gotlboys
Civil/Environmental
- May 31, 2015
- 61
Metric Units
Meyerholf's formula when B<=1.22m ----> qa = 12NFd where Fd=1+0.33D/B <=1.33 and B>=1.22m ---> qa=8N[(B+0.305)/B]^2(Fd).
Assume N=5 blows,D=2m and B>1.22m. The only variable we can play with is B. It could be larger or smaller but not less than 1.22m for second eq applicability.
Trial 1. Let B=1.5m, Fd=1+0.33(2/1.5)=1.44, use 1.33. Thus qa=8*5[(1.5+0.305)/1.5]^2(1.44)=77.03 kPa
Trial 2. Let B=2m, Fd=1.825, use 1.33. Thus qa=70.66 kPa
Can anyone explain what is the implication of the difference between the trials given that B=2m is larger in width but gives lesser allowable bearing capacity?
Meyerholf's formula when B<=1.22m ----> qa = 12NFd where Fd=1+0.33D/B <=1.33 and B>=1.22m ---> qa=8N[(B+0.305)/B]^2(Fd).
Assume N=5 blows,D=2m and B>1.22m. The only variable we can play with is B. It could be larger or smaller but not less than 1.22m for second eq applicability.
Trial 1. Let B=1.5m, Fd=1+0.33(2/1.5)=1.44, use 1.33. Thus qa=8*5[(1.5+0.305)/1.5]^2(1.44)=77.03 kPa
Trial 2. Let B=2m, Fd=1.825, use 1.33. Thus qa=70.66 kPa
Can anyone explain what is the implication of the difference between the trials given that B=2m is larger in width but gives lesser allowable bearing capacity?