Longitudinal forces, due to train braking (acting at the center of gravity of the live load) and locomotive tractive effort (acting at the freight equipment drawbars or couplers), are considerable for modern railway freight equipment. Longitudinal forces from railway live loads exhibit the following characteristics (Otter et al., 2000):
• Tractive effort and dynamic braking forces are greatest when accelerat-ing/decelerating at low train speeds.
• Span length does not affect the relative magnitude of braking forces, due to the distributed nature of emergency train braking systems.
• Traction forces from locomotives may affect a smaller length of the bridge.
• Participation of the rails is relatively small (particularly when the bridge and approaches are loaded) due to the relatively stiff elastic fastenings used in modern bridge deck construction.
• The ability of the approach embankments to resist longitudinal forces is reduced when the bridge and approaches are loaded.
• Grade-related traction is relatively insignificant for modern high adhesion locomotives.
The locomotive and car wheels may be modeled as accelerating or decelerating rolling∗ masses that do not slide (complete adhesion†) as they traverse the bridge superstructure. The forces created by the vertical, horizontal, and rotational translation of the rolling mass are shown inFigures 4.10a–c.The longitudinal traction forces applied to the superstructure may be determined by superposition of the vertical, horizontal, and rotational effects of the rolling mass for linear elastic structures.
Neglecting axle bearing and wheel rim friction,‡ the force equilibrium relating to the vertical effects of rolling motion, considering complete adhesion (no sliding), provides (Figure 4.10a)
W− mF
d2y(t)
dt2 − RV(t)= 0. (4.19)
∗Rolling is the superposition of translation and rotation (Beer and Johnston, 1976).
†Nonuniform speed (acceleration for starting and deceleration for braking) and adhesion must exist between the wheel and rail interface to start and stop trains.
‡Axle bearing and wheel rim friction are very small in comparison to rolling friction.
108 Design of Modern Steel Railway Bridges
x x = Vt
L y
y (t) W
ar αr
P
RV(t) RH(t) mF
–RV(t) = vertical force on beam from wheel rolling motion
r
FIGURE 4.10a Vertical effects of concentrated rolling mass on a simply supported span.
x x = Vt
L y
x (t)
–HLF(t) = horizontal force on beam from wheel rolling motion
r mF TF(t)
HLF(t)
RT(t)
FIGURE 4.10b Horizontal effects of concentrated rolling mass on a simply supported span.
x x = Vt
L y
r
ar Q (t)
HLF(t) RH(t) Ip
RV(t) MT(t)
FIGURE 4.10c Rotational effects of concentrated rolling mass on a simply supported span.
Loads and Forces on Steel Railway Bridges 109
The horizontal reaction at the wheel axle, P, is
P= RH(t)= War
r , (4.20)
where W= mFg is the weight of the concentrated force, mFis the mass of the concen-trated force, r is the wheel radius, and RV(t) and RH(t) are the vertical and horizontal components, respectively, of the reaction force due to rolling friction. The resultant reaction force, R(t), is located at a horizontal distance, ar, from the wheel centroid as a result of rolling friction (McLean and Nelson, 1962). The distance ar is often referred to as the coefficient of rolling resistance. Rolling friction is small at constant train speed and greater at nonuniform train speeds. The horizontal component of the reaction, RH(t), is generally small because the applied vertical forces greatly exceed applied horizontal forces. Neglecting axle bearing and wheel rim friction again, the force equilibrium relating to the horizontal effects of rolling motion, considering complete adhesion, yields(Figure 4.10b)
HLF(t)− RT(t)+ TF(t)− mF
d2x(t)
dt2 = 0, (4.21)
where HLF(t) is the longitudinal force transferred to rails and deck/superstructure and RT(t) is the resistance to horizontal movement (primarily air resistance or vehicle drag forces since axle bearing and wheel flange friction is considered negligible). RT(t) is generally relativity small in comparison to other horizontal forces and it is not too conservative to neglect this force. TF(t) is the locomotive traction force and is equal to (MT(t)c/r), where MT(t) is the driving torque applied to wheel, and cis a constant depending on locomotive engine characteristics and gear ratio.
Therefore, Equation 4.21 may be simplified to
HLF(t)+ TF(t)− mF
d2x(t)
dt2 = 0. (4.21a)
Also, neglecting axle bearing and wheel rim friction, the force equilibrium relating to the rotational effects of rolling motion, considering complete adhesion, provides (Figure 4.10c)
−rHLF(t)+ MT(t)− arRV(t)+ rRH(t)− Ip
d2θ(t)
dt2 = 0, (4.22) where Ipis the rotational moment of the inertia of mass.
Since the distance, ar, is small, the moment from rolling friction, arRV(t), may be neglected. In addition, because RH(t) is relatively small, Equation 4.22 may be simplified to
−rHLF(t)+ MT(t)− Ip
d2θ(t)
dt2 = 0. (4.23)
110 Design of Modern Steel Railway Bridges
Traction Braking
k Time HLF
FIGURE 4.11 Time history of braking and traction forces (at fixed bearing) from railroad equipment.
For the condition of no slippage (complete adhesion),θ(t) = (x(t)/r). Substitution of (d2θ(t)/dt2)= (d2x(t)/rdt2) into Equation 4.23 yields
d2x(t) dt2 = Ip
r(MT(t)− rHLF(t)). (4.24) Substitution of Equation 4.24 into Equation 4.21a provides
HLF(t)= 1 1− mFIp
mFIp
r (MT(t)+ TF(t))
≤ μRV(t), (4.25)
whereμ is the coefficient of adhesion between locomotive wheels and rail without slippage (can be as high as 0.35 for modern locomotives with software-controlled wheel slip).
Equation 4.25 allows the numerical solution for longitudinal force, HLF(t), which remains, however, too arduous for ordinary design. The longitudinal forces described by Equation 4.25 (including the effects of axle bearing, wheel rim friction, air resis-tance, rolling friction, and other effects) have been observed and recorded by field testing in both Europe and the United States. The longitudinal forces exhibit almost static behavior since maximum traction and braking forces occur at low speeds when starting and at the end of braking, respectively (Figure 4.11). Therefore, a static analysis can be performed with HLF = μRV= LF = μW.
For a static longitudinal analysis, the bridge may be modeled as a series of lon-gitudinal elastic bars (with independent lonlon-gitudinal and flexural deformations) on horizontal elastic foundations simply modeled∗as equivalent horizontal springs with stiffness, ki. The static longitudinal equilibrium equations for a system of i bars (spans and rails) on elastic foundations (elastic horizontal stiffness of bridge deck†
∗Other models that incorporate different longitudinal restraint at the rail-to-deck and deck-to-superstructure may be used to provide greater accuracy.
†Particularly appropriate for modern elastic rail fastening systems.
Loads and Forces on Steel Railway Bridges 111
and approach track) is (seeFigure E4.6)
−EiAid2xi(x)
dx2 + kixi(x)= qi(x), (4.26) where EiAiis the axial stiffness of the member (rail or span).
The resulting system of equations may be solved for the longitudinal displace-ments, xi(x), and forces, Ni(x)= EiAi(dxi(x)/dx), in the bars. The solution may be obtained using transformation methods (Fryba, 1996) and the appropriate boundary conditions (e.g.,Table E4.2in Example 4.7).
The longitudinal traction and braking forces transferred to the bearings may be determined from equilibrium following computation of the rail and span axial forces, Ni(x). However, as seen in Example 4.7, even the simplest bridge models will involve considerable calculation.
Example 4.7
Develop the equations of longitudinal forces and boundary conditions for the open deck steel railway bridge shown in Figure E4.6.
L1 L2
L5
Li ui(x) LFin
Sin
L6 L3
LF L4
k1
k2 k2 k3
Ei Ai
–Ei Aid2ui(x)+ kiui(x) = qi(x), i = 1, 4 dx2
–Ei Aid2ui(x)+ ki[ui(x) – ui+3(x)] = qi(x), i = 2, 3 dx2
–Ei Aid2ui(x)+ k2[ui(x) – ui–3(x)] = 0, i = 5, 6 dx2
qi(x) = S
Ni
n=1d(x–Sin)LFin FIGURE E4.6
112 Design of Modern Steel Railway Bridges
TABLE E4.2
Boundary Conditions Force and Displacement Conditions
Rails (i= 1, 2, 3, 4) N1(0)= N4(L4)= 0
N1(L1)− LFL1= N2(0) N2(L2)− LFL2= N3(0) N3(L3)− LFL3= N4(0) u1(L1)= u2(0) u2(L2)= u3(0) u3(L3)= u4(0)
Span (i= 5, 6) N5(L5)= N6(L6)= 0
u5(0)= u6(0)= 0 Particular
Expansion joints at end of bridge L1= L4= 0
CWR across bridge L1= L4
No longitudinal rail restraint (free rails) k2= 0 Rails fixed (direct fixation to deck) k2
The equations of longitudinal forces and boundary conditions are shown inFigure E4.6and Table E4.2, respectively.
Extensive testing and analytical work has been performed (see references Foutch et al., 1996, 1997; LoPresti et al., 1998; LoPresti and Otter, 1998; Otter et al., 1996, 1997, 1999, 2000; Tobias Otter and LoPresti, 1998; Tobias et al., 1999; Uppal et al., 2001) to overcome the theoretical model complexities and numerical modeling efforts.
This work has established relationships for braking and traction dependent on the length of the portion of the bridge under consideration. Testing in the United States has provided longitudinal forces for Cooper’s E80 design live load that are shown in Figure 4.12and Equations 4.27 and 4.28. It appears that, for loaded lengths less than about 350 ft, longitudinal force due to traction governs. However, locomotive traction occurs over a relatively small length and braking forces on a loaded length consisting of the entire bridge may exceed the tractive effort (see Examples 4.9 and 4.10).∗The force due to traction governs for short- and medium-length bridges.
LFB= 45 + 1.2L, (4.27)
LFT= 25√
L, (4.28)
where LFBis the longitudinal force due to train braking (kips), LFTis the longitudinal force due to locomotive traction (kips), and L is the length of the portion of the bridge under consideration (ft).
However, while an estimate of the magnitude of the applied longitudinal trac-tion and braking forces appropriate for design is readily available, the distributrac-tion of
∗As illustrated byFigure 4.13showing the ratio of the longitudinal force transmitted to the bearings, HB, to the applied longitudinal force for bridges with continuous welded rail and steel bearings (based on European tests reported by Fryba, 1996).
Loads and Forces on Steel Railway Bridges 113
FIGURE 4.12 AREMA design longitudinal forces.
200
FIGURE 4.13 Bearing forces from European testing. (After Fryba, L., 1996, Dynamics of Railway Bridges, Thomas Telford, London, UK.)
longitudinal forces for the design of span bracing, bearings, substructures, and foun-dations needs careful consideration. The distribution and path of longitudinal forces between their point of application and the bridge supports depend on the arrangement, orientation, and relative stiffness of
• Bridge members in the load path
• Bearing type (fixed or expansion)
• Substructure characteristics.
Example 4.8
The longitudinal design force for Cooper’s E80 loading is required for each track of the open deck steel multibeam railway bridge shown inFigures E4.1
114 Design of Modern Steel Railway Bridges
andE4.5.FromFigure 4.12,it is determined that
• The longitudinal force due to train braking is LFB= 153.0 kips per track on the entire bridge; because of relative span lengths and bearing arrangement, it may be equally distributed to each span as 76.5 kips.
• The longitudinal force due to train braking is LFB= 99.0 kips per track on one span. However, this is an unlikely scenario considering the bridge length, train length, and distributed nature of train brake application.
• The longitudinal force due to locomotive traction is LFT= 237.2 kips per track on the entire bridge; because of relative span lengths and bearing arrangement, it may be equally distributed to each span as 118.6 kips.
• The longitudinal force due to locomotive traction is LFT= 167.7 kips per track on one span.
The longitudinal force due to locomotive traction is LFT= 167.7 kips per track and may be used for superstructure design. The longitudinal forces are distributed through the superstructure to the bearings and substruc-tures. Bearing component and substructure design will require consideration of these longitudinal forces. However, in this multibeam span, longitudi-nal forces of this magnitude will result in only small axial stresses in the longitudinal beams or girders, which may be disregarded in the design.
Example 4.9
The longitudinal design force for Cooper’s E90 loading is required for the deck truss of the 1100 ft long single track 10 span steel bridge outlined in the data ofTable E4.3.Each span has fixed and expansion bearings. All substructures have spans with adjacent fixed and expansion bearings.
• The longitudinal force due to train braking is LFB= (9/8)1365 = 1536 kips on the entire bridge; it is distributed to the deck truss span as (400/1100)(1536)= 558 kips.
• The longitudinal force due to train braking is LFB= (9/8)525 = 591 kips on the deck truss span. However, this is an unlikely scenario considering the bridge length, train length, and distributed nature of train brake appli-cation. Therefore, other portions of the bridge should be investigated for train braking. For example, the longitudinal force due to train brak-ing, LFB= (9/8)(400/600)(765) = 574 kips on the deck truss span when the train is on spans 7–10 only and (9/8)(400/980)(1221)= 561 kip when the train is on spans 1–7 only.
TABLE E4.3
Span Type Length (ft)
1 Through plate girder 100
2–6 Deck plate girder 80
7 Deck truss 400
8–10 Deck plate girder 100
Total 1100
Loads and Forces on Steel Railway Bridges 115
• The longitudinal force due to locomotive traction is LFT= (9/8)829 = 933 kips on the entire bridge. However, this is not likely (unless a string of powered accelerating/decelerating locomotives traverses the bridge) and other portions of the bridge should be investigated. For example, the longitudinal force due to locomotive traction, LFT= (9/8)(400/600)(612) = 459 kips on the deck truss span when the train is on spans 7–10 only and (9/8)(400/980)(783)= 359 kips when the train is on spans 1–7 only.
• The longitudinal force due to locomotive traction is LFT= (9/8)500 = 563 kips on the 400 ft deck truss span.
The longitudinal force due to train braking, LFT= 574 kips, is likely to be used for design of the deck truss span.
As noted in Example 4.8, the distribution of longitudinal forces in the superstructure may be of little concern for some span types (e.g., multiple longitudinal beam and deck plate girder spans). However, for other types of superstructures, the longitudinal force path from rails to bearings is of considerable importance (e.g., floorbeams with direct fixation of track and span floor systems). The horizontal axial force resistance of deck plates from diaphragm behavior may preclude the need for bracing elements to carry longitudinal forces to the main girders or trusses. Nevertheless, in some open deck spans, specific consideration of the lateral bracing (traction bracing) requirements is necessary to adequately transfer longitudinal forces to the main girders or trusses for transfer to the substructures at the bearings. A typical instance where traction bracing may be required is within the panel adjacent to the fixed bearings in an open deck span with a stringer and floorbeam system supported each side of the track by long-span main girders or trusses. In order to preclude the torsional and/or lateral bending of floorbeams that might result from longitudinal forces transmitted by floor systems without connection to the lateral bracing(Figure 4.14a),traction bracing is used (Figure 4.14b). Traction bracing is provided through connection of the stringers to the lateral bracing and addition of a new transverse member (shown dashed in Figure 4.14b) between the stringers at the bracing connections. Provided the main girder or truss fixed bearings are adequate to transfer the longitudinal forces to the substructure, the traction bracing truss (Figures 4.14b and 4.15)will avoid lateral loading of floor beams (member 1–1 in Figure 4.15) since the stringers (members 2–3 in Figure 4.15) can carry no longitudinal force. Other traction bracing arrangements may be used in a similar manner at the fixed end of long single and multiple track spans to properly transmit longitudinal traction and braking forces to the bearings.