Civil Engineering Reference
In-Depth Information
10
−4
Substitution into Equation 4.43 yields
Δ
x
s
=−
6.50
×
[
(
1
/(
5.16
×
10
−4
))
+
(
1
/(
2.58
10
−4
))
]=−
3.78 in., which may be excessive, requiring a
longitudinally stiffer deck or track. For example, if a frozen ballast with
k
t
=
×
100 lb/in. (rapid strain rate) is considered, the rail gap is reduced to 3.0 in.
Relative displacement
:
Substitution into Equation 4.48 with
n
=
2 yields
0.34
1
e
−λ
L
)
e
−λ
L
Δ
x
=
+ λ
L
−
(
λ
L
+
Δ
x
=
0.34 [1
+
0.39
−
(
0.39
+
0.67
)
0.67]
=
0.23 in., which is likely OK and will
not cause deck fastener damage.
Fixed bearing force at the pier
:
Substitution into Equation 4.47 yields
t
α
0
Δ
C
2
)
T
X
F
=
N
4
(
0
)
−
N
3
(
0
)
=−
EA
r
αΔ
t
(C
3
−
2
αΔ
=−
487,500
(
0.385
)(C
3
−
C
2
)
l
d
L
=
0.39
e
−λ
L
C
1
=
=
0.67
C
1
)
e
−λ
L
C
2
=
(
λ
d
L
+
=
0.72
C
2
)
e
−λ
L
C
3
=
(
λ
d
L
+
=
0.75
X
F
=
187,688
(
0.75
−
0.72
)
=
5630 lb for both bearings.
For bridges with short spans, the amount of thermal movement per span is
smallandgenerallyeasilyaccommodatedbythenormaltolerancesofrailroad
track on bridges. It should be noted that the longitudinal stiffness values
assumed for the rail-to-deck and deck-to-superstructure interfaces may not
reflect values at a particular railroad location.
Example 4.18
An open deck steel deck truss bridge comprises a single 225 ft span. The CWR
with elastic rail fastenings is used on the friction-bolt fastened timber deck.
Determine the maximum stress in the CWR, relative displacement between
the rail and superstructure, rail separation, and longitudinal bearing force at
the abutment. The following are characteristics of the bridge:
70
◦
F
Δ
T
c
= Δ
t
c
=−
50
◦
F
Δ
t
h
=
40
◦
F
Δ
T
h
=
10
−4
(bridge)
α
0
Δ
T
c
=−
3.50
×
10
−4
(CWR)
αΔ
t
c
= αΔ
t
c
=−
4.55
×
10
−4
(bridge)
α
0
Δ
T
h
=
2.00
×
10
−4
(CWR)
αΔ
t
h
=
3.25
×
10
6
(
26
)
=
10
8
lb (for two typical CWRs)
EA
r
=
29
×
7.5
×
k
d
=
400 lb/in. (normal strain rate)
k
t
=
100 lb/in. (normal strain rate)
