Month: November 2020

Ceramic Materials – MCQ || Construction Materials

Ceramic Materials – MCQ

1. The word ceramic comes from which word?

a) Keramos
b) Kamos
c) Ceramos
d) None of these

View Answer

a) Keramos

2. Which of the following is not a ceramic material

a) Glass
b) Clay
c) Lime
d) Abrasives

View Answer

c) Lime

3. Which of the following properties ceramics do not possess

a) Hardness
b) Brittleness
c) Elasticity at low temperature
d) Malleability

View Answer

d) Malleability

4. Which of the following accelerator added for the fusion of glass

a) Nickel
b) Silica
c) Soda
d) Lime

View Answer

c) Soda

5. Which of the following compounds are used in ceramics to improve the shock resistance?

a) Lithium
b) Silicon nitride
c) Iron
d) Only (a) and (b)

View Answer

d) Only (a) and (b)

6. Ceramic materials have

a) Low electric conductivity
b) High electric conductivity
c) Very high electric conductivity
d) None

View Answer

a) Low electric conductivity

7. Carbonaceos refractories are manufactured from

a) Chromous iron
b) Chromous sulfate
c) Graphite
d) None of these

View Answer

c) Graphite

8. Which of the following statement is true

a) Refractories are able to withstand high temperature.
b) Refractories are able to resist thermal shocks.
c) Expansion and contraction of these materials are minimum.
d) It has low electric conductivity.
e) All of these.

View Answer

e) All of these.

9. Which of the following components make the glass infusible and fire-resistant

a) Soda
b) Lead oxide
c) Potash
d) Cullets

View Answer

c) Potash

10. Lead glass is also know as

a) Soft glass
b) Flint glass
c) Soda-ash glass
d) Boro glass

View Answer

b) Flint glass

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Structural Clay Products – MCQ

Ferrous Metals – MCQ

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Ferrous Metals – MCQ || Building Materials ||

Ferrous Metals – MCQ

1. The ratio of Young’s modulus of high tensile steel to that of mild steel is about

a) 0.5
b) 1.0
c) 1.5
d) 2.0

View Answer

b) 1.0

2. In mild steel, the iron content is about

a) 50 %
b) 80 %
c) 90 %
d) 99 %

View Answer

d) 99 %

3. The ultimate tensile strength of high carbon steel in N/mm2 may be as high as

a) 600
b) 1000
c) 1500
d) 2000

View Answer

d) 2000

4. Which of the following type of iron is capable of dissolving carbon and non-magnetic

a) Alpha iron
b) Beta iron
c) Gamma iron
d) Delta iron

View Answer

c) Gamma iron

5. High-strength deformed steel bars with a yield stress of 500 Mpa should have a minimum elongation of

a) 6 %
b) 8 %
c) 13.5 %
d) 20 %

View Answer

a) 6 %

6. Which of the following is the crudest form of iron?

a) Dead mild steel
b) Wrought iron
c) Cast iron
d) Pig iron

View Answer

d) Pig iron

7. Which of the following kind of steel is used in case of manufacturing of rails?

a) Cast steel.
b) Bessemer steel.
c) Manganese steel.
d) Mild steel.

View Answer

c) Manganese steel.

8. Molybdenum steel is normally used in the manufacture of

a) Gears
b) Delicated instruments
c) Magnets
d) Point and crossing

View Answer

a) Gears

9. Tungsten steel which is generally used for the manufacturing of

a) drilling machines.
b) Heavy earth equipment.
c) Heavy mining equipment.
d) Delicate instrument.

View Answer

a) drilling machines.

10. Which of the following type of iron is non-magnetic and absorbed very little carbon

a) Alpha iron
b) Beta iron
c) Gamma iron
d) Delta iron

View Answer

d) Delta iron

11. Which of the following type of iron is strongly magnetic

a) Alpha iron
b) Beta iron
c) Gamma iron
d) Delta iron

View Answer

b) Beta iron

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Principal Properties of Building Materials – MCQ

Timber – MCQ || Building Materials ||

Timber – MCQ

1. How many months are required for natural seasoning of timber

a) 1 to 3 months.
b) 4 to 6 months.
c) 8 to 11 months.
d) 12 to 14 months.

View Answer

b) 4 to 6 months.

2. The drawback of electric seasoning of timber is

a) Checks.
b) Splitting.
c) Cracks.
d) Reduced strength.

View Answer

b) Splitting.

3. Lumber

a) Implies a living tree.
b) is a part of felled tree.
c) is log of timber sawn into pieces of desired shape.
d) is used to denote standing timber.

View Answer

c) is log of timber sawn into pieces of desired shape.

4. The kiln seasoning of timber causes

a) Increase in strength.
b) Decrease in shear strength.
c) Case hardening.
d) No internal stress.

View Answer

c) Case hardening.

5. As per the I.S code, the weight of the timber is to be considered at a moisture content of

a) 12 %
b) 6 %
c) 2 %
d) zero

View Answer

a) 12 %

6. Seasoning of timber results in

a) Increase in the tendency to decay.
b) Increase in the tendency to split.
c) Increase in the tendency to warp.
d) The moisture present in timber approximately same as that of surroundings.

View Answer

d) The moisture present in timber approximately same as that of surroundings.

7. As a natural material timber is an

a) Isotropic material.
b) Heterogeneous material.
c) Homogeneous material.
d) None of these.

View Answer

a) Isotropic material.

8. Timber can be made fire resistant by

a) Kiln seasoning.
b) Sir Abel’s process.
c) Charging.
d) Mc. Neill’s process.

View Answer

b) Sir Abel’s process.

9. The defect in timber causes by fungus is

a) Checks
b) Foxiness
c) Dry rot
d) Wane

View Answer

c) Dry rot

10. By reducing the moisture content of timber above fibre saturation points which one of the following is not correct

a) Tensile strength is increased
b) Compressive strength is increased
c) Stiffness is increased
d) Shear strength is increased

View Answer

a) Tensile strength is increased

11. To obtain strong timber pieces from logs against shrinkage and swelling the best method is

a) Machine sawing
b) Tangential sawing
c) Quarter sawing
d) Radial sawing

View Answer

b) Tangential sawing

12. Fireproofing of timber

a) Makes it difficult to ignite and support its own combustion.
b) Does not allow the fire to come closer to the wood.
c) Extinguishes the fire and dissipate the heat generated.
d) All of these.

View Answer

a) Makes it difficult to ignite and support its own combustion

Read More:

Characteristics of Good Timber

Stone – MCQ

Precautions in Timbering

Reasons To Use Lumber Takeoff Services

Structural Clay Products – MCQ || Building Materials

Structural Clay Products – MCQ

1. The Compressive strength of burnt clay bricks as per IS 1077 is

a) 75 kg/cm2
b) 100 kg/cm2
c) 75 – 100 kg/cm2
d) 35 – 350 kg/cm2

View Answer

d) 35 – 350 kg/cm2

2. Glazing of clay products is achieved by throwing sodium chloride in kiln at a temperature of

a) 600° – 800° C
b) 700° – 1000° C
c) 900° – 1100° C
d) 1200° – 1300° C

View Answer

d) 1200° – 1300° C

3. Plasticity to mould clay bricks in proper shape is achieved by

a) Alumina
b) Lime
c) Alkalis
d) Ferric oxide

View Answer

a) Alumina

4. The raw bricks shrink during drying and warp during burning because of

a) Less silica in brick earth
b) Less silica and excess magnesia in brick earth
c) Excess of alumina and silica in brick earth
d) Less lime and excess silica in brick earth

View Answer

c) Excess of alumina and silica in brick earth.

5. Which of the following in the clay used for making bricks decomposes the brick on burning?

a) Iron oxide
b) Iron pyrite
c) Alkali
d) Magnesia

View Answer

b) Iron pyrite

6. When carbonaceous materials in the form of bituminous matter of carbon are presented in the clay, the brick will

a) be spongy
b) have black core
c) be porus
d) have cracks

View Answer

b) have black core

7. The defect which is seen in the clay products due to imprisoned air during their moulding is called

a) Spots
b) Lamination
c) Blister
d) Cracks

View Answer

b) Lamination

8. Considering the following three stages in the manufacturing of clay bricks:

i. Weathering
ii. Moulding
iii. Tempering

The correct sequence of these stages in the manufacturing of the bricks is

a) (i), (ii), (iii)
b) (ii), (iii), (i)
c) (iii), (ii), (i)
d) (i), (iii), (ii)

View Answer

d) (i), (iii), (ii)

9. Pick up the correct statement with regards to clay brick

a) Efflorescence is due to the soluble salts present in the clay
b) High porosity of bricks if, it is there, will make stress distribution uniform.
c) Excessive burning of bricks causes efflorescence
d) Magnesia in brick earth causes brick to warp

View Answer

a) Efflorescence is due to the soluble salts present in the clay

10. In the steel industry the brick used for lining furnaces should be

a) Acid refractory
b) Basic refractory
c) Neutral refractory
d) Heavy duty

View Answer

a) Acid refractory

Read More:

Principal Properties of Building Materials – MCQ

Structures in Clay Minerals – MCQ

Ceramic Materials – MCQ

Different Types of Bricks Used in Construction

Principal Properties of Building Materials – MCQ

Principal Properties of Building Materials – MCQ

1. Low melting point materials are those which can withstand maximum temperatures up to about

a) 34° C
b) 690° C
c) 1350° C
d) 1550° C

View Answer

c) 1350° C.

2. Density of wood (in g/cm3) ranges between

a) 0.5 – 0.95
b) 1.5 to 1.6
c) 2.5 to 2.8
d) 2.9 to 3.1

View Answer

b) 1.5 to 1.6

3. Which of the following is an example of brittle material

a) Copper
b) Zinc
c) Cast iron
d) Wrought iron

View Answer

c) Cast iron

4. Which of the following is an example of plastic materials

a) Cast iron
b) Zinc
c) Brick
d) Concrete

View Answer

b) Zinc

5. High melting point materials are those which can withstand maximum temperatures up to about

a) 34° C
b) 690° C
c) 1350° C
d) 1550° C, or More.

View Answer

d) 1550° C, or More.

6. For materials subjected to combined bending and torsion, the best theory to estimate breaking load will be

a) Maximum shear stress theory.
b) Strain energy theory.
c) Shear strain energy theory.
d) Maximum stress theory.

View Answer

a) Maximum shear stress theory.

7. What will be the shape of the failure surface of a standard cast iron specimen subjected to torque?

a) Cup and cone.
b) Plane surface perpendicular to the axis of the specimen.
c) Helicoid surface at 45 to the axis of the specimen.
d) None of these

View Answer

c) Helicoid surface at 45 to the axis of the specimen.

8. Consider following for a material under compressive load

⇰ It can be crushed to powder.
⇰ It crumbles to pieces.
⇰ It fails under definite angles.

The material can be

a) Ductile
b) Brittle
c) Viscous
d) Plastic

View Answer

b) Brittle

9. Consider the following stresses with regard to beams.

i. Bending tensile stress.
ii. Bending compressive stress.
iii. Shear stress.

Of the above, a cross-section of the beam will be subjected to

a) (i) and (ii) only
b) (ii) and (iii) only
c) (iii) and (i) only
d) All (i), (ii) and (iii).

View Answer

d) All (i), (ii) and (iii).

10. For a material tested under compression, the plane of rupture was at 58° with horizontal. The best shape of the specimen would be.

a) Cube.
b) of figure 8.
c) Cylinder with height equal to the diameter.
d) Prism with height two times the diameter.

View Answer

d) Prism with height two times the diameter.

11. For coarse aggregate to be tested for their use in making pavements, the most useful information can be gathered from.

a) Shearing test.
b) Tension test.
c) Compression test.
d) Abrasion test.

View Answer

d) Abrasion test.

12. The modulus of elasticity of a material is 208 GPa and its Poisson’s ratio is 0.3. What is the value of shear modulus?

a) 74 GPa
b) 80 GPa
c) 100 GPa
d) 128.5 Gpa

View Answer

b) 80 GPa.

The relation between Shear modulus(G), Elastic modulus(E) and Poisson’s ratio(μ) is

G = E / 2(1+μ)

G = 208 / 2(1+0.3) = 208/2.6 = 80 Gpa.

Read More:

Stone – MCQ

Rocks – MCQ

What is Cement?

Qualities of Good Building Stone

Ultimate Compressive Load for Axially Loaded Short Column

Ultimate Compressive Load for Axially Loaded Short Column

The ultimate compressive load for axially loaded short column is determined by the following assumption

a) The maximum compressive strain in concrete is 0.002
b) Strain in concrete is equal to strain in steel
c) Stress-strain relation for steel is the same in compression or tension

For an absolutely axially loaded short column, at the ultimate stage, the ultimate compressive load is resisted partly by concrete and partly by steel. Thus, at the ultimate stage,

Ultimate load = Pu = Puc + Pus

Where,

  • Puc = Ultimate load on concrete = 0.45 fck Ac
  • Pus = Ultimate load on steel = 0.75 fy Asc
  • Ac = Area of concrete
  • Asc = Area of longitudinal steel

After putting the value of Puc and Pus in the above equation

Pu = 0.45 fck Ac + 0.75 fy Asc

This relation is applicable for the ideal condition of axial loading. In the practical conditions, the loading is never absolutely axial and there will always be some eccentricity that cannot be avoided. Hence we may consider the possibility of a minimum eccentricity of 0.05 times the lateral dimension and assume an 11% reduction in the ultimate strength of the column.

After an 11% reduction, we can write as,

Pu = 0.40 fck Ac + 0.67 fy Asc

Assume, Ag = Gross sectional area of the column

Therefore, Ag = Ac + Asc

Now,

Pu = 0.40 fck (Ag – Asc) + 0.67 fyAsc [ Note, Ag = Ac + Asc So, Ac = Ag – Asc].

Or, Pu = 0.40 fck Ag – 0.40 fck Asc + 0.67 fy Asc ……………….(a)

Or, Pu = 0.40 fck Ag + (0.67 fy – 0.40 fck) Asc

If ‘p’ = percentage of steel provided = [latex] \frac{A_{sc}}{A_{g}}\times 100 [/latex]

Or, Asc = [latex] \frac{p}{100}\times A_{g} [/latex]

After putting the value of Asc in equation (a)

Pu = 0.40 fck Ag – 0.40 fck [latex] \frac{p}{100}\times A_{g} [/latex] + 0.67 fy [latex] \frac{p}{100}\times A_{g} [/latex]

Or, Pu = 0.40 fck (Ag – [latex] \frac{p}{100}\times A_{g} [/latex]) + 0.67 fy [latex] \frac{p}{100}\times A_{g} [/latex]

Or, Pu/Ag = 0.40 fck + [latex] \frac{p}{100} [/latex] (0.67 fy – 0.40 fck)

Read More:

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11 Situation Which Demands Pile Foundation for a Structure

11 Situation Which Demands Pile Foundation For a Structure

The Pile foundation can be used for any type of soil and for any type of structure. In fact, they provide a common solution to all the difficult problems of the foundation. But it doesn’t mean that you can use the pile foundation wherever you want, because it is very costly than other types of foundations.

11 Situation Which Demands Pile Foundation for a Structure
11 Situation Which Demands Pile Foundation For a Structure

If we provide this foundation without knowing whether it is needed, then the cost of construction will be very high. Following are the situations which demand the adoption of the pile foundation for the structure:

1. If the load coming from the structure is very high, then the pile foundation is the best solution in this situation.

2. When loads coming from the structure is not uniform, and also very heavy.

3. Where it is not possible to provide a grillage or raft foundation due to local difficulties.

4. When the layer of soil is compressible or very weak.

5. If the foundation is to be built up in a water-logged area then pile foundation is preferred.

6. If hard-bearing strata is located at a large depth, then it is necessary to consider the pile foundation.

7. If the other types of foundations are not possible to adopt due to the presence of the groundwater table in a very shallow depth, then the adoption of the pile foundation will be the best solution.

8. They are also used as anchors.

9. When the structure is to be constructed on a rive-bed or sea shore.

10. If it is not possible to keep the foundation trenches in dry condition by pumping due to a very strong inflow of seepage or capillary water.

11. They are also used for the construction of docks, piers, and other marine structures.

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Maximum Permissible Concentration of Various Inorganic Elements Present in Potable Water As per WHO & BIS Standards

Maximum Permissible Concentration of Various Inorganic Elements Present in Potable Water

Following are the values of the maximum permissible concentration of various inorganic elements present in potable water as per the World Health Organization (WHO) and Bureau of Indian Standards (BIS).

S.NoInorganic ElementMaximum Permissible Concentration in Potable Water as per WHO in mg/LMaximum Permissible Concentration in Potable Water as per BIS in mg/L
1.Arsenic (As)0.010.01
2. Boron (B)0.31.0
3. Antimony (Sb)0.05Not Specified (N.S)
4.Cadmium (Cd)0.0030.01
5. Barium (Ba)0.70N.S
6.Beryllium (Be)N.SN.S
7.Lead (Pb)0.010.05
8. Cobalt (Co)N.SN.S
9.Lithium (Li)N.SN.S
10.Chromium (Cr)0.050.05
11.Copper (Cu)N.S0.05
12.Thallium (Tl)N.SN.S
13. Fluoride (F)1.51.0
14.Sodium (Na)N.SN.S
15.Molybdenum (Mo)N.SN.S
16.Cyanide (CN–)0.070.05
17.Mercury (Hg)0.0010.001
18.Ammonia (NH3)N.SN.S
19. Nickel (Ni)0.02N.S
20.Nitrates (NO3–)11.345.0
21.Nitrites (NO2 –)0.91N.S
22.Silver (Ag)N.SN.S
23.Zinc (Zn)N.SN.S
24.Selenium (Se)0.010.01
25.Phenolic compounds as phenol (C6H5OH)N.S0.001

Read More:

Water Treatment – MCQ

Secondary Sedimentation Tank

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Water Treatment – MCQ || Water Supply

Water Treatment – MCQ

1. The rate of filtration of a slow sand filter is

a) 200 to 400 lit/m2/hr
b) 100 to 200 lit/m2/hr
c) 200 to 300 lit/m2/hr
d) 100 to 300 lit/m2/hr

View Answer

b) 100 to 200 lit/m2/hr

2. A city supply of 15 MLD is to be disinfected with chlorine doses of 0.5 ppm. The bleaching powder contains 25 % of chlorine. The requirement of bleaching powder per day is

a) 300 kg
b) 75 kg
c) 30 kg
d) 7.5 kg

View Answer

c) 30 kg

3. The expression for settling velocity for a sedimentation tank is given by

a) Q/BH
b) Q/BL
c) Q/LH
d) None of these.

View Answer

b) Q/BL

4. What is the maximum permissible velocity in horizontal flow type sedimentation tank?

a) 1.2 m/sec
b) 0.9 m/ sec
c) 0.6 m/sec
d) 0.3 m/sec

View Answer

d) 0.3 m/sec

5. What is the normal detection period for a plain sedimentation water tank?

a) 2 to 4 hours
b) 4 to 8 hours
c) 6 to 12 hours
d 8 to 16 hours

View Answer

b) 4 to 8 hours

6. What is the effective size of sand particles in the case of a slow sand filter?

a) 0.7 mm to 1.3 mm
b) 0. 53 mm to 0.7 mm
c) 0. 4 mm to 0.53 mm
d) 0.2 mm to 0.4 mm

View Answer

d) 0.2 mm to 0.4 mm

7. Which of the following factors affects the settlement of particles in the sedimentation tank?

a) Viscosity of water
b) Velocity of flow
c) Specific gravity of solid
d) Size and shape of solid
e) All of these

View Answer

e) All of these

8. Settling velocity is proportional to the square of the diameter of the particle( according to Stoke’s law) when

a) Diameter of particles is between 0.1 and 10 mm
b) Diameter of particles is less than 0.1 mm
c) Diameter of particles is more than 1.0 mm
d) For all diameter

View Answer

b) Diameter of particles is less than 0.1 mm

9. The maximum permissible concentration of cadmium in potable water as per who guide lines

a) 0.01 mg/l
b) 0.009 mg/l
c) 0.006 mg/l
d) 0.003 mg/l

View Answer

d) 0.003 mg/l

10. Which of the following condition indicates the 100 % removal of particles from the bottom of a sedimentation tank?

a) Settling velocity < Liquide rising velocity
b) Settling velocity ≤ Liquide rising velocity
c) Settling velocity ≥ Liquide rising velocity
d) None of these

View Answer

c) Settling velocity ≥ Liquide rising velocity

11. If the liquid rising velocity is greater than the settling velocity, then one would expect

a) 100 % removal of particles
b) 50 % removal of particles
c) 0 % removal of particles
d) None

View Answer

c) 0 % removal of particles

Read More:

Water Quality – MCQ

Sources of Water – MCQ

Water Demand – MCQ

Mud Plaster – How to Prepare & Apply Mud Plaster & Benefits

What is Mud Plaster?

Mud plaster is another type of plaster that is mostly used in cases of village and temporary kutcha construction. This is the cheapest type of plaster. Mud plaster is also known as earth plaster.

Mud Plaster - How to Prepare & Apply Mud Plaster & Benefits

Process of Preparing Mud Plaster

It is made up of earth with an adequate percentage of clay and sand content. The earth should be free from any types of unwanted elements like grass, stone pebbles, roots, organic matter, etc. If clods are found in the excavated earth, they should be broken into the fine earth.

The soil is mixed with sufficient water and left to season for 5 to 7 days. To increase the strength of mud plaster, chopped straw or hemp is added to the seasoning mixture. Around 30 kg of chopped straw or hemp is used for 1 m3 of earth.

During the seasoning period, the mixture is worked up from time to time under cattle feet and converted into a homogeneous mass. The mud mortar is now ready for plastering.

Process of Preparing Surface Before Applying Mud Plaster?

The surface which is to be plastered should be prepared in the same way as for cement plaster or lime plaster. Loose dust should be removed from surface joints. Before beginning the plastering work, the surface which is to be plastered should be wetted thoroughly.

Excessive projections extending from the surface should be knocked off and screeds formed, so as to act as thickness gauges. Mud plaster is now applied between screeds by dashing the mortar against the prepared surface.

Process of Applying Mud Plaster?

Mud plaster is usually done in two coats. The second coat is applied after the first coat has set but not fully dried. Usually, the thickness of the first coat is 13 mm and the thickness of the second coat is 6 to 7 mm. The mud plaster may finally be swept with a mixture of clay and cow dung.

Benefits of Mud Plaster

1. This plaster gives a smooth and decorative look to the mud walls.

2. Some designs may be created over the wall through mud plastering.

3. This type of plaster is environment friendly

4. They are less toxic than other types of plaster.

5. The repairing work is also very easy and inexpensive.

Read More:

Plastering – MCQ

Remedies of Plaster Defects

Types of Pointing used In Masonry Work