Chapter 6
6.1 The following table can now be prepared. Note: B = 2 m (diameter); z = 3 m;
qo = 150 kN/m2
2
B
z
r (m)
2
B
r
a
o
q
 (kN/m2)
3
3
3
3
0
0.4
0.8
1
0
0.4
0.8
1
0.146
0.141
0.127
0.118
21.9
21.15
19.05
17.7
a Table 6.1
6.2 See Eq. (6.1).
Load
@
r
(m)
z
(m)
z
r
2/5
2
12
3
z
r
5.2
2
2
12
3
z
r
z
P
z
(kN/m2)
B
C
D
6
(62 + 62)0.5 = 8.48
(62+32) = 6.708
6
6
6
1.0
1.41
1.118
0.0844
0.0309
0.0629
0.234
0.172
0.699
z = 1.105 kN/m2
6.3 Eq. (6.4):
50
50
6.4 B = 36 ft; q = 900 lb/ft2; x = 27 ft; z = 15 ft
6.5
. :2) Eq.(6.1; :Eq.(6.11) z
L
n
z
B
m
Refer to areas in Figure 6.6.
2
lb/ft 1365.66
6.6 Refer to Figure 6.6.
Chapter 6
51
6.7 L = L1 + L2 = 8 + 10 = 18 ft; B = B1 + B2 = 4 + 6 = 10 ft; z = 10 ft
10
18 11
z
L
6.8 Referring to the figure shown and Eq. (6.28):
av
Chapter 6
52
52
6.9 Eq. (6.14):
coIq
. qo = 4000 lb/ft2.
ft 5 ;
2
;1 1 1
B
B
z
n
B
L
m
z (m)
qo (lb/ft2)
m1
n1
Ic (Table 6.5)
 (lb/ft2)
3
4000
1
1.2
0.606
2424 = t
8
4000
1
3.2
0.165
660 = m
13
4000
1
5.2
0.068
272 = n
Eq. (6.29):
2
lb/ft 889
6
272)660)(4(2424
av
6.10
2
2lb/ft 5.5261
)35(
200050
t
6.11 Point A:
Chapter 6
53
Point B:
6.12 Eq. (6.3):
2/3
2
2
1
1
1
z
B
qo
. qo = 150 kN/m2; B = 2m
Chapter 6
54
54
2
1
av
6.13 Referring to Figure 6.15:
6
2
2
6
2
;3
2
2
3
2
21
B
H
B
H