Title PSAD 44 - Construction Engineering.pdf ✅

PSAD 44: Construction Engineering Video Transcript SITUATION. The table below shows the schedule of works necessary to carry out a construction project. 1. What is the number of days required to carry out the entire project? 2. What is the critical path of the project? 3. What is the total float, free float, and interfering float of activity H? Solution: Shown is the PERT-CPM diagram (forward values are in blue and backward values are in red) From forward computation, the days required to finish the project is 65. From the PERT-CPM diagram, the critical path is A-B-D-G-I-K. The total float is the total number of days that activity H can be delayed so as not to change the duration of the project. TFH = Late Start − Early Start = 34 − 27 = 7 The free float is the total number of days activity H can be delayed so as not to disrupt the next activity FFH = Late Finish − Early Finish = 46 − 45 = 1 The interfering float is the time that task H can be delayed where it has no bearing on the early finish date of the project. IFH = TF − FF = 7 − 1 = 6
SITUATION. It is required to produce 30 m3 of class B (1:2:4) concrete using the shown properties. Assume 1 bag of cement has a volume of 0.028 m3. Use 25 liters of water per bag of cement. Material SG Unit weight Cement 3.10 1506 kg/m3 Sand 2.65 1680 kg/m3 Gravel 2.50 1525 kg/m3 1. Determine the number of bags of cement required. 2. Determine the quantity of sand. 3. Determine the quantity of gravel. 4. Determine the quantity of water. Solution: Shown is the table that shows the computations and the amounts of each material, per bag of cement. Material Loose Volume Absolute Volume Weight, kg Cement 0.028 42.168 3.10 × 1000 = 0.0136 0.028 × 1506 = 42.168 Sand 0.028 × 2 = 0.056 94.08 2.65 × 1000 = 0.0355 0.056 × 1680 = 94.08 Gravel 0.028 × 4 = 0.112 170.8 2.50 × 1000 = 0.06832 0.112 × 1525 = 170.8 Water - 25 1000 = 0.025 0.025 × 1000 = 25 Total 0.221 0.14242 332 For the number of bags of cement, n = 30 m3 0.14242 m3/bag = 210.638 bags, say 211 bags For the volume of sand, Vs = (210.638)(0.056) Vs = 11. 796 m3 For the volume of gravel, Vg = (210.638)(0.112) Vg = 23. 592 m3 For the volume of water, Vw = (210.638)(0.025) Vw = 5. 266 m3