Question

In: Civil Engineering

Find the weight (in kg) of cement, water, flyash, fine and coarse aggregate to produce a...

Find the weight (in kg) of cement, water, flyash, fine and coarse aggregate to produce a cement/flyash mix for pumped concrete, using the British Method that has a characteristic 28-day compressive strength of 45 MPa. One hundred laboratory test results on the controlling mix show a standard deviation of 5.3 MPa (k=1.65). The exposure classification for durability purposes is A2 and the conditions require a fly ash/ high early strength cement blend (with a 40% proportion of fly ash). The specific gravity of the crushed coarse 20 mm aggregate to be used is 2850 kg/m3 and its dry rodded density is 1600 kg/m3. The specific gravity of fine aggregate is 2750 kg/m3 and its fineness modulus is 2.5. Assume the high early strength cement density is 3150 kg/m3 and fly ash density to be 2000 kg/m3

Solutions

Expert Solution

1. Target mean strength & standard deviation.

fck' = fck + kS ...... k = constant = 1.65 , Standard deviation = 5.3 MPa

fck' = 45 + 5.3*1.65 = 53.745 MPa. = 54 MPa.

2. water-cement ratio

exposure condition is A2 as per British standard, max. w/c ratio = 0.5

3. selection off water content

for crushed 20mm aggregate & concrete of slump 25-50mm water content = 186kg/m3

For pumped concrete slump of 75-100 mm

therefore,

add extra 3% water for evvery 25mm increse in slump, 2 * 3 % = 6 %

water content = 1.06 * 186 = 197.16 kg/m3.

4. cement conternt = 197.16 / 0.5 = 394.32 kg/m3 of concrete ( less than minimum required of 400kg/m3 for M55) therefore 400kg/m3

flyash = 400 * 0.4 = 160 kg/m3

cement = 400-160 = 240 kg/m3

w/c = 197.15/400 = 0.49....ok

5. CA & FA

for w/c ratio of 0.5, percentage of CA of 20mm is 0.64 of total aggregaates.

VCA = 0.64 * total aggregates

VFA = 0.36 * total aggregates

The volume of total aggregate = 1 - { [ ( C/Sc) + (F ash/ Sflyash) + (W/1) ] / 1000 }

= 1 - { [ ( 240/3.15) + (160/ 2) + (197.16/1) ] / 1000 }

= 0.6466 m3

VCA = 0.64* 0.65 = 0.415 m3

VFA = 0.36 *0.65 = 0.2335 m3

MCA = 0.415 m3 * 2850 kg/m3 = 1,182.75 kg

MFA = 0.2335 * 2750 = 642.125‬ kg

Therefore,

water = 197.16 kg/m3

cement = 240 kg/m3

flyash = 160 kg/m3

CA = 1182.75 kg/m3

FA = 642.125 kg/m3


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