HW2 Solution.pdf

# HW2 Solution.pdf - Q No 1(Solution Nfield = Penetration...

This preview shows pages 1–3. Sign up to view the full content.

Q No. 1 (Solution) Nfield = Penetration against one feet i.e Neglect number of blow for first 6”. The Number of blows for nex 1 ft. Is N field At depth of 8.5 ft Nfield = 6+3 =9 ? 60 = 𝐸 ? ? ? ? ? ? ? ? ? ? ? ? 0.6 Where Em= Hammer Efficiency = 0.74 (Supposed Value) CB= Bore Hole Correction Factor If Borehole diameter = 2.5-in to 4-in C B = 1.0 Borehole diameter = 6-in C B = 1.05 Borehole diameter = 8-in C B = 1.15 So CB= 1.15 Cs= Sample Correction factor Cs= 1.0 for SPT and 1.2 for RING CR= Rod Length Correction Factor if C R = 0.8; 0ft < L ≤ 12ft C R = 0.015L + 0.61; 12ft < L ≤ 20ft C R = -0.000037L 2 + 0.005L + 0.83 20ft < L ≤ 65ft C R = 1.0; L > 65f So CR= 0.8 Putting all Values ? 60 = 0.73 x 1.15 x 1.0 x 0.8 x 9 0.6 = 9 ? 1 60 = ? 60 ∗ ( 2000 ? 𝑍 ) 0.5 ? ? = ϒ ? (1 + 𝑤%) ????ℎ ? ? = 98.5 (1 + 6.6%) 8.5 = 892.51 ? 1 60 = 9 ∗ ( 2000 892.51 ) 0.5 = 14 Relative Density/ Consistency = Fine Grained Soil So Consistency is Stiff See Table 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Table No. 1 Depth ft Sample Type USCS W% Dry Unit Weight Stress blows per 6" Nfield Em CB Cs CR N 60 N1) 60 Relative Density/ Consistency 2 SPT SM 5.6 90 190.08 2 9 0.74 1.15 1 0.8 10 33 Stiff 3 6 5 RING SC 7.9 92.7 500.1165 7 16 0.74 1.15 1.2 0.8 22 44 Stiff 8 8 8.5 SPT SC 6.6 98.5 892.5085 1 9 0.74 1.15 1 0.7375 9 14 Stiff 3 6 13.5 RING SC 10.7 99.4 1485.483 8
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern