Civil Engineering Reference
In-Depth Information
With
NP
=
=
13, the first wheel on the span
=
N1
=
9
x
1
=
(S
1
)
13
−
(S
1
)
9
=
74
−
48
=
26 ft
support B is 30
+
26
=
56 ft from N1
=
9
With
NP
=
13, the last wheel on the span is
NL
=
18
=
(L/
2
)
−[
(S
1
)
18
−
(S
1
)
13
]=
30
−
(
104
−
74
)
=
0 ft and N18 is over
xL
support B
NE
is the last wheel on the span and is equal to
NL
−
1
=
17 when
NL
is
over support B
M
17,9
+
P
9,17
(xL)
60
M
B
L
8448
+
(
310
−
52
)(
0
)
R
B
=
=
=
=
140.8 kips,
60
R
B
L
2
−
M
12,9
=
M
C
=
140.8
(
30
)
−
1636
=
2588 ft-kips.
With
NP
14 (Cooper's load configuration wheel number 14)
From Table 5.1:
With
NP
=
=
14, the first wheel on the span
=
N1
=
10
x
1
=
(S
1
)
14
−
(S
1
)
10
=
79
−
56
=
23 ft
support B is 30
+
23
=
53 ft from N1
=
10
14, the last wheel on the span,
NL
, is the beginning of the
uniform load,
w
xL
With
NP
=
=
(L/
2
)
−[
(S
1
)
w
−
(S
1
)
14
]=
30
−
(
104
+
5
−
79
)
=
0 ft from the begin-
ning of the uniform load,
w
, to support B
NE
is the last wheel on the span and is equal to the beginning of the
uniform load,
w
M
B
L
M
18,10
+
P
10,w
(xL)
60
8412
+
(
284
)(
0
)
60
R
B
=
=
=
=
140.2 kips,
R
B
L
2
−
M
13,10
=
M
C
=
140.2
(
30
)
−
1660
=
2546 ft-kips.
With
NP
=
=
15, the first wheel on the span
=
N1
=
11
x
1
=
(S
1
)
15
−
(S
1
)
11
=
88
−
64
=
24 ft
support B is 30
+
24
=
54 ft from N1
=
11
15, the last wheel on the span,
NL
, is the end of 9 ft of the
uniform load,
w
xL
With
NP
=
=
(L/
2
)
−[
(S
1
)
w
−
(S
1
)
15
]=
30
−
(
104
+
5
−
88
)
=
9 ft from the begin-
ning of the uniform load,
w
, to support B
NE
is the last wheel on the span and is equal to 9 ft of the uniform load,
w
Σ
M
18,11
+ Σ
Σ
M
B
L
P
11,
w
(xL)
7352
+[
(
264
)(
9
)
+
4
(
9
)(
9
/
2
)
]
R
B
=
=
=
60
60
=
164.8 kips,
R
B
L
2
M
14,11
=
M
C
=
−
164.8
(
30
)
−
2640
=
2305 ft-kips.
The maximum bending moment is 2588 ft-kips.
(NP
=
13
)
