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eric08
I think I'm losing my mind! This problem gives the total unit weight as 127 lb/ft3 and the unit weight of the solids as 165.4. Is this possible?
PLEASE help!
PLEASE help!
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Guest
Eric --
The short answer is yes. How is this possible?
Let's first consider the specific gravity for each of the three possible phases:
So, putting that all together,
Mathematically this means Unit Wt = Mass/Volume, where the volumetric contributions of water and void space (are) are RELATIVELY weighted more which means that the unit weight of solids will typically be LESS THAN the total unit weight.
Total Unit Weight = (Mass of Solids + Water + Void Space)/(Volume of Solids + Water + Void Space)
Therefore, while the RELATIVE mass contributions of the water and void space may not substantially affect the total unit weight calculation, the volumetric contribution certainly will depending on how much saturation and void space is contained within your soil sample.
I hope this helps - please let me know if a sketch might be more useful for illustrating the concept.
Regards,
JR
The short answer is yes. How is this possible?
Let's first consider the specific gravity for each of the three possible phases:
- Specific Gravity for Solids ~ 2.7 (give or take)
- Specific Gravity for Water = 1.0 (by definition)
- Specific Gravity for Void Space (Air) ~ 0 (Infintely small)
So, putting that all together,
Mathematically this means Unit Wt = Mass/Volume, where the volumetric contributions of water and void space (are) are RELATIVELY weighted more which means that the unit weight of solids will typically be LESS THAN the total unit weight.
Total Unit Weight = (Mass of Solids + Water + Void Space)/(Volume of Solids + Water + Void Space)
Therefore, while the RELATIVE mass contributions of the water and void space may not substantially affect the total unit weight calculation, the volumetric contribution certainly will depending on how much saturation and void space is contained within your soil sample.
I hope this helps - please let me know if a sketch might be more useful for illustrating the concept.
Regards,
JR
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eric08
I think we're making the same point. because their giving me a total unit wieght that is smaller than the unit wieght of the solids (total=127, solids=165.4)
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jartgo
I don't have the problem to look at, but I think you missed jr's point above.
I'm assuming that what you're reporting as "total unit weight" is the in-situ or dry unit weight (127), either way it includes void spaces, whether it be water or air...the matter filling those void spaces has a unit weight less than that of the solids (165.4). So, a cubic foot of the soil weighs less than what a cubic foot of the solids would weigh.
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eric08
Problem #6 ( 6 min. sol.) Given the phase diagram shown, what is most nearly the porosity?
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jartgo
yeah, so the unit weight of the solids is 165.4 #/c.f. and the total unit weight of the soil is 127 #/c.f., as explained above. In one cubic foot of soil, you have less than a cubic foot of solids. If you still don't understand, you'd probably benefit from reading a little from a soil mechanics book. This topic can be confusing.
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eric08
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Guest
^^^ Does it make sense now?
JR
JR
squishles10
CEO
woohoo! i'm not the only one studying!!!!!
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eric08
owillis28
Well-known member
In the solution, it states "By assuming a unit value for the total volume, an equation for the void ratio can be simplified to determine the porosity." How did they do that?????????????????????
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I don't know what they mean, but I can show you the steps I would use to solve this problem.
To solve three-phase problems, you need to know 6 factors: the weight of air, water, and solids; and the volume of air, water, and solids. You can assume that the weight of air is zero. That leaves five factors.
The problem states that the density of solids is 165.4 lb/ft3. Assume that you have a sample with 1 ft3 of solids. Now you know the weight of solids (165.4 lb) and the volume of solids (1 ft3) in the sample. That leaves three factors.
The problem states that the water content is 10%. Since we know the weight of solids (165.4 lb), we can calculate the weight of water in the sample (16.54 lb). Now we know the weights of all three phases in the sample: soil (165.4 lb), water (16.54 lb), and air (0 lb). Total weight of the sample is 181.9 lb.
Since we know the weight of water (16.54 lb) and the density of water (62.4 lb/ft3), we can calculate the volume of water (0.265 ft3). Now we know five of the six factors, but we are still missing the volume of air.
The problem states that the total density is 127.0 lb/ft3. We know that our sample has a total weight of 181.9 lb. Since density is weight/volume, we can calculate the total volume (1.432 ft3). We can subtract the volume of solids (1 ft3) and the volume of water (0.265 ft3) to get the volume of air (0.167 ft3). Now we know the weights and volumes of all three phases. The total volume of the sample is the sum of the volumes of soil (1 ft3), water (0.265 ft3) and air (0.167 ft3), or 1.432 ft3.
The porosity is the volume of voids divided by the total volume. The volume of voids is the volume of water (0.265 ft3) plus the volume of air (0.167 ft3), or 0.432 ft3. The porosity is 0.432 ft3 / 1.432 ft3, or 0.30 (30%).
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