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159
Chapter 16
16.1
.28
°
=
′
φ
From Table 16.1, =
c
N31.61;
=
q
N
17.81;
=
γ
N
13.70
2
kN/m 499.4=
++=
++
′
==
2
1
5.3
1
2
1
5.3
1
all γqc
s
u
γBNqNNc
F
q
q
16.2 For square footing, Eq. (16.12) is used.
16.3
.35
°
=
′
φ
From Table 16.1, =
c
N57.75;
=
q
N
41.44;
=
γ
N
45.41
16.4
.0
°
=
′
φ
From Table 16.1, =
c
N5.7;
=
q
N
1;
=
γ
N
0
2
1
1
u
q
160
16.5
.28
°
=
′
φ
From Table 16.2, =
c
N25.80;
=
q
N
14.72;
=
γ
N
16.72
For a continuous foundation with vertical loading, all inclination factors and
shape factors are equal to one. So,
16.6
.28
°
=
′
φ
From Table 16.2, =
c
N25.80;
=
q
N
14.72;
γ
N
= 16.72
s
F
For vertical loading, all inclination factors are equal to one.
γs
161
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1=
γd
λ
2
kN/m 745.5
=
+
+
=
)1)(6.0)(72.16)(0.2)(19(
2
1
)224.1)(531.1)(72.14)(5.1)(19()3.1)(57.1)(80.25)(31(
5.3
1
all
q
16.7 γ = 115 lb/ft
3
;
1100
=
′
c
lb/ft
2
;
°
=
′
35
φ
; D
f
= 3.5 ft; B = 5 ft; factor of safety = 4
+
)1)(03.48)(0.5)(115(
2
1
4
all
16.8 γ = 16.5 kN/m
3
; 41=
u
c kN/m
3
;
0
=
′
φ
; D
f
= 1.5 m; factor of safety = 5
162
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
3.1
0.2
5.1
4.014.01 =
+=
+= B
D
λ
f
cd
1)sin1(tan21 2=
′
−
′
+=
B
D
λf
qd
φφ
1=
γd
λ
2
kN/m 59.7=
++= )1)(0)(0.2)(5.16(
2
1
)1)(1)(5.1)(5.16()3.1)(14.5)(41(
5
1
all
q
16.9 Eq. (16.12):
γqcu
BNγqNNcq
′
++
′
=4.03.1
16.10 Since the groundwater table is 0.25 m below the footing, it represents Case III in
Figure 16.6.
163
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4.18=++= )]84.9)(5.1)(66.10)(4.0()21.14)(17()09.27)(32)(3.1[(
771
5.1
2
s
F
16.11
.29
°
=
′
φ
From Table 16.1, =
c
N34.24;
=
q
N
19.98;
=
γ
N
16.18
all
s
F
16.12
°
=
′
27
φ
;
38
=
′
c
kN/m
2
; B = 2.25 m; D
f
= 2 m; α = 12°
Using Eq. (16.31) which provides the vertical component of the inclined loading,
++
′
=
γiγdγsγqiqdqsqcicdcsc
s
λλλγBNλλλqNλλλNc
F
q2
11
all
90
90
qi
°
ci
164
509.127tan1tan1 =+=
′
+=
φ
L
B
λ
qs
27.1
25.2
2
)27sin1(27tan21)sin1(tan21
22
=
−+=
′
−
′
+= B
D
λ
f
qd
φφ
12
cos
all(i)
16.13 γ = 17 kN/m
3
;
48
=
′
c
kN/m
2
;
°
=
′
31
φ
; factor of safety = 4.5
1
165
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
B
Bq
22.3435.751
)]65.22)()(17)(4.0()28.25)(17)(2()41.40)(48)(3.1[(
5.4
1
all
+=
++=
(a)
2
all
1160
B
q=
(b)
From Eqs. (a) and (b),
B
B
22.3435.751
1160
2
+=
By trial and error, B ≈
1.21 m
16.14 γ = 121 lb/ft
3
;
0
=
′
c
;
°
=
′
26
φ
; factor of safety = 2.5
1
16.15 γ = 17 kN/m
3
;
48
=
′
c
kN/m
2
;
°
=
′
31
φ
; factor of safety = 4.5
++
11
Trial 1: Assume B = 1.25 m
631.1
67.32
63.20
11 =
+=
+=
c
q
cs
N
N
L
B
λ
404.1
25.1
2
tan)4.0(1tan)4.0(1
11
=
+=
+=
−−
B
D
λ
f
cd
6.131tan1tan1 =+=
′
+=
φ
L
B
λ
qs
286
.
1
25.1
2
tan)31sin1(31tan21tan)sin1(tan21
1212
=
−+=
′
−
′
+=
−−
B
D
λ
f
qd
φφ
6.04.01 =
−= L
B
λ
γs
γd
167
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1176)1)(5.1176())((
22
allall
≈== BqQ
kN, which is slightly higher than 1160 kN,
and therefore acceptable for design.
Design B
= 1 m
16.16 γ = 16 kN/m
3
;
0
=
′
c
;
°
=
′
29
φ
, D
f
= 1.3 m; B = 1.75 m; and x = 0.25 m
From Table 16.2:
=
q
N
16.44;
=
γ
N
19.34
1
all
s
F
16.17 γ = 118 lb/ft
3
,
700
=
′
c
lb/ft
2
,
°
=
′
33
φ
, D
f
= 4.5 ft, B = 6.25 ft, and x = 0.75 ft
168
For vertical load [Eq. (16.44)]:
1
4
all
16.18 γ = 16 kN/m
3
;
0
=
′
c
;
°
=
′
29
φ
, D
f
= 1.3 m; B = 1.75 m; and x = 0.25 m
From Table 16.2:
=
q
N
16.44;
=
γ
N
19.34
1
Eq. (16.48):
( )
(eccentric) (centric) (centric)
1 1
k
u u u k
e
Q Q a Q R
B
= − = −
(eccentric) (centric)
u u k
16.19 γ = 17 kN/m
3
;
0
=
′
c
;
°
=
′
34
φ
, D
f
= 1.2 m; B = 1.8 m; e = 0.3 m; and α = 10°
170
B′ = 1.8 – (2)(0.3) = 1.2 m
Since it is a continuous foundation, all shape factors are equal to one.
1
)(
ueiu
B
′
φ
171
u
q
is the ultimate bearing capacity with vertical centric load. Since it is a
continuous foundation, all shape factors are equal to one.
1
CRITICAL THINKING PROBLEM
C.16.1 The footing is placed at a depth of 1.5 m.
a.
B = 1 m
Eq. (16.49):
=25
)mm(
05.0
60
net
e
d
S
F
N
q
net
25
05.0
172
kN 71
3
)1)(8.212(
3
))((
2
net
net-all
≈== Aq
Q
B = 1.5 m
+
)mm(3.0
2
60
e
S
BN
3
3
net-all
B = 2 m
+
)mm(3.0
2
60
e
S
BN
net
25
2
08.0
173
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kN 229
3
)2)(7.114(
3
))((
2
net
net-all
≈== Aq
Q
B = 3 m
Eq. (16.50):
+
=25
)mm(3.0
08.0
2
60
net
e
d
S
F
B
BN
q
3
3
net-all
The design chart is shown.
174
b. For any design footing size B,
appliednetall
QQ ≥
−
(1)
c.
φ
′ = 33º; cʹ = 0;
=
q
N
32.33;
=
γ
N
31.94 (Table 16.1); B = 2.25 m
−
netallnet-all
d. The net allowable column load calculated by Terzaghi’s bearing capacity
equation (Part c) is significantly higher than that calculated by the method
