14.310 ENGINEERING MATERIALS Assignment #8: Concrete Mix Design PROBLEM #1: GIVEN: Concrete is required for a pavement slab that will be exposed to severe freeze-thaw climate as well as deicing chemicals. A specified compressive strength of 3,000 psi is required at 28 days (use Type I cement with a specific gravity of 3.15). Coarse Aggregates: 1.5 Nominal Maximum Size; Bulk Specific Gravity (oven-dry) = 2.67; Dry-Rodded Unit Weight = 103 lb/ft 3 ; Total Moisture = 2.1%; Absorption = 1.6%. Fine Aggregates: Sand - Bulk Specific Gravity (oven-dry) = 2.65; Fineness Modulus (F.M.) = 2.80; Total Moisture = 4.8%; Absorption = 1.4%. REQUIRED: Determine the materials required for one cubic yard of concrete using the absolute volume method. SOLUTION: Step 1: Determine Strength (f cr ) No statistical data available. Using Table 12-10 (PCA Design and Control of Concrete Mixtures): f cr = f c + 1,200 psi Therefore f cr = 4,200 psi. Step 2: Determine w/c Ratio Using Table 12-1 (PCA Design and Control of Concrete Mixtures), you have Exposure Category F3 (concrete exposed to freezing and thawing with deicers). Therefore, Maximum w/c = 0.45, minimum f c = 4,500 psi. You should increase f c (in other words, you should not design the pavement for the specified 3,000 psi compressive strength). You will need to revise f cr. Revised f cr = f c + 1,200 psi (Table 12-10) = 4,500 psi + 1,200 psi = 5,700 psi. Therefore f cr = 5,700 psi. Using Table 12-3 (PCA Design and Control of Concrete Mixtures): w/c = 0.34 Use this w/c instead of maximum value from Table 12-1. 14.310 2015 Assignment 8 Solution Page 1 of 8
Step 3: Determine Coarse Aggregate Size: This is given (1½ inches). There is no additional data provided to check if this is acceptable (i.e. no reinforcing steel spacing or pavement thickness information is provided). Step 4: Determine Air Content: Using Table 12-5 (PCA Design and Control of Concrete Mixtures), for 1 ½ inch coarse aggregate and severe exposure (F3), use an air content of 5.5%. Add an additional 1% to cover -1% to + 2% range (see class example). Therefore, use air content = 6.5% (although 5.5% is also acceptable no calculations are provided for this air content). Using 6.5% air content, the volume of air = 0.065 x (27ft 3 ) = 1.76 ft 3. Step 5: Determine Slump: For pavements placed by mechanical pavers, ranges between 1-3 inches. The higher the, the more workability of the concrete, which is typically desired. So therefore, select a = 3 inches (although 1-2 inches is also acceptable). Step 5: Determine Water Content: Using Table 12-5 (PCA Design and Control of Concrete Mixtures) and a of 3 inches, air entrained concrete, and 1½ inch aggregate, water = 275 lb/yd 3. This takes up a volume of 275 lb/(62.4 lb/ft 3 ) = 4.41 ft 3. If you were using a 1-2 inches, water = 250 lb/yd 3. This takes up a volume of 250 lb/(62.4 lb/ft 3 ) = 4.01 ft 3. As discussed in class, you do not need to reduce water content for subangular aggregate or crushed particles. Step 6: Determine Cement Content: For a w/c = 0.34 (see Step 2), 3 inch, and 275 lb/yd 3 of water, you will need 809 lb/yd 3 of cement. This takes up a volume of 809 lb/(3.15 x 62.4 lb/ft 3 ) = 4.12 ft 3. For a w/c = 0.34 (see Step 2), 1-2 inch, and 250 lb/yd 3 of water, you will need 735 lb/yd 3 of cement. This takes up a volume of 735 lb/(3.15 x 62.4 lb/ft 3 ) = 3.74 ft 3. Step 6: Determine Coarse Aggregate Content: Using Table 12-4 (PCA Design and Control of Concrete Mixtures), 1½ inch aggregate, and a FM = 2.8, the bulk volume of coarse aggregate per unit volume of concrete = 0.71. Therefore, the weight of the course aggregate = 0.71 x (27ft 3 /yd 3 ) x (103 lb/ft 3 ) 14.310 2015 Assignment 8 Solution Page 2 of 8
= 1,975 lb/yd 3. NOTE: This is for the same for both selections. This takes up a volume of 1,975 lb/(2.67 x 62.4 lb/ft 3 ) = 11.85 ft 3. Step 6: Determine Fine Aggregate Content: A summary of the volumes of Air, Water, Cement, and Coarse Aggregates from Steps 1-5 is presented in Table A. The fine aggregates = 27 ft 3 (the sum of air, water, cement, and coarse aggregate volumes). Table A. Summary of Steps 1-5. Mix Part Selection of 1-2 inch Selection of 3 inch Air 1.76 1.76 Water 4.01 4.41 Cement 3.74 4.12 Coarse Aggregate (C.A.) 11.85 11.85 Fine Aggregate (F.A.) 5.64 4.86 The weight of the fine aggregate = (Volume)x(Specific Gravity)x(Unit Weight of Water). For 3 inch : Weight (Fine Aggregate) = (4.86 ft 3 )(2.65)(62.4lb/ft 3 ) = 803 lb. For 1-2 inch : Weight (Fine Aggregate) = (5.64 ft 3 )(2.65)(62.4lb/ft 3 ) = 933 lb. Step 7: Apply Moisture Correction: For 3 inch : Weight of Wet C.A. required for 1 yd 3 of concrete = 1975 (1.021) = 2016 lb Weight of Wet F.A. required for 1 yd 3 of concrete = 803 (1.048) = 842 lb Free Moisture in C.A. = (0.021 0.016) 1975 = 10 lb Free Moisture in F.A. = (0.048 0.014) 842 = 29 lb Water contributed by the aggregates = 39 lb Actual amount of water to be added = 275 lb 39 lb = 236 lb 14.310 2015 Assignment 8 Solution Page 3 of 8
For 1-2 inch : Weight of Wet C.A. required for 1 yd 3 of concrete = 1975 (1.021) = 2016 lb Weight of Wet F.A. required for 1 yd 3 of concrete = 933 (1.048) = 978 lb Free Moisture in C.A. = (0.021 0.016) 1975 = 10 lb Free Moisture in F.A. = (0.048 0.014) 978 = 33 lb Water contributed by the aggregates = 43 lb Actual amount of water to be added = 250 lb 43 lb = 207 lb A summary of the final batch weight corrected for moisture is presented in Table B. Table B. Summary of Final Batch Weight corrected for Moisture. Mix Part Selection of 1-2 inch Selection of 3 inch Air 6.5% 6.5% Water 207 lb 236 lb Cement 735 lb 809 lb Coarse Aggregate (C.A.) 2016 lb 2016 lb PROBLEM #2: Fine Aggregate (F.A.) 978 lb 842 lb GIVEN: The project will be in Columbia, SC and you will be using the concrete for a bridge abutment. You have the following coarse and fine aggregates available: Coarse Aggregates: ¾ inch Nominal Maximum Size (ASTM C33) gravel, Bulk Specific Gravity (oven-dry) = 2.69; Absorption = 0.5% (SSD Saturated Surface Dry), and an oven-dry rodded bulk density (unit weight) = 102 lb/ft 3 ; Lab trail batching moisture content = 2.30%. Fine Aggregates: Sand - Bulk Specific Gravity (oven-dry) = 2.66; Fineness Modulus (F.M.) = 2.70; Total Moisture = 4.1%; Absorption = 1.0%. 14.310 2015 Assignment 8 Solution Page 4 of 8
REQUIRED: Determine the materials required for one cubic yard of concrete for SCDOT Class 4000 structural concrete using the absolute volume method. SOLUTION: Read the Specifications! SCDOT Standard Specifications, Table from Section 701.2.12.2: SCDOT Standard Specifications, Section 701.2.5 (Air Entrained Concrete): Unless otherwise specified, use a design mix for air-entrained concrete based on 4.5% (± 1.5%) entrained air, except for prestressed concrete. If the concrete is pumped, then the entrained air will be acceptable at 5.5% (± 1.5%) measured at the truck. Now use the PCA Design and Control of Concrete Mixtures Book. Step 1: Determine Strength (f cr ) For SCDOT Class 4000 structural concrete: f c = 4,000 psi No statistical data available. Using Table 12-10 (PCA Design and Control of Concrete Mixtures): f cr = f c + 1,200 psi Therefore f cr = 5,200 psi. 14.310 2015 Assignment 8 Solution Page 5 of 8
Step 2: Determine w/c Ratio Using Table 12-1 (PCA Design and Control of Concrete Mixtures), you have Exposure Category F0 (does it freeze often in Columbia, SC?). Therefore, Maximum w/c = NA, minimum f c = NA. You don t have minimums for this exposure level. Using Table 12-3 (PCA Design and Control of Concrete Mixtures): w/c = 0.38. Step 3: Determine Coarse Aggregate Size: This is given (¾ inches). There is no additional data provided to check if this is acceptable (i.e. no reinforcing steel spacing or pavement thickness information is provided). Step 4: Determine Air Content: As previously noted, SCDOT Standard Specifications, Section 701.2.5 (Air Entrained Concrete) states to use air content of 5.5% (± 1.5%). Use 6.5% (5.5% + 1%) for batching (+1% same as previous and in class example). Using 6.5% air content, the volume of air = 0.065 x (27ft 3 ) = 1.76 ft 3. Step 5: Determine Slump: SCDOT Standard Specifications, Section 701.4.6 (Slump) notes a maximum of 4 inches for all concrete except Class 2500 or unless otherwise specified. Use a = 4 inches (although 3 inches is acceptable). Step 5: Determine Water Content: Using Table 12-5 (PCA Design and Control of Concrete Mixtures) and a of 4 inches, air entrained concrete, and ¾ inch aggregate, water = 305 lb/yd 3. This takes up a volume of 305 lb/(62.4 lb/ft 3 ) = 4.89 ft 3. As discussed in class, you do not need to reduce water content for subangular aggregate or crushed particles. Step 6: Determine Cement Content: For a w/c = 0.38 (see Step 2), 4 inch, and 305 lb/yd 3 of water, you will need 803 lb/yd 3 of cement. Select Type I cement (no reason to select another Type). Use a specific gravity of 3.15 for Type I (See Problem #1). This takes up a volume of 803 lb/(3.15 x 62.4 lb/ft 3 ) = 4.09 ft. Step 6: Determine Coarse Aggregate Content: Using Table 12-4 (PCA Design and Control of Concrete Mixtures), ¾ inch aggregate, and a FM = 2.7, the bulk volume of coarse aggregate per unit volume of concrete = 14.310 2015 Assignment 8 Solution Page 6 of 8
0.63. Therefore, the weight of the course aggregate = 0.63 x (27ft 3 /yd 3 ) x (102 lb/ft 3 ) = 1,735 lb/yd 3. This takes up a volume of 1,735 lb/(2.69 x 62.4 lb/ft 3 ) = 10.34 ft 3. Step 6: Determine Fine Aggregate Content: A summary of the volumes of Air, Water, Cement, and Coarse Aggregates from Steps 1-5 is presented in Table C. The fine aggregates = 27 ft 3 (the sum of air, water, cement, and coarse aggregate volumes). Table C. Summary of Steps 1-5. Mix Part Selection of 4 inch Air 1.76 Water 4.89 Cement 4.09 Coarse Aggregate (C.A.) 10.34 Fine Aggregate (F.A.) 5.92 The weight of the fine aggregate = (Volume)x(Specific Gravity)x(Unit Weight of Water). For 3 inch : Weight (Fine Aggregate) = (5.92 ft 3 )(2.66)(62.4lb/ft 3 ) = 803 lb. Step 7: Apply Moisture Correction: Weight of Wet C.A. required for 1 yd 3 of concrete = 1735 (1.023) = 1775 lb Weight of Wet F.A. required for 1 yd 3 of concrete = 979 (1.041) = 1019 lb Free Moisture in C.A. = (0.023 0.005) 1735 = 31 lb Free Moisture in F.A. = (0.041 0.010) 979= 30 lb Water contributed by the aggregates = 61 lb Actual amount of water to be added = 305 61 = 244 lb A summary of the batch weight corrected for moisture is presented in Table D. 14.310 2015 Assignment 8 Solution Page 7 of 8
Table D. Summary of Batch Weight corrected for Moisture. Does it meet specifications? Mix Part Selection of 4 inch Air 6.5% Water Cement Coarse Aggregate (C.A.) Fine Aggregate (F.A.) Cement Content: 803 lbs/yd 3 > 611 lb/yd 3 (minimum) CHECKS! Water to Cement Ratio: w/c ratio = 0.38 < 0.40 (maximum) CHECKS! 244 lb 803 lb 1775 lb 1019 lb Percent Fine to Coarse Aggegate: SCDOT Specification Section 701.2.12.1 Base the sand/stone ratio on volume. 36% Fine: 64% Coarse 34% Fine: 66% Coarse DOES NOT CHECK Notes SCDOT Specifications Section 701.2.12.1 states that you can vary this ratio to obtain good workability. Try a trial batch BEFORE adjusting! 14.310 2015 Assignment 8 Solution Page 8 of 8