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Construction Materials & Management Test 1

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Construction Materials & Management Test 1
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  • Question 1
    1 / -0

    High percentage of C3S and low percentage of C2S in a cement mixture will result in:

    i) Rapid hardening.

    ii) High early strength with high heat generation.

    iii) More resistance to chemical attack.

    The correct answer is:

    Solution

    Concept

    There are four compounds (Called Bogue's Compounds) formed as a result of hydration of cement:

    Alite: C3S, or Tricalcium Silicate

    Belite: C2S, or Dicalcium Silicate

    Aluminate phase: C3A, or Tricalcium Aluminate

    Ferrite phase: C4AF, or Tetracalcium Aluminoferrite

    Bogue Compounds

    Dicalcium Silicate (C2S): This compound will undergo reaction slowly. It is responsible for the progressive strength of concrete. This is also called as Belite. 

    A higher percentage of C2S results in slow hardening, less heat of hydration, and great resistance to chemical attack.

    The heat of hydration is 260 J/Cal.

    Tricalcium silicate (C3S): This is also called as Alite. It undergoes hydration within one week and helps in the development of strength in the early stages of concrete (aka hardening).

    It has the best cementitious property among all the other Bogue's Compounds. Tricalcium Silicate (C3S) hardens rapidly and is largely responsible for the initial set and early strength.

     The cement that has more C­­­­3S content is good for cold weather concreting. The heat of hydration is 500 J/Cal.

    Tricalcium aluminate (C3A): Celite is the quickest one to react when the water is added to the cement. It is responsible for the flash setting. The increase of this content will help in the manufacture of Quick Setting Cement. 

    It provides weak resistance against sulphate attack and contribution to the development of strength is significantly less than above two bogue compounds.

    The heat of hydration is 865 J/Cal.

    Tetra calcium Alumino ferrite (C4AF): This is called as Felite. The heat of hydration is 420 J/Cal. It has the poorest cementing value but it responsible for long term gain of strength of the cement.

     

  • Question 2
    1 / -0
    Creep strains in concrete structure are usually ______.
    Solution

    Concept:

    Under sustained compressive loading, deformation in concrete increases with time even though the stress level remains constant.

    This time-dependent component of strain is called creep strain.

    Creep occurs due to:

    • Internal movement of absorbed water
    • Moisture loss
    • Sliding between gel particles
    • Growth in microcracks.

    Hence creep is caused due to dead load and sustained live load. It depends on the stress level.

    Important Point:-

    Defects of concrete:-

    1. Crazing:- It is a development of fine random cracks on the concrete surface due to difference in shrinkage between surface and interior.

    2. Segregation:- It is a separation of coarse aggregate from fine aggregate, paste from coarse aggregate or water from the mix and hence reducing the strength.

    There are 5 types of shrinkage:-

    a) Chemical shrinkage, b) Plastic shrinkage, c) Drying shrinkage, d) Autogenous shrinkage, and e) Carbonation shrinkage

    3. Shrinkage:- The reduction in the volume of concrete when it changes phase from plastic to solid-state. The shrinkage strain is independent of stresses.

    4. Bleeding:- It is an autogenous flow of mixing water within or emergence to the surface from freshly prepared concrete usually due to excessive vibrations imparted to concrete to achieve full compaction. It leads to formation of pores inside the concrete.
  • Question 3
    1 / -0
    Stainless steel is an alloy of________.
    Solution

    Concept

    Option 1 is correct, i.e. Iron, chromium and nickel.

    • Steel (Carbon steel) is composed of Iron and carbon, which is the main component of stainless steel.
    • Stainless steel differs from carbon steel by the amount of chromium present, which is added to make it resistant to rust.
    • Stainless steel, also known as inox steel or inox is a steel alloy of iron, nickel, and a minimum of 10.5% chromium.
    • Stainless steel is notable for its corrosion resistance, and it is widely used for food handling and cutlery among many other applications.

    Note:

    An alloy is a substance made by melting two or more elements together, at least one of the metal.

    Name of the alloyMade up of
    BrassCopper and Zinc
    BronzeCopper and Tin
    Stainless steelIron, chromium, Nickle, Carbon
    German SilverCopper, Zinc, and Nickle
    Nickel SteelIron and Nickel
  • Question 4
    1 / -0

    The number of cement bags required in making plain cement concrete flooring in an area of size 3 m ×  3 m and 100 mm thick using cement concrete of mix 1: 3: 6 is

    Solution

    Concept

    Weight of one cement bag = 50 kg

    1 m3 of dry cement = 1440 kg

    Concrete is a mixed proportion of Cement: Fine aggregate: Coarse Aggregate

    Calculation

    Given,

    Area of floor = 3 m × 3 m

    Thickness of floor = 100 mm = 0.1 m

    Concrete mix proportion = 1: 3: 6

    Total wet quantity of concrete required = Area × Thickness

    i.e. = (3 m × 3 m) × ( 0.1 m)

    = 0.9 m3

    Total dry Quantity = 1.54 × Wet Quantity

    = 1.54 × 0.9 = 1.386 m3

    \(Quantity\;of\;one\;proportion = \frac{{1.386}}{{1 + 3 + 6}}\)

    = 0.1386 m3

    Total Quantity of cement = 1 × 0.1386 = 0.1386 m3

    = 0.1386 × 1440 = 199.584 kg

    Number of cement bags = 199.58/50

    = 3.99 bags

    ∴ ≈ 4 bags of cement

  • Question 5
    1 / -0

    The fineness modulus of fine aggregate is 2.78 and of coarse aggregate is 7.82 and the desired fineness modulus of mixed aggregate is 6.14. What is the amount of fine aggregate to be mixed with one part of coarse aggregate?

    Solution

    Concept:

    Fineness modulus (F.M)

    It is a numerical index of fineness or grading of aggregate. It gives some idea of the average size of the aggregate. The value of fineness modulus is higher for coarse aggregate.

    \({\rm{F}}.{\rm{M}} = \frac{{{\rm{Cummulative\;percentage\;of\;material\;retained\;on\;each\;sieve}}}}{{100}}\)

    Classification based on Fineness modulus

    Sr no

    Type of aggregate

    Fineness modulus

    1

    Fine aggregate

    2.2 – 2.6

    2

    Medium aggregate

    2.6 – 2.9

    3

    Coarse aggregate

    2.9 – 3.2

    The percentage of fine aggregate (X) to be combined with coarse aggregate is determined by,

    \({\rm{X}} = \frac{{{{\left( {{\rm{FM}}} \right)}_{{\rm{CA}}}} - {{\left( {{\rm{FM}}} \right)}_{{\rm{MA}}}}}}{{{{\left( {{\rm{FM}}} \right)}_{{\rm{MA}}}} - {{\left( {{\rm{FM}}} \right)}_{{\rm{FA}}}}}} \times 100\)

    (FM)CA and (FM)FA = Fineness modulus of coarse aggregate and fine aggregate respectively

    (FM)MA = Fineness modulus of mixed aggregate

    Calculation:

    Given,

    (FM)CA  = 7.82, (FM)FA = 2.78, (FM)MA = 6.14

    \({\rm{X}} = \frac{{{{\left( {{\rm{FM}}} \right)}_{{\rm{CA}}}} - {{\left( {{\rm{FM}}} \right)}_{{\rm{MA}}}}}}{{{{\left( {{\rm{FM}}} \right)}_{{\rm{MA}}}} - {{\left( {{\rm{FM}}} \right)}_{{\rm{FA}}}}}} \times 100\)

    \({\rm{X}} = \frac{{7.82 - 6.14}}{{6.14 - 2.78}} \times 100 = 50{\rm{\% }}\)

    Hence the percentage of fine aggregate (X) to be combined with coarse aggregate is 50%

  • Question 6
    1 / -0

    Consider the following statements:

    1. If more water is added to concrete for increasing its workability, it results into concrete of low strength.

    2. No slump is an indication of a good workable concrete.

    3. Higher the slump of concrete, lower will be its workability.

    4. Workability of concrete is affected by water content as well as water-cement ratio.

    Which of the above statements are correct?

    Solution

    Statement I- True

    Water Cement ratio (W/C) is the ratio of weight of water to weight of cement in a concrete mix. This ratio decides the strength and workability of concrete.  According to Abram’s law, the strength of a concrete mix is inversely related to the weight ratio of water to cement. Lower W/C ratio, higher the strength but lower is the workability.

    Statement II & III- False

    When a standard seized mould filled completely with concrete, it is lifted in vertically upward direction that causes the concrete to subsidise and this subsidence is referred as slump value and it is further used to indicate the workability of concrete. 

    Higher Subsidence ⇒ Higher the Slump value ⇒  Higher will be workability

    Statement-IV – True

    Factors affecting the workability of Concrete are:

    1.  Water Content

    2. Water –Cement Ratio

    3. Size of Aggregates

    4. Admixtures

    5. Texture

    6. Grading of Aggregates

    7. Cement –aggregate ratio

  • Question 7
    1 / -0

    Consider the following particulars in respect of a concrete mix design:

     

    Weight

    Specific Gravity

    Cement

    400 kg/m3

    3.2

    Find aggregates

    -

    2.5

    Coarse aggregates

    1040 kg/m3

    2.6

    Water

    200 kg/m3

    1.0

     

    What shall be the weight of the Fine aggregates?

    Solution

    Concept

    Volume of concrete = Volume of F.A + Volume of C.A + Volume of Cement + Volume of Water + Volume of air

    Where,

    F.A is fine aggregate

    CA is coarse aggregate

    Calculation

    Let volume of Concrete is 1 m3 and assume that there is not air content.

    Composition

    Weight per

    1 m3  Concrete

    Actual weight

    Specific Gravity

    Density = specific  gravity × 1000

    Volume = mass/density

    Cement

    400 kg/m3

    400 kg

    3.2

    3200 kg/m3

    0.125 m3

    Find aggregates(FA)

    -

    ‘x’ kg

    2.5

    2500 kg/m3

    x/2500 m3

    Coarse aggregates(CA)

    1040 kg/m3

    1040 kg

    2.6

    2600 kg/m3

    0.4 m3

    Water

    200 kg/m3

    200 kg

    1.0

    1000 kg/m3

    0.2 m3

    Air

    NIL (assumed as it is  not given)

     

    Now,

    V = VFA + VCA + Vcement + Vwater + Vair

    1 = x/2500 + 0.4 + 0.125 + 0.2 + 0

    Weight of Fine aggregates, x = 687.5 Kg  ≈ 690 kg

  • Question 8
    1 / -0

    Out of the following test methods for Ordinary Portland cement (OPC) and concrete which of them are correctly matched.

    Sr.No.

    Tests

    Properties

    a)

    Vee-Bee Test

    Workability of concrete

    b)

    Blaine air permeability test

    Fineness of OPC

    c)

    Vicats apparatus

    Soundness of OPC

    d)

    Le Chatelier test

    Consistency and Setting time of OPC

    Solution

    Concept:

    1) Vee-Bee consistometer test is used for the concrete which possess low workability, slump for which is limited to 50 mm.

    The measurement of the effort is done by time measurement in seconds. The amount of work measured in seconds is called as the remolding effort. The time required for the complete remolding is a measure of the workability and is expressed in the Vee-Bee seconds.

    2) Air permeability method (Blaine) 

    The fineness of cement is measured as the specific surface. Specific surface is expressed as the total surface area in square meters of all the cement particles in one kilogram of cement. The higher the specific surface is, the finer cement will be. 

    Fineness test is performed on the Blaine apparatus. It is practically a manometer in the U-tube form. One arm of the manometer is provided at the top with a conical socket to form an airtight fit with the conical surface of the cell.

    3) Standard consistency test:

    This test determines the percentage of water required to make workable cement paste.

    Vicat’s apparatus is used to perform this test

    Temperature during test = 27 ± 2°C, Relative humidity = 90%

    As per Vicat’s test ‘the percentage of water added to the cement at which the needle cannot penetrate 5-7 mm from bottom of the mould is called consistency.

    For OPC consistency is around 30%

    In order to make a cement paste of normal consistency the percentage of water varies from 25 to 35%.

    4) Le Chatelier Test:

    This test is used to measure the soundness of OPC due to lime. Lime & Magnesia are two primary compounds responsible for the soundness of cement.

  • Question 9
    1 / -0

    For the following Strength of concrete statements. Identify the correct statements.

    Solution

    Concept

    1) Split Tensile strength

    It is the standard test to determine the tensile strength of concrete indirectly as per IS: 5816-1970

    The magnitude of the tensile stress (fct) is obtained by –

    \({f_{ct}} = \frac{{2P}}{{\pi \times D \times L}}\)

    Where, P = Applied load, D = Diameter of the cylinder, L = Length of the cylinder

    Split tensile strength is about 2/3 of Modulus of Rupture

    Due to the difficulty in applying uni-axial tension to a concrete specimen, the tensile strength is determined by indirect methods.

    2) Strength of concrete

    It is determined by the compressive strength test on a standard 150 mm concrete cube in a compressive testing machine as per IS 516: 1959. The test specimens are generally tested after 28 days of casting and continuous curing.

    In USA standard cylinder of height to diameter ratio of 2 is taken.  (150 mm diameter, 300 mm height) for determining.

    It is observed that the Cube strength of concrete is nearly 1.25 times the cylinder strength.

    3) Influence of Specimen Size on Strength of concrete.

    A standard test cylinder of a concrete specimen of 300 mm × 150 mm diameter is placed horizontally between the loading surfaces of the compression testing machine.

    The compression load is applied diametrically and uniformly along the length of the cylinder until the failure of the cylinder along vertical diameter.

    The loading plates and the top/bottom surface of the concrete specimen offer some frictional resistance called platen restraint which introduces shear stress at the top and bottom surfaces of the specimen and this effect diminish as the distance between the platen surface increases.

    ∴ For this reason, standard concrete cube (height/diameter ratio = 1) specimen gives higher compressive strength then the cylindrical specimen (height/diameter ratio = 2).

    Split tensile strength < modulus of rupture < Cylinder strength < Cube strength

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