (unit cost in $/CCM) considering optimal production? A. Wheel tractor towed with tamping foot, $0.48/CCM B. Wheel tractor towed with tamping foot, $0.70/CCM C. Heavy pneumatic with tamping foot, $0.37/CCM D. Heavy pneumatic with tamping foot, $1.30/CCM Solution: The correct answer is B The answer is the Wheel tractor towed with tamping foot will be the most economic to choose as the price would be $ 0.70 per m3 compacted. Step 1: Calculate the density. = 1.7 gr/cm3 13

Step 2: Number of passes required for the compactor. Number of passes = 11 Step 3: Estimate the Compactor Production Wheel tractor towed with tamping foot: = 698.27 CCM /h Heavy pneumatic: = 620.68 CCM /h 14

Heavy pneumatic: = $ 0.74 / CCM N. N ON -C ALCULATIVE Q UESTION Which compaction equipment is recommended for a wide range of materials, except for sand, clean, or silty materials? E. Pneumatic F. Steel wheel G. Tamping Foot H. Vibratory Solution: The correct answer is C Tamping Foot is the only machine that can be used in varied materials such as rock, gravel, clean or silty, gravel, clayey, sand, clayey silt, clay, sandy or silty and clay heavy; however, this equipment is least effective on sand, clean or silty materials. [ S.W. Nunnally, “Compaction and Finishing” in Construction Methods & Management, Prentice-Hall Pearson, 8th Edition, 2014, P.85, Figure 10] 16

8) AGGREGATE PRODUCTION O. C ALCULATIVE Q UESTION Concrete is to be produced at 200 cum per hour with a mix design of 1:2:4 (cement: fine aggregate: coarse aggregate) by weight with a minimum cement content of 400 kg per cum. The gradation of coarse aggregate is 25mm to 19mm - 70%, 19 to 13mm – 15%, 13 to 10mm – 15%. Considering open circuit, which of the following crusher closed side setting requires lowest total feed and yields the above gradation, so that the concrete production is continuous. A. 38mm B. 25mm C. 32mm D. 19mm Sol. The correct answer is B. Step – 1: Calculate the aggregate required to produce 200 cum of concrete per hour. Mix design is 1:2:4 by weight and minimum cement quantity is 400 kg per cum Total quantity of cement = 200 cum x 400 kg per cum => 80000 kg => 80 tonnes per hour Quantity of coarse aggregate = 4 x Total quantity of cement => 4 x 80 tonnes => 320 tonnes per hour 320 tonnes per hour of coarse aggregate is required to produce 200 cum of concrete per hour. Step – 2: Calculate the quantity of different aggregate sizes required. 25 – 19mm = 70% x 320 tonnes per hour => 224 tonnes per hour 19 – 13mm = 15% x 320 tonnes per hour => 48 tonnes per hour 13 – 10mm = 15% x 320 tonnes per hour => 48 tonnes per hour Step – 3: Select screen sizes required based on aggregate sizes required. The screen sizes required to yield the gradation in step-2 in the order of their placement in screen deck are 25mm, 19mm, 13mm, and 10mm. Step – 4: Select the crusher closed side setting with maximum retainage. 17

13mm -10mm: The percentage of aggregate passing 13mm screen and retained on 10mm screen gives this grading; the percentage retained on 10mm screen from step – 5 is 12% Output = Total feed x percentage retained. 48 tonnes per hour = total feed x 12% Total feed = 48 tonnes per hour / 12% => 400 tonnes per hour Every other crusher closed side setting requires total feed more than 1244.45 tonnes per hour, as the percentage of 25mm to 19mm retained is lower than 18%. P. N ON -C ALCULATIVE Q UESTION Which of the following is the correct sequence of producing aggregate? A. Excavation, loading, hauling, crushing, screening, washing, storage. B. Excavation, crushing, screening, loading, hauling, washing, storage. C. Excavation, crushing, screening, washing, loading, hauling, storage. D. Excavation, screening, crushing, washing, loading, storage. Sol. The correct answer is A. To produce aggregate, rock or gravel is excavated, loaded into haul trucks, and transported to the aggregate processing plant. Then the rock or gravel is crushed using rock crushers, screened, washed using log washers, scrubber drums, etc. and then stored. 9) CONCRETE FORM DESIGN Q. C ALCULATIVE Q UESTION You are the project engineer of a multiple story building project. while preparing the casting of the first typical floor slab designed for 200 lb/sq. ft, you have decided to lay out the joists at minimum 30 inches spacing due to the limited formwork material available at site. You have the choice between actual 1-inch plyform class 1 or actual 1-inch Eastern spruce for the decking material spanning a single span, which will be satisfactory for the 30-inch joist spacing? (Assume grains across the supports for the plyform calculations and l/360 for deflection limit). 19

A. Plyform class 1 B. Eastern Spruce C. Both of them D. None of them Sol: The correct answer is B Step-1: Calculate the design load for joist spacing We design joist for 1ft X 1ft sheathing. So, W (lb/ft)= 200 (lb/sq. ft)x 1ft = 200 lb/ft Step-2: Calculate joist spacing for plyform class 1 Considering the plyform 1” , from table 13-6: d=1” EI = 0.641 × 10 6 F b KS = 1.422 × 10 3 F s lb Q = 0.675 × 10 3 From table 13-5: choosing the plywood equations for 1 span condition: Bending: l = 9.8 ( F b KS W ) 1 2 = 9.8 ( 1.422 × 10 3 200 ) 1 2 = 26.13 in Shear: l = 24 ( F s lb Q W ) + 2 d = 24 ( 0.675 × 10 3 200 ) + 2 × 1 = 83 in Deflection (l/360): l = 1.37 ( EI W ) 1 3 = 1.37 ( 0.641 × 10 6 200 ) 1 3 = 20.19 in From these three values we can choose minimum value 20.19 in but it less than our minimum requirement 30 in. So, we cannot use plyform class 1 for above conditions. Step-3: Calculate joist spacing for Eastern spruce Considering the Eastern spruce 1 in From table 13-8: d=1 in, b=12 in (As we assume 1ft x 1ft strip) E = 1.2 × 10 6 × 0.97 × 1 = 1.164 × 10 6 F b = 1050 × 0.86 × 1.25 = 1128.75 F v = 140 × 0.97 × 1.25 = 169.75 and from cross section, A = 12 × 1 = 12 i n 2 20