The heights of two towers are recorded as (10.2 ± 0.4) m and (14.1 ± 0.3) m. The difference in heights of the towers with error limits is
(i)2.5 ms⁻² (ii) 0 (iii) 6 ms⁻² (iv) 2100 m
(i)1.5 ms⁻² (ii) 0 (iii) 6 ms⁻² (iv) 2100 m
(i)2.5 ms⁻² (ii) 0 (iii) 6 ms⁻² (iv) 1100 m
(i)1.5 ms⁻² (ii) 0 (iii) 8 ms⁻² (iv) 2100 m
Answers
Answer:
sampling tube has an outside diameter Do of 3.00 inches, a tip diameter De of 2.84 inches, and a wall thickness of 0.065 inches. Calculate the clearance ratio, area ratio, and indicate if the sampling tube meets the criteria for undisturbed soil sampling.
Solution:
Di = Do − 2 t (where t = wall thickness)
Di = 3.00 − (2)(0.065) = 2.87 in.
Do = 3.0 in.
From Table 2.8, inside clearance ratio = 100 (Di −De)/De = 100 (2.87 − 2.84)/2.84 = 1.06%
From Table 2.8, area ratio = 100 (Do2 −De2)/De2 = 100 (32 − 2.842)/2.842 = 11.6%
These values are close to meeting the criteria for undisturbed soil sampling.
2.2 A standard penetration test SPT was performed on a near surface deposit of clean sand where the number of blows to drive the sampler 45 cm was 5 for the first 15 cm, 8 for the second 15 cm, and 9 for the third 15 cm. Assume that Em = 60% (i.e., Em = 0.60), the borehole diameter is 100 mm, and the drill rod length is 5 m. Calculate the measured SPT N-value (blows per foot), N60, and (N1)60 assuming that the vertical effective stress σ′vo = 50 kPa. Also indicate the density condition of the sand.
Solution:
N-value = 8 + 9 = 17
For 100 mm borehole diameter,Cb = 1.0
For drill rod length = 5 m,Cr = 0.85
From Table 2.11: N60 = 1.67 Em Cb Cr N = (1.67)(0.60)(1.0)(0.85)(17) = 14.5
From Table 14.22: (N1)60 = CN N60 = (100/σ′vo)0.5N60 = (100/50)0.5 (14.5) = 20.4
Per Table 2.12, for N60 = 14.5, the sand is in a medium condition
2.3 A field vane shear test was performed on a clay, where the rectangular vane had a length H of 4.0 in. and a diameter D of 2.0 in. The maximum torque Tmax required to shear the soil was 8.5 ft-lb. Calculate the undrained shear strength su of the soil.
Solution:
H = 4 in. = 0.333 ft D = 2 in. = 0.167 ft
From Table 2.13: su = Tmax/[π (0.5 D2H + 0.167 D3)] = 8.5/[π (0.5 (0.167)2 0.333 + 0.167 (0.167)3)] = 500 psf
2.4 A construction site in New England requires excavation of rock. The geologist has determined that the rock is granite and from geophysical methods (i.e., seismic refraction), the seismic velocity of the in situ granite is 12,000 to 15,000 feet per second. A Caterpillar D11R tractor/ripper is available. Can the granite be ripped apart?
Solution:
D11R tractor/ripper, therefore use Fig. 2.17. For granite with a seismic velocity = 12,000 to 15,000 feet per second, it is non-rippable.
2.5 A plate load test is performed where a 1-foot wide square steel plate is subjected to a vertical stress of 2,000 psf and the recorded depth of penetration of the steel plate is 0.25 inch. If the width of the actual square footing will be 10 feet, calculate the settlement of the footing if it is subjected to a vertical stress of 2,000 psf.
Solution:
S1 = 0.25 in., D1 = 12 in., D = (10)(12) = 120 in.
From Table 2.13: S = 4 S1/(1 +D1/D)2 = (4)(0.25)/[1 + (12/120)]2 = 0.83 in