Question

To evaluate the effect of a certain admixture on the flexure strength of concrete, two mixes were prepared, one without admixture and one with admixture. Three beams were prepared of each mix. All the beams had a cross-section of 0.15 m by 0.15 m and a span of 0.45 m. The third-point loading flexure strength test was performed on each beam after 7 days of curing. The loads at the failure of the beams without admixture were 36.1, 38.0, and 34.9 KN, while the loads at the failure of beams with admixture were 43.3, 39.2, 38.5 KN.

Determine:
a. The modulus of rupture of each beam in MPa.
b. The average moduli of rupture of the beams without and with admixture.
c. The percent of the increase of the average modulus of rupture due to adding the admixture.

Answer 1

(a) Modulus of rupture without admixture:

R = 4813.33 MPa, R =5066.67 MPa, R = 4653.33 MPa

Modulus of rupture with admixture:

R =5773.33 MPa, R = 5226.67 MPa, R = 5133.33 MPa

(b) Average modulus of rupture without mixture = 4844.43MPa

Average modulus of rupture with mixture = 5377.78MPa

(c) % of increase = 11%

Step-by-step explanation:

Given Data;

Specimen with =0.15m

Specimen height = 0.15 m

Span length = 0.45 m

The loads of failure without admixture = 36.1 KN, 38.0 KN and 34.9 KN

The loads of failure with admixture = 43.3 KN, 39.2 KN and 38.5 KN.

The flexure strength of a concrete beam is determined by calculating the modulus of rupture using the formula;

R = PL/bd² -------------------------------------------1

Where;

P= Load in KN

L = span length

b =specimen width

d= height of specimen

R = modulus of rupture

Note; The modulus of rupture will be calculated for each of the load with admixture and without admixture.

Calculating Modulus of rupture without admixture:

For load 36.1 KN:

R = PL/bd²

= (36.1 *0.45)/(0.15 *0.15²)

= 16.245/0.003375

R = 4813.33 MPa

For load 38.0 KN:

R = PL/bd²

R = (38.0 *0.45)/(0.15 *0.15²)

R =17.1/0.003375

R =5066.67 MPa

For load 34.9 KN:

R = PL/bd²

R = (34.9 *0.45)/(0.15 *0.15²)

R =15.705/0.003375

R = 4653.33 MPa

Calculating Modulus of rupture with admixture:

For load 43.3 KN:

R = PL/bd²

R = (43.3 *0.45)/(0.15 *0.15²)

R = 19.485/0.003375

R =5773.33 MPa

For load 39.2 KN:

R = PL/bd²

R = (39.2 *0.45)/(0.15 *0.15²)

R =17.64/0.003375

R = 5226.67 MPa

For load 38.5 KN:

R = PL/bd²

R = (38.5 *0.45)/(0.15 *0.15²)

R = 17.325/0.003375

R = 5133.33 MPa

(b)

For rupture of beam without mixture,

Average modulus = sum of rupture without admixture/number of beam

= (4813.33 MPa+ 5066.67 MPa+4653.33 MPa)/3

= 14533.3/3

= 4844.43MPa

For rupture of beam with mixture,

Average modulus = sum of rupture with admixture/number of beam

= (5773.33 MPa +5226.67 MPa +5133.33 MPa)/3

=16133.33/3

= 5377.78MPa

(c)

% of increase =

(average modulus with admixture - average modulus without admixture)/

average modulus without admixture *100

% of increase = (5377.78 - 4844.43)/4844.43 *100

= 533.35/4844.43 * 100

= 0.110 *100

= 11%