Aggregates II
Definitions
• Unit Weight
– Weight of a material divided by its volume
• Density
– Mass of a material divided by its volume
• Specific Gravity
– Ratio of a solid’s density to the density of
water
Specific Gravity
M
V
G
w
G
M
V w
1
Voids and Moisture
Oven Dry
SSD Dry
Moist
Air Dry
Phase Diagram
VA
Air
MA
Vw
Water
Mw
MT
VT
Solids
Vs
Ms
Apparent Specific Gravity
Oven Dry Mass
Total Aggregate Volume
Gsa
Vs
Ms
VA
w
Includes solids and impermeable voids in the
total aggregate volume
2
Bulk SSD Specific Gravity
SSD Mass
Total Aggregate Volume
GSSD
Ms Mw
Vs VA Vw
w
Ms Mw
VT w
Includes solids, impermeable voids, and water
permeable voids in the total aggregate volume
Bulk Specific Gravity
Oven Dry Mass
Total Aggregate Volume
Gsb
Vs
Ms
VA Vw
w
Ms
VT w
Includes solids, impermeable voids, and water
permeable voids in total aggregate volume
Specific Gravity
• Apparent
• Bulk SSD
• Bulk
Largest
Smallest
3
Volume Measurement
Water
Solid
Volume of Water
Equal Volume of Solid
Archimedes’ Principle
The buoyant force on a submerged object is equal
to the mass of the water displaced.
Displaced Water Mass = 2g
Displaced Water Volume = 2cm3
Therefore Volume of Rock = 2cm3
G
5g
2cm3
cm3
1g
2.500
Mass in Air = 5g
Mass in Water = 3g
Specific Gravity
• Coarse Aggregate (AASHTO T85, ASTM
C127
• Fine Aggregate (AASHTO T84, ASTM C128
• Soak Aggregate For 24 Hours
– Mass in Water
– Mass in SSD Condition
– Oven Dry Mass
4
Coarse Aggregate Specific Gravity
• 3 Values Determined
A = Oven-dry mass
B = SSD mass
C = Mass in water
SG Values should
Have 3 decimal
places!
Coarse Aggregate Specific Gravity
Gsa
A
A C
Gsb
GSSD
%Absorptio n
A
B C
B
B C
B- A
A
100
Coarse Aggregate Example
Given: Oven-dry mass = 1252.0g
SSD Mass = 1259.6g
Mass in Water = 787.1g
Find: Apparent, Bulk SSD, and Bulk specific
gravities, and the percent water absorption
Solution:
5
Fine Aggregate Specific Gravity
• 4 Values Determined
A = Oven-dry mass
B = Mass of flask + water
C = Mass of flask + sample + water
D = SSD mass
Fine Aggregate Specific Gravity
Gsa
A
B A C
Gsb
GSSD
%Absorptio n
A
B D C
D
B D C
D- A
A
100
Fine Aggregate Example
Given: Oven-dry mass = 1250.0g
SSD Mass = 1258.8g
Flask + Water Mass = 7784.5g
Flask + Sample + Water Mass = 8575.0
Find: Apparent, Bulk SSD, and Bulk specific gravities, and
the percent water absorption
Solution:
6
Aggregate Blends
• Aggregates are typically blended
G
P P ... P
P P
P
...
G G
G
1
1
1
2
2
n
n
2
n
• Pn = Percent of aggregate n
• Gn = Aggregate n specific gravity
• You must use this equation!
Dry-Rodded Unit Weight
• Used when aggregate will be used to fill a
volume
• Method Steps:
– Calibrate measure
– Rod each layer (3) 25 times
– Strike off excess and weigh
– Calculate DRUW
Dry-Rodded Unit Weight
M
•
•
•
•
G- T
V
M = DRUW (pcf)
G = Weight of Aggregate + Measure (lbs)
T = Weight of Measure (lbs)
V = Volume of measure (ft3)
7
Fineness Modulus
• Check uniformity of graded material
• Cannot be used to compare aggregates
from different sources
• Typically used with fine aggregates
• Smaller the number, finer the gradation
• For PCC, value should not change by
more than 0.2
Fineness Modulus
• Calculation
FM
Cumulative % Retained on Standard Sieves
100
• Standard sieve sizes used
– #100, #50, #30, #16, #8, #4, 3/8 in., 3/4
in., 1.5 in., etc., increasing at a 2:1 ratio
Fineness Modulus Example
Sieve
Size
(mm)
No. 4
Retained Percent
Mass (g) Retained
9
2
Cumulative
Percent
Percent
Retained Passing
2
98
No. 8
46
9
11
89
No. 16
97
19
30
70
No. 30
99
20
50
50
No. 50
120
24
74
26
No. 100
91
18
92
8
Sample
Mass (g)
500
∑=
8