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Chapter 14
14.1 a. Refer to Figure 14.9 in the text.
L1 = 4 m; L2 = 8 m; = 32
)948.2)(39.8(
2A
134
14.2 a. Refer to Figure14. 9.
333.01545tan
2
45tan 22
a
K
;
31545tan2
p
K
)667.2)(81.94.19(
)( 2
3
ap KK
Chapter 14
135
43 LLD
Chapter 14
14.3
3 ;333.0
2
45tan2
pa KK
2
2kN/m 16.68)333.0)(3)(7.16(a
LK
Chapter 14
137
)15.28)(2(
2
P
14.4 a. Refer to the following figure:
Chapter 14
138
)4.26()3.2)(78.57()4.5)(07.15(
14.5
271.0
2
35
45tan
2
45tan 22
a
K
3
3
1 L
Chapter 14
139
)12(
2
zcPP
14.6 a. Refer to the figure. = 34.
Chapter 14
©2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
140
283.0
2
45tan2
a
K
;
537.3
2
45tan2
p
K
2
11 kN/m 19.244)283.0)(4)(17(a
KL
2
212 kN/m 42.65)283.0)](9)(81.919()4)(17[()(
a
KLL
m 1.43
)254.3)(19.9(
65.42
)( 2
3
ap KK
L
kN/m 7.514349.3033.1052.17349.384321 of Areas
1P
m 5.68
)]95.0)(49.30()33.4)(33.105()93.5)(2.173()76.11)(49.38[(
51.347
1
z
Eq. (14.67):
0
)(
)()[(3
)(5.1 1321
322
2
4
3
4
ap KK
lzLLLP
LLlLL
0
)254.3)(19.9(
)268.5()43.194)[(51.347)(3(
)43.192(5.1 2
4
3
4
LL
m 3.3 ;032.235645.18 4
2
4
3
4 LLL
m 4.73 1.433.33.3 3
LD
14.7 a. Dactual = (1.3)(Dtheory) = (1.3)(4.75) 6.15 m
Chapter 14
141
1
12
1 Lxz
Taking the moment about the point of zero shear,
1
4
12
2
x
Mmax 759 kN-m/m
b. H = L1 + L2 + Dactual = 4 + 9 + 6.15 = 19.15 m 20 m
14.8 a.
333.0
84
4
;35 ;167.0
84
2
21
1
21
1
LL
L
LL
l
Chapter 14
©2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
142
From Figure 14.19, GD = 0.18; from Figure 14.22, CDL1 = 1.144
Eq. (14.70): D = (L1 + L2)(GD)(CDL1) = (12)(0.18)(1.144) = 2.47 m
b. From Figure 14.20, GF 0.059. From Figure 14.23, CFL1 = 1.071
2
22
2
21
21
2
2
2
1
)12(
)8)(4)(16)(2()8)(81.95.18()4)(16(
)(
2
LL
LLLL
a
c. From Figure 14.21, GM = 0.018. From Figure 14.24, CML1 = 1.026
14.9
3
sat kN/m 9.6981.95.19 w
Chapter 14
143
2 DD
m/m-kN 519.1
11
11
11
14.10 a. Refer to the figure.
Chapter 14
©2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
144
2
212 kN/m 26.5)307.0)](6)(81.986.18()2)(16[()(
a
KLL
2
1kN/m 76.1185094.58824.9321 of Areas P
m 2.88
76.118
10082.17653.65
76.118
)2)(50()3)(94.58()67.6)(824.9(
1
z
Eq. (14.83):
0)(2)(2 112111216
2
6 zlLLPlLLDD
2
216 kN/m 21.73.861083.86)27)(4()(4
LLc
So
0)88.2162)(76.118)(2()7)(7.21)(2(7.21 2 DD
. Or
;01.4514
2 DD
D 2.8 m
b.
FDP 61
118.76 – (21.7))(2.8) = 58 kN/m
14.11 From Figure 14.38(a), for = 30, the value of Ka 0.31.
W = H t concrete = (1.52)(0.076)(23.58) = 2.72 kN/m
Eq. (14.87):
Chapter 14
©2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Eq. (14.88):
3
kN/m 59.9)59.62(
91.0 52.1
14
114
1
u
ov
ov
us P
h
H
C
C
P
375.0
91.052.1
22.113.2
hH
BS
Referring to Figure 14.40(b),
m 1.6822.1)43.2)(19.0( ;19.0
e
eB
hH
BB
.
kN 100.6
)68.1)(9.59(
eusuBPP
14.12 Eq. (11.92):
9.0
4.5
tan
4.5
28.0
2
28.0
2
BhH
Bh
H
P
u
Chapter 14
146
