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5.2.1.1.1 Criteria for Maximum Shear Force (with Loads Applied Directly to the Superstructure)
A series of concentrated loads applied directly to the steel beam or girder is typically assumed in the design of both open and ballasted deck spans.
The maximum shear force, VC, at a location, C, in a simply supported span of length, L, traversed by a series of concentrated loads with resultant force at a distance, xT, from one end of the span is(Figure 5.2)
VC= PT
xT
L − PL, (5.1)
∗ Some reasons for this are given in Chapter 3.
† The equations of static equilibrium suffice to determine forces in the structure.
‡ Typical of continuous and some movable steel superstructures.
§ The influence lines for simple span shear force and bending moment are useful for the construction of influence lines for axial force in truss web and chord members, respectively.
∗∗Two-hinged arches (hinged at bases) are statically indeterminate and many steel railway arch superstruc-tures are designed and constructed as three-hinged arches to create a statically determinate structure.
For statically determinate arches, influence lines for axial forces in members may be constructed by superposition of horizontal and vertical effects. The influence lines for simple span bending moment are useful for the construction of the influence lines for the vertical components of axial force in arch members.
152 Design of Modern Steel Railway Bridges
PT
L/2 L/2
©
A B
C
a xT
PL
Pn
bn xL
Direction of movement
FIGURE 5.2 Concentrated moving loads applied directly to the superstructure.
where PT is the total load on the span and PL is the load to left of location C.
Equation 5.1 indicates that VCwill be a maximum at a location where PT(xT/L) is a maximum and PLa minimum. If PL= 0 the absolute maximum shear in the span occurs at the end of the span and is
VA= PT
xT
L . (5.2)
For any span length, L, the maximum end shear, VA, will be largest when the product PTxTis greatest. Therefore, for a series of concentrated loads (such as the Cooper’s E loading), the maximum end shear, VA, must be determined with the heaviest loads included in PT and these heavy loads should be close to the end of the beam (to maximize the distance, xT).
This information assists in determination of the absolute maximum value of end shear force, which can be determined by a stepping the load configuration across the span (by each successive load spacing) until PTxTcauses a decrease in VA. With the exception of end shear in spans between L= 23 and 27.3 ft, this occurs when the second axle∗ of the Cooper’s design load configuration is placed at the end of the beam (location A inFigure 5.2).For spans between L= 23 and 27.3 ft, the maximum shear occurs with the fifth axle at the end of the span.
The maximum shear force at other locations, C, may be determined in a simi-lar manner by considering a constant PT moving from xT to xT+ bn(where bn is successive load spacing). In that case, the change in shear force,ΔVC, at location C is
ΔVC= PT
bn
L − PL. (5.3)
The relative changes in shear given by Equation 5.3 can be examined to determine the location of the concentrated loads for maximum shear at any location, C, in the span.
∗ The first driving wheel of the configuration.
Structural Analysis and Design of Steel Railway Bridges 153
5.2.1.1.2 Criteria for Maximum Shear Force (with Loads Applied to the Superstructure through Transverse Members)
A series of concentrated loads applied through longitudinal members (stringers) to transverse members (floorbeams) to the steel beam or girder is typically assumed in the design of open deck through spans. Ballasted deck superstructures that transfer load to the beams or girders by closely spaced transverse members without stringers may be treated as outlined in Section 5.2.1.1.1.
The maximum shear force, VBC, in panel BC in a simply supported span of length, L, traversed by a series of concentrated loads (with resultant force at a distance, xT) is(Figure 5.3)
The maximum shear force in panel BC may be determined by considering a constant PTmoving from xT to xT+ ΔxT, whereΔxTis a small increment of movement of load assuming no change in concentrated forces on the span or within panel BC. In that case, the change in shear force,ΔVBC, in panel BC, is
ΔVBC= changes sign (positive to negative). Therefore, the maximum shear occurs in panel BC when the average distributed load on the span, PT/L, equals the average distributed load in panel BC, PBC/sp.
When sp= L/n (where n is the number of equal length panels) ΔVBC=
FIGURE 5.3 Concentrated moving loads applied to the superstructure at transverse members.
154 Design of Modern Steel Railway Bridges
and PT/n− PBC= 0 for maximum shear in the panel. Therefore, the maximum shear in spans with equal length panels occurs in panel BC when the average panel load on the span, PT/n, equals the load in panel BC, PBC.
The relative changes in shear given by Equation 5.6 can be examined to determine the location of the concentrated loads for maximum shear in any panel on the span.
5.2.1.1.3 Criteria for Maximum Bending Moment (with Loads Applied Directly to the Superstructure)
A series of concentrated loads applied directly to the steel beam or girder is typically assumed in the design of both open and ballasted deck spans.
The maximum bending moment, MC, at a location, C, in a simply supported span, of length, L, traversed by a series of concentrated loads with resultant at a distance, xT, from one end of the span is (Figure 5.2)
The change in bending moment, ΔMC, at location C as the constant force PT moves from xT+ ΔxTis
ΔMC= PTa
L − PL
ΔxT, (5.8)
where ΔxT is a small increment of movement of load assuming no change in concentrated forces on the span.
When PT/L= PL/a, maximum bending moment at location C occurs as the change in bending moment changes sign (positive to negative). Therefore, the maximum bending moment occurs at location C when the average distributed load on the span, PT/L, equals the average distributed load to the left of location C, PL/a.
The relative changes in bending moment given by Equation 5.8 can be examined to determine the location of the concentrated loads for maximum bending moment at any location, C, in the span.
5.2.1.1.4 Criteria for Maximum Bending Moment (with Loads Applied to the Superstructure through Transverse Members)
A series of concentrated loads applied through longitudinal members (stringers) to transverse members (floorbeams) to the steel beam or girder is typically assumed in the design of open deck through spans. Ballasted deck superstructures that transfer load to the beams or girders by closely spaced transverse members without stringers may be treated as outlined in Section 5.2.1.1.3.
The maximum bending moment, MBC, in panel BC in a simply supported span of length, L, traversed by a series of concentrated loads (with resultant force at xT) transferred to the span by stringers and transverse floorbeams is (Figure 5.3)
MBC=
Structural Analysis and Design of Steel Railway Bridges 155
The change in bending moment,ΔMBC, in panel BC is
ΔMBC=
= 0, maximum bending moment occurs in panel BC.
The relative changes in bending moment given by Equation 5.10 can be examined to determine the location of the concentrated loads for maximum bending moment at any panel in the span. The maximum bending moment will occur at the panel point nearest the center of the span.
5.2.1.1.5 Maximum Bending Moment with Cooper’s E80 Load
The criteria for maximum shear force and bending moment in a simply supported span illustrate that loads must be stepped across the span and their effects investigated at the location of interest. In particular, the load position for maximum bending moment is of interest to engineers.
For live load configurations, such as Cooper’s E80, that are expressed as a series of concentrated loads (with or without uniform load segments), the wheel load under which the maximum bending moment occurs may not be readily known by inspection, particularly on longer spans. In such cases, the development of a moment table or chart is of benefit for determining the maximum bending moments at any location along the span. For example, to determine the maximum moment at location C in Figure 5.4, the load configuration would have to be moved in many successive increments across the span. The construction of a moment table, for the particular live load configuration, makes such an iterative analysis unnecessary.
x
FIGURE 5.4 Bending moment at location C for a series of concentrated moving loads.
156 Design of Modern Steel Railway Bridges
The bending moment, MC, at any location, C, due to moving concentrated and uniform loads as shown inFigure 5.4is
MC= RA
whereΣPixiis the sum of moments due to loads to the left of C. The left reaction, RA, is
RA=
Pizi+ (wlw2/2)
L . (5.12)
Substitution of Equation 5.12 into Equation 5.11 yields
MC=
From Figure 5.4, the sum of the moments of concentrated loads about B is
Pizi= PT(xP)+ Pi
lP. (5.14)
Substitution of Equation 5.14 into Equations 5.12 and 5.13 yields
RA= PT(xP)+
Equations 5.15 and 5.16 illustrate that to determine the end shear force and bending moment at any location in the simple span due to moving concentrated and uniform loads (such as the Cooper’s E80 load), the following is required:
• The sum of the bending moments of all concentrated loads in front of, and about, the last concentrated load (at lPfrom B inFigure 5.4)on the span, PT(xP).
• The sum of all concentrated loads on the span,ΣPi.
• The negative bending moment or the sum of the moments about C of all concentrated loads in front of C,ΣPixi.
Since the Cooper’s load pattern is constant, it is possible to develop charts and tables to readily determine the bending moment for various simple span lengths using Equation 5.16. Table 5.1 is developed for the wheel load (1/2 of axle load) of the Cooper’s E80 live load. The legend to Table 5.1 outlines the methods used to
StructuralAnalysisandDesignofSteelRailwayBridges157 TABLE 5.1
Moment Table for Cooper’s E80 Wheel Load
N 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
P 20 40 40 40 40 26 26 26 26 20 40 40 40 40 26 26 26 26
S1 0 8 13 18 23 32 37 43 48 56 64 69 74 79 88 93 99 104
Sw 109 101 96 91 86 77 72 66 61 53 45 40 35 30 21 16 10 5
SP1 20 60 100 140 180 206 232 258 284 304 344 384 424 464 490 516 542 568
SP18 568 548 508 468 428 388 362 336 310 284 264 224 184 144 104 78 52 26
SM 0 320 840 1560 2480 3313 4273 5387 6640 7760 10,320 13,080 16,040 19,200 21,467 23,907 26,467 29,200
SM18 32,728 30,548 26,508 22,668 19,028 15,588 13,586 11,714 9998 8412 7352 5552 3952 2552 1352 806 390 130
SM17 29,888 27,808 23,698 20,328 16,888 13,648 11,776 10,034 8448 6992 6032 4432 3032 1832 832 416 130
SM16 27,178 25,198 21,558 18,118 14,878 11,838 10,096 8484 7028 5702 4842 3442 2242 1242 442 156
SM15 24,082 22,222 18,822 15,622 12,622 9822 8236 6780 5480 4310 3570 2410 1450 690 130
SM14 21,632 19,872 16,672 13,672 10,872 8272 6816 5490 4320 3280 2640 1680 920 360
SM13 17,456 15,876 13,036 10,396 7956 5716 4494 3402 2466 1660 1200 600 200
SM12 15,336 13,856 11,216 8776 6536 4496 3404 2442 1636 960 600 200
SM11 13,416 12,036 9596 7356 5316 3476 2514 1682 1006 460 200
SM10 11,696 10,416 8176 6136 4296 2656 1824 1122 576 160
SM9 9264 8144 6224 4504 2984 1664 1040 546 208 SM8 6992 6032 4432 3032 1832 832 416 130 SM7 5702 4842 3442 2242 1242 442 156 SM6 4310 3570 2410 1450 690 130 SM5 3280 2640 1680 920 360 SM4 1660 1200 600 200 SM3 960 600 200 SM2 460 200 SM1 160
N= Wheel number; P = Coopers E80 wheel load, kips (1/2 axle load); S1= Distance, ft, from wheel N to wheel 1; Sw= Distance, ft, from wheel N to uniform train load of 4000 lb/ft; S1+ Sw= 109 ft;
SP1= Sum of wheel loads, kips, between and including wheel 1 to wheel N; SP18= Sum of wheel loads, kips, between and including wheel N to wheel 18; SM = Sum of the moments, ft-kips, about wheel 1 of wheel loads between and including wheel 2 to wheel N ; SM18= Sum of the moments, ft-kips, about the beginning of the uniform load of wheel loads between and including wheel N to wheel 18; SMk= Sum of the moments, ft-kips, about wheel load (k + 1) of wheel loads between and including wheel N to wheel k.
158 Design of Modern Steel Railway Bridges
determine the values shown. The use ofTable 5.1for determining maximum bending moments due to Cooper’s E80 load is outlined in Examples 5.1 and 5.2.
Example 5.1
Determine the maximum bending moment per rail for Cooper’s E80 load on a 60 ft long deck plate girder (DPG) span. The moment at the center, C, is assumed to be near the maximum bending moment in both location and magnitude.
A review of the Cooper’s load configuration indicates that the maximum moment will likely occur under axles, NP= 3, 4, 5, 12, 13, or 14 (Figure E5.1).
With NP= 3 (Cooper’s load configuration wheel number 3) From Table 5.1:
x1= (S1)3= 13 ft
since x1≤ L/2 ≤ 30 ft; N1 = 1
support B is (L/2+ x1) = 30 + 13 = 43 ft from N1 since (S1)8= 43 ft, NL = 8 and is over support B xL= 43–43 = 0
NE is the last wheel on the span and is equal to NL− 1 = 7 when NL is over support B
RB=
MB
L =
M(NL−1),N1+
PN1,NE(xL)
60 =
M7,1+ P1,7(xL) 60
= 5702+ 232(0)
60 = 95.03 kips, MC= RB(L/2)−
M(NP−1),N1= RB(L/2)− M2,1
= 95.03(30) − 460 = 2391 ft-kips.
L/2 L/2
©
A B
N1 NP NL
x1 xL
C
FIGURE E5.1
Structural Analysis and Design of Steel Railway Bridges 159
With NP= 4 (Cooper’s load configuration wheel number 4) FromTable 5.1:
With NP= 5 (Cooper’s load configuration wheel number 5) From Table 5.1:
With NP= 12 (Cooper’s load configuration wheel number 12) FromTable 5.1:
160 Design of Modern Steel Railway Bridges
With NP= 13 (Cooper’s load configuration wheel number 13) FromTable 5.1:
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 x1= (S1)14− (S1)10= 79 − 56 = 23 ft
support B is 30+ 23 = 53 ft from N1 = 10
With NP= 14, the last wheel on the span, NL, is the beginning of the uniform load, w
xL= (L/2) − [(S1)w− (S1)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
With NP= 15 (Cooper’s load configuration wheel number 15) FromTable 5.1: 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 RB= ΣMB
The maximum bending moment is 2588 ft-kips. (NP= 13)
Structural Analysis and Design of Steel Railway Bridges 161
Example 5.2
Determine the bending moment per rail at location C under axles NP= 3, 4, and 13 of the Cooper’s E80 load on a 60 ft long through plate girder span with a floor system comprising floorbeams and 20 ft long stringers (Figure E5.2).
With NP= 3 (Cooper’s load configuration wheel number 3) From Table 5.1:
The superstructure with transverse floorbeams has a 100[1 − (2344/2391)] = 2.0% decrease in bending moment with NP = 3 at the location of maximum moment.
With NP= 4 (Cooper’s load configuration wheel number 4) FromTable 5.1:
162 Design of Modern Steel Railway Bridges
The superstructure loaded with transverse floorbeams has a 100[1 − (2331/2536)] = 8.1% decrease in bending moment with NP = 4 at the location of maximum moment.
With NP= 13 (Cooper’s load configuration wheel number 13) FromTable 5.1: begin-ning of the uniform load, w, to support B
NE is the last wheel on the span and is equal to 5 ft of the uniform load, w
RB=
The superstructure with transverse floorbeams has a 100[1 − (2334/2588)] = 9.8% decrease in bending moment with NP = 13 at the loca-tion of maximum moment.
5.2.1.2 Influence Lines for Maximum Effects of Moving Loads on