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7 Pages
BC-General-methods
Course: CEG 4012, Spring 2010School: FIU
Rating:
Word Count: 359
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lb/ft3) ( = sat ( lb/ft3) = DB (ft) = Dw (ft) = c ( lb/ft2) = Bearing Capacity Methods - Comparisons Given Variables 105 ()= 25 FS = 3.000 eB (ft) = 0.060 118 = 0 4 Q (lb) = 50000 eL (ft) = 0.060 MB (lb-ft) = eL / L = 2 3000 0.015 ML (lb-ft) = eB / B = 0.015 350 3000 B' (ft) = 2.329 L' (ft) = 3.400 A' (ft) = 7.917 B= 4 L= 4 Hansen Method Calculated Variables c ( lb/ft2) = Nc Fcs Fcd Fci q (lb/ft2) = Nq Fqs...
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lb/ft3) ( =
sat ( lb/ft3) =
DB (ft) =
Dw (ft) =
c ( lb/ft2) =
Bearing Capacity Methods - Comparisons
Given Variables
105
()=
25
FS =
3.000
eB (ft) = 0.060
118
=
0
4
Q (lb) =
50000 eL (ft) = 0.060
MB (lb-ft) =
eL / L =
2
3000
0.015
ML (lb-ft) =
eB / B = 0.015
350
3000
B' (ft) = 2.329
L' (ft) = 3.400
A' (ft) = 7.917
B=
4
L=
4
Hansen Method
Calculated Variables
c ( lb/ft2) =
Nc
Fcs
Fcd
Fci
q (lb/ft2) =
Nq
Fqs
Fqd
Fqi
350.00
20.79
1.52
1.40
1.00
321.20
10.73
1.47
1.31
1.00
B
N
Fs
Fd
Fi
129.47
6.85
0.60
1.00
1.00
Calculation of the ultimate bearing capacity (qu)
qu = c Nc Fcs Fcd Fci + qNq Fqs Fqd Fqi + (sat - w) B N Fs Fd Fi
qu =
350.00
321.20
x
+
55.60
qu =
x
x
4.00
20.79
10.73
x
x
x
6.85
1.52
1.47
x
x
x
0.60
1.40
1.31
x
x
1 +
x
1 +
1.00 x 1 =
22983
122574
Q=
OK
Terzaghi Method
Calculated & Obtained Variables
c ( lb/ft2) =
Nc
350.00
25.10
q (lb/ft2) =
Nq
321.20
12.70
B
N
129.47
9.20
Calculation of the ultimate bearing capacity (qu)
qu = 1.3(c)(Nc) + q(Nq) + 0.4()(B)( N)
qu =
17045
Q=
44983
Not OK
Meyerhof Method
Calculated & Obtained Variables
c ( lb/ft2) =
Nc
Fcs
Fcd
Fci
350.00
20.72
1.49
1.31
1.00
q (lb/ft2) =
Nq
Fqs
Fqd
Fqi
321.20
10.66
1.00
1.00
1.00
B
N
Fs
Fd
Fi
129.47
6.77
1.00
1.00
1.00
Kp =
2.46
x
x
6.77
1.31
1.00
x
Calculation of the ultimate bearing capacity (qu)
qu = c Nc Fcs Fcd Fci + qNq Fqs Fqd Fqi + 0.5 B Ng Fgs Fgd Fgi
qu =
+
0.50
1.00
qu =
Q=
350.00
321.20
x
x
19070
50326
OK
x
x
105.00
1
20.72
10.66
x
=
x
x
4.00
1.49
1.00
x
x
1 +
x
1 +
1.00 x
DETERMINATION OF () BASED ON SPT DATA
Standard Penetration number
Depth
Nf
'
5
10
15
20
25
4
6
6
10
5
0.188
0.327
0.466
0.605
0.744
N corr =
= 20 N corr + 20
= 34 o
N corr = N f
9
10
9
13
6
9
1
'v
B=
L=
4 Eb
4 El
Case I
0.2000 >= 1/6
0.1000 >= 1/6
el/L
eb/B
Case 2
1 < 0.5
0 <1/6
B1=
L1=
A'=
L'=
B'=
L'
B'
A'
0.060
0.060
Case 3
1 < 1/6
1 < 0.5
5.820 L1
5.820 L2
16.94 A'
4.000 L'
4.234 B'
0.000
3.400 B1
0.559 B2
7.92 A'
3.400 L'
2.329 B'
1.000
3.400
2.329
7.917
3.400
2.329
7.917
Eb/B
b
0.17
0.1
0.08
0.06
0.04
0.02
0.01
0
Eb/B
m
1.58
1.43
1.4
1.34
1.23
1.1
1.05
1
b
0.17
0.14
0.12
0.10
0.08
0.06
0.04
0.02
0.01
0
eL/L
m
b
0.17
0.1
0.08
0.06
0.04
0.02
0.01
0
0.92
0.85
0.84
0.79
0.72
0.7
0.63
0.6
1
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0.024
0.081
0.140
0.214
0.291
0.361
0.436
0.533
0.560
0.600
1
1
1
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
0
0
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
1.25
1.14
1.12
1.07
0.98
0.9
0.84
0.8
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
L2/L
-0.1
-0.19
-0.45
-0.71
-0.99
-1.27
-1.56
-1.74
-1.94
-2
m
1.58
1.43
1.4
1.34
1.23
1.1
1.05
1
#DIV/0! B2
#DIV/0! L2
#DIV/0! A'
#DIV/0! L'
#DIV/0! B'
0.000
L1/L
-3.3
-2.9
-2.82
-2.75
-2.55
-2.00
-2.10
-2
0.044
0.119
0.230
0.356
0.489
0.615
0.748
0.881
0.948
1
Case 4
0 < 1/6
1 < 1/6
L1/L
-3.3
-2.9
-2.82
-2.75
-2.55
-2.00
-2.10
-2
eL/L
b
0.17
0.14
0.12
0.10
0.08
0.06
0.04
0.02
0.01
0
eL/L
m
0.044
0.119
0.230
0.356
0.489
0.615
0.748
0.881
0.948
1
eB/B
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
L2/L
-0.1
-0.19
-0.45
-0.71
-0.99
-1.27
-1.56
-1.74
-1.94
-2
0.034
0.100
0.185
0.285
0.390
0.488
0.592
0.707
0.754
0.800
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
-0.01
0.03
0.1
0.2
0.37
0.52
0.67
1.3
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
-0.01
B2/B
0.100
0.100
0.100
0.100
0.100
0.100
0.100
0.100
0
#DIV/0!
eL/L
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
eB/B
B1/B
0.1000
0.1000
0.1000
0.1000
0.1000
0.1000
0.1000
0.1000
0.14
0.14
0.14
0.13
0.12
0.1
0.08
0.04
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0.14
0
#DIV/0!
0
1
#DIV/0!
#DIV/0!
#DIV/0!
4.000
#DIV/0!
0.000
0.92
0.85
0.17
0.1
0.07
0.85
0.07
0.14
0.21
0.12
0.1
0.07
0.14
0.02
1.25
0.17
0
#DIV/0!
0
0
#DIV/0!
0
0.16
0
0.16
0
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Finance 320 Problem Set #11. The rate on a banker's acceptance, quoted on a bank discount basis, is 6 percent. The securityhas a face value of$100,000 and will mature in 71 days. What is its price?2. AT-bill with 45 days to maturity has an Ask Yield of
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Finance 320 Problem Set #11. The rate on a banker's acceptance, quoted on a bank discount basis, is 6 percent. The securityhas a face value of$100,000 and will mature in 71 days. What is its price?2. AT-bill with 45 days to maturity has an Ask Yield of
Iowa State - FINANCE - 320
Expected Values);. Random variables are often described in terms of theirmeans and variances, which in turn are defined in terms ofthe expectations operator E.Elementary Statistics:A Review);. Assume that X" X2, , XN represent the N possibleoutcome
Iowa State - FINANCE - 320
Examplel>Example ContinuedSuppose we have two random variables, X and Y. X has amean of 0 and a variance of I. Y has a mean of2 and avariance of 4. The covariance between X and Yis-1.From Result 3, the mean of a sum is just the sum of themeans:l>
Iowa State - FINANCE - 320
,.Inflation Defined Inflation is an increase in the general price level.Risk and Return I The general price level is periodically measured from a standardbasket of goods. The Inflation Rate is the annual percent change in the price level. The most
Iowa State - FINANCE - 320
,L.~AnnualizationDollar- Weighted Returns Variance is proportional to time, t. What is the annualvariance of ABIT returns?cr~"1= Consider two portfolio managers (A and B) who haveidentical investment styles and skill.cr;"k!)' x 52 Assume over a