11 Simple . 1:13 PM faculty.uml.edu 1 of 1 University of CIVE.3100 ENGINEERING M

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11 Simple . 1:13 PM faculty.uml.edu 1 of 1 University of CIVE.3100 ENGINEERING MATERIALS ASSIGNMENT #8: CONCRETE PROPORTION!NG Massachusetts UMASS Lowell Learning with Purpose PROBLEM #1 (40 Points) Concrete is required for an exterior building column in Gloucester, MA. A specified compressive strength of 3,500 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) 2.68; Dry-Rodded Unit weight = 102.5 lb/ft*, Total Moisture = 2.8%; Absorption = 1.2%. Fine Aggregates: Sand-Bulk Specific Gravity (oven-dry) = 2.65; Fineness Modulus (F.M.) = 2.60; Total Moisture = 4.2%; Absorption = 1.2%. Determine the materials required for one cubic yard of concrete using the absolute volume method. Also, determine the amount of materials required for a trail batch.

Explanation / Answer

Step 1. Choice of slump

Value of Slump ( Building Column):

Maximum: 4" & Minimum : 1”

Slump Value = 2”

Step 2. Choice of maximum size of aggregate

Maximum Size of aggregates = 1.5" (given)

Step 3. Estimation of mixing water and air content

The approximate amount of water required for average aggregates is given in Table 10.2.

For Slump value of 2" and aggregate size of 1.5", the amount of water = 275 lb/yd3

Step 4. Selection of water/cement ratio

The required water/cement ratio is determined by strength, durability and finishability.

From Table 10.3: Water-Cement Ratio and Compressive Strength Relationship, the value for 3500 psi and non air entrained concrete is

For Compressive stength of 3000 psi the value is 0.68

For Compressive stength of 4000 psi the value is 0.57

Hence, For Compressive stength of 3500 psi the value is 0.625 (By method of interpolation)

Maximum water cement ratio is limited to 0.50 as per Table 11.5, ACI 211.1-91

Water cement ratio = 0.5

Step 5. Calculation of cement content

Weight of cement = Weight of water /water cement ratio = 275/0.5 = 550 lb/yd3

Step 6. Estimation of coarse aggregate content

The CA content is estimated from Table 10.5: Volume of Coarse Aggregate per Unit Volume for Different Fine aggregate Fineness Moduli

For Fine aggregate moduli = 2.60 and nominal maximum aggregate size 1.5", the value is 0.73

Dry rodded density = 102.5 lb/ft3

Therefore the OD weight of CA = 0.73*26*102.5=1945.45 lb/ft3

Step 7. Estimation of Fine Aggregate Content by Absolute Volume method

Water = 275/62.4 = 4.41 lb/ft3

Cement =550/(3.15*62.4) = 2.80 lb/ft3

CA = 1945.45/(2.68*62.4) = 11.63 lb/ft3

Air = 0 lb/ft3

Therefore, the fine aggregate must occupy a volume = 27- 4.41 - 2.80-11.63 = 8.56 lb/yd3

Step 8. Adjustments for Aggregate Moisture

Fine aggregate

The SSD weight of FA = 11.63*2.65*62.4 = 1923.14 lb

Considering Moisture content of 4.2%

Final value = 1923.14(1+0.042) = 2004 lb/yd3

Coarse aggregate

The SSD weight of CA = 1945.45 lb

Considering Moisture content of 2.8 %

Final value = 1945.45(1+0.028) = 2000 lb/yd3

Adjust the amount of water based on moisture content for FA = 1.2 and CA =1.2

The required mixing water required = 275-(1923.14*0.012) - (1945.45*0.012) = 227.65 lb/yd3

Step 9. Trial Batch Adjustments

Thus the estimated batch weights per yd.3 are:

Water = 227.65 lb.

Cement = 550 lb.

Coarse aggregate (wet) = 2000 lb.

Fine aggregate (wet) = 1945.45 lb.

Total = 4723 lb/yd3 = 174.93 lb/ft3

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