Problem 1: For a coarse aggregate sample air dry weight of sample is 1790 gm, weight when it is immersed in water is 1180 gm and SSD weight is 1850 gm.
Calculate bulk specific gravity and apparent specific gravity and % of absorption?
Solution:
Apparent specific gravity,
= (MD / VB)/w
Here, VB = (1850 - 1180) /1
= 670 cm3
MD = 1790 gm
w = 1gm/ cm3
∴Apparent specific gravity GA= (1790/ 670)/ 1
= 2.67
Bulk Specific Gravity,
GB= (MD / VB)/w
Here, VB = 670 + (1850 - 1790)
= (670 + 60) cm3
= 730 cm3
MD = 1790 gm
w = 1gm/ cm3
∴Bulk Specific Gravity,
GB= (1790/730)/ 1
= 2.45
Percentage of absorption = {(1850-1790)/1790}*100
= 3.35 %
Answer: Apparent specific gravity 2.67, Bulk Specific Gravity 2.45, Percentage of absorption 3.35%
Problem 2: Dry weight of sample is 1206 gm, SSD weight is 1226.4 gm. The volume of water exerted by the sample is 440.6 cm3. Find bulk specific weight and apparent specific weight. Also find the absorption capacity of sample?
Solution:
Apparent specific gravity,
= (MD / VB)/w
Here, VB = 440.6 cm3
MD = 1206 gm
w = 1gm/ cm3
∴Apparent specific gravity, GA= (1206/ 440.6)/ 1
= 2.74
Bulk Specific Gravity,
GB= (MD / VB)/w
Here, VB = 440 + (1226.4 - 1206)
= 460.4 cm3
MD = 1206 gm
w = 1gm/ cm3
∴Bulk Specific Gravity,
GB=( 1206/730)/ 1
= 2.62
Percentage of absorption = {(1226.4 -1206)/1206}*100
= 1.69 %
Answer: Apparent specific gravity 2.74, Bulk Specific Gravity 2.62, Percentage of absorption 1.69%
Apparent specific gravity is error... formula wrong
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