Question

Question 1. A concrete mix is to be designed for use in a 250 mm thick road pavement. The desired properties are provided below. Use the ACI method to calculate the required amounts per unit volume (/1m) of water, cement, course aggregate and fine aggregate. Slump 25 mm Average 28-day flexural strength of at least 4.5 MPa Coarse aggregate: nominal maximum size = 37.5 mm, dry-rodded weight 1600 kg/m3, specific gravity 2.68, moisture content 1.0 percent, absorption 0.5 percent Fine aggregate: fineness modulus = 2.80, specific gravity 2.64, moisture content percent, absorption = 0.7 percent 5 Air content: 4.5-6.5 percent Maximum allowable water-cement ratio = 0.44 Minimum cement content 335 kg Density of water 1000 kg/m3 Specific gravity of portland cement 3.15 (typically assumed value) Slump, specified as 25 mm. This is typical for concrete pavement.

Answer 1

Ans) According to ACI ,for air content concrete,

Water required per m³ for maximum aggregate size of 37.5mm and slump of 25mm = 150 kg/m³

For 28 days strength of 4.5Mpa, water cement ratio as per ACI = 0.48

But according to question, maximum allowable water cement ratio = 0.44

=> Weight of cement per m³ = 150 / 0.44 = 340.90 kg/m³ > 335 kg (Hence OK)

Now , as per ACI, for fineness modulus of 2.80 and aggregate size of 37.5mm, volume of coarse aggregate = 0.71 m³

Density of Coarse Aggregate = 1600 kg/m³

Weight of 0.71m³ aggregate = 1600 x 0.71 = 1136 kg

Hence, coarse aggregate required = 1136 kg/m³

Now, volume of concrete = 1m³

Volume of Fine aggregate = Total Volume - volume of coarse aggregate + cement + water + air

= 1 - (1136/2680) - (340.90 / 3150) + (150/1000) + 0.045

= 1 - 0.727

= 0.272 m³

=>Amount of fine aggregate required = Specific gravity x density of water x volume

=   718.08 kg /m³

Since, coarse aggregate and fine aggregate absorb moisture,

Therefore water absorbed = (0.50/100 x 1136) + (0.70/100 x 718.08 )

= 10.70 kg

Water present in aggregates = (0.01 x 1136) + (0.05 x 718.08)

= 46.56 kg

=> Net water provided by aggregate = 46.56 - 10.7 = 36.56 kg

Therefore actual water to be added = 150 - 36.56 = 113.43 kg

Actual weight of cement = 113.43 / 0.44 = 257.80 kg

According to question, minimum quantity of cement = 335 kg so provide cement as 335 kg/m³

So, final quantities are as required for 1m³ :

Water = 113.43 kg

Cement = 335 kg

Coarse aggregate = 1136 kg

Fine aggregate = 718.08 kg