Civil Engineering Reference
In-Depth Information
1
2
3
4
5
0.95
2.44
3.32
3.50
1.74
2.35
2.35
2.35
2.35
2.35
42.9
110.1
149.8
157.9
78.5
-10
4
20
35
57
0.985
0.998
0.940
0.819
0.545
1.015
1.002
1.064
1.221
1.836
0.654
1.958
2.440
2.020
0.246
0.352
0.410
0.376
0.295
0.072
0.985
0.587
0.540
0.459
0.412
15.4
23.5
29.4
26.4
11.8
-0.174
0.070
0.342
0.574
0.839
-7.4
7.7
51.2
90.6
65.8
∑
106.5
∑
207.9
(a) Conventional method
(1)
(2)
(3)
(4)
(5)
(6)
sec
α
α
(4)
×
(5)
1
+
tan
φ
′
tan
α
F
φ
α
F
=
1.5 F
=
1.43
F
=
1.5
F
=
1.43
1
2
3
4
5
0.95
2.44
3.32
3.50
1.74
2.35
2.35
2.35
2.35
2.35
42.9
110.1
149.8
157.9
78.5
-10
4
20
35
57
-0.17
0.07
0.34
0.57
0.84
-7.4
7.7
51.2
90.6
65.8
∑
207.9
22.6
22.6
22.6
22.6
22.6
10.1
23.6
34.1
40.5
26.5
38.3
51.8
62.2
68.7
54.7
1.015
1.002
1.064
1.221
1.836
-0.18
0.07
0.36
0.70
1.54
1.061
0.986
0.978
1.043
1.337
1.063
0.985
0.974
1.036
1.319
40.6
51.1
60.9
71.7
73.1
∑
297.5
40.7
51.1
60.6
71.2
72.2
295.8
(b) Rigorous method
Fig. 13.20
Example
13.4.
The calculations for the conventional method are set out in Fig.
13.20a
:
θ
= °
89
π
⇒ ′ = ′ = ×
c l
c R
θ
12 9 15
.
×
× =
89
170 6
.
kN
180
106 5 170 6
207 9
.
+
.
F
=
=
1 33
.
.
The rigorous method calculations are set out in Fig.
13.20b
. With the first
approximation:
297 5
207 9
.
.
F
=
=
1 43
.
This value was obtained by assuming a value for F of 1.5 in the expression:
sec
tan
α
φ
′
tan
α
1
+
F
