Question

Using the following data for an unreinforced PCC Pavement slab: ■ Design strength f′ c = 28 MPa ■ Slab thickness = 300 mm ■ Standard deviation of f′ c obtained from 20 samples = 1.4 MPa ■ Ignore any exposure requirement ■ Use air-entrained concrete ■ Fineness modulus of fine aggregate = 2.60 ■ Maximum aggregate size = 50 mm and nominal maximum aggregate size = 37.5 mm ■ Bulk oven-dry specific gravity of coarse aggregate = 2.6 ■ Oven-dry rodded density of coarse aggregate = 2002 kg/m3 Find the following: a. Required compressive strength b. w/c ratio c. Coarse aggregate amount (kg/m3 ) d. If the w/c ratio is 10% reduced, will the quantity of coarse aggregate increase, decrease or remain the same? Explain your answer.

Answer 1

Question:

Using the following data for an unreinforced PCC Pavement slab:

■ Design strength f′ c = 28 MPa

■ Slab thickness = 300 mm

■ Standard deviation of f′ c obtained from 20 samples = 1.4 MPa

■ Ignore any exposure requirement

■ Use air-entrained concrete

■ Fineness modulus of fine aggregate = 2.60

■ Maximum aggregate size = 50 mm and nominal maximum aggregate size = 37.5 mm

■ Bulk oven-dry specific gravity of coarse aggregate = 2.6

■ Oven-dry rodded density of coarse aggregate = 2002 kg/m3

Find the following: a. Required compressive strength b. w/c ratio c. Coarse aggregate amount (kg/m3 ) d. If the w/c ratio is 10% reduced, will the quantity of coarse aggregate increase, decrease or remain the same? Explain your answer.

Answer:

The following are the given data:

f′ c = 28 MPa

Slab thickness = 300 mm

Standard deviation of f′ c obtained from 20 samples = 1.4 MPa

Ignore any exposure requirement

Use air-entrained concrete

Fineness modulus of fine aggregate = 2.60

Maximum aggregate size = 50 mm and nominal maximum aggregate size = 37.5 mm

Bulk oven-dry specific gravity of coarse aggregate = 2.6

Oven-dry rodded density of coarse aggregate = 2002 kg/m3

a. Required compressive strength :

f_c = Max of { (f′ c + 1.34 s) or ( f′ c + 2.33 s -3.45)}

s= Modification factor * given standard deviation

Modification facor = 1.08 ( for 20 sample)

f_c = Max of { ( 28+ (1.34*1.08*1.4) ) or ( 28+(2.33*1.08*1.4) - 3.45)}

f_c = Max of { (30.026 MPa ) or ( 28.0726 MPa) }

Hence f_c is 30.026 MPa

b. w/c ratio:

For AIr Entrained Concrete and Compressive strength of 30.026 MPa, w/c ratio is found by using the following two figures

Hence interpolating from the above table :

w/c ratio = 0.48 - { [(0.48 - 0.4) * (30.026 - 28)] / (35-28) }

= 0.48 - 0.02315 = 0.456

Hence, w/c ratio = 0.456 ( Air Entrained concrete)

c. Coarse aggregate amount (kg/m3 ):

Slab thickness = 300 mm

max. aggregate size < [(1/3) * (Slab thickness)]

50 mm < [ (1/3) * (300)]

50 mm < 100 mm

Therefore, the aggregate size is OK.

Given: Nominal Max Size of aggregate = 37.5 mm

Fineness modulus of Fine Aggregate = 2.6

using the following figure the Coarse aggregate factor is determined

From the above table, the Coarse aggregate factor = 0.73,

Oven dry weight of Coarse aggregate is found as follows,

Coarse aggregate weight = (0.73)* (2002) = 1461.46 kg/m³

Therefore, the Coarse aggregate amount = 1461.46 kg/m³

d. If the w/c ratio is 10% reduced, will the quantity of coarse aggregate increase, decrease or remain the same? Explain your answer.

The quantity of coarse aggregate will remain the same because it is not affected by the w/c ratio.