CLE204 - Construction Materials and Technology - Unit 1

CLE204 - Construction Materials and TechnologyUnit I – Sources of Aggregates & Properties

Ramesh KannanAssistant Professor

School of Mechanical and Building SciencesVIT Chennai, Chennai - 600127

1 / 24 Ramesh Kannan, VIT Chennai CLE204 - Construction Materials and Technology - Unit I

Syllabus

Unit I - Sources of Aggregates & PropertiesPhysical and Mechanical properties of construction materials - commonlyused types of stones - Tests for stones, road aggregates and concreteaggregates, properties of sand, BIS specification for testing of aggregates- Bricks - Properties and testing methods for Bricks.


Physical Properties of Construction Materials

Physical Properties

The measurable properties which describes the physical state(without altering its characteristics) of the materials are commonlytermed as Physical Properties.

The physical properties of construction materials includes specificmass (density), specific gravity (relative density), specificweight (unit weight), porosity, water absorption and so on.


1. Density (or) Specific Mass

Density (ρ) is defined as the “ratio of mass per unit volume of ahomogeneous material”. This is commonly referred to as ‘volumetricmass density’ or ‘specific mass’.

Density can also be represented in terms of weight as weight densitywhich is the “ratio of weight per unit volume of a material”.

ρ = mv and ρw = w

v = m.gv (1)

Where,ρ = Density of the material, in kg/m3

ρw = Weight density of the material, in kg/m3

m = Mass of the material, in kgw = Weight of the material, in kgg = Acceleration due to gravity, in m/s2 (g = 9.81 m/s2 on Earth)v = Volume of the material, in m3


The density of some of the important building and construction materialsare shown below.

0 2,000 4,000 6,000 8,000

AluminiumBrick

ConcreteGlass

LimestonePolystyrene Foam

SteelWater

Wood (Pine)Wood (Oak)

2,7081,971

2,4002,467

1,65033

7,8331,000

513721

Density (in kg/m3)


1.1. Bulk Density

Bulk Density (ρb) is defined as the ratio of mass per unit volume ofa material in its natural state (with extrinsic voids and pores).

The intrinsic voids or pores, cracks, gaps and flaws, etc., present inthe materials are generally not considered for the determination ofbulk density.

For example, for a fired clay brick, the bulk density is calculatedbased on its density of the constituent materials and macroscopicpores developed due to the entrapped air only but not the mesoscopicand microscopic pores, which is an intrinsic characteristic of brittlematerials.

ρb = mv (2)

Where,ρb = Bulk density of the material, in kg/m3

m = Mass of the material, in kgv = Volume of the material, in m3


1.1. Bulk Density – Dry & Wet Bulk Density

Bulk density is a property of granular materials and it is commonlyreferred as ‘Density of Granual Materials’ .

Bulk density can be categorized into Dry bulk density ρb(dry) andWet bulk density ρb(wet).

Dry bulk density is normally the bulk density of the material

Wet bulk density is the ratio of total mass of material (mass ofmaterial + mass of water) to the total volume of the material (volumeof material + volume of water).

ρb(dry) = mv and ρb(wet) = mt

vt= m + mw

v + vw(3)

Where,mt = Total or apparent mass of the material, in kgmw = Mass of the water, in kgvt = Volume of the material, in m3

vw = Volume of the water, in m3


Example 1: Calculation of Bulk DensityCalculate the dry and wet bulk density of a structural fired clay brick ofdimension 19 × 9 × 9 cm with a standard frog of size 10 × 4 × 2 cm asshown below. The weight of the brick which is dried naturally at 27± 1oCtemperature is 2.76 kg and the weight of brick removed after immersedand soaked in cold water for 24 hours is found to be 3.12 kg. The volumeof water consumed by the brick is 0.0009 m3.


Solution: Given Data:-

¶ Dimension of brick, (L×B×D) = 19 × 9 × 9 cm = 0.19 × 0.09× 0.09 m

=⇒ Total volume of brick, V = 0.19 × 0.09 × 0.09 cm =0.001539 m3

· Dimension of frog, (l×b×d) = 10 × 4 × 2 cm = 0.10 × 0.04× 0.02 m

=⇒ Volume of frog, vf = 0.10 × 0.04 × 0.02 m =0.00008 m3

=⇒ Volume of brick (dry), v = Total volume of brick, V –Volume of frog, vf

= 0.001539 – 0.00008 m3 =0.001459 m3 ≈ 0.0015 m3

¸ Mass of brick (dry), m = 2.76 kg (Initial Mass)

¹ Total mass of brick (wet), mt = 3.12 kg (Final Mass)


Solution: Given Data (Cont...):-

º Mass of water, mw = Total mass of brick, mt – Massof brick, m

= 3.12 – 2.76 kg = 0.36 kg

» Volume of water, vw = 0.0009 m3

=⇒ Volume of brick (wet), vt = Volume of brick (dry), v +Volume of water, vw

= 0.0015 + 0.0009 m3 = 0.0024m3


Dry bulk density and wet bulk density of a material is given by

ρb(dry) = ρb = mv and ρb(wet) = mt

vt

Dry bulk density, ρb(dry) = ρb = mv = 2.76

0.0015 = 1840 kg/m3

Wet bulk density, ρb(wet) = mt

vt= 3.12

0.0024 = 1300 kg/m3

Answer∴ Dry density of structural fired clay brick, ρb(dry) = 1840 kg/m3 and

Wet density of structural fired clay brick, ρb(wet) = 1300 kg/m3


1.2. Density IndexDensity Index (ρ0) is defined as the ratio of bulk density to the density ofsame material.

ρo = Bulk DensityDensity = ρb

ρ(4)

Where,ρo = Density Index of the material (Constant)ρb = Bulk density of the material, in kg/m3

ρ = Density of the material, in kg/m3

In general, the density index of almost all the building and constructionmaterials are less than or nearly equal to 1 but not exactly equal to1 since there isn’t any perfect or ideal material exists whose density andbulk density are the same.


Example 2: Calculation of Density IndexCalculate the density index of a structural fired clay brick whose densityand bulk density are 1971 kg/m3 and 1840 kg/m3 respectively.

Solution: Given Data:-

¶ Density of brick, ρ = 1971 kg/m3

· Bulk density of brick, ρb = 1840 kg/m3

Density Index (ρo) of a material is given by ρo = ρbρ

ρo = ρbρ

= 18401971 = 0.933536 ≈ 0.94 (Constant)

Answer∴ The density index of this structural fired clay brick is 0.94


2. Unit Weight (or) Specific Weight

The unit weight (γ) of the material is defined as the weight perunit volume of a material. This is commonly referred to as ‘weightdensity’ or ‘specific weight’.

γ = ρw = wv = m.g

v (5)

The relationship between density and unit weight of a material isgiven by the following expression.

γ = ρ.g (6)

Where,γ = Unit weight of the material, in kN/m3

ρ = Density of the material, in kg/m3

g = Acceleration due to gravity, in m/s2 (g = 9.81 m/s2 on Earth)


2. Unit Weight (or) Specific Weight (Cont...)

The unit weight places an important role in the structural engineeringas it is used for the determination of loads that could possibly beencounter in a building along with the angle of internal friction.

The comprehensive list of weight, density and unit weight ofcommonly used building and construction materials are given inBureau of Indian Standard, IS 875-1 (1987): Code of Practice ForDesign Loads (Other Than Earthquake) For Buildings And Structures,Part 1: Dead Loads - Unit Weights of Building Material And StoredMaterials.https://law.resource.org/pub/in/bis/S03/is.875.1.1987.pdf


https://law.resource.org/pub/in/bis/S03/is.875.1.1987.pdf

https://law.resource.org/pub/in/bis/S03/is.875.1.1987.pdf

Example 3: Calculation of Unit WeightCalculate the unit weight of a hardened Plain Cement Concrete (P.C.C)cube whose density is found to be 2400 kg/m3.

The unit weight (γ) of a material is given by γ = ρ.g

γ = ρ.g = 2400× 9.81 = 23544 N/m3

= 23.544 kN/m3 ≈ 24 kN/m3

Answer∴ The unit weight of hardened Plain Cement Concrete (P.C.C) cube is 24kN/m3


The unit weight of some of the commonly used building and constructionmaterials are shown below.

0 20 40 60 80

AluminiumBrick

ConcreteGlass


SteelWater


2719.3

2424.2

16.23 · 10−2

76.89.8

57.1

Unit Weight (in kN/m3)


3. Specific Gravity (or) Relative Density

Specific gravity Gs is defined as the “ratio of mass or weight of avolume (density or unit weight) of the material to the mass or weightof a volume (density or unit weight) of the reference material”.

Specific gravity is commonly referred to as ‘relative density’.

Generally, water is taken as the reference material whose density andunit weight at 4oC are 1000 kg/m3 and 9.8 kN/m3 respectively.

Gs = γmγw

= ρm.�gρw.�g

= ρmρw

∵ γ = ρ.g (7)

Where,Gs = Specific Gravity of the material (Constant)γm = Unit weight of the material, in kN/m3

γw = Unit weight of the water, in kN/m3

ρm = Density of the material, in kg/m3

ρw = Density of the water, in kg/m3

g = Acceleration due to gravity, in m/s2 (g = 9.81 m/s2 on Earth)


3. Specific Gravity (or) Relative Density (Cont...)– Apparent Specific Gravity & Absolute Specific Gravity

Apparent Specific Gravity is defined as the ratio of mass or weightincluding both intrinsic and extrinsic voids in a volume (density orunit weight) of the material to the mass or weight (density or unitweight) of equal volume of the reference material.

This is also referred to as ‘mass specific gravity’ .

Absolute Specific Gravity is defined as the ratio of mass or weightexcluding voids in a volume (density or unit weight) of the materialto the mass or weight (density or unit weight) of equal volume of thereference material.

This is also referred to as ‘true specific gravity’ .


Example 4: Calculation of Apparent & Absolute Specific GravityCalculate the apparent and absolute specific gravity of mineral admixturesample whose density at natural state (loose, dry) and when densified(compacted, dry) is 1442 kg/m3 and 1922 kg/m3 respectively.

Solution: Given Data:-Density of mineral admixture sample at natural state (loose, dry),ρm(undensified) = 1442 kg/m3

Density of mineral admixture sample after densification (compacted, dry),ρm(densified) = 1922 kg/m3

We know that, density of water at 4oC, ρw = 1000 kg/m3

Apparent Specific Gravity =ρm(undensified)

ρw= 1442

1000 = 1.442 (Constant)

Absolute Specific Gravity =ρm(densified)

ρw= 1922

1000 = 1.922 (Constant)


The specific gravity of some of the commonly used building andconstruction materials are shown below.

0 2 4 6 8

AluminiumBrick

ConcreteGlass


SteelWater


2.781.97

2.42.47

1.653 · 10−2

7.831

0.510.72

Specific Gravity


Relative Volume Fraction– Porosity, Compactness & Void Ratio

Porosity (p) is defined as the “ratio of volume of voids (ve) to theapparent volume (V ) of the material”.

p = veV = ve

vm + ve(8)

Compactness (c) is defined as the “ratio of volume of material (vm)(excluding, volume of voids) to the apparent volume (V ) of thematerial”.

c = vmV = vm

vm + ve(9)

=⇒ c = vmV = V − ve

V = VV −

veV = 1− ve

V = 1− p

∴ c = (1− p)


Relative Volume Fraction (Cont...)– Porosity, Compactness & Void Ratio

Void Ratio (e) is defined as the “ratio of volume of voids (ve) to thevolume of the material (vm) (excluding, volume of voids)”.

e = vevm

(10)

=⇒ e = vevm

= veV − ve

= ve

ve ×(

Vve− 1

)

e = ve

ve ×(

Vve− 1

) = 1(1p − 1

) = 1(1− p

p

) = p1− p

∴ e = p(1− p) = p

c


Relative Volume Fraction (Cont...)– Porosity, Compactness & Void Ratio

A relationships exists between these three quantities as shown inEquation 10 and is also expressed as follows.

p = veV = e × vm

V = e ×(

vmV

)= e × c

∴ p = e . c

Knowing of one of the quantities enables us to calculate the othertwo quantities.

Porosity, compactness and void ratio are the three quantities thatcharacterize the relative volume fractions of voids and materials.


CLE204 - Construction Materials and Technology - Unit 1

Engineering

Transcript of CLE204 - Construction Materials and Technology - Unit 1