SELECTION-AND-COMPUTATION-OF-MATERIALS.pptx

Intended for ABE STUDENTS

1. SELECTION AND COMPUTATION OF MATERIALS ABE 155 – AB STRUCTURES ENGINEERING Prepared by: Engr. CFM

2. CEMENT, SAND, GRAVEL For floor/ floor slabs and sidewalks – (4-inches thick), use CLASS A concrete proportion For wall posts – Use Class B concrete proportion (posts do not have CHB) For reinforced concrete walls, use CLASS A concrete proportion; for walls with only vertical reinforcement, use Class B concrete proportion For footings and foundations – use Class B concrete proportion

3. TAGAYUN’S APPROACH

4. Sample Computation: Floor/sidewalk Determine the number of bags of cement (94 lbs /bag), sand, and gravel of a proposed swine housing solid floor whose width and length are 12 meters x 15 meters. The thickness is 4 in. Use T agayun’s Approach. Given: Width - 12 meters Length - 15 meters Thickness - 4 in. = 0.1016 Plus 10% allowance for wastage – no form Concrete Proportion – 1:2:4 Required: No. of bags of cement, sand, and gravel Solution: Volume of concrete = 12 m x 15 m x  0.1016 m    = 18.288 m 3  18.288 x 0.10 allowance = 1.8288 m 3  18.288 + 1.8288 = 20.1168 m 3  Cement = Vol of concrete  x  7.85 = no. of bags Sand = Vol of concrete  x 0.42 = m 3 Gravel = Vol of concrete  x 0.84 = m 3 Cement = 20.1168 m 3  x  7.85 = 157.92 or 158 bags Sand = Vol of concrete  x 0.42 = 8.45 or 9 m 3 Gravel = Vol of concrete  x 0.84 = 16.90 or 17 m 3

5. SAMPLE COMPUTATION: POSTS AND FOOTINGS A p roposed farmhouse will be designed with 12 posts, each having a dimension of 0.2 meters x 0.2 meters x 2.45-m high.   The square footing depth for each pole is 1.5 meters (0.6 m x 0.6 m).  Determine the number of bags of cement and the volume of sand and gravels needed for the farmhouse. Given: Width – 0.2 meters Length – 0.2 meters Height – 2.45 meters Footing Depth - 1.5 meters Plus 5% allowance for wastage – with form Mixing proportion – Class B (1:2.5:5) Required:  number of bags of cement, the volume of sand and gravel Solution: Volume of footing = 0.6m x 0.6m x 1.5m = 0.54 m 3 x 12 footings = 6.48 m 3 6.48 m 3 x 0.5 allowance = 0.324 m 3 6.46 m 3 + 0.324 m 3 = 6.804 m 3 Volume of posts = 0.2m x 0.2m x 2.45 m = 0.098 m 3 x 12 posts = 1.176 m 3 1.176 m 3 x 0.05 allowance = 0.0588 m 3 1.176 m 3 + 0.0588 m 3 = 1.2348 m 3 Bags of Cement = 8.0388 m 3 x  6.49 = 52.17 or 53 bags of cement Sand = 8.0388 m 3 x 0.44 = 3.54 or 4 m 3 Gravel = 8.0388 m 3 x 0.87 =  6.99 or 7 m 3   Total Volume = 6.804 + 1.2348 = 8.0388 m 3

6. SAMPLE COMPUTATION: CHB A farmstead will be enclosed with a concrete hallow block wall with a height of 2 meters.  The total length of fence will be 1,000 meters.   Using the class B mixture, what is the total number of 6in x 8in x 16in  hollow blocks needed for the wall?   Given:   Structure - concrete hallow block wall Height of wall - 2 meters   Total length of wall - 1000 meters Mixture - class B Hollow block size – 6in x 8in x 16in (T x H x L) Required: Number of hollow blocks Solution: Total surface Area =  2 m x 1000 m =  2000 square meters No. of hollow blocks =  2000 sq m x 12.5 CHB/sq m = 25000 pieces of CHB (using the rule of thumb formula) No of CHB (column) = 2 /0.2032 = 9.8425 column No of CHB (row) = 1000/0.4064 = 2,460.6299 rows Answer: Column x row = 9.8425 x 2,460.6299 = 24,218.75 pcs of CHB or 24, 219 pcs of CHB

7. PLASTER

8. Given the previous problem, what is the volume of mortar and plaster needed for the hollow blocks as well for the wall surface on both sides of the wall.  Also compute the number of bags of cement, lime and sand needed for the construction. Use 1:3 proportion ( cement:sand ) Required: Number of bags of cement Volume of lime and sand Volume of plaster and mortar SOLUTION: Volume of concrete (mortar joint requirement) = 24,219 pieces x 0.0008 m 3 /block = 19.3752 or 20 m 3 (rule of thumb) Vol. of concrete per CHB (per 4 cell) = 24,219 pieces x 0.004 = 116.876 or 117 m 3 Total volume = 19.3725 + 116.876 = 136.2512 or 137 m 3 Volume of Cement mortar (1:3 proportion) = 137 m 3 x 12.04 bags/m 3 = 1,649.48 or 1,650 bags of cement Volume of sand = 137 m 3 x 0.97 = 132.89 or 133 m 3 SAMPLE COMPUTATION: mortar & plaster, cement, LIME and sand Volume of plaster = 24,219 pieces x 0.002 m 3 /pc = 48.438 m 3 (double face) Bags of cement = 48.438 m 3 x 5.5 = 266.409 or 267 bags of cement Bags of Lime = 48.438 m 3 x 5.5 = 266.409 or 267 bags of lime Volume of sand = 48.438 m 3 x 0.90 = 43.59 or 44 m 3

9. REINFORCEMENT BARS

10. American Society for Testing and Materials (ASTM)

11. The common rebar size used for footing: Ø 16mm x 6 m Rebar spacing in footings is 150mm Minimum depth of footings below the surface of undisturbed soil, compacted fill material or controlled low strength material (CLSM) shall be 12 inches or 305 mm) For example: Determine the number of rebars needed for 5 footings with 0.305 m depth and 0.6m x 0.6m cross-sectional area. Formula: 0.6m / 150mm spacing  there will be 8 rebars ( with 0.6 m long) per cross battens x 2 = 16 rebars x 5 footings = 80 rebars 6 meters/0.6 = 10 rebars with 0.6m long will be produced in 6m rebar (80 rebars /10 = 8 pcs rebars are needed. SAMPLE COMPUTATION: rebar for footings or foundations

12. Sample computation: rebar for walls Rebar recommended for concrete wall measures 5-6 inches in depth. The common rebar size used for CHB WALL: A proposed farm machinery shed has one concrete wall (rear) with dimensions of 5m x 9m (height x length). Det ermine the number of rebars (horizontal and vertical) needed for the concrete wall. CHB used is 6” x 8” x 16”. (see next slide for the solution) Ø 10mm x 6 m

13. Given: Height of wall = 5meters Length of wall = 9meters Horizontal rebars will be placed per 2 layers of CHB Rebar size = Ø 10mm x 6 m height of chb = 8 in = 0.2032 m le ngth of chb = 16 in = 0.4064 m Solution: Horizontal Rebar: 5 meters /(0.2032m x 2 layers) = 12.3031 or 13 layers 2 pieces of horizontal rebars per layer x 13 layers = 26 pcs of horizontal rebars Splice allowance = lapping length = 50 time Ø 10mm = 500 mm or 0.5 m 6 meters long (standard size of rebar) – 0.5 m lapping length = 5.5 m x 2 = 11 meters 11 meters x 2 pcs rebars per layer x 13 layers = 286 pcs of rebars with dims of Ø 10mm x 6 m

14. Vertical rebars