e. Net Bearing Capacity if GWL = 3 feet deep
Here, the GWL is below the footing, therefore
Tult = yDfNq + 0.5By'Ny
= (120 pcf)(3.0 feet)(22.5) + (0.5)(3.0 feet)(120 pcf - 62.4 pcf)(19.7)
= 9802.08 psf
Tnet = Tult - yDf = 9802.08 - (120 pcf)(3.0 feet) = 9442.08 psf or 9.4 ksf
f. Net Bearing Capacity if GWL = 4 feet deep. For a water table located so that 0 =< d <= B, we use an equivalent unit weight i.e., yeff = [yd + y'(B-d)]/B in the part of the equation that represents the failure zone below the footing, where d is the depth of GWT measured from the base of the footing. In this case, d = 1.0'
Tult = yDfNq + (0.5)B(yeff)Ny
Tult = (120 pcf)(3.0')(22.5) + (0.5)[(120 pcf)(1.0') + (120 - 62.4)(1)](19.7)
= 8100 psf + 2316.72 psf
= 10416.72 psf or 10.4 ksf
Tnet = Tult - yDf = 10416.72 - (120 pcf)(3.0') = 10056.72 psf or 10.1 ksf
I am sure you'd be able solve part g and h without much help. If you still have doubts, I'll be more than happy to help you.
Also, I'd always draw a figure (if possible) before solving a problem. It makes things a bit clearer.
The 6 minute solutions uses a different equation for circular foundations than the DAS book. Can you reconcile these two equations?
ReplyDeleteDAS: 1.3cNc + qNq + 0.3yBNy
6 min: (1.2)cNc + qNq + 0.5yBNy(0.7)
Is it something that will allow me to get the correct answer on the test or are they different enough that the solutions could be similar and I would have to decide between to very close sol'ns?
I'll post a blog very soon to clarify all this. Thanks for raising this question tymr.
ReplyDeleteRegarding the following:
ReplyDeleteyeff = [yd + y'(B-d)]/B in the part of the equation that represents the failure zone below the footing, where d is the depth of GWT measured from the base of the footing. In this case, d = 1.0'
Tult = yDfNq + (0.5)B(yeff)Ny
Tult = (120 pcf)(3.0')(22.5) + (0.5)[(120 pcf)(1.0') + (120 - 62.4)(1)](19.7)
= 8100 psf + 2316.72 psf
Is the value of B missing from the second term?
I seem to be missing something.
I am sorry, I skipped a step there:
ReplyDeleteTult = yDfNq + (0.5)B(yeff)Ny
from
yeff = [yd + y'(B-d)]/B
we can write
B(yeff) = [yd + y'(B-d)]
by bringing B to the left side of the equation
substitute the value of B(yeff) from above in Tult equation:
Tult = (120 pcf)(3.0')(22.5) + (0.5)[(120 pcf)(1.0') + (120 - 62.4)(1)](19.7)
= 8100 psf + 2316.72 psf
I hope that helps.
I'm confused in calculating the effective vertical stress in a sandy gravel soil layer below the graound water level and it is between a sand layer on the top and a clay layer on the bottom.
ReplyDeleteWhat unit weight i have to consider the sand unit waight or I need the porosity(n)of the sandy gravel to get the unit waight?
Thanks
suppose we have a situation like the following:
ReplyDelete_____________________________________________
Sand dry unit weight = Ys,
layer thickness = d1
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
water table @ depth dw, unit weight of water = Yw
______________________________________________
Sandy Gravel
saturated unit weight = Ysg
layer thickness = d2
______________________________________________
Clay
saturated unit weight = Yc
layer thickness = d3
______________________________________________
Effective Stress = Total Stress - Pore Water Pressure
Total Stress = (Ys)(d1) + (Ysg)(d2) + (Yc)(d3)
Pore water pressure = (Yw)(d1-dw) + (Yw)(d2) + (Yw)(d3)
Therefore, Effective Stress =
(Ys)(d1) + (Ysg)(d2) + (Yc)(d3) - (Yw)(d1-dw) - (Yw)(d2) - (Yw)(d3)
= (Ys-Yw)(d1)+(Ysg-Yw)(d2)+(Yc-Yw)(d3)+(Yw)(dw)
Let me know if it helps (or if it does not).