Lecture_3-Aggregate_Civil Engineering.pdf

CE 201: Engineering Materials
Lecture# 3
Aggregates
Dr. Sheikh Mokhlesur Rahman
Associate Professor, Dept. of CE
Contact: mokhles@gmail.com

Aggregate
Aggregate is an aggregation of non-metallic minerals obtained in particulate form and can
be processed and used for civil and highway engineering constructions.
• Non-metallic minerals
• Granular in form
• Occupies 60 – 75% volume of concrete
• Affects
▪ Workability of fresh concrete
▪ Properties of hardened concrete
• Sources:
▪ Mining of mineral aggregate deposits (e.g., sand, gravel and stone)
▪ Use of waste slag from the manufacture of iron/steel
▪ Recycling of concrete
July 2023 Semester - SMR CE 201_Aggregates
2

Classification of Aggregate
➢ Based on source:
• Natural: crushed stone, crushed boulder, crushed gravels, shingles, sands
• Artificial: crushed brick, blast furnace slag, synthetic
➢ Based on size:
• Coarse aggregate: Retaining on sieve no. 4 (opening 4.75 mm)
• Fine aggregate: Passing no. 4 sieve and retaining on no. 200 sieve
• Silt and Clay: Passing no. 200 sieve (opening 0.075 mm)
➢ Based on weight:
• Normal weight: bulk unit weight 1520-1680 kg/m3
• Light weight: bulk unit weight <1120 kg/m3
• Heavy weight: bulk unit weight > 2080 kg/m3
3

Grading Requirements for Fine Aggregates
ASTM C33/33M-13
4
Sand

Grading Requirements for Coarse Aggregates
5
ASTM C33/33M-13
Crushed Stone/Stone Chips
Crushed Brick/Brick Chips

Characteristics of Aggregates
6

Properties of Aggregate
➢ Gradation
➢ Water Absorption of Aggregate
➢ Specific Gravity of Aggregate
➢ Unit weight of Aggregate
➢ Voids in Aggregate
Will be covered in this lecture
➢ Shape of Particles
➢ Strength of Aggregate
➢ Abrasion Test of Aggregate
Will be covered in Transportation Engineering Courses
7

Gradation
➢ Gradation means distribution of particle sizes, usually expressed as percent of
total weight
➢ Gradation of an aggregate (sand or coarse aggregate) is one of the most key
characteristics that determine the performance of aggregate in a concrete mix.
➢ An aggregate mix should contain particles of varying sizes that means it should
be well graded.
➢ Sieve Analysis is usually performed to get the gradation of an aggregate (ASTM
C136, E11).
➢ Sieve analysis is conducted by passing the aggregates through a series of
sieves, ordered as in the descending order of their opening with so that the
sieves with larger opening remain at the top of sieves with smaller opening.
➢ The sieve retains particles larger than the opening, while smaller ones pass
through.
8

Sieve Analysis
9
➢ Sieves are numbered based on their
opening size.
➢ ASTM Standard Sieves:
No. 100, No. 50, No. 30, No. 16, No. 8, No. 4,
⅜″, ¾″, 1½″ and at a ratio of 1:2 beyond that
Standard Sieve
Designation
Nominal Sieve Opening
mm in
Coarse Sieves
75.0 mm 3″ 75 3
63.0 mm 2½″ 63 2.5
50.0 mm 2″ 50 2
37.5 mm 1½″ 37.5 1.5
25.0 mm 1″ 25 1
19.0 mm ¾″ 19 0.75
12.5 mm ½″ 12.5 0.5
9.5 mm ⅜″ 9.5 0.375
Fine Sieves
4.75 mm No. 4 4.75 0.187
2.36 mm No. 8 2.36 0.0937
1.18 mm No. 16 1.18 0.0469
600 µm No. 30 0.6 0.0234
300 µm No. 50 0.3 0.0117
150 µm No. 100 0.15 0.0059
75 µm No. 200 0.075 0.0029

Sieve Analysis: Data processing
➢ The weight of materials retained on each sieve
is determined, from which percentage of
materials retained is estimated.
➢ Cumulative percentage retained on a specific
sieve is determined by summing up the
percentage retained on sieves of higher sizes
and percentage retained on that particular
sieve.
➢ Percentage finer (% finer) for a specific sieve
is calculated as
% finer = 100 – Cumulative percentage retained
➢ Gradation curve is usually plotted as % finer
versus sieve opening (mm), where sieve
opening is plotted on log scale.
10

Properties of Gradation Curve
➢ Fineness Modulus (FM) is an index of the fineness of an aggregate
• the higher the FM, the coarser the aggregate.
• Different aggregate grading may have the same FM.
➢ FM is obtained by adding the cumulative % retained on specified sieve (ASTM standard
sieves) and dividing the sum by 100.
𝐹𝑀 =
σ 𝐶𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 % 𝑟𝑒𝑡𝑎𝑖𝑛𝑒𝑑 𝑜𝑛 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑠𝑖𝑒𝑣𝑒𝑠
100
➢ D10 = Effective size, 10% materials are finer than D10
➢ D50 = Mean diameter, 50% materials are finer than D50
➢ 30% materials are finer than D30 and 60% materials are finer than D60
➢ Coefficient of Uniformity, 𝐶𝑢 =
𝐷60
𝐷10
➢ Coefficient of Curvature or Coefficient of Gradation (𝐶𝑐 𝑜𝑟 𝐶𝑍), =
𝐷30
2
𝐷10𝐷60
11

Example Problem 1: Sieve Analysis of Fine Aggregate
Sieve Size
(mm)
Mass Retained
(g)
Individual Retained
(%)
Cumulative
Retained (%)
% Finer/
Total Passing (%)
4.75 (No. 4) 3
2.36 (No. 8) 29
1.18 (No. 16) 65
0.60 (No. 30) 165
0.30 (No. 50) 131
0.15 (No. 100) 97
0.075 (No. 200) 10
Pan 1
12

Example Problem 1: Solution
Sieve Size
(mm)
Mass Retained
(g)
Individual Retained
(%)
Cumulative
Retained (%)
% Finer/
Total Passing (%)
4.75 (No. 4) 3 1 1 99
2.36 (No. 8) 29 6 7 93
1.18 (No. 16) 65 13 20 80
0.60 (No. 30) 165 33 53 47
0.30 (No. 50) 131 26 79 21
0.15 (No. 100) 97 19 98 2
0.075 (No. 200) 10 2 100 0
Pan 1 0 100 0
Total 501 100 --- ---
13
𝐹𝑀 =
1 + 7 + 20 + 53 + 79 + 98
100
= 2.58

𝐶𝑢 =
𝐷60
𝐷10
=
0.8
0.2
= 4
𝐶𝑍 =
𝐷30
2
𝐷10𝐷60
=
(0.38)2
0.2∗0.8
= 0.90
Gradation Curve
𝐶𝑢 =
𝐷60
𝐷10
𝐶𝑍 =
𝐷30
2
𝐷10𝐷60
14
0
20
40
60
80
100
0.01 0.1 1 10
%
Finer
Sieve opening (mm)
Gradation Curve
Also known as grain size distribution curve

Example Problem 2: Sieve Analysis of Coarse Aggregate
15
Draw gradation curve and Determine the FM of aggregates from the following
particle distribution.
Sieve Size
(mm)
Mass Retained
(g)
25.0 (1in.) 0
19.0 (3/4 in.) 405
12.5 (1/2 in.) 2850
9.5 (3/8 in.) 2435
4.75 (No. 4) 2030
2.36 (No. 8) 375
Pan 35

Sieve Analysis of Coarse Aggregate
Sieve Size
(mm)
Mass Retained
(g)
Individual
Retained
(%)
Cumulative
Retained (%)
Total Passing (%)
25.0 (1 in.) 0 0 0 100
19.0 (3/4 in.) 405 5 5 95
12.5 (1/2 in.) 2850 35 40 60
9.5 (3/8 in.) 2435 30 70 30
4.75 (No. 4) 2030 25 95 5
2.36 (No. 8) 375 5 100 0
Pan 35 0 100 0
Total 8130 100 --- ---
16
Solution:

Gradation Curve of Coarse Aggregate
17
0
20
40
60
80
100
0.1 1 10 100
%
Finer
Sieve opening (mm)
Gradation Curve

Fineness Modulus of Coarse Aggregate
Sieve Size
(mm)
Cumulative
Retained (%)
25.0 (1in.) 0
19.0 (3/4 in.) 5
12.5 (1/2 in.) 40
9.5 (3/8 in.) 70
4.75 (No. 4) 95
2.36 (No. 8) 100
𝐹𝑀 =
5 + 70 + 95 + 100 + 100 + 100 + 100 + 100
100
𝐹𝑀 = 6.70
▪ Add No. 16, 30, 50 and 100
sieves
▪ Take only the standard sieves
18
1.18 (No. 16) 100
0.6 (No. 30) 100
0.3 (No. 50) 100
0.15 (No. 100) 100

Types of Gradation
➢ Well graded/smooth grading
➢ Poor Graded
▪ Open graded
▪ Gap graded
▪ Uniform graded
19

Gradation of Aggregate
Well graded/ Dense graded
• Sample contains certain amount of
almost all sizes of particles
• Provides good interlocking
• Contain minimum amount of voids
• Low permeability
• Good Compaction
• Leads to less cement requirement
20
Open graded
• Contain very little amount of fine
particles
• High void content
• Poor compaction
• Grain-to-grain contact
• Provides high permeability

Gradation of Aggregate
Gap graded
• Very few middle size particles
• Provides good interlocking
(No grain-to-grain contact)
• Moderate void content
• Moderate to low permeability
• Low stability
• Easy to compact
• Requires large amount of cement to
coat the surface
21
Uniformly graded
• Narrow range of sizes
• Grain-to-grain contact
• High void content
• High permeability
• Low stability
• Difficult to compact

Different Types of Gradation Curves
22
0
20
40
60
80
100
0.01 0.1 1 10 100
%
Finer
Sieve opening (mm)
Gradation Curve

Blending of Aggregate
➢ A single aggregate source is generally unlikely to meet gradation
requirements for cement or asphalt concrete mixes.
➢ Blending of aggregates from two or more sources would be required to
satisfy the specifications.
➢ Reasons behind blending
• Obtain desirable gradation
• Single natural or quarried material not enough
• Economical to combine natural and process materials
➢ Methods
• Trial and Error method
• Equation solution method
• Graphical method
23

Trial and Error Method
Example Problem 3: Find the mix ratio of aggregate A and B to meet the
specification.
Material
Aggregate
A
Aggregate
B
Target
Specification
Sieve % Passing % Passing % Passing
3/8 100 100 100
No. 4 90 100 90 to 100
No. 8 30 100 36 to 76
No. 16 7 88 -
No. 30 3 47 -
No. 50 1 32 -
No. 100 0 24 -
No. 200 0 10 2 to 10
24
The basic equation for blending:
𝑃𝑖 = 𝑎 ∗ 𝐴𝑖 + 𝑏 ∗ 𝐵𝑖 + 𝑐 ∗ 𝐶𝑖 + … …
Here,
𝑃𝑖 = percent of blended aggregate
passing sieve i
𝐴𝑖, 𝐵𝑖, 𝐶𝑖= percent of aggregates A,
B, C, . . . passing sieve i
𝑎, 𝑏, 𝑐 = proportion of aggregates A,
B, and C used in the blend, where
the total is 1.00.

Trial 01:
Material
Aggregate
A
Aggregate
B Aggregate
Blend
Target
Specification
Proportion 30.0% Proportion 70.0%
Sieve % Passing % Batch % Passing % Batch
% Passing % Passing
3/8 100 30 100 70 100 100
No. 4 90 27 100 70 97 90 to 100
No. 8 30 9 100 70 79 36 to 76
No. 16 7 2 88 62 64
No. 30 3 1 47 33 34
No. 50 1 0 32 22 22
No. 100 0 0 24 17 17
No. 200 0 0 10 7 7 2 to 10
25

Trial 01:
Material
Aggregate
A
Aggregate
B Aggregate
Blend
Target
Specification
% Passing % Passing
3/8 100 30 100 70 100 100
No. 4 90 27 100 70 97 90 to 100
No. 8 30 9 100 70 79 36 to 76
No. 16 7 2 88 62 64
No. 30 3 1 47 33 34
No. 50 1 0 32 22 22
No. 100 0 0 24 17 17
No. 200 0 0 10 7 7 2 to 10
26

Trial 02:
Material
Aggregate
A
Aggregate
B Aggregate
Blend
Target
Specification
% Passing % Passing
3/8 100 40 100 60 100 100
No. 4 90 36 100 60 96 90 to 100
No. 8 30 12 100 60 72 36 to 76
No. 16 7 3 88 53 56
No. 30 3 1 47 28 29
No. 50 1 0 32 19 220
No. 100 0 0 24 14 14
No. 200 0 0 10 6 6 2 to 10
27

Trial 03:
The solution may not be unique
Material
Aggregate
A
Aggregate
B Aggregate
Blend
Target
Specification
% Passing % Passing
3/8 100 50 100 50 100 100
No. 4 90 45 100 50 95 90 to 100
No. 8 30 15 100 50 65 36 to 76
No. 16 7 4 88 44 48
No. 30 3 2 47 24 26
No. 50 1 1 32 16 17
No. 100 0 0 24 12 12
No. 200 0 0 10 5 5 2 to 10
28

Equation Solution Method
Example Problem 4: Find the mix ratio of aggregate A, B, and C to meet the
specification.
Material
Aggregate
A
Aggregate
B
Aggregate
C
Target
Specification
Sieve % Passing % Passing % Passing % Passing
1.5 100 100 100
¾ 100 99 13 50%
3/8 100 33 8
No. 4 99 5 2 24%
No. 8 76 0 0
No. 16 58 - -
No. 30 40 - - 9%
No. 50 12 - -
No. 100 2 - -
29
The basic equation for blending:
𝑃𝑖 = 𝑎 ∗ 𝐴𝑖 + 𝑏 ∗ 𝐵𝑖 + 𝑐 ∗ 𝐶𝑖 + … …
Here,
𝑃𝑖 = percent of blended aggregate
passing sieve i
𝐴𝑖, 𝐵𝑖, 𝐶𝑖= percent of aggregates A,
B, C, . . . passing sieve i
𝑎, 𝑏, 𝑐 = proportion of aggregates A,
B, and C used in the blend, where
the total is 1.00.

Solution:
For ¾” sieve: 100 ∗ 𝑎 + 99 ∗ 𝑏 + 13 ∗ 𝑐 = 50 (1)
For No. 4 sieve: 99 ∗ 𝑎 + 5 ∗ 𝑏 + 2 ∗ 𝑐 = 24 (2)
For No. 30 sieve: 40 ∗ 𝑎 = 9 (3)
Constraint: 𝑎 + 𝑏 + 𝑐 = 1 (4)
Solving equation 1, 2, and 4
𝑎 = 0.22, 𝑏 = 0.21, 𝑐 = 0.57
30

Alternate Equation Framework:
For ¾” sieve: 100 ∗ 𝑎 + 99 ∗ 𝑏 + 13 ∗ 𝑐 = 50 ∗ (𝑎 + 𝑏 + 𝑐) (1)
For No. 4 sieve: 99 ∗ 𝑎 + 5 ∗ 𝑏 + 2 ∗ 𝑐 = 24 ∗ (𝑎 + 𝑏 + 𝑐) (2)
For No. 30 sieve: 40 ∗ 𝑎 = 9 ∗ (𝑎 + 𝑏 + 𝑐) (3)
Constraint: 𝑎 + 𝑏 + 𝑐 = 1
With 𝑎 = 1 from equation 1 and 2
𝑏 = 0.94 and 𝑐 = 2.60
𝑎 =
1
1 + 0.94 + 2.60
= 0.22
𝑏 =
0.94
1 + 0.94 + 2.60
= 0.21
𝑐 =
2.60
1 + 0.94 + 2.60
= 0.57
31

The solution may not be unique
Material
Aggregate
A
Aggregate
B
Aggregate
C Aggregate
Blend
Target
Specification
Proportion 22.0% Proportion 21.0% Proportion 57.0%
Sieve % Passing % Batch % Passing % Batch % Passing % Batch
% Passing % Passing
1.5 100 22 100 21 100 57 100
¾ 100 22 99 21 13 7 50 50%
3/8 100 22 33 7 8 5 34
No. 4 99 22 5 1 2 1 24 24%
No. 8 76 17 0 0 0 0 17
No. 16 58 13 0 0 0 0 13
No. 30 40 9 0 0 0 0 9 9%
No. 50 12 3 0 0 0 0 3
No. 100 2 0 0 0 0 0 0
32

Properties of Blended Aggregate
33
➢ When two or more aggregates from different sources are blended, some of the
properties of the blend can be calculated from the properties of the individual
components.
➢ With the exception of specific gravity and density, the properties of the blend are
the simple weighted averages of the properties of the components:
𝑋 = 𝑃𝐴 ∗ 𝑋𝐴 + 𝑃𝐵 ∗ 𝑋𝐵 + 𝑃𝐶 ∗ 𝑋𝐶 + … …
where, 𝑋 = composite property of the blend
𝑋𝐴, 𝑋𝐵, 𝑋𝐶, ...... = Property of aggregates A, B, C, ..........
𝑃𝐴, 𝑃𝐵, 𝑃𝐶, ...... = Proportion of aggregates used in the blend in fractions
➢ This equation applies to properties such as angularity, absorption, strength, and
fineness modulus.

Properties of Blended Aggregate
34
➢ Fineness modulus of the blend:
𝐹𝑀 = 𝑃𝐴 ∗ 𝐹𝑀𝐴 + 𝑃𝐵 ∗ 𝐹𝑀𝐵 + 𝑃𝐶 ∗ 𝐹𝑀𝐶 + … …
where, 𝐹𝑀 = Fineness modulus of the blend
𝐹𝑀𝐴, 𝐹𝑀𝐵, 𝐹𝑀𝐶, ...... = Fineness modulus of aggregates A, B, C, ..........
➢ Specific gravity of the blend:
𝐺 =
1
𝑃𝐴
𝐺𝐴
+
𝑃𝐵
𝐺𝐵
+
𝑃𝐶
𝐺𝐶
+ ……
where, 𝐺 = Specific gravity of the blend
𝐺𝐴, 𝐺𝐵, 𝐺𝐶, ...... = specific gravity of aggregates A, B, C, ..........

Water Absorption of Aggregates
➢ Pores
• Impermeable pores
• Permeable pores – Interconnected and extended up to surface
➢ Water Absorption of Aggregate
• Four moisture conditions of aggregate
▪ Oven dry/ Bone dry (0% moisture)
▪ Air Dry (Moisture % less than capacity)
▪ Saturated and Surface Dry, SSD (Moisture % = capacity)
▪ Moist/Damp/Wet (Moisture % more than capacity)
35

Oven dry
Absorption Capacity
Different Stages of Absorption
𝑀𝑜𝑖𝑠𝑡𝑢𝑟𝑒 𝐶𝑜𝑛𝑡𝑒𝑛𝑡 𝑀𝐶 =
𝑤𝑒𝑖𝑔ℎ𝑡 − 𝑂𝐷
𝑂𝐷
∗ 100 (%)
𝐴𝑏𝑠𝑜𝑟𝑝𝑡𝑖𝑜𝑛 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 (𝐴𝐶) =
𝑆𝑆𝐷 − 𝑂𝐷
𝑂𝐷
∗ 100 (%)
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑀𝑜𝑖𝑠𝑡𝑢𝑟𝑒 = 𝑀𝐶 − 𝐴𝐶
36
Free/ surface Moisture
Effective Absorption
Moisture Content

Example Problem
Problem 5: Moist mass of sand sample is 625.2 g and dry mass is 589.9 g.
If the absorption capacity is 1.6%, Determine (a) total moisture content and
(b) free moisture content.
Solution:
a) 𝑀𝐶 =
𝑤𝑒𝑖𝑔ℎ𝑡 −𝑂𝐷
𝑂𝐷
∗ 100% =
625.2−589.9
589.9
= 5.98%
b) 𝐹𝑟𝑒𝑒 𝑀𝑜𝑖𝑠𝑡𝑢𝑟𝑒 = 𝑀𝐶 − 𝐴𝐶 = 5.98 − 1.6 = 4.38%
37

Example Problem
Problem 6: 50 kg of gravel is mixed with 30 kg of sand. The gravel has a
moisture content of 3.9% and absorption of 4.7%, whereas the sand has a
moisture content of 3.5% and absorption of 4.9%. What is the amount of
water required to reach the SSD condition for both gravel and sand?
Solution:
38

Specific Gravity of Aggregate
➢ Absolute Specific Gravity, S
All pores are excluded from V
➢ Apparent Specific Gravity, Sa
• Impermeable pores are included
• Permeable pores are excluded from V
➢ Bulk Specific Gravity (OD basis), Sd
• All pores are included in V
• OD weight is used
➢ Bulk Specific Gravity (SSD basis), Ss
• All pores are included in V
• SSD weight is used
39
𝑆𝑎 =
𝑂𝐷𝑤𝑒𝑖𝑔ℎ𝑡
𝑤𝑡. 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 occupying equal volume excluding permeable pores
𝑆𝑑 =
𝑂𝐷𝑤𝑒𝑖𝑔ℎ𝑡
𝑤𝑡. 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 occupying equal volume including permeable pores
𝑆𝑠 =
𝑆𝑆𝐷𝑤𝑒𝑖𝑔ℎ𝑡
𝑤𝑡. 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 occupying equal volume including permeable pores
S > Sa > Ss > Sd

Unit Weight/Bulk Density
➢ Mass/Volume, V
• Volume includes all kind of voids (inter-particle, permeable and impermeable
voids)
➢ Loose unit weight
• By Shoveling
➢ Compacted unit weight
• By Rodding
• By Jigging
➢ Voids in aggregate is also depends on degree of compaction. Voids are
lower for the more compact aggregate.
➢ The void content between particles affects the amount of cement paste
required for the mix. Angularity increases void content while well-graded
aggregate decreases void content
40

Deleterious Substances in Aggregate
Deleterious substances Effects on Cement Concrete
Organic impurities
Delay settling and hardening, may reduce strength gain, may
cause deterioration
Coatings on aggregate
Silt present in the aggregate may form a surface coatings on
the aggregate and may prevent development of good bond
between aggregate and cement paste
Particles < 0.075 mm Weaken bond, may increase water materials requirements
Coal, lignite or other low-density
materials
Reduce durability, may cause popouts or stains
Clay lumps and friable particles Popouts, reduce durability and wear resistance
Soft particles Reduce durability and wear resistance, popouts
41
Deleterious substances and their effects on Portland cement concrete

Alkali-Aggregate Reactivity
➢ Though aggregate is considered as inert materials earlier, some aggregates may
react with cement harming the concrete structure.
➢ The most common reaction, particularly in humid and warm climates, is between
the active silica constituents of an aggregate and the alkalis in cement (sodium
oxide, Na2O, and potassium oxide, K2O).
➢ The alkali–silica reaction results in excessive expansion, cracking, or popouts in
concrete.
➢ Other constituents in the aggregate, such as carbonates, can also react with the
alkali in the cement (alkali–carbonate reactivity); however, their reaction is less
harmful.
➢ The alkali–aggregate reactivity is affected by the amount, type, and particle size
of the reactive material, as well as by the soluble alkali and water content of the
concrete.
42

Alkali-Aggregate Reactivity
➢ Control of Alkali-Aggregate Reactivity
• By selecting non-reactive aggregates
• By using low alkali cement (0.4 ~ 0.6 alkali content), if alkali-reactive aggregate
must be used
• By using fly ash, ground granulated blast furnace slag, silica fume, or natural
pozzolans
• By using admixtures such as pozzolana
• By keeping the concrete structure as dry as possible
• By controlling void space in concrete
• By replacing about 30% of a reactive sand–gravel aggregate with crushed
limestone (limestone sweetening)
43

Practice Problems
Problem 1: Estimate the mix ratio of Aggregates 1, 2, 3, and 4 to obtain the
aggregate blend to meet the specification. Also determine the % passing for
the combined aggregate in each sieve and the FM of the combined aggregate.
Sieve Size % passing Specification
mm inch Aggregate 01 Aggregate 02 Aggregate 03 Aggregate 04 % Passing
175 7 100 - - -
150 6 98 - - -
100 4 30 100 - -
75 3 10 92 - - 55
50 2 2 30 100 -
37.5 1.5 0 6 94 - 28
25 1 0 4 36 100
19 3/4 0 0 4 92 13
9.5 3/8 0 0 2 30
4.76 No. 4 0 0 0 2
44

Practice Problems
Problem 2: Aggregates from three sources having bulk specific gravities of
2.753, 2.649, and 2.689 were blended at a ratio of 70:20:10 by weight,
respectively. What is the bulk specific gravity of the aggregate blend?
Problem 3: 20, 30, and 45 kg of three aggregates with fineness modulus of
2.6, 2.8, and 2.9, respectively, are mixed together to get a blended aggregate.
Determine the fineness modulus of the aggregate blend.
45

Lecture_3-Aggregate_Civil Engineering.pdf

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