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**Q.1 When a specimen of M25 concrete is loaded to a stress level of 12.5 MPa, a strain of 500 x 10 ^{-6} is recorded. If this load is allowed to stand for a long time, the strain increases to 1000 x 10^{-6}. In accordance with the provisions of IS:456-2000, considering the long-term effects, the effective modulus of elasticity of the concrete (in MPa) is_______ . [2019 : 2 Marks, Set-II]**

Solution:

Initial strain = 500 x 10

stress = 12.5 N/mm

Total strain after long time

Effective modulus of elasticity

Target mean strength

Water content required

Say 46%(round off to the lower integer)

(a) 30 mm

(b) 50 mm

(c) 42 mm

(d) 36 mm

Ans.

Nominal cover = Effective cover

Solution:

From clause 26.5.1.1 of IS 456 : 2000, we know minimum reinforcement is given by

∴ Min. % of steel

For Fe500, min. % of steel

If μ is the mean strength of the specimens and o is the standard deviation, the number of specimens (out of 20) with compressive strength less than μ - 3σ is _______ . [2018 : 1 Mark, Set-I]

Solution:

Average strength,

= 3.7

Now, μ - 3σ = 26.575 - 3 x 3.7 = 15.48

Thus, no specimen is having compressive strength less than μ - 3σ.

No. of specimen = 0.

(a) creep

(b) hydration

(c) segregation

(d) shrinkage

Ans.

Creep is inelastic deformation with time due to sustained loading.

(a) 18.42

(b) 21.00

(c) 25.00

(d) 31.58

Ans.

If f

f

[Mean strength = Characteristics strength]

So target mean strength of concrete to be considered in design mix.

= Mean strength

f

(a) x

(b) x

(c) x

(d) x

Ans.

So it depends upon grade of steel only.

Solution:

Long term static modulus of elasticity,

Solution:

The structure may fail when,

(i) the load exceeds the design load

(ii) the strength is less than the characteristic strength

(iii) both the load exceeds the design load and strength is less than the characteristic strength. The probability of load exceeding design load = 5%

(iv) The probability of strength less than characteristic strength = 5%.

The probability of failure is

= 0.05 x 0.95 + 0.05 x 0.95 + 0.05 x 0.05

= 0.0975

Solution:

∴ C = T

(i) Air-entrainment reduces the water demand for a given level of workability.

(ii) Use of air-entrained concrete is required in environments where cyclic freezing and thawing is expected.

Which of the following is TRUE? [2015 : 1 Mark, Set-I]

(a) Both (i) and (ii) are True

(b) Both (i) and (ii) are False

(c) (i) is True and (ii) is False

(d) (i) is False and (ii) is True

Ans.

Solution:

(i) An air-entraining agent introduces air in the form of bubbles that occupy upto 5% of the volume of concrete distributed uniformly throughout the cement paste. Thus, for the same amount of water/cement ratio, we get higher workability.

(ii) Since water takes up the air voids in the concrete and its freezing and melting causes cracks in concrete, hence using air entraiment agents improves the freezing and throwing resistance.

At cross-section XX, which of the following statements is TRUE at the limit state? [2015 : 1 Mark, Set-I]

(a) The variation of stress is linear and that of strain is non-linear

(b) The variation of strain is linear and that of stress is non-linear

(c) The variation of both stress and strain is linear

(d) The variation of both stress and strain is nonlinear

Ans.

(i) As the slump increases, the Vee-bee time increases.

(ii) As the slump increases, the compaction factor increases.

(a) Both (i) and (ii) are True

(b) Both (i) and (ii) are False

(c) (i) is True and (ii) is False

(d) (i) is False and (ii) is True

Ans.

As slump increase, water content increases means workability increases.

Compaction factor

∴ Compaction factor will also increase

The time required for complete remoulding in seconds is considered as a measure of workability and is expressed as the number of Vee-Beeseconds

∴ As workability increases, Vee-Bee time decreases.

(a) 2.0 x 10

(b) 3.0 x 10

(c) 4.0 x 10

(d) 5.0 x 10

Ans.

Solution:

Modulus of elasticity of concrete,

⇒ Curvature = 4 x 10

Solution:

Flexural strength =

Ans.

The learner or less stiller the material, lesser is the slope of stress-strain curve, which is a measure of modulus of elasticity. Hence the diagram is self explanatory.

Solution:

Permissible bearing stress = 0.45 f

= 0.45 x 20 = 9 N/mm

(a) 0.5

(b) 1.0

(c) 1.5

(d) 2.0

Ans.

Compaction factor

Hence, its maximum value for fresh concrete can be 1.

Ans.

Solution:

(a) soft ferrite-pearlite throughout.

(b) hard martensite throughout.

(c) a soft ferrite-pearlite core with a hard martensitic rim.

(d) a hard martensitic core with a soft pearlite- bainitic rim.

Ans.

TMT or thermo mechanicaly treated bars are high strength reinforcement bars having a tough outer core and a soft inner core.

They provide strength is well as enough ductility to the structure.

List-I

A. Resonant frequency test

B. Rebound hammer test

C. Split cylinder test

D. Compacting factor test

List-ll

1. Tensile strength

2. Dynamic modulus of elasticity

3. Workability

4. Compressive strength

Ans.

Solution:

A. Resonant frequency test is used to determine dynamic modulus of elasticity whereas destructive test like cube test etc measures the static modulus of elasticity.

B. Rebound hammer test is a non-destructive test used for compressive strength of concrete.

C. Split cylinder test is used to determine the direct tensile strength of concrete.

D. Compaction factor test is used to measure workability of concrete.

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