Implementation Example for I-Girder System under Vertical Load
Legal Load Rating Example
A numerical example is provided to illustrate the procedure for using the system factor tables
during the rating of an existing bridge.
In this example, we assume a hypothetical case where the bending moment capacity of a simple
span 120-ft prestressed concrete bridge with six beams at 8-ft spacing was found to be Rn=7200
kip-ft. The dead load effect is Dn=3500 kip-ft. The moment due to the AASHTO truck load
alone is LLHS20=1880 kip-ft. The moment for the AASHTO 3S-2 Legal Load is 1682 kip-ft.
The distribution factor from the AASHTO LRFD tables is D.F.=0.75 and the impact factor is
IM=1.33.
The LRFR Operating Rating for a site where the average daily truck traffic is ADTT>5000 is
obtained as:
R.F . 
Rn   D Dn
Rn   D Dn
1.0  7200  1.25  3500


 0.94
 L Ln
 L LL3S 2  IM  D.F . 1.80 1682 1.33  0.75
This rating factor value based on individual member capacity implies that the bridge should be
closed, posted, rehabilitated, or replaced. Given the redundant configuration of the bridge, the
bridge owners may choose to delay such actions if the bridge system capacity is found to be
sufficiently high so that the bridge would be able to withstand the potential overloading of a
main girder.
To assess the entire system’s load carrying capacity, the system factor is calculated based on
Table 1.3.6.1-1 in the proposed Specifications. A first step would require the calculation of the
unfactored live load carrying capacity of the member, LF1:
LF1 
RD
7200  3500

 2.62
D.F .  LLHS 20 0.75 1880
The bridge’s six beams are spaced at more than 4-ft and the system factor is a function of the
 0.49 .
dead load to resistance ratio D  3500
R
7200
1  1.5  D / R 
1  1.5  0.49 
s  1 

1

 1.08
2
1  LF12
1   2.62 
2
2
A.3-1
The adjusted System Legal Load Rating of the bridge is then executed using:
R.F . 
s Rn   D Dn 1.08 1.0  7200  1.25  3500

 1.13
 L Ln
1.80 1682 1.33  0.75
Where the system factor s is 1.08 and the other factors and variables are those that are normally
used during the usual rating process.
The adjusted rating factor R.F.=1.13, which is higher than 1.0, implies that, because of the
bridge’s redundant configuration, the bridge system will not collapse and will still be able to
support the applied legal loads even if one member reaches its limiting capacity. The bridge’s
redundancy is making the system’s capacity significantly higher than the capacities of the
individual members.
A.3-2
Implementation Example for a Continuous Three-Span Steel Tub Girder System under
Vertical Load
This Example is adopted from Example 4 in the Highway Structures Design Handbook, Vol. II, Chapter 1.A,
entitled: “Four LRFD Design Examples of Steel Highway Bridges,” Published by the American Iron and Steel
Institute (AISC) in cooperation with HDR Engineering, Inc. 1996.
This example illustrates the procedure for calculating a system factor for a continuous steel box
girder with spans of 58 000 mm (190 ft) - 72 000 mm (236 ft) -58 000 mm (190 ft). The bridge
cross section consists of two trapezoidal box sections with top flanges spaced at 3400 mm
(11.2 ft) on centers, 3400 mm (11.2 ft) between the centerlines of adjacent top flanges, and 1340
mm (4.40 ft) overhangs for a roadway width equal to 12 000 mm (39.37 ft). The concrete deck is
245 mm (9.65 in) thick including a 13 mm (0.5 in) integral wearing surface.
In the example, the positive moment capacity of the entire composite box was found to be
Rn+= 38.78 x 109 N-mm (28.60 x 103 kip-ft), the negative bending capacity near the interior
supports is Rn- = 62.47 x 109 N-mm (46.08 x 103 kip-ft) for the steel section including the
longitudinal reinforcement in the deck or Rn- = 58.86 x 109 N-mm (43.41 x 103 kip-ft) for steel
section only. The dead load effect for the box is Dn+=14.035 x 109 N-mm (10.35 x 103 kip-ft) for
the maximum positive bending region and Dn- =30.95 x 109 N-mm (22.83 x 103 kip-ft) for the
section over the pier. The maximum positive moment due to the AASHTO truck load alone (one
design truck is placed at the center of the middle span) is LLHS20+=3272 x 106 N-mm (2413.3
kip-ft). The maximum negative moment due to the AASHTO truck load (two design trucks are
placed in adjacent spans) is LLHS20- = 2324 x 106 N-mm (1714.1 kip-ft).
For multiple steel box girders with a concrete deck, the live load flexural moment can be
determined using the distribution factor determined by the following expression:
D.F .  0.05  0.85
N L 0.425

Nb
NL
(Table 4.6.2.2.2b-1, AASHTO LRFD 2012, 6th edition)
where:
N L = Number of design lanes
N b = Number of beams or girder
Therefore, the distribution factor for the two tub-girder bridge loaded by two lanes is:
N
0.425
D.F .  0.05  0.85 L 
Nb
NL
3 0.425
 0.05  0.85 
2
3
 1.467
To assess the entire system’s load carrying capacity, the system factor is calculated based on
Table 1.3.6.1-2. A first step would require the calculation of the unfactored live load carrying
capacity of the member, LF1:
A.3-3
LF1 for positive bending region (moments in N.mm as provided in the AISC design Handbook)
R  D
37.78 109  14.035 109
LF 

 4.95
D.F .  LLHS 20
1.467  3272 106

1
LF1 for negative flexure (stresses in MPa as provided in the AISC design Handbook)
Compression flanges (Steel section+Long.Reinforcement)
f r  ( f DC1  f DC2  f DW ) 330.9  (138.78  18.36  10.81)
LF1 

 8.97
D.F .  f LLHS 20
1.467 12.31
Tension flanges (Steel section+Long.Reinforcement)
f y  ( f DC1  f DC2  f DW ) 345  (149.12  17.2  10.13)
LF1 

 9.97
D.F .  f LLHS 20
1.467 11.52
The compression flange controls the negative bending load factor LF1-.
The above calculations use the distribution factors from the AASHTO LRFD tables. One can
also find actual maximum live load effects on the external beam (half box) using a structural
analysis program with a grillage model. To find the maximum positive moment, we place one
design truck at the center of the middle span. The two design trucks are placed in adjacent spans
to produce the maximum negative effects. The results for LF1 are calculated as follows:
LF1 
R   D  37.78 109 / 2  14.035 109 / 2

 4.79
L
2481106
LF1 
R   D  58.86 109 / 2  30.95 109 / 2

 9.45
L
1476 106
Using the AASHTO LRFD Tables for the distribution factors, the controlling load factor is
LF1  LF1  4.95 . The dead load to resistance ratio:
D
 14.035 10
9
R

37.78 109
=0.37
Given that the ratio for LF1 / LF1  1.81  1.75 , the system factor is calculated as:
1  1.5  D / R 
1  1.5  0.37 
s  1  4 
 1 4
 1.12
2
2
1  LF1
1   4.95
2
2
The calculated system factor indicates that the bridge system is highly redundant whereby even if
one section reaches its limiting capacity, the system will be able to withstand significant
A.3-4
additional load. Therefore, it would still be safe to design the members using lower safety levels.
Specifically, the moment capacity of the positive bending region which was designed for Rn =
38.78 x 109 N-mm (28.60 x 103 kip-ft) can be reduced to:
RnN  
Rn 37.78 109

 33.73 109 N  mm  24.88 103 Kip  ft
s
1.12
Similarly, the moment capacity of the negative bending region which was designed for Rn =
62.47 x 109 N-mm (46.08 x 103 kip-ft) can be reduced to:
RnN  
Rn 62.47 109

 55.78 109 N  mm  41.14 103 Kip  ft
s
1.12
Comments
The system factor s =1.12 is obtained because the moment capacity of the steel section in the
negative bending region is significantly larger than that in positive bending. Even though the
negative bending section is non compact, the system will still be able to undergo significant
nonlinear deformations before the negative bending section fails. If the negative bending section
did not satisfy the condition LF1 / LF1  1.75 , then the lack of ductility in negative bending
would lead to a significantly lower system capacity and the system factor would have been
reduced to s =1.03, which implies a reduced level of redundancy.
A.3-5
Implementation Example for Bridge System under Lateral Load
Weak Cap Beam
This example is for a three-span continuous bridge with two three-column bents where the lateral
confinement reinforcement ratio of each column is s=0.3% (detail category B). Each column’s
height is 32.6-ft with a diameter of 7-ft. The bridge columns are based on stiff foundations that
are assumed fixed. A cross-section analysis of the cap beam shows that the plastic moment
capacity Mp beam = 202,000 kip-in. The ultimate curvature for the cap beam is u beam = 9.03 ×10-4
in-1. The ultimate moment, plastic moment and ultimate curvature of the middle column
subjected to an axial load of 214 kips which includes the dead load and 20% of design live load
are Mu column 214,600 kip-in, Mp column=198,600 kip-in, and u column = 5.74×10-4 in-1, respectively.
The pushover analysis shows that the first column reaches its plastic capacity when the lateral
force is Pp1= 5,244.8 kip.
Because the cap beam strength is higher than the plastic moment of the column, but less than the
ultimate moment capacity of the column, [Mcol.ultimate>Mbeam.plastic.>Mcol.plastic], then the reduction
in the ultimate curvature of the beam-column connection,   , is obtained using Equation
(1.3.6.2-4).
 
M available  M p column
M u column  M p column

M p beam  M p column
M u column  M p column

202,000  198，
600
 0.21
214,600  198，
600
Compare the reduced column curvature to that of the beam and use the lower of the two:
 u column  u beam  0.21 5.74 104 in1  1.21104 in1  9.03 104 in1
The bridge members with the relatively weak cap beam are evaluated using the system factor in
Eq. (1.3.6.2-2):

s    Fmc  C

 u  tunc 

0.21(5.74 104 )  3.64 104 
 0.82
  0.75 1.16  0.24
tconf  tunc 
1.55 103  3.64 104 

Although this bridge is formed by two 3-column bents, the weakness in the cap beam reduces the
ability of the columns to deform until they crush. The failure of the cap beam before concrete
column crushing reduces the redundancy of the system that will be able to sustain an applied
load PEQ lower than the calculated PP1 while maintaining an adequate level of system reliability.
The maximum applied lateral load PEQ is found using Eq. (1.3.2.1-1):
PEQ  s Pp1  0.82  5, 244.8 kip  4300 kip
A.3-6