LRFD Steel Design
AASHTO LRFD Bridge Design
Specifications
Example Problems

Created July 2007
This material is copyrighted by
The University of Cincinnati
and
Dr. James A Swanson.
It may not be reproduced, distribute...
LRFD Steel Design
AASHTO LRFD Bridge Design Specification
Example Problems
Case Study: 2-Span Continuous Bridge..............
Ad-Hoc Web Strength and Stiffener Examples
Web Strength Example #1 ..........................................................
James A Swanson
Associate Professor
University of Cincinnati
Dept of Civil & Env. Engineering
765 Baldwin Hall
Cincinnati,...
1. PROBLEM STATEMENT AND ASSUMPTIONS:
A two-span continuous composite I-girder bridge has two equal spans of 165’ and a 42...
Negative Bending Section (Section 2)

Positive Bending Section (Section 1)

2. LOAD CALCULATIONS:
DC dead loads (structura...
Deck Self-Weight (DC1):

Wdeck

⎛
⎜ 150 pcf
= (9.5")(144") ⎜
2
⎜ 12 in
ft
⎝

(

)

⎞
⎟
Lb
⎟ = 1, 425 ft per girder
⎟
⎠

Ha...
Unfactored Dead Load Moment Diagrams from SAP
4,000
DC1

3,000
2,000
DW

1,000

Moment (kip-ft)

0
DC2

-1,000
-2,000
-3,0...
The following Dead Load results were obtained from the FE analysis:
•

The maximum positive live-load moments occur at sta...
2.2: Live Load Calculations
The following design vehicular live load cases described in AASHTO-LRFD are considered:

1) Th...
Unfactored Moving Load Moment Envelopes from SAP
6,000

Single Truck

4,000
Tandem

Moment (kip-ft)

2,000

Fatigue

0
Fat...
The following Live Load results were obtained from the SAP analysis:
•

The maximum positive live-load moments occur at st...
2.2.1: Verify the Maximum Positive Live-Load Moment at Station 73.3’:
25kip

25kip

Tandem:
32kip 32kip
8kip

Single Truck...
2.2.2: Verify the Maximum Negative Live-Load Moment at Station 165.0’:
25kip

25kip

Tandem:
32kip 32kip
8kip

Single Truc...
Based on the influence line analysis, we can say that the moments obtained from SAP appear to be
reasonable and will be us...
2.4: Centrifugal Force

A centrifugal force results when a vehicle turns on a structure. Although a centrifugal force does...
2.5: Wind Loads

For the calculation of wind loads, assume that the bridge is located in the “open country” at an elevatio...
Vertical Wind Load on Structure: (WS)
When no traffic is on the bridge, a vertical uplift (a line load) with a magnitude e...
3. SECTION PROPERTIES AND CALCULATIONS:
3.1: Effective Flange Width, beff:

For an interior beam, beff is the lesser of:
⎧...
3.2: Section 1 Flexural Properties
Bare Steel
Ix
Ay
791.72
0.53
1,377.84 15,398.86
10.50
1.75

Ad2

b
15.00
69.00
21.00

A...
3.3: Section 2 Flexural Properties
Bare Steel
Ix
Ay
1,512.00
1.75
1,436.06 15,398.86
65.63
27.34

Ad2

b
21.00
69.00
21.00...
4. DISTRIBUTION FACTOR FOR MOMENT
4.1: Positive Moment Region (Section 1):
Interior Girder –
One Lane Loaded:
0.4
0.3
⎛ S ...
Two or More Lanes Loaded:

DFM2,Ext+ = e DFM2,Int+

e = 0.77 +
= 0.77 +

de
9.1
1.5
9.1

= 0.9348

DFM2,Ext+ = (0.9348) (0...
Exterior Girder –
One Lane Loaded:

Same as for the positive moment section: DFM1,Ext- = 0.8500
Two or More Lanes Loaded:
...
Two Lanes Loaded:
14.5'
2.5'
2'

3'

3'

2'

3'

3'

P1

P2

DF

M 2 , Ext , Min

=

2
4

+

(18.0 ')(14.5 '+ 2.5 ')
(2) ⎡...
Nominal Girder Moments for Design
Station
(ft)
0.0
14.7
29.3
44.0
58.7
73.3
88.0
102.7
117.3
132.0
135.7
139.3
143.0
146.7...
5. DISTRIBUTION FACTOR FOR SHEAR

The distribution factors for shear are independent of the section properties and span le...
5.4: Shear Distribution Factor Summary
Strength and Service Shear Distribution:

1 Lane Loaded:
2 Lanes Loaded:

Shear Dis...
Nominal Girder Shears for Design
Station
(ft)
0.0
14.7
29.3
44.0
58.7
73.3
88.0
102.7
117.3
132.0
135.7
139.3
143.0
146.7
...
6. FACTORED SHEAR AND MOMENT ENVELOPES

The following load combinations were considered in this example:
Strength I:
Stren...
Strength Limit Moment Envelopes
20,000
15,000
Strength I

10,000
Strength IV

Moment (kip-ft)

5,000
0
-5,000
Strength IV
...
Service II Moment Envelope
12,500
10,000
7,500
5,000

Moment (kip-ft)

2,500
0
-2,500
-5,000
-7,500
-10,000
-12,500
-15,00...
Factored Fatigue Moment Envelope
1,500

1,000

Moment (kip-ft)

500

0

-500

-1,000

-1,500
0

30

60

90

120

150

180
...
Factored Girder Moments for Design
Station
(ft)
0.0
14.7
29.3
44.0
58.7
73.3
88.0
102.7
117.3
132.0
135.7
139.3
143.0
146....
Factored Girder Shears for Design
Station
(ft)
0.0
14.7
29.3
44.0
58.7
73.3
88.0
102.7
117.3
132.0
135.7
139.3
143.0
146.7...
7. FATIGUE CHECKS
7.1: Check transverse stiffener to flange weld at Station 73.3:

Traffic information:
ADTT given as 2400...
Check Bottom Flange Weld:
The permanent loads at Station 73.3 cause tension in the bottom flange, thus by inspection
fatig...
Calculate the design life of the part under consideration:
Since γ ( Δf ) is greater than

( ΔF )TH
2

, solve for N in th...
8. CHECK CROSS_SECTION PROPORTION LIMITS
Web Proportions

•

69"

D
≤ 150
tw

9 "
16

= 122.7 ≤ 150

O.K.

Flange Proporti...
9. CHECK SERVICE LIMIT STATE
9.1: Check Absolute Deflection of the Bridge: Section 6.10.4.1
Section 1

The cross section o...
Section 2

The cross section of Section 2 that is used for computing deflections is shown above.
The transformed width of ...
The following model, which represents the stiffness of a single girder, was used to compute absolute liveload deflections ...
9.2: Check the Maximum Span-to-Depth Ratio: Section 6.10.4.1
From Table 2.5.2.6.3-1, (1) the overall depth of a composite ...
9.3: Permanent Deformations - Section 1
At the Service Limit State, the following shall be satisfied for composite section...
Bottom Flange, Positive Moment

⎛ (2, 979

f f ,58.7 = 1.00 ⎜

⎝

k-ft

)(12 in ) ⎞
ft

1,733 in

⎛ (549.7

⎟ + 1.00 ⎜
⎠

...
9.4: Permanent Deformations - Section 2
Top Flange, Negative Moment

⎛ (7,109 )(12 in ) ⎞
⎛ (1, 250 )(12 in ) ⎞
⎛ (2, 292 ...
9.5: Bend Buckling Checks
At the Service Limit State, all sections except composite sections in positive flexure shall sat...
10. CHECK STRENGTH LIMIT STATE
10.1: Section 1 Positive Flexure
Section Classification (§6.10.6.2, Pg. 6.98 – 6.99)

Check...
?

M u + 13 f l S xt ≤ φ f M n

?

(13,568k-ft ) + (0) ≤(1.00)(13,060k-ft )

No Good.

Note that the check of M n ≤ 1.3Rh ...
Hybrid Girder Factors Will Now be Required:

Compute the Hybrid Girder Factor, Rh, for Section 1:
Per AASHTO Commentary Pg...
11. RECHECK SERVICE LIMIT STATE WITH 70KSI FLANGES
11.1: Permanent Deformations - Section 1
At the Service Limit State, th...
11.3: Bend Buckling Checks
At the Service Limit State, all sections except composite sections in positive flexure shall sa...
12. RECHECK STRENGTH LIMIT STATE WITH 70KSI FLANGES
12.1: Section 1 - Positive Flexure
Section Classification (§6.10.6.2, ...
(The haunch is not included in Dt, as per ODOT Exceptions)
Dp ⎞
⎛
Since Dp = 8.804” > 0.1Dt = 7.925”, M n = M p ⎜1.07 − 0....
12.2: Section 2 - Negative Flexure
Section Classification (§6.10.6.2, Pg. 6.98 – 6.99)

Check

2 Dc
E
≤ 5.70
tw
Fyc

Dc is...
As it stands here, this girder is clearly not adequate over the pier. The compression flange is
overstressed as per the pr...
12.3 Vertical Shear Capacity
At the strength Limit, the following must be satisfied

Vu ≤ φVn
For an unstiffened web,

Vn ...
Try adding transverse stiffeners spaced at do = 8’ = 96”

k =5+

5
⎛ do ⎞
⎜D⎟
⎝ ⎠

2

=5+

5
⎛ 96" ⎞
⎜ 69" ⎟
⎝
⎠

= 7.583
...
With TFA:
Since

2 Dtw
(2)(69")( 916 ")
=
= 1.056 ≤ 2.5 ,
( b fct fc + b ft t ft ) ( (21")(2 12 ") + (21")(1") )

⎡
⎤
⎡
⎤
...
Strength Limit Shear Capacity
800
Tension Field Action

600

Strength I

400

Strength IV

Shear (kip)

200

0

-200

-400...
12.4: Horizontal Shear Strength
Per ODOT Standard practice, shear studs will be used to transfer horizontal shear between ...
Zr = α d 2 ≥

5.5d 2
2

(6.10.10.2-1)

α = 34.5 − 4.28Log( N )

(6.10.10.2-2)

α = 34.5 − 4.28Log(55.84 ×106 ) = 1.343ksi
...
Strength Limit:

Qr = φsc Qn
Qn = 0.5 Asc

φsc = 0.85
f c' Ec ≤ Asc Fu

(6.10.10.4.3-1)

2
⎛π ⎞
Asc = ⎜ ⎟ ( 7 8 ") = 0.601...
Positive moment - Section 1: Station 0.0’ - 73.3’

Pp = Min ( PConcrete , Psteel )

PConcrete = 0.85 f c'bets

= ( 0.85 ) ...
Negative Moment - Section 2: Station 73.3’ - 165.0’

Pn = Min ( Psteel , PCrack )

PCrack = 0.45 f c'bets

= ( 0.45 ) ( 4....
Shear Stud Summary:

This table represents that pitch of shear studs required for either 3 or 4 studs per row based on loc...
ONE-SPAN INELASTIC I-GIRDER BRIDGE DESIGN EXAMPLE

1. PROBLEM STATEMENT AND ASSUMPTIONS:
A single span composite I-girder ...
172' - 4" Total Girder Length

G
2

G
3

G
4

G
5

G
6

166' - 4" cc Bearings

Cross Frames Spaced @ 22' - 0" cc
G
1

Sing...
Positive Bending Section (Section 2)

Positive Bending Section (Section 1)

Positive Bending Section (Section 3)

Single-S...
2. LOAD CALCULATIONS:
DC dead loads (structural components) include:
• Steel girder self weight (DC1)
• Concrete deck self...
Haunch Self-Weight (DC1):
Average width of haunch: 14’’

⎛ 150pcf ⎞
⎟ = 94.33 lbs per girder
Whaunch = (14 )( 4 ) + 2 ( ( ...
The maximum shear forces at the ends of the girder are also easily computed.
lbs
⎛ wL ⎞ ⎡ ( 413.3 ft ) (166.3' ) ⎤
kip
VDC...
2b. Live Load Calculations
The following design vehicular live load cases described in AASHTO-LRFD are considered:
1) The ...
Unfactored HL-93 Moment Envelopes from SAP
6,000
Single Truck

4,000
Tandem

Moment (kip-ft)

2,000

0

-2,000

-4,000

-6...
2c. Verify the Maximum Positive Live-Load Moment at Station 83.15’:
kip

kip

25

25

Tandem:
kip

32
8

kip

32

kip

:
S...
Based on the influence line analysis, we can say that the moments obtained from SAP appear to be
reasonable and will be us...
3b. Section 1 Flexural Properties

Single Span Bridge Example - Section 1
Bare Steel
t

b

A

y

Ay

d

Ix

Ad2

IX

Top F...
3c. Section 2 Flexural Properties

Single Span Bridge Example - Section 2
Bare Steel
t

b

A

y

Ay

d

Ix

Ad2

IX

Top F...
3d. Section 3 Flexural Properties

Single Span Bridge Example - Section 3
Bare Steel
t

b

A

y

Ay

d

Ix

Ad2

IX

Top F...
4. DISTRIBUTION FACTOR FOR MOMENT
4a. Section 1:
Interior Girder - One Lane Loaded:

DFM1,Int,Sec1

0.4
0.3
⎛ S ⎞ ⎛ S ⎞ ⎛ ...
Exterior Girder – One Lane Loaded:
The lever rule is applied by assuming that a
hinge forms over the first interior girder...
Interior Girder – Two or More Lanes Loaded:
0.4
0.3
⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎞
DFM2,Int,Sec2 = 0.075 + ⎜
⎟ ⎜ ⎟ ⎜
3 ⎟
⎝ 9.5 ⎠ ⎝ L ⎠...
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
LRFD Steel Design example
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LRFD Steel Design example
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LRFD Steel Design example

  1. 1. LRFD Steel Design AASHTO LRFD Bridge Design Specifications Example Problems Created July 2007
  2. 2. This material is copyrighted by The University of Cincinnati and Dr. James A Swanson. It may not be reproduced, distributed, sold, or stored by any means, electrical or mechanical, without the expressed written consent of The University of Cincinnati and Dr. James A Swanson. July 31, 2007
  3. 3. LRFD Steel Design AASHTO LRFD Bridge Design Specification Example Problems Case Study: 2-Span Continuous Bridge.......................................................................................1 Case Study: 1-Span Simply-Supported Bridge .........................................................................63 Case Study: 1-Span Truss Bridge...............................................................................................87 Ad-Hoc Tension Member Examples Tension Member Example #1 ..........................................................................................105 Tension Member Example #2 ..........................................................................................106 Tension Member Example #3 ..........................................................................................108 Tension Member Example #4 ..........................................................................................110 Ad-Hoc Compression Member Examples Compression Member Example #1 .................................................................................111 Compression Member Example #2 .................................................................................112 Compression Member Example #3 .................................................................................114 Compression Member Example #4 .................................................................................116 Compression Member Example #5 .................................................................................119 Compression Member Example #6 .................................................................................121 Compression Member Example #7 .................................................................................123 Ad-Hoc Flexural Member Examples Flexure Example #1 ..........................................................................................................127 Flexure Example #2 ..........................................................................................................129 Flexure Example #3 ..........................................................................................................131 Flexure Example #4 ..........................................................................................................134 Flexure Example #5a ........................................................................................................137 Flexure Example #5b ........................................................................................................141 Flexure Example #6a ........................................................................................................147 Flexure Example #6b ........................................................................................................152 Ad-Hoc Shear Strength Examples Shear Strength Example #1 .............................................................................................159 Shear Strength Example #2 .............................................................................................161
  4. 4. Ad-Hoc Web Strength and Stiffener Examples Web Strength Example #1 ...............................................................................................165 Web Strength Example #2 ...............................................................................................168 Ad-Hoc Connection and Splice Examples Connection Example #1....................................................................................................175 Connection Example #2....................................................................................................179 Connection Example #3....................................................................................................181 Connection Example #4....................................................................................................182 Connection Example #5....................................................................................................185 Connection Example #6a..................................................................................................187 Connection Example #6b .................................................................................................189 Connection Example #7....................................................................................................190
  5. 5. James A Swanson Associate Professor University of Cincinnati Dept of Civil & Env. Engineering 765 Baldwin Hall Cincinnati, OH 45221-0071 Ph: (513) 556-3774 Fx: (513) 556-2599 James.Swanson@uc.edu
  6. 6. 1. PROBLEM STATEMENT AND ASSUMPTIONS: A two-span continuous composite I-girder bridge has two equal spans of 165’ and a 42’ deck width. The steel girders have Fy = 50ksi and all concrete has a 28-day compressive strength of f’c = 4.5ksi. The concrete slab is 91/2” thick. A typical 2¾” haunch was used in the section properties. Concrete barriers weighing 640plf and an asphalt wearing surface weighing 60psf have also been applied as a composite dead load. HL-93 loading was used per AASHTO (2004), including dynamic load allowance. 42' - 0" Out to Out of Deck 39' - 0" Roadway Width 9½” (typ) 23/4" Haunch (typ) 3'-0" 3 spaces @ 12' - 0" = 36' - 0" 3'-0" References: Barth, K.E., Hartnagel, B.A., White, D.W., and Barker, M.G., 2004, “Recommended Procedures for Simplified Inelastic Design of Steel I-Girder Bridges,” ASCE Journal of Bridge Engineering, May/June Vol. 9, No. 3 “Four LRFD Design Examples of Steel Highway Bridges,” Vol. II, Chapter 1A Highway Structures Design Handbook, Published by American Iron and Steel Institute in cooperation with HDR Engineering, Inc. Available at http://www.aisc.org/ 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 1 of 62 -- 1 --
  7. 7. Negative Bending Section (Section 2) Positive Bending Section (Section 1) 2. LOAD CALCULATIONS: DC dead loads (structural components) include: • Steel girder self weight (DC1) • Concrete deck self weight (DC1) • Haunch self weight (DC1) • Barrier walls (DC2) DW dead loads (structural attachments) include: • Wearing surface (DW) 2.1: Dead Load Calculations Steel Girder Self-Weight (DC1): (Add 15% for Miscellaneous Steel) (a) Section 1 (Positive Bending) A = (15”)(3/4”) + (69”)(9/16”) + (21”)(1”) = 71.06 in2 ⎛ 2 ⎜ 490 Wsec tion1 = 71.06 in ⎜ ⎜ ⎝ ⎞ pcf ⎟ ( 12 in ft ) 2 ⎟ (1.15 ) = 278.1 ft ⎟ ⎠ Lb per girder (b) Section 2 (Negative Bending) A = (21”)(1”) + (69”)(9/16”) + (21”)(2-1/2”) = 112.3 in2 ⎛ 2⎜ Wsec tion 2 = 112.3 in ⎜ ⎜ ⎝ ⎞ 490 pcf ⎟ ( 12 in ft ) 2 ⎟ (1.15 ) = 439.5 ⎟ ⎠ 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel Lb ft per girder AASHTO-LRFD 2007 Page 2 of 62 -- 2 --
  8. 8. Deck Self-Weight (DC1): Wdeck ⎛ ⎜ 150 pcf = (9.5")(144") ⎜ 2 ⎜ 12 in ft ⎝ ( ) ⎞ ⎟ Lb ⎟ = 1, 425 ft per girder ⎟ ⎠ Haunch Self-Weight (DC1): ⎛ 21"(66') + 15"(264') ⎞ ⎟ = 16.2" 66'+ 264' ⎝ ⎠ Average width of flange: ⎜ Average width of haunch: Whaunch ( 1 2 ) ⎡(16.2"+ (2)(9") ) + 16.2"⎤ = 25.2" ⎣ ⎦ ⎛ ⎞ ⎜ ( 2")( 25.2") ⎟ (150 pcf ) = 52.5 Lb per girder =⎜ ft 2 ⎟ ⎜ 12 in ⎟ ft ⎝ ⎠ ( ) Barrier Walls (DC2): ⎛ (2 each) ( 640 plf ) ⎞ ⎟ = 320.0 Lb ft per girder ⎜ ⎟ 4 girders ⎝ ⎠ Wbarriers = ⎜ Wearing Surface (DW): W fws = (39')(60 psf ) 4 girders = 585 Lb per girder ft The moment effect due to dead loads was found using an FE model composed of four frame elements. This data was input into Excel to be combined with data from moving live load analyses performed in SAP 2000. DC1 dead loads were applied to the non-composite section (bare steel). All live loads were applied to the short-term composite section (1n = 8). DW (barriers) and DC2 (wearing surface) dead loads were applied to the long-term composite section (3n = 24). 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 3 of 62 -- 3 --
  9. 9. Unfactored Dead Load Moment Diagrams from SAP 4,000 DC1 3,000 2,000 DW 1,000 Moment (kip-ft) 0 DC2 -1,000 -2,000 -3,000 -4,000 -5,000 -6,000 -7,000 -8,000 0 30 60 90 120 150 180 210 240 270 300 330 Station (ft) Unfactored Dead Load Shear Diagrams from SAP 200 DC1 150 100 DW Shear (kip) 50 DC2 0 -50 -100 -150 -200 0 30 60 90 120 150 180 210 240 270 300 330 Station (ft) 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 4 of 62 -- 4 --
  10. 10. The following Dead Load results were obtained from the FE analysis: • The maximum positive live-load moments occur at stations 58.7’ and 271.3’ • The maximum negative live-load moments occur over the center support at station 165.0’ DC1 - Steel: DC1 - Deck: DC1 - Haunch: DC1 - Total: DC2: DW Max (+) Moment Stations 58.7’ and 271.3’ 475k-ft 2,415k-ft 89k-ft 2,979k-ft 553k-ft 1,011k-ft 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel Max (-) Moment Station 165.0’ -1,189k-ft -5,708k-ft -210k-ft -7,107k-ft -1,251k-ft -2,286k-ft AASHTO-LRFD 2007 Page 5 of 62 -- 5 --
  11. 11. 2.2: Live Load Calculations The following design vehicular live load cases described in AASHTO-LRFD are considered: 1) The effect of a design tandem combined with the effect of the lane loading. The design tandem consists of two 25kip axles spaced 4.0’ apart. The lane loading consists of a 0.64klf uniform load on all spans of the bridge. (HL-93M in SAP) 2) The effect of one design truck with variable axle spacing combined with the effect of the 0.64klf lane loading. (HL-93K in SAP) 3) For negative moment between points of contraflexure only: 90% of the effect of a truck-train combined with 90% of the effect of the lane loading. The truck train consists of two design trucks (shown below) spaced a minimum of 50’ between the lead axle of one truck and the rear axle of the other truck. The distance between the two 32kip axles should be taken as 14’ for each truck. The points of contraflexure were taken as the field splices at 132’ and 198’ from the left end of the bridge. (HL-93S in SAP) 4) The effect of one design truck with fixed axle spacing used for fatigue loading. All live load calculations were performed in SAP 2000 using a beam line analysis. The nominal moment data from SAP was then input into Excel. An Impact Factor of 1.33 was applied to the truck and tandem loads and an impact factor of 1.15 was applied to the fatigue loads within SAP. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 6 of 62 -- 6 --
  12. 12. Unfactored Moving Load Moment Envelopes from SAP 6,000 Single Truck 4,000 Tandem Moment (kip-ft) 2,000 Fatigue 0 Fatigue Tandem -2,000 Contraflexure Point Contraflexure Point -4,000 Single Truck Two Trucks -6,000 0 30 60 90 120 150 180 210 240 270 300 330 270 300 330 Station (ft) Unfactored Moving Load Shear Envelopes from SAP 200 Single Truck 150 Tandem 100 Fatigue Shear (kip) 50 0 -50 -100 -150 -200 0 30 60 90 120 150 180 210 240 Station (ft) 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 7 of 62 -- 7 --
  13. 13. The following Live Load results were obtained from the SAP analysis: • The maximum positive live-load moments occur at stations 73.3’ and 256.7’ • The maximum negative live-load moments occur over the center support at station 165.0’ HL-93M HL-93K HL-93S Fatigue Max (+) Moment Stations 73.3’ and 256’ 3,725k-ft 4,396k-ft N/A 2,327k-ft Max (-) Moment Station 165’ -3,737k-ft -4,261k-ft -5,317k-ft -1,095k-ft Before proceeding, these live-load moments will be confirmed with an influence line analysis. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 8 of 62 -- 8 --
  14. 14. 2.2.1: Verify the Maximum Positive Live-Load Moment at Station 73.3’: 25kip 25kip Tandem: 32kip 32kip 8kip Single Truck: 0.640kip/ft Lane: Moment (k-ft / kip) 40 30 20 10 0 -10 -20 0 15 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240 255 270 285 300 315 330 Station (ft) Tandem: Single Truck: Lane Load: ( 25 ) ( 33.00 ) + ( 25 ) ( 31.11 ) = 1, 603 ( 8 ) ( 26.13 ) + ( 32 ) ( 33.00 ) + ( 32 ) ( 26.33 ) = 2,108 kip kip k-ft kip kip kip k-ft k-ft kip ( 0.640 ) ( 2, 491 ) = 1,594 ft kip k-ft kip kip k-ft k-ft kip k-ft 2 k-ft kip k-ft kip (IM)(Tandem) + Lane: (1.33 ) (1, 603 (IM)(Single Truck) + Lane: (1.33 ) ( 2,108 k-ft k-ft ) + 1,594 k-ft ) + 1,594 k-ft = 3,726 k-ft = 4,397 k-ft GOVERNS The case of two trucks is not considered here because it is only used when computing negative moments. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 9 of 62 -- 9 --
  15. 15. 2.2.2: Verify the Maximum Negative Live-Load Moment at Station 165.0’: 25kip 25kip Tandem: 32kip 32kip 8kip Single Truck: 32kip 32kip 32kip 32kip 8kip 8kip Two Trucks: 0.640kip/ft Lane: Station (ft) 0 15 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240 255 270 285 300 315 330 Moment (k-ft / kip) 0 -5 -10 -15 -20 Tandem: Single Truck: ( 25 ) (18.51 ) + ( 25 ) (18.45 ) = 924.0 ( 8 ) (17.47 ) + ( 32 ) (18.51 ) + ( 32 ) (18.31 ) = 1,318 kip kip k-ft k-ft kip kip kip k-ft kip k-ft kip k-ft kip kip k-ft kip kip k-ft kip kip kip k-ft ft k-ft kip k-ft 2 k-ft kip k-ft kip (IM)(Tandem) + Lane: (1.33 ) ( 924.0 (IM)(Single Truck) + Lane: (1.33 ) (1,318 (0.90){(IM)(Two Trucks) + Lane}: kip k-ft ( 0.640 ) ( 3,918 ) = 2,508 kip k-ft kip kip Lane Load: k-ft kip ( 8 ) (17.47 ) + ( 32 ) (18.51 ) + ( 32 ) (18.31 ) + ... ... + ( 8 ) (16.72 ) + ( 32 ) (18.31 ) + ( 32 ) (18.51 ) = 2, 630 kip Two Trucks: kip k-ft k-ft k-ft ) + 2,508 ) + 2,508 ( 0.90 ) ⎡(1.33 ) ( 2,630 ⎣ 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel k-ft k-ft k-ft = 3,737 k-ft = 4, 261 ) + 2,508 k-ft k-ft ⎤ = 5, 405k-ft GOVERNS ⎦ AASHTO-LRFD 2007 Page 10 of 62 -- 10 --
  16. 16. Based on the influence line analysis, we can say that the moments obtained from SAP appear to be reasonable and will be used for design. Before these Service moments can be factored and combined, we must compute the distribution factors. Since the distribution factors are a function of Kg, the longitudinal stiffness parameter, we must first compute the sections properties of the girders. 2.3: Braking Force The Breaking Force, BR, is taken as the maximum of: A) 25% of the Design Truck BRSingle Lane = ( 0.25 ) ( 8kip + 32kip + 32kip ) = 18.00kip B) 25% of the Design Tandem BRSingle Lane = ( 0.25 ) ( 25kip + 25kip ) = 12.50kip C) 5% of the Design Truck with the Lane Load. ( ) BRSingle Lane = ( 0.05 ) ⎡( 8kip + 32kip + 32kip ) + ( 2 )(165') 0.640 kip ⎤ = 14.16kip ft ⎦ ⎣ D) 5% of the Design Tandem with the Lane Load. ( ) BRSingle Lane = ( 0.05 ) ⎡( 25kip + 25kip ) + ( 2 )(165' ) 0.640 kip ⎤ = 13.06kip ft ⎦ ⎣ Case (A) Governs: BRNet = ( BRSingle Lane ) ( # Lanes )( MPF ) = (18.00kip ) ( 3)( 0.85 ) = 45.90kip 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel This load has not been factored… AASHTO-LRFD 2007 Page 11 of 62 -- 11 --
  17. 17. 2.4: Centrifugal Force A centrifugal force results when a vehicle turns on a structure. Although a centrifugal force doesn’t apply to this bridge since it is straight, the centrifugal load that would result from a hypothetical horizontal curve will be computed to illustrate the procedure. The centrifugal force is computed as the product of the axle loads and the factor, C. C= f v2 gR (3.6.3-1) where: ft ( sec ) v - Highway design speed f - 4/3 for all load combinations except for Fatigue, in which case it is 1.0 g - The acceleration of gravity R - The radius of curvature for the traffic lane (ft). ( ) ft sec 2 Suppose that we have a radius of R = 600’ and a design speed of v = 65mph = 95.33ft/sec. ft 2 ⎤ ⎛ 4 ⎞ ⎡ ( 95.33 sec ) ⎥ = 0.6272 C = ⎜ ⎟⎢ ft ⎝ 3 ⎠ ⎢ 32.2 sec2 ( 600 ') ⎥ ⎣ ⎦ ( ) CE = ( Axle Loads )( C )( # Lanes )( MPF ) = ( 72kip ) ( 0.6272 )( 3)( 0.85 ) = 115.2kip This force has not been factored… The centrifugal force acts horizontally in the direction pointing away from the center of curvature and at a height of 6’ above the deck. Design the cross frames at the supports to carry this horizontal force into the bearings and design the bearings to resist the horizontal force and the resulting overturning moment. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 12 of 62 -- 12 --
  18. 18. 2.5: Wind Loads For the calculation of wind loads, assume that the bridge is located in the “open country” at an elevation of 40’ above the ground. Take Z = 40’ V o = 8.20mph Z o = 0.23ft Open Country Horizontal Wind Load on Structure: (WS) Design Pressure: 2 ⎛V ⎞ VDZ 2 PD = PB ⎜ D Z ⎟ = PB 2 10, 000mph ⎝ VB ⎠ (3.8.1.2.1-1) - Base Pressure - For beams, PB = 50psf when VB = 100mph. - Base Wind Velocity, typically taken as 100mph. - Wind Velocity at an elevation of Z = 30’ (mph) - Design Wind Velocity (mph) PB VB V30 VDZ (Table 3.8.1.2.1-1) Design Wind Velocity: ⎛V ⎞ ⎛ Z ⎞ VDZ = 2.5Vo ⎜ 30 ⎟ ln ⎜ ⎟ ⎝ VB ⎠ ⎝ Z o ⎠ = ( 2.5 ) ( 8.20 PD = ( 50 psf ) (105.8 ) mph (3.8.1.1-1) ft ⎛ 100 ⎞ ⎛ 40 ⎞ ) ⎜ 100 ⎟ Ln ⎜ 0.23ft ⎟ = 105.8mph ⎝ ⎠ ⎝ ⎠ mph 2 (10, 000 ) mph 2 PD = 55.92psf The height of exposure, hexp, for the finished bridge is computed as hexp hexp = 71.5"+ 11.75"+ 42" = 125.3" = 10.44 ' The wind load per unit length of the bridge, W, is then computed as: W = ( 55.92psf ) (10.44 ' ) = 583.7 lbs ft = ( 583.7 lbs ) ( 2 )(165') = 192.6kip ft Total Wind Load: WS H ,Total For End Abutments: WS H , Abt = ( 583.7 lbs ) ( 1 ) (165' ) = 48.16kip ft 2 For Center Pier: WS H , Pier = ( 583.7 lbs ) ( 2 ) ( 1 ) (165' ) = 96.31kip ft 2 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 13 of 62 -- 13 --
  19. 19. Vertical Wind Load on Structure: (WS) When no traffic is on the bridge, a vertical uplift (a line load) with a magnitude equal to 20psf times the overall width of the structure, w, acts at the windward quarter point of the deck. P = ( 20psf ) ( w ) = ( 20psf ) ( 42 ' ) = 840 lbs V ft Total Uplift: (840 lbs ) ( 2 )(165') = 277.2kip ft For End Abutments: (840 lbs ) ( 1 ) (165') = 69.30kip ft 2 For Center Pier: ( 840 lbs ) ( 2 ) ( 1 ) (165') = 138.6kip ft 2 Wind Load on Live Load: (WL) The wind acting on live load is applied as a line load of 100 lbs/ft acting at a distance of 6’ above the deck, as is shown below. This is applied along with the horizontal wind load on the structure but in the absence of the vertical wind load on the structure. WL PD 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 14 of 62 -- 14 --
  20. 20. 3. SECTION PROPERTIES AND CALCULATIONS: 3.1: Effective Flange Width, beff: For an interior beam, beff is the lesser of: ⎧ Leff 132' = = 33' = 396" ⎪• 4 ⎪ 4 bf ⎪ 15" = (12)(8.5") + = 109.5" ⎨• 12ts + 2 2 ⎪ ⎪• S = (12')(12 in ft ) = 144" ⎪ ⎩ For an exterior beam, beff is the lesser of: ⎧ Leff 132' = = 33' = 198.0" ⎪• 4 ⎪ 4 bf ⎪ 15" = (12)(8.5") + = 109.5" ⎨• 12ts + 2 2 ⎪ ⎪ S ⎛ 12' ⎞ + 3' ⎟ (12 in ) = 108.0" ⎪• + d e = ⎜ ft ⎝ 2 ⎠ ⎩ 2 Note that Leff was taken as 132.0’ in the above calculations since for the case of effective width in continuous bridges, the span length is taken as the distance from the support to the point of dead load contra flexure. For computing the section properties shown on the two pages that follow, reinforcing steel in the deck was ignored for short-term and long-term composite calculations but was included for the cracked section. The properties for the cracked Section #1 are not used in this example, thus the amount of rebar included is moot. For the properties of cracked Section #2, As = 13.02 in2 located 4.5” from the top of the slab was taken from an underlying example problem first presented by Barth (2004). 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 15 of 62 -- 15 --
  21. 21. 3.2: Section 1 Flexural Properties Bare Steel Ix Ay 791.72 0.53 1,377.84 15,398.86 10.50 1.75 Ad2 b 15.00 69.00 21.00 A 11.25 38.81 21.00 Y= ITotal = 53,157 30.68 71.06 IX 17,729 16,301 19,127 2,180.06 y 70.38 35.50 0.50 17,728 902 19,125 SBS1,top = 1,327 SBS1,bot = Top Flange Web Bot Flange t 0.7500 0.5625 1.0000 1,733 d -39.70 -4.82 30.18 Short-Term Composite (n = 8) b 109.50 15.00 15.0000 69.0000 21.0000 n: Slab Haunch Top Flange Web Bot Flange t 8.5000 0.0000 0.7500 0.5625 1.0000 A 116.34 0.00 11.25 38.81 21.00 187.41 y 75.00 70.75 70.38 35.50 0.50 Ix Ay 8,725.78 700.49 0.00 0.00 791.72 0.53 1,377.84 15,398.86 10.50 1.75 10,905.84 Ad2 IX 8.00 Y= d -16.81 -12.56 -12.18 22.69 57.69 33,562 0 1,670 35,387 69,901 140,521 SST1,top = SST1,bot = 58.19 32,862 0 1,669 19,988 69,900 ITotal = 11,191 2,415 Long-Term Composite (n = 24) b 109.50 15.00 15.0000 69.0000 21.0000 n: Slab Haunch Top Flange Web Bot Flange t 8.5000 0.0000 0.7500 0.5625 1.0000 A 38.78 0.00 11.25 38.81 21.00 109.84 y 75.00 70.75 70.38 35.50 0.50 Ix Ay 2,908.59 233.50 0.00 0.00 791.72 0.53 1,377.84 15,398.86 10.50 1.75 5,088.66 Ad2 31,885 0 6,506 4,549 44,101 ITotal = IX 32,119 0 6,507 19,948 44,103 102,676 SLT1,top = SLT1,bot = 4,204 2,216 24.00 Y= d -28.67 -24.42 -24.05 10.83 45.83 46.33 Cracked Section Rebar Top Flange Web Bot Flange t 4.5000 0.7500 0.5625 1.0000 b 15.0000 69.0000 21.0000 A 13.02 11.25 38.81 21.00 84.08 y 75.25 70.38 35.50 0.50 Y= Ix Ay 979.76 791.72 0.53 1,377.84 15,398.86 10.50 1.75 3,159.82 37.58 Ad2 73,727 55,717 48,913 5 ITotal = IX 73,727 55,718 64,312 7 193,764 SCR1,top = SCR1,bot = 5,842 5,156 d -75.25 -70.38 -35.50 -0.50 These section properties do NOT include the haunch or sacrificial wearing surface. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 16 of 62 -- 16 --
  22. 22. 3.3: Section 2 Flexural Properties Bare Steel Ix Ay 1,512.00 1.75 1,436.06 15,398.86 65.63 27.34 Ad2 b 21.00 69.00 21.00 A 21.00 38.81 52.50 Y= ITotal = 96,642 26.83 112.31 IX 42,843 19,411 34,388 3,013.69 y 72.00 37.00 1.25 42,841 4,012 34,361 SBS2,top = 2,116 SBS2,bot = Top Flange Web Bot Flange t 1.0000 0.5625 2.5000 3,602 d -45.17 -10.17 25.58 Short Term Composite (n = 8) b 109.50 21.00 21.0000 69.0000 21.0000 n: Slab Haunch Top Flange Web Bot Flange t 8.5000 0.0000 1.0000 0.5625 2.5000 A 116.34 0.00 21.00 38.81 52.50 228.66 y 76.75 72.50 72.00 37.00 1.25 Ix Ay 8,929.38 700.49 0.00 0.00 1,512.00 1.75 1,436.06 15,398.86 65.63 27.34 11,943.07 Ad2 IX 8.00 Y= d -24.52 -20.27 -19.77 15.23 50.98 70,641 0 8,208 24,403 136,481 239,734 SST2,top = SST2,bot = 52.23 69,941 0 8,207 9,005 136,454 ITotal = 11,828 4,590 Long-Term Composite (n = 24) b 109.50 15.00 21.0000 69.0000 21.0000 n: Slab Haunch Top Flange Web Bot Flange t 8.5000 0.0000 1.0000 0.5625 2.5000 A 38.78 0.00 21.00 38.81 52.50 151.09 y 76.75 72.50 72.00 37.00 1.25 Ix Ay 2,976.46 233.50 0.00 0.00 1,512.00 1.75 1,436.06 15,398.86 65.63 27.34 5,990.15 Ad2 53,393 0 21,983 272 77,395 ITotal = IX 53,626 0 21,985 15,670 77,423 168,704 SLT2,top = SLT2,bot = 5,135 4,255 24.00 Y= d -37.10 -32.85 -32.35 2.65 38.40 39.65 Cracked Section Rebar Top Flange Web Bot Flange t 4.5000 1.0000 0.5625 2.5000 b 21.0000 69.0000 21.0000 A 13.02 21.00 38.81 52.50 125.33 y 77.00 72.00 37.00 1.25 Y= Ix Ay 1,002.54 1,512.00 1.75 1,436.06 15,398.86 65.63 27.34 4,016.23 32.04 Ad2 26,313 33,525 953 49,786 ITotal = IX 26,313 33,527 16,352 49,813 126,006 SCR2,top = SCR2,bot = 3,115 3,932 d -44.96 -39.96 -4.96 30.79 These section properties do NOT include the haunch or sacrificial wearing surface. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 17 of 62 -- 17 --
  23. 23. 4. DISTRIBUTION FACTOR FOR MOMENT 4.1: Positive Moment Region (Section 1): Interior Girder – One Lane Loaded: 0.4 0.3 ⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎟ ⎜ ⎟ ⎜ ⎝ 14 ⎠ ⎝ L ⎠ ⎜ 12 Lt s3 ⎝ DFM 1, Int + = 0.06 + ⎜ ⎞ ⎟ ⎟ ⎠ 0.1 2 K g = n( I + Aeg ) 4 2 2 K g = 8(53,157 in + (71.06 in )(46.82") ) K g = 1, 672, 000 in 4 0.4 0.3 4 ⎛ 12 ' ⎞ ⎛ 12 ' ⎞ ⎛ 1, 672, 000 in ⎞ ⎟ ⎟ ⎜ ⎟ ⎜ ⎝ 14 ⎠ ⎝ 165 ' ⎠ ⎝ (12)(165 ')(8.5")3 ⎠ 0.1 DFM 1, Int + = 0.06 + ⎜ DFM 1, Int + = 0.5021 In these calculations, the terms eg and Kg include the haunch and sacrificial wearing surface since doing so increases the resulting factor. Note that ts in the denominator of the final term excludes the sacrificial wearing surface since excluding it increases the resulting factor. Two or More Lanes Loaded: ⎞ ⎟ ⎟ ⎠ 0.1 DFM 2, Int + 0.6 0.2 ⎛ S ⎞ ⎛ S ⎞ ⎛ Kg = 0.075 + ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ 9.5 ⎠ ⎝ L ⎠ ⎜ 12 Lt s3 ⎝ DFM 2, Int + 0.6 0.2 4 ⎛ 12 ' ⎞ ⎛ 12 ' ⎞ ⎛ 1, 672, 000 in ⎞ = 0.075 + ⎜ ⎟ ⎟ ⎜ ⎟ ⎜ ⎝ 9.5 ⎠ ⎝ 165 ' ⎠ ⎝ 12(165 ')(8.5")3 ⎠ 0.1 DFM 2, Int + = 0.7781 Exterior Girder – One Lane Loaded: The lever rule is applied by assuming that a hinge forms over the first interior girder as a truck load is applied near the parapet. The resulting reaction in the exterior girder is the distribution factor. DFM 1, Ext + = 8.5 12 = 0.7083 Multiple Presence: DFM1,Ext+ = (1.2) (0.7083) = 0.8500 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 18 of 62 -- 18 --
  24. 24. Two or More Lanes Loaded: DFM2,Ext+ = e DFM2,Int+ e = 0.77 + = 0.77 + de 9.1 1.5 9.1 = 0.9348 DFM2,Ext+ = (0.9348) (0.7781) = 0.7274 4.2: Negative Moment Region (Section 2): The span length used for negative moment near the pier is the average of the lengths of the adjacent spans. In this case, it is the average of 165.0’ and 165.0’ = 165.0’. Interior Girder – One Lane Loaded: DFM 1, Int − 0.4 0.3 ⎛ S ⎞ ⎛ S ⎞ ⎛ Kg = 0.06 + ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ 14 ⎠ ⎝ L ⎠ ⎜ 12 Lts3 ⎝ ⎞ ⎟ ⎟ ⎠ 0.1 2 K g = n( I + Aeg ) K g = 8(96, 642 in 4 + (112.3 in 2 )(52.17 ") 2 ) K g = 3, 218, 000 in 4 0.4 0.3 4 ⎛ 12 ' ⎞ ⎛ 12 ' ⎞ ⎛ 3, 218, 000 in ⎞ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 14 ⎠ ⎝ 165 ' ⎠ ⎝ (12)(165 ')(8.5")3 ⎠ 0.1 DFM 1, Int − = 0.06 + ⎜ DFM 1, Int − = 0.5321 Two or More Lanes Loaded: ⎞ ⎟ ⎟ ⎠ 0.1 DFM 2, Int − 0.6 0.2 ⎛ S ⎞ ⎛ S ⎞ ⎛ Kg = 0.075 + ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ 9.5 ⎠ ⎝ L ⎠ ⎜ 12 Lt s3 ⎝ DFM 2, Int − 0.6 0.2 4 ⎛ 12 ' ⎞ ⎛ 12 ' ⎞ ⎛ 3, 218, 000 in ⎞ = 0.075 + ⎜ ⎟ ⎟ ⎜ ⎟ ⎜ ⎝ 9.5 ⎠ ⎝ 165 ' ⎠ ⎝ (12)(165 ')(8.5")3 ⎠ 0.1 DFM 2, Int − = 0.8257 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 19 of 62 -- 19 --
  25. 25. Exterior Girder – One Lane Loaded: Same as for the positive moment section: DFM1,Ext- = 0.8500 Two or More Lanes Loaded: DFM2,Ext- = e DFM2,Intd e = 0.77 + e 9.1 = 0.77 + 1.5 9.1 = 0.9348 DFM2,Ext- = (0.9348) (0.8257) = 0.7719 4.3: Minimum Exterior Girder Distribution Factor: NL DF Ext , Min = NL Nb + X Ext ∑ e Nb ∑x 2 One Lane Loaded: DF M 1, Ext , Min = 1 4 + (18.0 ')(14.5 ') (2) ⎡(18 ') 2 + (6 ') 2 ⎤ ⎣ ⎦ = 0.6125 Multiple Presence: DFM1,Ext,Min = (1.2) (0.6125) = 0.7350 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 20 of 62 -- 20 --
  26. 26. Two Lanes Loaded: 14.5' 2.5' 2' 3' 3' 2' 3' 3' P1 P2 DF M 2 , Ext , Min = 2 4 + (18.0 ')(14.5 '+ 2.5 ') (2) ⎡(18 ') 2 + (6 ') 2 ⎤ ⎣ ⎦ = 0.9250 Multiple Presence: DFM2,Ext,Min = (1.0) (0.9250) = 0.9250 Lane 1 (12') 3' Lane 2 (12') 12' 6' Three Lanes Loaded: The case of three lanes loaded is not considered for the minimum exterior distribution factor since the third truck will be placed to the right of the center of gravity of the girders, which will stabilize the rigid body rotation effect resulting in a lower factor. 4.4: Moment Distribution Factor Summary Strength and Service Moment Distribution: 1 Lane Loaded: 2 Lanes Loaded: Positive Moment Interior Exterior 0.5021 0.8500 ≥ 0.7350 0.7781 0.7274 ≥ 0.9250 Negative Moment Interior Exterior 0.5321 0.8500 ≥ 0.7350 0.8257 0.7719 ≥ 0.9250 For Simplicity, take the Moment Distribution Factor as 0.9250 everywhere for the Strength and Service load combinations. Fatigue Moment Distribution: For Fatigue, the distribution factor is based on the one-lane-loaded situations with a multiple presence factor of 1.00. Since the multiple presence factor for 1-lane loaded is 1.2, these factors can be obtained by divided the first row of the table above by 1.2. 1 Lane Loaded: Positive Moment Interior Exterior 0.4184 0.7083 ≥ 0.6125 Negative Moment Interior Exterior 0.4434 0.7083 ≥ 0.6125 For Simplicity, take the Moment Distribution Factor as 0.7083 everywhere for the Fatigue load combination Multiplying the live load moments by this distribution factor of 0.9250 yields the table of “nominal” girder moments shown on the following page. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 21 of 62 -- 21 --
  27. 27. Nominal Girder Moments for Design Station (ft) 0.0 14.7 29.3 44.0 58.7 73.3 88.0 102.7 117.3 132.0 135.7 139.3 143.0 146.7 150.3 154.0 157.7 161.3 165.0 168.7 172.3 176.0 179.7 183.3 187.0 190.7 194.3 198.0 212.7 227.3 242.0 256.7 271.3 286.0 300.7 315.3 330.0 (LL+IM)+ (k-ft) 0.0 1605.1 2791.4 3572.6 3999.4 4066.7 3842.5 3310.8 2509.4 1508.6 1274.6 1048.4 828.6 615.8 463.3 320.5 185.5 76.4 0.0 76.4 185.5 320.5 463.3 615.8 828.6 1048.4 1274.6 1508.6 2509.4 3310.8 3842.5 4066.7 3999.4 3572.6 2791.4 1605.1 0.0 (LL+IM)(k-ft) 0.0 -280.7 -561.3 -842.0 -1122.7 -1403.4 -1684.0 -1964.7 -2245.4 -2547.5 -2660.0 -2793.3 -2945.6 -3115.6 -3371.3 -3728.6 -4105.0 -4496.9 -4918.1 -4496.9 -4105.0 -3728.6 -3371.3 -3115.6 -2945.6 -2793.3 -2660.0 -2547.5 -2245.4 -1964.7 -1684.0 -1403.4 -1122.7 -842.0 -561.3 -280.7 0.0 Nominal Moments Fat+ Fat(k-ft) (k-ft) 0.2 0.0 645.6 -68.9 1127.9 -137.9 1449.4 -206.8 1626.1 -275.8 1647.9 -344.7 1599.4 -413.7 1439.3 -482.6 1148.6 -551.6 763.6 -620.5 651.3 -637.8 539.1 -655.0 425.3 -672.2 310.8 -689.5 221.9 -706.7 158.6 -724.0 98.8 -741.2 49.4 -758.4 0.1 -775.6 49.4 -758.4 98.8 -741.2 158.6 -724.0 221.9 -706.7 310.8 -689.5 425.3 -672.2 539.1 -655.0 651.3 -637.8 763.2 -620.6 1148.6 -551.6 1439.3 -482.6 1599.4 -413.7 1647.9 -344.7 1626.1 -275.8 1449.4 -206.8 1127.9 -137.9 645.6 -68.9 0.2 0.0 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel DC1 (k-ft) 0.0 1309.9 2244.5 2799.9 2978.6 2779.3 2202.1 1248.4 -84.8 -1793.1 -2280.8 -2794.0 -3333.2 -3898.1 -4488.6 -5105.1 -5747.2 -6415.3 -7108.8 -6415.3 -5747.2 -5105.1 -4488.6 -3898.1 -3333.2 -2794.0 -2280.8 -1793.1 -84.8 1248.4 2202.1 2779.3 2978.6 2799.9 2244.5 1309.9 0.0 DC2 (k-ft) 0.0 240.0 412.0 515.0 549.7 515.8 413.2 242.3 2.5 -305.4 -393.2 -485.2 -581.5 -682.1 -787.0 -896.2 -1009.7 -1127.5 -1249.5 -1127.5 -1009.7 -896.2 -787.0 -682.1 -581.5 -485.2 -393.2 -305.4 2.5 242.3 413.2 515.8 549.7 515.0 412.0 240.0 0.0 DW (k-ft) 0.0 440.3 755.6 944.7 1008.3 946.1 757.9 444.4 4.6 -560.2 -721.2 -890.0 -1066.7 -1251.3 -1443.7 -1643.9 -1852.1 -2068.1 -2291.9 -2068.1 -1852.1 -1643.9 -1443.7 -1251.3 -1066.7 -890.0 -721.2 -560.2 4.6 444.4 757.9 946.1 1008.3 944.7 755.6 440.3 0.0 AASHTO-LRFD 2007 Page 22 of 62 -- 22 --
  28. 28. 5. DISTRIBUTION FACTOR FOR SHEAR The distribution factors for shear are independent of the section properties and span length. Thus, the only one set of calculations are need - they apply to both the section 1 and section 2 5.1: Interior Girder – One Lane Loaded: S 25.0 12 ' = 0.36 + = 0.8400 25.0 DFV 1,Int = 0.36 + Two or More Lanes Loaded: DFV 2 ,Int S ⎛ S ⎞ = 0.2 + − ⎜ ⎟ 12 ⎝ 35 ⎠ 2 2 = 0.2 + 12 ' ⎛ 12 ' ⎞ −⎜ ⎟ = 1.082 12 ⎝ 35 ⎠ 5.2: Exterior Girder – One Lane Loaded: Lever Rule, which is the same as for moment: DFV1,Ext = 0.8500 Two or More Lanes Loaded: DFV2,Ext = e DFV2,Int de 10 1.5' = 0.60 + = 0.7500 10 e = 0.60 + DFV2,Ext = (0.7500) (1.082) = 0.8115 5.3: Minimum Exterior Girder Distribution Factor - The minimum exterior girder distribution factor applies to shear as well as moment. DFV1,Ext,Min = 0.7350 DFV2,Ext,Min = 0.9250 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 23 of 62 -- 23 --
  29. 29. 5.4: Shear Distribution Factor Summary Strength and Service Shear Distribution: 1 Lane Loaded: 2 Lanes Loaded: Shear Distribution Interior Exterior 0.8400 0.8500 ≥ 0.7350 1.082 0.6300 ≥ 0.9250 For Simplicity, take the Shear Distribution Factor as 1.082 everywhere for Strength and Service load combinations. Fatigue Shear Distribution: For Fatigue, the distribution factor is based on the one-lane-loaded situations with a multiple presence factor of 1.00. Since the multiple presence factor for 1-lane loaded is 1.2, these factors can be obtained by divided the first row of the table above by 1.2. 1 Lane Loaded: Shear Distribution Interior Exterior 0.7000 0.7083 ≥ 0.6125 For Simplicity, take the Shear Distribution Factor as 0.7083 everywhere for the Fatigue load combination. Multiplying the live load shears by these distribution factors yields the table of “nominal” girder shears shown on the following page. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 24 of 62 -- 24 --
  30. 30. Nominal Girder Shears for Design Station (ft) 0.0 14.7 29.3 44.0 58.7 73.3 88.0 102.7 117.3 132.0 135.7 139.3 143.0 146.7 150.3 154.0 157.7 161.3 165.0 168.7 172.3 176.0 179.7 183.3 187.0 190.7 194.3 198.0 212.7 227.3 242.0 256.7 271.3 286.0 300.7 315.3 330.0 (LL+IM)+ (kip) 144.9 123.5 103.5 85.0 68.1 52.8 39.4 27.8 18.0 10.0 8.3 6.7 5.5 4.3 3.2 2.2 1.3 0.0 0.0 170.1 166.2 162.3 158.4 154.5 150.5 146.5 142.5 138.6 122.3 105.7 89.1 72.7 56.7 41.4 26.8 20.3 19.7 (LL+IM)(kip) -19.7 -20.3 -26.8 -41.4 -56.7 -72.7 -89.1 -105.7 -122.3 -138.6 -142.5 -146.5 -150.5 -154.5 -158.4 -162.3 -166.2 -170.1 -173.9 -0.5 -1.3 -2.2 -3.2 -4.3 -5.5 -6.7 -8.3 -10.0 -18.0 -27.8 -39.4 -52.8 -68.1 -85.0 -103.5 -123.5 -144.9 Nominal Shears Fat+ Fat(kip) (kip) 50.8 -4.7 44.6 -4.7 38.5 -6.4 32.6 -11.1 26.9 -17.2 21.4 -23.2 16.3 -29.0 11.5 -34.6 7.3 -39.9 3.9 -44.9 3.4 -46.0 2.8 -47.2 2.3 -48.3 1.8 -49.4 1.4 -50.4 1.0 -51.5 0.6 -52.4 0.3 -53.4 54.3 -54.3 53.4 -0.3 52.4 -0.6 51.5 -1.0 50.4 -1.4 49.4 -1.8 48.3 -2.3 47.2 -2.8 46.0 -3.4 44.9 -3.9 39.9 -7.3 34.6 -11.5 29.0 -16.3 23.2 -21.4 17.2 -26.9 11.1 -32.6 6.4 -38.5 4.7 -44.6 4.7 -50.8 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel DC1 (kip) 115.0 88.8 62.5 36.3 10.1 -16.1 -42.3 -68.6 -94.8 -121.0 -127.6 -134.1 -140.7 -147.2 -153.8 -160.3 -166.9 -173.4 -180.0 173.4 166.9 160.3 153.8 147.2 140.7 134.1 127.6 121.0 94.8 68.6 42.3 16.1 -10.1 -36.3 -62.5 -88.8 -115.0 DC2 (kip) 20.6 15.9 11.2 6.5 1.8 -2.9 -7.6 -12.3 -17.0 -21.7 -22.8 -24.0 -25.2 -26.4 -27.5 -28.7 -29.9 -31.0 -32.2 31.0 29.9 28.7 27.5 26.4 25.2 24.0 22.8 21.7 17.0 12.3 7.6 2.9 -1.8 -6.5 -11.2 -15.9 -20.6 DW (kip) 37.6 29.0 20.5 11.9 3.3 -5.3 -13.9 -22.4 -31.0 -39.6 -41.7 -43.9 -46.0 -48.2 -50.3 -52.5 -54.6 -56.8 -58.9 56.8 54.6 52.5 50.3 48.2 46.0 43.9 41.7 39.6 31.0 22.4 13.9 5.3 -3.3 -11.9 -20.5 -29.0 -37.6 AASHTO-LRFD 2007 Page 25 of 62 -- 25 --
  31. 31. 6. FACTORED SHEAR AND MOMENT ENVELOPES The following load combinations were considered in this example: Strength I: Strength IV: 1.75(LL + IM) + 1.25DC1 + 1.25DC2 + 1.50DW 1.50DC1 + 1.50DC2 + 1.50DW Service II: 1.3(LL + IM) + 1.0DC1 + 1.0DC2 + 1.0DW Fatigue: 0.75(LL + IM) (IM = 15% for Fatigue; IM = 33% otherwise) Strength II is not considered since this deals with special permit loads. Strength III and V are not considered as they include wind effects, which will be handled separately as needed. Strength IV is considered but is not expected to govern since it addresses situations with high dead load that come into play for longer spans. Extreme Event load combinations are not included as they are also beyond the scope of this example. Service I again applies to wind loads and is not considered (except for deflection) and Service III and Service IV correspond to tension in prestressed concrete elements and are therefore not included in this example. In addition to the factors shown above, a load modifier, η, was applied as is shown below. Q = ∑ηiγ i Qi η is taken as the product of ηD, ηR, and ηI, and is taken as not less than 0.95. For this example, ηD and ηI are taken as 1.00 while ηR is taken as 1.05 since the bridge has 4 girders with a spacing greater than or equal to 12’. Using these load combinations, the shear and moment envelopes shown on the following pages were developed. Note that for the calculation of the Fatigue moments and shears that η is taken as 1.00 and the distribution factor is based on the one-lane-loaded situations with a multiple presence factor of 1.00 (AASHTO Sections 6.6.1.2.2, Page 6-29 and 3.6.1.4.3b, Page 3-25). 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 26 of 62 -- 26 --
  32. 32. Strength Limit Moment Envelopes 20,000 15,000 Strength I 10,000 Strength IV Moment (kip-ft) 5,000 0 -5,000 Strength IV -10,000 -15,000 Strength I -20,000 -25,000 0 30 60 90 120 150 180 210 240 270 300 330 270 300 330 Station (ft) Strength Limit Shear Force Envelope 800 Strength I 600 400 Strength IV Shear (kip) 200 0 -200 -400 -600 -800 0 30 60 90 120 150 180 210 240 Station (ft) 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 27 of 62 -- 27 --
  33. 33. Service II Moment Envelope 12,500 10,000 7,500 5,000 Moment (kip-ft) 2,500 0 -2,500 -5,000 -7,500 -10,000 -12,500 -15,000 -17,500 -20,000 0 30 60 90 120 150 180 210 240 270 300 330 270 300 330 Station (ft) Service II Shear Envelope 600 400 Shear (kip) 200 0 -200 -400 -600 0 30 60 90 120 150 180 210 240 Station (ft) 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 28 of 62 -- 28 --
  34. 34. Factored Fatigue Moment Envelope 1,500 1,000 Moment (kip-ft) 500 0 -500 -1,000 -1,500 0 30 60 90 120 150 180 210 240 270 300 330 270 300 330 Station (ft) Factored Fatigue Shear Envelope 50 40 30 20 Shear (kip) 10 0 -10 -20 -30 -40 -50 0 30 60 90 120 150 180 210 240 Station (ft) 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 29 of 62 -- 29 --
  35. 35. Factored Girder Moments for Design Station (ft) 0.0 14.7 29.3 44.0 58.7 73.3 88.0 102.7 117.3 132.0 135.7 139.3 143.0 146.7 150.3 154.0 157.7 161.3 165.0 168.7 172.3 176.0 179.7 183.3 187.0 190.7 194.3 198.0 212.7 227.3 242.0 256.7 271.3 286.0 300.7 315.3 330.0 Strength I Total + Total (k-ft) (k-ft) 0.0 0.0 5677.1 -515.7 9806.0 -1031.5 12403.3 -1547.2 13567.8 -2062.9 13287.4 -2578.7 11687.1 -3094.4 8740.0 -3610.2 4621.6 -4237.1 2772.1 -8317.5 2342.0 -9533.2 1926.4 -10838.2 1522.6 -12230.6 1131.6 -13707.1 851.2 -15392.8 588.9 -17317.3 340.9 -19328.3 140.4 -21420.1 0.0 -23617.1 140.4 -21420.1 340.9 -19328.3 588.9 -17317.3 851.2 -15392.8 1131.6 -13707.1 1522.6 -12230.6 1926.4 -10838.2 2342.0 -9533.2 2772.1 -8317.5 4621.6 -4237.1 8740.0 -3610.2 11687.1 -3094.4 13287.4 -2578.7 13567.8 -2062.9 12403.3 -1547.2 9806.0 -1031.5 5677.1 -515.7 0.0 0.0 Strength IV Total + Total (k-ft) (k-ft) 0.0 0.0 3134.6 0.0 5374.1 0.0 6708.8 0.0 7145.1 0.0 6679.8 0.0 5312.9 0.0 3047.7 0.0 11.2 -133.5 0.0 -4187.3 0.0 -5347.3 0.0 -6566.4 0.0 -7845.7 0.0 -9184.5 0.0 -10582.9 0.0 -12041.3 0.0 -13559.1 0.0 -15137.1 0.0 -16774.1 0.0 -15137.1 0.0 -13559.1 0.0 -12041.3 0.0 -10582.9 0.0 -9184.5 0.0 -7845.7 0.0 -6566.4 0.0 -5347.3 0.0 -4187.3 11.2 -133.5 3047.7 0.0 5312.9 0.0 6679.8 0.0 7145.1 0.0 6708.8 0.0 5374.1 0.0 3134.6 0.0 0.0 0.0 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel Service II Total + Total (k-ft) (k-ft) 0.0 0.0 4280.7 -383.1 7393.0 -766.2 9349.1 -1149.4 10222.6 -1532.5 10004.2 -1915.6 8787.0 -2298.7 6551.1 -2681.8 3432.8 -3153.9 2059.3 -6268.9 1739.8 -7195.8 1431.1 -8190.4 1131.1 -9251.2 840.6 -10375.8 632.3 -11657.1 437.4 -13117.1 253.3 -14642.7 104.3 -16229.6 0.0 -17895.9 104.3 -16229.6 253.3 -14642.7 437.4 -13117.1 632.3 -11657.1 840.6 -10375.8 1131.1 -9251.2 1431.1 -8190.4 1739.8 -7195.8 2059.3 -6268.9 3432.8 -3153.9 6551.1 -2681.8 8787.0 -2298.7 10004.2 -1915.6 10222.6 -1532.5 9349.1 -1149.4 7393.0 -766.2 4280.7 -383.1 0.0 0.0 Fatigue Total + Total (k-ft) (k-ft) 0.2 0.0 484.2 -51.7 845.9 -103.4 1087.1 -155.1 1219.6 -206.8 1235.9 -258.6 1199.5 -310.3 1079.5 -362.0 861.5 -413.7 572.7 -465.4 488.5 -478.3 404.3 -491.3 318.9 -504.2 233.1 -517.1 166.5 -530.0 119.0 -543.0 74.1 -555.9 37.1 -568.8 0.1 -581.7 37.1 -568.8 74.1 -555.9 119.0 -543.0 166.5 -530.0 233.1 -517.1 318.9 -504.2 404.3 -491.3 488.5 -478.3 572.4 -465.4 861.5 -413.7 1079.5 -362.0 1199.5 -310.3 1235.9 -258.6 1219.6 -206.8 1087.1 -155.1 845.9 -103.4 484.2 -51.7 0.2 0.0 AASHTO-LRFD 2007 Page 30 of 62 -- 30 --
  36. 36. Factored Girder Shears for Design Station (ft) 0.0 14.7 29.3 44.0 58.7 73.3 88.0 102.7 117.3 132.0 135.7 139.3 143.0 146.7 150.3 154.0 157.7 161.3 165.0 168.7 172.3 176.0 179.7 183.3 187.0 190.7 194.3 198.0 212.7 227.3 242.0 256.7 271.3 286.0 300.7 315.3 330.0 Strength I Total + Total (kip) (kip) 479.5 -34.5 390.5 -35.5 304.0 -46.9 220.1 -72.4 138.9 -99.3 92.5 -158.9 68.9 -239.1 48.6 -319.7 31.5 -400.1 17.5 -480.2 14.5 -500.0 11.7 -519.8 9.6 -539.7 7.6 -559.6 5.7 -579.3 3.9 -599.0 2.2 -618.7 0.0 -638.3 0.0 -657.9 638.3 -0.9 618.7 -2.2 599.0 -3.9 579.3 -5.7 559.6 -7.6 539.7 -9.6 519.8 -11.7 500.0 -14.5 480.2 -17.5 400.1 -31.5 319.7 -48.6 239.1 -68.9 158.9 -92.5 99.3 -138.9 72.4 -220.1 46.9 -304.0 35.5 -390.5 34.5 -479.5 Strength IV Total + Total (kip) (kip) 272.8 0.0 210.6 0.0 148.4 0.0 86.2 0.0 24.0 0.0 0.0 -38.2 0.0 -100.4 0.0 -162.6 0.0 -224.8 0.0 -287.0 0.0 -302.6 0.0 -318.1 0.0 -333.7 0.0 -349.2 0.0 -364.8 0.0 -380.3 0.0 -395.9 0.0 -411.4 0.0 -427.0 411.4 0.0 395.9 0.0 380.3 0.0 364.8 0.0 349.2 0.0 333.7 0.0 318.1 0.0 302.6 0.0 287.0 0.0 224.8 0.0 162.6 0.0 100.4 0.0 38.2 0.0 0.0 -24.0 0.0 -86.2 0.0 -148.4 0.0 -210.6 0.0 -272.8 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel Service II Total + Total (kip) (kip) 379.7 -26.9 309.0 -27.7 240.2 -36.6 173.4 -56.5 108.9 -77.5 72.1 -124.8 53.8 -188.6 37.9 -252.7 24.6 -316.8 13.7 -380.5 11.3 -396.3 9.2 -412.1 7.5 -427.9 5.9 -443.7 4.4 -459.4 3.0 -475.1 1.7 -490.8 0.0 -506.4 0.0 -522.0 506.4 -0.7 490.8 -1.7 475.1 -3.0 459.4 -4.4 443.7 -5.9 427.9 -7.5 412.1 -9.2 396.3 -11.3 380.5 -13.7 316.8 -24.6 252.7 -37.9 188.6 -53.8 124.8 -72.1 77.5 -108.9 56.5 -173.4 36.6 -240.2 27.7 -309.0 26.9 -379.7 Fatigue Total + Total (kip) (kip) 38.1 -3.5 33.5 -3.5 28.9 -4.8 24.5 -8.3 20.2 -12.9 16.1 -17.4 12.2 -21.8 8.6 -26.0 5.5 -29.9 3.0 -33.7 2.5 -34.5 2.1 -35.4 1.7 -36.2 1.4 -37.0 1.0 -37.8 0.8 -38.6 0.5 -39.3 0.2 -40.0 40.7 -40.7 40.0 -0.2 39.3 -0.5 38.6 -0.8 37.8 -1.0 37.0 -1.4 36.2 -1.7 35.4 -2.1 34.5 -2.5 33.7 -2.9 29.9 -5.5 26.0 -8.6 21.8 -12.2 17.4 -16.1 12.9 -20.2 8.3 -24.5 4.8 -28.9 3.5 -33.5 3.5 -38.1 AASHTO-LRFD 2007 Page 31 of 62 -- 31 --
  37. 37. 7. FATIGUE CHECKS 7.1: Check transverse stiffener to flange weld at Station 73.3: Traffic information: ADTT given as 2400. Three lanes are available to trucks. (ADTT)SL = (0.80) (2,400) = 1,920 N = (ADTT)SL (365) (75) n = (1,920) (365) (75) (1) = 52.56M Cycles Check Top Flange Weld: Fatigue need only be checked when the compressive stress due to unfactored permanent loads is less than twice the maximum tensile stress due to factored fatigue loads. ? Check f comp , DL ≤ 2 f Fat Distance from bottom of section to the detail under investigation y = tf,bottom + D = 1.00” + 69.00” = 70” (Pg 16) (Pg 24) ( 2, 779 ) (12 ) ( 70"− 30.68") = 24.67 k-ft M DC1 = 2, 779k-ft f DC1 = in ft ksi 53,157 in 4 (Pg 16) (Pg 24) (Pg 16) ( 515.8 ) (12 ) ( 70"− 46.33") = 1.427 k-ft M DC 2 = 515.8k-ft f DC 2 = in ft 102, 676 in ksi 4 (Pg 16) f comp , DL = 24.67 ksi + 1.427 ksi = 26.09ksi (Pg 16) (Pg 30) ( 258.6 ) (12 ) ( 70"− 58.19") = 0.261 k-ft M Fat , Neg = 258.6k-ft f Fat = in ft 140,521 in ksi 4 (Pg 16) ? Check f comp , DL ≤ 2 f Fat 26.09ksi ≤ ( 2 ) ( 0.261ksi ) = 0.521ksi , No. ? Fatigue need not be checked on the top flange at Station 73.3. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 32 of 62 -- 32 --
  38. 38. Check Bottom Flange Weld: The permanent loads at Station 73.3 cause tension in the bottom flange, thus by inspection fatigue needs to be checked. γ ( Δf ) ≤ ( ΔF )n 1 ( ΔF ) n ⎛ A ⎞ 3 ( ΔF )TH =⎜ ⎟ ≥ 2 ⎝N⎠ γ is a load factor of 0.75, which is already included in the fatigue moments. γ ( Δf ) = (1, 236 ) (12 ) ( 58.19"− 1.00") = 6.036 k-ft in ft 140,521 in ksi 4 The detail under consideration is a Category C’ detail. A = 44.0 x 108 ksi3 and (ΔF)TH = 12.0 ksi ( ΔF )TH 2 = 12.0ksi = 6.00ksi 2 The stress in the detail is almost less than the infinite life threshold 1 1 8 3 ⎛ A ⎞ 3 ⎛ 44 × 10 ksi ⎞ 3 =⎜ = 4.375ksi ⎜ ⎟ 6 ⎟ ⎝ N ⎠ ⎝ 52.56 × 10 ⎠ 1 ( ΔF )TH ⎛ A ⎞3 = 6.00ksi , the infinite life governs. Since ⎜ ⎟ = 4.375ksi is less than 2 ⎝N⎠ ( ΔF )n = 6.00ksi Since γ ( Δf ) = 6.036ksi > ( F )n = 6.00ksi , the detail is not satisfactory. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 33 of 62 -- 33 --
  39. 39. Calculate the design life of the part under consideration: Since γ ( Δf ) is greater than ( ΔF )TH 2 , solve for N in the following equation. 1 ⎛ A ⎞3 γ ( Δf ) ≤ ⎜ ⎟ ⎝N⎠ N≤ A γ ( Δf ) 3 = 44 × 108 ksi3 ( 6.036 ) ksi 3 = 20.01×106 cycles 20.01×106 cycles = 10, 420 days 1,920 10, 420 days = 28.55 years 365 ( = 28y, 6m, 19d, 2h, 38min...) 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel ☺ AASHTO-LRFD 2007 Page 34 of 62 -- 34 --
  40. 40. 8. CHECK CROSS_SECTION PROPORTION LIMITS Web Proportions • 69" D ≤ 150 tw 9 " 16 = 122.7 ≤ 150 O.K. Flange Proportions • bf ≤ 12 2t f 15" = 10.00 ≤12 (2)( 3 4 ") O.K. • bf ≤ 12 2t f 21" = 10.50 ≤12 (2)(1") O.K. • bf ≤ 12 2t f 21" = 4.200 ≤12 (2)(2 12 ") O.K. Check ODOT Criteria for Flange Width ? ⎛D ⎞ b f ≥ ⎜ + 2.5 ⎟ ≥ 12" ⎝6 ⎠ • b f ,min = → ⎛ 69" ⎞ + 2.5 ⎟ = 14" ⎜ ⎝ 6 ⎠ D 69" = =11.50" 6 6 • t f ,min = 1.1tw =(1.1)( 916 ") O.K. O.K. 5 O.K. 8" • 0.1 ≤ I yc ≤ 10 I yt 0.1 ≤ ( 3 4 ")(15")3 = 0.2733 ≤ 10 (1")(21")3 O.K. • 0.1 ≤ I yc ≤ 10 I yt 0.1 ≤ (2.5")(21")3 = 2.500 ≤ 10 (1")(21")3 O.K. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 35 of 62 -- 35 --
  41. 41. 9. CHECK SERVICE LIMIT STATE 9.1: Check Absolute Deflection of the Bridge: Section 6.10.4.1 Section 1 The cross section of Section 1 that is used for computing deflections is shown above. The entire deck width is used (as opposed to just the effective width that was used earlier) and the haunch and sacrificial wearing surface have been neglected. AASHTO permits the use of the stiffness of parapets and structurally continuous railing but ODOT does not. The transformed width of the bridge deck is w ' = ( 42 ') (12 in ) ft 8 = 63.00" Using the bottom of the steel as a datum, the location of the CG of the deck can be found as: yc = 1"+ 69"+ 3 4 "+ 8.5" = 75.00" 2 The CG of this composite cross section is found as: Y ( 63")(8.5")i( 75.00") + ( 4 ) ( 71.06 in 2 )i( 30.68") = = 59.63" ( 63")( 8.5") + ( 4 ) ( 71.06 in 2 ) Now the moment of inertia of the section can be found as: Concrete → Steel → ( 63")( 8.5") 12 3 + ( 63")( 8.5") [ 75.00"− 59.63"] = 2 129, 700 in 4 ( 4 ) ( 53,160 in 4 ) + ( 4 ) ( 71.06 in 2 ) [30.68"− 59.63"] 2 = 450,900 in 4 I1,total = 580, 600 in 4 I1 = 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel 580, 600 in 4 in 4 = 145,100 Girder 4 Girders AASHTO-LRFD 2007 Page 36 of 62 -- 36 --
  42. 42. Section 2 The cross section of Section 2 that is used for computing deflections is shown above. The transformed width of the bridge deck is w ' = ( 42 ') (12 in ) = 63.00" ft 8 Using the bottom of the steel as a datum, the location of the CG of the deck can be found as: yc = 2 1 2 "+ 69"+ 1"+ 8.5" = 76.75" 2 The CG of this composite cross section is found as: Y ( 63")(8.5")i( 76.75") + ( 4 ) (112.3 in 2 )i( 26.83") = = 53.98" ( 63")( 8.5") + ( 4 ) (112.3 in 2 ) Now the moment of inertia of the section can be found as: Concrete → Steel → ( 63")( 8.5") 12 3 + ( 63")( 8.5") [ 76.75"− 53.98"] = 2 280,900 in 4 ( 4 ) ( 96, 640 in 4 ) + ( 4 ) (112.3 in 2 ) [ 26.83"− 53.98"] 2 = 717, 700 in 4 I total = 998, 600 in 4 I2 = 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel 998, 600 in 4 in 4 = 249, 700 Girder 4 Girders AASHTO-LRFD 2007 Page 37 of 62 -- 37 --
  43. 43. The following model, which represents the stiffness of a single girder, was used to compute absolute liveload deflections assuming the entire width of the deck to be effective in both compression and tension. The live load component of the Service I load combination is applied. Based on AASHTO Section 3.6.1.3.2, the loading includes (1) the design truck alone and (2) the lane load with 25% of the design truck. The design truck and design lane load were applied separately in the model and will be combined below. The design truck included 33% impact. I = 145,100 in4 I = 249,700 in4 I = 145,100 in4 From the analysis: Deflection due to the Design Truck with Impact: ΔTruck = 2.442” Deflection due to the Design Lane Load: ΔLane = 0.8442” These deflections are taken at Stations 79.2’ and 250.8’. The model was broken into segments roughly 25’ long in the positive moment region and 7’ long in the negative moment region. A higher level of discretization may result in slightly different deflections but it is felt that this level of accuracy was acceptable for deflection calculations. Since the above results are from a single-girder model subjected to one lane’s worth of loading, distribution factors must be applied to obtain actual bridge deflections. Since it is the absolute deflection that is being investigated, all lanes are loaded (multiple presence factor apply) and it is assumed that all girders deflect equally. Given these assumptions, the distribution factor for deflection is simply the number of lanes times the multiple presence factor divided by the number of girders. Looking at the two loading criteria described above: ⎛ ( 0.85 )( 3) ⎞ Δ1 = ⎜ ⎜ ( 4 ) ⎟ ( 2.442") = 1.558" ⎟ ⎝ ⎠ ← Governs ⎛ ( 0.85 )( 3) ⎞ Δ2 = ⎜ ⎦ ⎜ ( 4 ) ⎟ ⎡( 0.8442") + ( 0.25 )( 2.442") ⎤ = 0.9274" ⎟⎣ ⎝ ⎠ The limiting deflection for this bridge is: Δ Limit = (165') (12 in ) L ft = = 2.475" 800 800 ← OK 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 38 of 62 -- 38 --
  44. 44. 9.2: Check the Maximum Span-to-Depth Ratio: Section 6.10.4.1 From Table 2.5.2.6.3-1, (1) the overall depth of a composite I-beam in a continuous span must not be less than 0.032L and (2) the depth of the steel in a composite I-beam in a continuous span must not be less than 0.027L. (1) ( 2) → 0.032 L = ( 0.032 )(165') (12 in ) = 63.36" ft OK → 0.027 L = ( 0.027 )(165') (12 in ) = 53.46" ft OK 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 39 of 62 -- 39 --
  45. 45. 9.3: Permanent Deformations - Section 1 At the Service Limit State, the following shall be satisfied for composite sections Top Flange: f f ≤ 0.95Rh Fyf Bottom Flange ff + fl ≤ 0.95Rh Fyf 2 Per §6.10.1.1.1a, elastic stresses at any location in a composite section shall consist of the sum of stresses caused by loads applied separately to the bare steel, short-term composite section, and long-term composite section. ⎛ M DC 1 ⎞ ⎛ M DC 2 ⎞ ⎛ M DW ⎟ + 1.00 ⎜ S ⎟ + 1.00 ⎜ S ⎝ LT ⎠ ⎝ LT ⎝ S BS ⎠ fc = 1.00 ⎜ ⎛ M LL + IM ⎞ ⎞ ⎟ + 1.30 ⎜ S ⎟ ⎠ ⎝ ST ⎠ Top Flange, Positive Moment It is not immediately evident to me whether the factored stress at 58.7’ or 73.3’ will govern. ⎛ (2, 979 f f ,58.7 = 1.00 ⎜ ⎝ k-ft )(12 in ) ⎞ ft 1,327 in 3 ⎛ (549.7 ⎟ + 1.00 ⎜ ⎠ ⎝ k-ft )(12 in ) ⎞ ft 4,204 in 3 ⎛ (1, 008 ⎟ + 1.00 ⎜ ⎠ ⎝ k-ft )(12 in ) ⎞ ft 4,204 in 3 ⎛ (3, 999 ⎟ + 1.30 ⎜ ⎠ ⎝ k-ft )(12 in ) ⎞ ft 11,191 in 3 ⎟ ⎠ f f ,58.7 = 36.96ksi ⎛ (2, 779 f f , 73.3 = 1.00 ⎜ ⎝ k-ft )(12 in ) ⎞ ft 1,327 in 3 ⎛ (515.8 ⎟ + 1.00 ⎜ ⎠ ⎝ k-ft )(12 in ) ⎞ ft 4,204 in 3 ⎛ (946.1 k-ft ⎟ + 1.00 ⎜ ⎠ ⎝ )(12 in ) ⎞ ft 4,204 in 3 ⎛ (4, 067 ⎟ + 1.30 ⎜ ⎠ ⎝ k-ft )(12 in ) ⎞ ft 11,191 in 3 ⎟ ⎠ f f ,73.3 = 34.97 ksi The stress at 58.7’ governs. ff = 36.96ksi. f f ≤ 0.95Rh Fyf → 36.96ksi ≤ (0.95)(1.00)(50ksi ) = 47.50ksi O.K. Note: The bending moments in the above calculations come from page 22 while the moments of inertia are found on page 16. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 40 of 62 -- 40 --
  46. 46. Bottom Flange, Positive Moment ⎛ (2, 979 f f ,58.7 = 1.00 ⎜ ⎝ k-ft )(12 in ) ⎞ ft 1,733 in ⎛ (549.7 ⎟ + 1.00 ⎜ ⎠ 3 ⎝ k-ft )(12 in ) ⎞ ft 2,216 in 3 ⎛ (1, 008 ⎟ + 1.00 ⎜ ⎠ ⎝ k-ft )(12 in ) ⎞ ft 2,216 in 3 ⎛ (3, 999 ⎟ + 1.30 ⎜ ⎠ ⎝ k-ft )(12 in ) ⎞ ft 2,415 in 3 ⎟ ⎠ f f ,58.7 = 54.90ksi ⎛ (2, 779 )(12 in ) ⎞ ⎛ (515.8 )(12 in ) ⎞ ⎛ (946.1 )(12 in ) ⎞ ⎛ (4, 067 )(12 in ) ⎞ ft ft ft ft = 1.00 ⎜ + 1.00 ⎜ + 1.00 ⎜ + 1.30 ⎜ ⎟ ⎟ ⎟ ⎟ ⎝ 1,733 in ⎠ ⎝ 2,216 in ⎠ ⎝ 2,216 in ⎠ ⎝ 2,415 in ⎠ k-ft f f , 73.3 k-ft 3 k-ft 3 k-ft 3 3 f f ,73.3 = 53.43ksi The stress at 58.7’ governs. ff = 54.90ksi. The load factor for wind under Service II is 0.00, ∴ fl = 0ksi ff + fl ≤ 0.95Rh Fyf 2 → 54.90ksi + 0ksi ≤ (0.95)(1.00)(50ksi ) = 47.50ksi 2 No Good. Note: The bending moments in the above calculations come from page 22 while the moments of inertia are found on page 17. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 41 of 62 -- 41 --
  47. 47. 9.4: Permanent Deformations - Section 2 Top Flange, Negative Moment ⎛ (7,109 )(12 in ) ⎞ ⎛ (1, 250 )(12 in ) ⎞ ⎛ (2, 292 )(12 in ) ⎞ ⎛ (4, 918 )(12 in ) ⎞ ft ft ft ft = 1.00 ⎜ + 1.00 ⎜ + 1.00 ⎜ + 1.30 ⎜ ⎟ ⎟ ⎟ ⎟ ⎝ 2,116 in ⎠ ⎝ 5,135 in ⎠ ⎝ 5,135 in ⎠ ⎝ 11,828 in ⎠ k-ft f f ,165 k-ft k-ft 3 k-ft 3 3 3 f f ,165 = 55.08ksi f f ≤ 0.95 Rh Fyf ? → 55.08ksi ≤(0.95)(1.00)(50ksi ) = 47.50ksi No Good. Bottom Flange, Negative Moment ⎛ (7,108 f f = 1.00 ⎜ ⎝ k-ft )(12 in ) ⎞ ft 3,602 in 3 ⎛ (1, 250 ⎟ + 1.00 ⎜ ⎠ ⎝ k-ft )(12 in ) ⎞ ft 4,255 in 3 ⎛ (2, 292 ⎟ + 1.00 ⎜ ⎠ ⎝ k-ft )(12 in ) ⎞ ft 4,255 in 3 ⎛ (4, 918 ⎟ + 1.30 ⎜ ⎠ ⎝ k-ft )(12 in ) ⎞ ft 4,590 in 3 ⎟ ⎠ f f = 50.39 ksi The load factor for wind under Service II is 0.00, ∴ fl = 0ksi ff + fl ≤ 0.95 Rh Fyf 2 → 50.39 ksi + 0ksi ≤ (0.95)(1.00)(50ksi ) = 47.50ksi 2 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel No Good. AASHTO-LRFD 2007 Page 42 of 62 -- 42 --
  48. 48. 9.5: Bend Buckling Checks At the Service Limit State, all sections except composite sections in positive flexure shall satisfy: f c ≤ Fcrw where: Fcrw = 0.9 Ek ⎛D⎞ ⎜ ⎟ ⎝ tw ⎠ k= and 2 9 ( Dc / D ) 2 Section 1 Not Applicable Section 2 ⎛ (7,108 )(12 ⎝ 3,602 in k-ft fc = 1.00 ⎜ )⎞ in ⎛ (1, 250 )(12 ⎝ 4,255 in k-ft ⎟ + 1.00 ⎜ ft ⎠ 3 in ft )⎞ ⎠ 3 ⎛ (2, 292 )(12 ⎝ 4,255 in k-ft ⎟ + 1.00 ⎜ in ft )⎞ ⎠ 3 ⎛ (4, 918 )(12 ⎝ 4,590 in k-ft ⎟ + 1.30 ⎜ ft )⎞ ft )⎞ in 3 ⎟ ⎠ f c = 50.39 ksi ⎛ (7,108 )(12 ⎝ 2,116 in k-ft ft = 1.00 ⎜ in ft )⎞ ⎛ (1, 250 )(12 ⎝ 5,135 in k-ft ⎟ + 1.00 ⎜ ⎠ 3 3 in ft )⎞ ⎛ (2, 292 )(12 ⎝ 5,135 in ⎟ + 1.00 ⎜ ⎠ k-ft 3 in ft )⎞ ⎛ (4, 918 )(12 ⎝ 11,828 in k-ft ⎟ + 1.30 ⎜ ⎠ 3 in ⎟ ⎠ ft = 55.08ksi ⎛ − fc ⎞ 0 Dc = ⎜ ⎜ f + f ⎟ d − tcf ≥ ⎟ t ⎠ ⎝ c ⎛ ⎞ 50.39ksi 72.5") − 2.5" ≥ 0 =⎜ ksi ksi ⎟ ( ⎝ 50.39 + 55.08 ⎠ = 32.14′′ k= 9 ( Dc / D ) Fcrw = 2 = 9 ⎛ 32.14" ⎞ ⎜ ⎟ ⎝ 69" ⎠ 2 = 41.49 (0.90)(29, 000ksi )(41.49) ⎛ 69" ⎞ ⎜ 9 "⎟ ⎝ 16 ⎠ 2 = 71.96 ksi This is larger than fc…O.K. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 43 of 62 -- 43 --
  49. 49. 10. CHECK STRENGTH LIMIT STATE 10.1: Section 1 Positive Flexure Section Classification (§6.10.6.2, Pg. 6.98 – 6.99) Check 2 Dcp E ≤ 3.76 tw Fyc Find Dcp, the depth of the web in compression at Mp (compression rebar in the slab is ignored). Pt = Fyt bt tt = (50ksi ) ( 21")(1") = 1,050kip Pw = Fyw Dtw = (50ksi )(69")(916 ") = 1,941kip Pc = Fyc bc tc = (50ksi )(15")( 3 4 ") = 562.5kip Ps = 0.85 f c'bs ts = (0.85)(4.5ksi )(109.5")(8.5") = 3,560kip 3,554kip < 3,560kip, the PNA lies in the slab. Since Pt + Pw +Pc < Ps ⎡ P + Pw + Pt ⎤ ⎡ 3,554kip ⎤ Y = ( ts ) ⎢ c = ( 8.5") ⎢ ⎥ kip ⎥ Ps ⎣ 3,560 ⎦ ⎣ ⎦ Y = 8.486 " ↓ from top of slab ∴ D p = Y = 8.486 " Since none of the web is in compression, Dcp = 0 and the web is compact. For Composite Sections in Positive Flexure, (§6.10.7.1, Pg. 6.101 – 6.102) Mu + 1 f S ≤ φf Mn 3 l xt Mu = 13,568k-ft from Page 30; take fl = 0 Dt = 1” + 69” + 3/4” + 8.5” = 79.25” 0.1Dt = 7.925” (The haunch is not included in Dt, as per ODOT Exceptions) Dp ⎞ ⎛ Since Dp =8.486 > 0.1Dt = 7.925, M n = M p ⎜ 1.07 − 0.7 ⎟ Dt ⎠ ⎝ ⎛ M p = (3,554kip ) ⎜ 79.25"− ⎝ = 157,500 k-in 8.486" ⎞ − 30.68" ⎟ 2 ⎠ = 13,130k-ft ⎡ ⎛ 8.486" ⎞ ⎤ k-ft M n = 13,130k-ft ⎢1.07 − ( 0.7 ) ⎜ ⎟ ⎥ = 13,060 ⎝ 79.25" ⎠ ⎦ ⎣ ( ) 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 44 of 62 -- 44 --
  50. 50. ? M u + 13 f l S xt ≤ φ f M n ? (13,568k-ft ) + (0) ≤(1.00)(13,060k-ft ) No Good. Note that the check of M n ≤ 1.3Rh M y has not been made in the above calculations. This section would satisfy the Article B6.2 so this check doesn’t need to be made. Check the ductility requirement to prevent crushing of the slab: ? D p ≤ 0.42 Dt ? → 8.486" ≤ ( 0.42 )( 79.25") = 33.29" O.K. The Section is NOT Adequate for Positive Flexure at Stations 58.7’ and 271.3’ The Girder failed the checks for service limits and has failed the first of several checks at the strength limit state. At this point I will investigate the strength of a section with 70ksi steel in the top and bottom flanges. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 45 of 62 -- 45 --
  51. 51. Hybrid Girder Factors Will Now be Required: Compute the Hybrid Girder Factor, Rh, for Section 1: Per AASHTO Commentary Pg 6-95, Dn shall be taken for the bottom flange since this is a composite section in positive flexure. Dn , Bottom = 58.19"− 1" = 57.19" Rh = ( 12 + β 3ρ − ρ 3 ) β= 12 + 2 β ρ= Rh , Section 1 = 2 Dn tw (2) ( 57.19")( 916 ") = = 3.064 Afn (1")( 21") Fyw fn ≤ 1.0 → ρ= 12 + ( 3.064 ) ⎡(3)(0.7143) − (0.7143)3 ⎤ ⎣ ⎦ 12 + ( 2 )( 3.064 ) 50ksi = 0.7143 70ksi = 0.9626 Compute the Hybrid Girder Factor, Rh, for Section 2: For the short-term composite section, Dn ,Top = 2 1 2 "+ 69"− 52.23" = 19.27" Dn , Bottom = 52.23"− 2 1 2 " = 49.73" Rh = ( 12 + β 3ρ − ρ 3 ) Governs β= 12 + 2 β ρ= Rh , Section 2 = 2 Dn tw (2) ( 49.73")( 916 ") = = 1.066 Afn ( 2 12 ")( 21") Fyw fn ≤ 1.0 → ρ= 12 + (1.066 ) ⎡(3)(0.7143) − (0.7143)3 ⎤ ⎣ ⎦ 12 + ( 2 )(1.066 ) 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel 50ksi = 0.7143 70ksi = 0.9833 AASHTO-LRFD 2007 Page 46 of 62 -- 46 --
  52. 52. 11. RECHECK SERVICE LIMIT STATE WITH 70KSI FLANGES 11.1: Permanent Deformations - Section 1 At the Service Limit State, the following shall be satisfied for composite sections Top Flange: f f ≤ 0.95Rh Fyf Bottom Flange ff + fl ≤ 0.95Rh Fyf 2 Top Flange, Positive Moment From before: f f ,58.7 = 36.96ksi f f ≤ 0.95Rh Fyf ? 36.96ksi ≤(0.95)(0.9626)(70ksi ) = 64.01ksi → O.K. Bottom Flange, Positive Moment f f ,58.7 = 54.90ksi ff + The load factor for wind under Service II is 0.00, ∴ fl = 0ksi fl ≤ 0.95Rh Fyf 2 → 54.90ksi + 0ksi ? ≤(0.95)(0.9626)(70ksi ) = 64.01ksi 2 O.K. 11.2: Permanent Deformations - Section 2 Top Flange, Negative Moment From before: f f ,165 = 55.08ksi f f ≤ 0.95Rh Fyf ? → 55.08ksi ≤(0.95)(0.9833)(70ksi ) = 65.39ksi O.K. Bottom Flange, Negative Moment f f = 50.39 ksi The load factor for wind under Service II is 0.00, ∴ fl = 0ksi ff + fl ≤ 0.95 Rh Fyf 2 → 50.39ksi + 0ksi ? ≤(0.95)(0.9833)(70ksi ) = 65.39ksi 2 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel O.K. AASHTO-LRFD 2007 Page 47 of 62 -- 47 --
  53. 53. 11.3: Bend Buckling Checks At the Service Limit State, all sections except composite sections in positive flexure shall satisfy: f c ≤ Fcrw where: Fcrw = 0.9 Ek ⎛D⎞ ⎜ ⎟ ⎝ tw ⎠ k= and 2 9 ( Dc / D ) 2 Not Applicable Section 1 - Section 2 ⎛ (7,108 )(12 ⎝ 3,602 in k-ft fc = 1.00 ⎜ )⎞ in ⎛ (1, 250 )(12 ⎝ 4,255 in k-ft ⎟ + 1.00 ⎜ ft ⎠ 3 in ft )⎞ ⎠ 3 ⎛ (2, 292 )(12 ⎝ 4,255 in k-ft ⎟ + 1.00 ⎜ in ft )⎞ ⎠ 3 ⎛ (4, 918 )(12 ⎝ 4,590 in k-ft ⎟ + 1.30 ⎜ ft )⎞ ft )⎞ in 3 ⎟ ⎠ f c = 50.39ksi ⎛ (7,108 )(12 ⎝ 2,116 in k-ft ft = 1.00 ⎜ in ft )⎞ ⎛ (1, 250 )(12 ⎝ 5,135 in k-ft ⎟ + 1.00 ⎜ ⎠ 3 3 in ft )⎞ ⎛ (2, 292 )(12 ⎝ 5,135 in ⎟ + 1.00 ⎜ ⎠ k-ft 3 in ft )⎞ ⎛ (4, 918 )(12 ⎝ 11,828 in k-ft ⎟ + 1.30 ⎜ ⎠ 3 in ⎟ ⎠ ft = 55.08ksi ⎛ − fc ⎞ 0 Dc = ⎜ ⎜ f + f ⎟ d − tcf ≥ ⎟ t ⎠ ⎝ c ⎛ ⎞ 50.39ksi 72.5") − 2.5" ≥ 0 =⎜ ksi ksi ⎟ ( ⎝ 50.39 + 55.08 ⎠ = 32.14′′ k= 9 ( Dc / D ) Fcrw = 2 = 9 ⎛ 32.14" ⎞ ⎜ ⎟ ⎝ 69" ⎠ 2 = 41.49 (0.90)(29, 000ksi )(41.49) ⎛ 69" ⎞ ⎜ 9 "⎟ ⎝ 16 ⎠ 2 = 71.96 ksi This is larger than fc…O.K. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 48 of 62 -- 48 --
  54. 54. 12. RECHECK STRENGTH LIMIT STATE WITH 70KSI FLANGES 12.1: Section 1 - Positive Flexure Section Classification (§6.10.6.2, Pg. 6.98 – 6.99) Check 2 Dcp E ≤ 3.76 tw Fyc Find Dcp, the depth of the web in compression at Mp (compression rebar in the slab is ignored). Pt = Fyt bt tt = (70ksi ) ( 21")(1") = 1, 470kip Pw = Fyw Dtw = (50ksi )(69")(916 ") = 1,941kip Pc = Fyc bc tc = (70ksi )(15")( 3 4 ") = 787.5kip Ps = 0.85 f c'bs ts = (0.85)(4.5ksi )(109.5")(8.5") = 3,560kip 4,199kip > 3,560kip, the PNA is NOT in the slab. Since Pt + Pw +Pc > Ps ? Check Case I Pt + Pw ≥ Pc + Ps ? 1, 470kip + 1,941kip ≥ 787.5kip + 3,560kip NO ? Check Case II Pt + Pw + Pc ≥ Ps ? 1, 470kip + 1,941kip + 787.5kip ≥ 3,560kip YES - PNA in Top Flange ⎞ ⎛ t ⎞ ⎛ P + P − Ps Y = ⎜ c ⎟⎜ w t + 1⎟ Pc ⎝ 2 ⎠⎝ ⎠ kip kip kip ⎞ ⎛ 0.750" ⎞ ⎛ 1,941 + 1, 470 − 3,560 =⎜ + 1⎟ = 0.3040" (from the top of steel) ⎟⎜ kip 787.5 ⎝ 2 ⎠⎝ ⎠ Dp = 8.5” + 0.3040” = 8.804” Since none of the web is in compression, Dcp = 0 and the web is compact. For Composite Sections in Positive Flexure, (§6.10.7.1, Pg. 6.101 – 6.102) Mu + 1 f S ≤ φf Mn 3 l xt Mu = 13,568k-ft from Page 30; take fl = 0 Dt = 1” + 69” + 3/4” + 8.5” = 79.25” 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel 0.1Dt = 7.925” AASHTO-LRFD 2007 Page 49 of 62 -- 49 --
  55. 55. (The haunch is not included in Dt, as per ODOT Exceptions) Dp ⎞ ⎛ Since Dp = 8.804” > 0.1Dt = 7.925”, M n = M p ⎜1.07 − 0.7 ⎟ Dt ⎠ ⎝ Determine Mp: The distances from the component forces to the PNA are calculated. 8.5" + 0.3040" = 4.554" 2 69" − ( 0.75"− 0.3040") = 34.05" dw = 2 1" dt = 70.75"− − 0.3040" = 69.95" 2 ds = The plastic moment is computed. ⎛P ⎞ 2 M p = ⎜ c ⎟ ⎡Y 2 + ( tc − Y ) ⎤ + [ Ps d s + Pw d w + Pdt ] t ⎦ ⎝ 2tc ⎠ ⎣ ⎛ 787.5kip ⎞ ⎡ 2 2 =⎜ ⎟ ⎣( 0.3040") + ( 0.750"− 0.3040") ⎤ + ... ⎦ ⎝ (2)(0.750") ⎠ ... + ⎡( 3,560kip ) ( 4.554") + (1,941kip ) ( 34.05") + (1, 470kip ) ( 69.95") ⎤ ⎣ ⎦ = ( 525 kip in ) ⎡0.2913 in 2 ⎤ + ⎡185,100k-in ⎤ ⎣ ⎦ ⎣ ⎦ = 185,300k-in = 15, 440k-ft ⎡ ⎛ 8.804" ⎞ ⎤ k-ft M n = 15, 440k-ft ⎢1.07 − ( 0.7 ) ⎜ ⎟ ⎥ = 15,320 ⎝ 79.25" ⎠ ⎦ ⎣ ( ) ? M u + 13 f l S xt ≤ φ f M n ? (13,568k-ft ) + (0) ≤(1.00)(15,320 k-ft ) O.K. Note that the check of M n ≤ 1.3Rh M y has not been made in the above calculations. This section would satisfy the Article B6.2 so this check doesn’t need to be made. Check the ductility requirement to prevent crushing of the slab: ? D p ≤ 0.42 Dt ? → 8.804" ≤ ( 0.42 )( 79.25") = 33.29" O.K. The Section is Adequate for Positive Flexure at Stations 58.7’ and 271.3’ with 70ksi Flanges 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 50 of 62 -- 50 --
  56. 56. 12.2: Section 2 - Negative Flexure Section Classification (§6.10.6.2, Pg. 6.98 – 6.99) Check 2 Dc E ≤ 5.70 tw Fyc Dc is the depth of the web in compression for the cracked section. Dc = 32.04” – 21/2” = 29.54” 2 Dc (2)(29.54") E 29,000ksi = = 105.0 < 5.70 = 5.70 = 137.3 tw (916 ") Fyc 50ksi The web is non-slender. Since the web is non-slender we have the option of using the provisions in Appendix A to determine the moment capacity. I will first determine the capacity using the provisions in §6.10.8, which will provide a somewhat conservative determination of the flexural resistance. For Composite Sections in Negative Flexure, (§6.10.8.1, Pg. 6.105 – 6.114) The Compression Flange must satisfy: fbu + 1 f ≤ φ f Fnc 3 l Per §6.10.1.1.1a, elastic stresses at any location in a composite section shall consist of the sum of stresses caused by loads applied separately to the bare steel, short-term composite section, and long-term composite section. In §6.10.1.1.1c, though, it states that for the Strength Limit, the short-term and long-term composite sections shall consist of the bare steel and the longitudinal rebar. In other words, for determining negative moment stresses over the pier, we can use the factored moment above with the properties for the cracked section. ⎛ M DC1 ⎞ ⎛ 1.25M DC 2 + 1.50 M DW + 1.75M LL − ⎞ ⎟+⎜ ⎟ SCR ⎝ S BS ⎠ ⎝ ⎠ fbu = 1.25 ⎜ k-ft k-ft k-ft ⎛ (7,109 k-ft )(12 in ) ⎞ ⎛ ⎡ (1.25)(1, 250 ) + (1.50)(2, 292 ) + (1.75)(4, 918 ) ⎤ (12 in ) ⎞ ⎣ ⎦ ft ft fbu = 1.25 ⎜ ⎟ ⎟+⎜ 3 3 3,932 in ⎝ 3,602 in ⎠ ⎝ ⎠ fbu = 71.13ksi Since fbu is greater than Fyc, it is obvious that a strength computed based on the provisions in §6.10.8 will not be adequate. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 51 of 62 -- 51 --
  57. 57. As it stands here, this girder is clearly not adequate over the pier. The compression flange is overstressed as per the provisions in §6.10.8. There are still other options to explore, though, before increasing the plate dimensions. 1. Since the web is non-slender for Section 2 in Negative Flexure, we have the option of using the provisions in Appendix A6 to determine moment capacity. This would provide an upper bound strength of Mp instead of My as was determined in §6.10.8. 2. The provisions in Appendix B6 allow for redistribution of negative moment from the region near the pier to the positive moment region near mid-span for sections that satisfy stringent compactness and stability criteria. If this section qualifies, as much as ~2,000k-ft may be able to be redistributed from the pier to mid-span, which could enable the plastic moment strength from Appendix A6 to be adequate. (This solution may even work with the flange strength at 50ksi, but I doubt it…) Despite the fact that the girder appears to have failed our flexural capacity checks, let’s look at the shear capacity. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 52 of 62 -- 52 --
  58. 58. 12.3 Vertical Shear Capacity At the strength Limit, the following must be satisfied Vu ≤ φVn For an unstiffened web, Vn = Vcr = CV p Check, D Ek , ≤ 1.12 tw Fyw D 69" = = 122.7 tw 916 " Since there are no transverse stiffeners, k = 5 (29, 000ksi )(5) 1.12 = 60.31 (50ksi ) Since C= (29, 000ksi )(5) 1.40 = 75.39 (50ksi ) D Ek , elastic shear buckling of the web controls. > 1.40 tw Fyw 1.57 ⎛ kE ⎞ 1.57 ⎛ (5)(29, 000ksi ) ⎞ ⎜ ⎟= ⎟ = 0.3026 2 2 ⎜ (50ksi ) ⎛ D ⎞ ⎜ Fyw ⎟ ⎛ 69" ⎞ ⎝ ⎠ ⎝ ⎠ ⎜ ⎟ ⎜ 9 "⎟ ⎝ 16 ⎠ ⎝ tw ⎠ V p = 0.58 Fyw Dt w = (0.58)(50 ksi )(69")( 9 16 ") = 1,126 kip Vn = CV p = (0.3026)(1,126 kip ) = 340.6 kip ( ) φVn = (1.00 ) 340.6 kip = 340.6 kip No Good. This strength is adequate from 16’ – 100’ and 230’ - 314’. This strength is not adequate near the end supports or near the pier, however. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 53 of 62 -- 53 --
  59. 59. Try adding transverse stiffeners spaced at do = 8’ = 96” k =5+ 5 ⎛ do ⎞ ⎜D⎟ ⎝ ⎠ 2 =5+ 5 ⎛ 96" ⎞ ⎜ 69" ⎟ ⎝ ⎠ = 7.583 2 D (29, 000ksi )(7.583) (29, 000ksi )(7.583) = 122.7 , 1.12 = 74.28 , 1.40 = 92.85 (50ksi ) (50ksi ) tw Since D Ek > 1.40 , elastic shear buckling of the web controls. tw Fyw 1.57 ⎛ Ek ⎞ 1.57 ⎛ (29, 000ksi )(7.583) ⎞ C= ⎟= ⎟ = 0.4589 2 ⎜ 2 ⎜ ⎜ ⎟ (50ksi ) ⎛ D ⎞ ⎝ Fyw ⎠ ⎛ 69" ⎞ ⎝ ⎠ ⎜ ⎟ ⎜ 9 "⎟ ⎝ 16 ⎠ ⎝ tw ⎠ φVn = φCV p = (1.00)(0.4589)(1,126 kip ) = 516.5kip O.K. This capacity is fine but we may be able to do better if we account for tension field action. Try adding transverse stiffeners spaced at do = 12’ = 144” k =5+ 5 ⎛ do ⎞ ⎜D⎟ ⎝ ⎠ 2 =5+ 5 ⎛ 144" ⎞ ⎜ 69" ⎟ ⎝ ⎠ 2 = 6.148 D (29, 000ksi )(6.148) (29, 000ksi )(6.148) = 122.7 , 1.12 = 66.88 , 1.40 = 83.60 (50ksi ) (50ksi ) tw Since C= D Ek , elastic shear buckling of the web controls. > 1.40 tw Fyw 1.57 ⎛ Ek ⎞ 1.57 ⎛ (29, 000ksi )(6.148) ⎞ = ⎟ ⎟ = 0.3721 2 ⎜ 2 ⎜ ⎜ ⎟ (50ksi ) ⎛ D ⎞ ⎝ Fyw ⎠ ⎛ 69" ⎞ ⎝ ⎠ ⎜ ⎟ ⎜ 9 "⎟ ⎝ 16 ⎠ ⎝ tw ⎠ Without TFA: Vn = CV p = (0.3721)(1,126 kip ) = 418.9 kip 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 54 of 62 -- 54 --
  60. 60. With TFA: Since 2 Dtw (2)(69")( 916 ") = = 1.056 ≤ 2.5 , ( b fct fc + b ft t ft ) ( (21")(2 12 ") + (21")(1") ) ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ 0.87(1 − C ) ⎥ (0.87)(1 − 0.3721) ⎥ ⎢ ⎢ = (1,126kip ) ⎢0.3721 + Vn = V p ⎢C + ⎥ 2 ⎥ 2 ⎛ do ⎞ ⎥ ⎛ 144" ⎞ ⎥ ⎢ ⎢ 1+ ⎜ 1+ ⎜ ⎟ ⎟ ⎢ ⎥ ⎢ ⎝ 69" ⎠ ⎥ ⎝D⎠ ⎦ ⎣ ⎦ ⎣ Vn = (1,126kip )(0.6082) = 684.8kip ( ) φVn = (1.00 ) 684.8kip = 684.8kip O.K. This TFA strength is adequate near the pier but TFA is not permitted in the end panels. The following stiffener configuration should provide adequate shear strength. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 55 of 62 -- 55 --
  61. 61. Strength Limit Shear Capacity 800 Tension Field Action 600 Strength I 400 Strength IV Shear (kip) 200 0 -200 -400 -600 -800 0 30 60 90 120 150 180 210 240 270 300 330 Station (ft) 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 56 of 62 -- 56 --
  62. 62. 12.4: Horizontal Shear Strength Per ODOT Standard practice, shear studs will be used to transfer horizontal shear between the concrete deck and top flange of the steel girder. ODOT prefers the use of 7/8”diameter studs. Ideally, the studs should extend to the mid-thickness of the deck. Using this criterion, the height of the studs can be determined. be ts + thaunch − t flange 2 9.5" = + 2.75"− 0.75" = 6.75" 2 h= ts thaunch tc bc Use 7/8” x 61/2” shear studs D tw h AASHTO requires that the ratio of /d be greater than or equal to 4.0. tt bt ? h ≥ 4.0 d 6 12 " = 7.429 ≥ 4.0 7 " 8 OK AASHTO requires a center-to-center transverse spacing of 4d and a clear edge distance of 1”. With 7/8” diameter studs, there is room enough transversely to use up to 4 studs in each row. With this in mind, I will investigate the option of either 3 or 4 studs per row. Fatigue Limit State: The longitudinal pitch of the shear studs based on the Fatigue Limit is determined as p≤ nZ r Vsr Vsr = Vf Q I (6.10.10.1.2-1 & 3) where: n Zr Vsr Vf Q I - Number of studs per row Fatigue resistance of a single stud Horizontal fatigue shear range per unit length Vertical shear force under fatigue load combination 1st moment of inertia of the transformed slab about the short-term NA 2nd moment of inertia of the short-term composite section 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 57 of 62 -- 57 --
  63. 63. Zr = α d 2 ≥ 5.5d 2 2 (6.10.10.2-1) α = 34.5 − 4.28Log( N ) (6.10.10.2-2) α = 34.5 − 4.28Log(55.84 ×106 ) = 1.343ksi 2 ⎛ 5.5 ⎞ 7 2 Z r = (1.343ksi ) ( 7 8 ") ≥ ⎜ ⎟ ( 8 ") ⎝ 2 ⎠ = 1.028kip ≥ 2.105kip → Z r = 2.105kip Q = Atc d c ⎡ (109.5")( 9.5") ⎤ ⎛ 9.5" ⎞ − 58.19" ⎟ = 2,511 in 3 QSection 1 = ⎢ ⎥ ⎜ 1"+ 69"+ 2.75"+ 8 2 ⎠ ⎣ ⎦⎝ ⎡ (109.5")( 9.5") ⎤ ⎛ 9.5" ⎞ − 52.23" ⎟ = 3, 481 in 3 QSection 2 = ⎢ ⎥ ⎜ 2.5"+ 69"+ 2.75"+ 8 2 ⎠ ⎣ ⎦⎝ I Section 1 = 140,500 in 4 I Section 2 = 239, 700 in 4 Since the fatigue shear varies along the length of the bridge, the longitudinal distribution of shear studs based on the Fatigue Limit also varies. These results are presented in a tabular format on a subsequent page. To illustrate the computations, I have chosen the shear at the abutment as an example. ( ) At the abutment, V f = 38.13kip − −3.53kip = 41.66kip ( 41.66 )( 2,511 in ) = 0.7445 = (140,500 in ) kip Vsr 4 p≤ ( 0.7445 ) kip inch = 8.482 kip inch For 4 studs in each row: For 3 studs in each row: ( 3) ( 2.105ksi ) 3 p≤ in row 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel ( 4 ) ( 2.105ksi ) ( 0.7453 ) kip inch in = 11.31 row AASHTO-LRFD 2007 Page 58 of 62 -- 58 --
  64. 64. Strength Limit: Qr = φsc Qn Qn = 0.5 Asc φsc = 0.85 f c' Ec ≤ Asc Fu (6.10.10.4.3-1) 2 ⎛π ⎞ Asc = ⎜ ⎟ ( 7 8 ") = 0.6013 in 2 ⎝4⎠ f c' = 4.5ksi Since n = 8, Ec = Es 29, 000ksi = = 3, 625ksi 8 n Fu = 60ksi Qn = ( 0.5 ) ( 0.6013 in 2 ) ( 4.5 )( 3, 625 ) ≤ ( 0.6013 in )( 60 ) ksi ksi 2 ksi kip kip = 38.40 stud ≤ 36.08 stud kip kip φscQn = ( 0.85 ) ( 36.08 stud ) = 30.67 stud n+ = Pp n− = Qr 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel Pp + Pn Qr AASHTO-LRFD 2007 Page 59 of 62 -- 59 --
  65. 65. Positive moment - Section 1: Station 0.0’ - 73.3’ Pp = Min ( PConcrete , Psteel ) PConcrete = 0.85 f c'bets = ( 0.85 ) ( 4.5ksi ) (109.5")( 9.5") = 3,979kip PSteel = Fyw Dtw + Fft b ft t ft + Ffc b fc t fc = ( 70ksi ) ⎡(15")( 0.75") + ( 21")(1") ⎤ + ( 50ksi ) ( 69")( 0.5625") = 4,198kip ⎣ ⎦ Pp = 3,979kip n+ = Pp Qr = 3,979kip = 129.7studs kip 30.67 stud 3 Studs per Row: 129.7studs = 44rows studs 3 row → p= ( 73.3'− 0 ') (12 in ) ft = 20.46 inch row ( 44 − 1) → Say 20" → p= ( 73.3'− 0 ') (12 in ) ft = 27.49 inch row ( 33 − 1) → Say 24" 4 Studs per Row: 129.7studs = 33rows studs 4 row 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 60 of 62 -- 60 --
  66. 66. Negative Moment - Section 2: Station 73.3’ - 165.0’ Pn = Min ( Psteel , PCrack ) PCrack = 0.45 f c'bets = ( 0.45 ) ( 4.5ksi ) (109.5")( 9.5") = 2,107 kip PSteel = Fyw Dtw + Fft b ft t ft + Ffc b fc t fc = ( 70ksi ) ⎡( 21")( 2.5") + ( 21")(1") ⎤ + ( 50ksi ) ( 69")( 0.5625") = 7, 086kip ⎣ ⎦ Pn = 2,107 kip n− = Pp + Pn Qr = 3,979kip + 2,107 kip = 198.4studs kip 30.67 stud 3 Studs per Row: 198.4studs = 67 rows studs 3 row → p= (165.0 '− 73.3') (12 in ) ft = 16.67 inch row ( 67 − 1) → Say 16" → p= (165.0 '− 73.3') (12 in ) ft = 22.48 inch row ( 50 − 1) → Say 20" 4 Studs per Row: 198.4studs = 50rows studs 4 row 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 61 of 62 -- 61 --
  67. 67. Shear Stud Summary: This table represents that pitch of shear studs required for either 3 or 4 studs per row based on location in the bridge. 3 Studs Per Row Station (ft) 0.0 14.7 29.3 44.0 58.7 73.3 88.0 102.7 117.3 132.0 135.7 139.3 143.0 146.7 150.3 154.0 157.7 161.3 165.0 168.7 172.3 176.0 179.7 183.3 187.0 190.7 194.3 198.0 212.7 227.3 242.0 256.7 271.3 286.0 300.7 315.3 330.0 Vf (kip) 41.66 37.01 33.68 32.79 33.04 33.46 33.98 34.59 35.38 36.62 37.07 37.53 37.98 38.42 38.88 39.34 39.81 40.26 81.44 40.26 39.81 39.34 38.88 38.42 37.98 37.53 37.07 36.62 35.38 34.59 33.98 33.46 33.04 32.79 33.68 37.01 41.66 Q V sr I 3 (in ) 2,511 2,511 2,511 2,511 2,511 2,511 2,511 2,511 2,511 2,511 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 2,511 2,511 2,511 2,511 2,511 2,511 2,511 2,511 2,511 2,511 140,521 140,521 140,521 140,521 140,521 140,521 140,521 140,521 140,521 140,521 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 140,521 140,521 140,521 140,521 140,521 140,521 140,521 140,521 140,521 140,521 (in ) 4 kip ( /in) 0.7444 0.6613 0.6018 0.5859 0.5904 0.5979 0.6071 0.6181 0.6323 0.6543 0.5383 0.5449 0.5514 0.5579 0.5645 0.5713 0.5780 0.5847 1.1826 0.5847 0.5780 0.5713 0.5645 0.5579 0.5514 0.5449 0.5383 0.6543 0.6323 0.6181 0.6071 0.5979 0.5904 0.5859 0.6018 0.6613 0.7444 p Fat p Str (in) 8.48 9.55 10.49 10.78 10.70 10.56 10.40 10.22 9.99 9.65 11.73 11.59 11.45 11.32 11.19 11.05 10.93 10.80 5.34 10.80 10.93 11.05 11.19 11.32 11.45 11.59 11.73 9.65 9.99 10.22 10.40 10.56 10.70 10.78 10.49 9.55 8.48 (in) 20.00 20.00 20.00 20.00 20.00 20.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 20.00 20.00 20.00 20.00 20.00 20.00 4 Studs Per Row p max (in) 8.48 9.55 10.49 10.78 10.70 10.56 10.40 10.22 9.99 9.65 11.73 11.59 11.45 11.32 11.19 11.05 10.93 10.80 5.34 10.80 10.93 11.05 11.19 11.32 11.45 11.59 11.73 9.65 9.99 10.22 10.40 10.56 10.70 10.78 10.49 9.55 8.48 p Fat p Str (in) 11.31 12.73 13.99 14.37 14.26 14.08 13.87 13.62 13.32 12.87 15.64 15.45 15.27 15.09 14.92 14.74 14.57 14.40 7.12 14.40 14.57 14.74 14.92 15.09 15.27 15.45 15.64 12.87 13.32 13.62 13.87 14.08 14.26 14.37 13.99 12.73 11.31 (in) 24.00 24.00 24.00 24.00 24.00 24.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 24.00 24.00 24.00 24.00 24.00 24.00 p max (in) 11.31 12.73 13.99 14.37 14.26 14.08 13.87 13.62 13.32 12.87 15.64 15.45 15.27 15.09 14.92 14.74 14.57 14.40 7.12 14.40 14.57 14.74 14.92 15.09 15.27 15.45 15.64 12.87 13.32 13.62 13.87 14.08 14.26 14.37 13.99 12.73 11.31 The arrangement of shear studs is shown below. 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Page 62 of 62 -- 62 --
  68. 68. ONE-SPAN INELASTIC I-GIRDER BRIDGE DESIGN EXAMPLE 1. PROBLEM STATEMENT AND ASSUMPTIONS: A single span composite I-girder bridge has span length of 166.3’ and a 64’ deck width. The steel girders have Fy = 50ksi and all concrete has a 28-day compressive strength of ksi f’c = 4.5 . The concrete slab is 9.5” thick. A typical 4” haunch was used in the section properties. Concrete barriers weighing 640plf and an asphalt wearing surface weighing 60psf have also been applied as a composite dead load. HL-93 loading was used per AASHTO (2004), including dynamic load allowance. Single-Span Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Created July 2007: Page 1 of 21 -- 63 --
  69. 69. 172' - 4" Total Girder Length G 2 G 3 G 4 G 5 G 6 166' - 4" cc Bearings Cross Frames Spaced @ 22' - 0" cc G 1 Single-Span Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Created July 2007: Page 2 of 21 -- 64 --
  70. 70. Positive Bending Section (Section 2) Positive Bending Section (Section 1) Positive Bending Section (Section 3) Single-Span Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Created July2007: Page 3 of 23 -- 65 --
  71. 71. 2. LOAD CALCULATIONS: DC dead loads (structural components) include: • Steel girder self weight (DC1) • Concrete deck self weight (DC1) • Haunch self weight (DC1) • Barrier (DC2) DW dead loads (structural attachments) include: • Wearing surface (DW), Including FWS 2a. Dead Load Calculations Steel Girder Self-Weight (DC1): (a) Section 1 A = (14”)(1.125”) + (68”)(0.6875”) + (22”)(1.5”) = 95.5 in2 ⎛ ⎛ 490pcf ⎞ ⎞ ⎟ ⎟ (1.15 ) = 373.7 lbs per girder Wsection1 = ⎜ 95.5 in 2 ⎜ ft ⎜ (12 in )2 ⎟ ⎟ ⎜ ft ⎝ ⎠⎠ ⎝ (b) Section 2 A = (14”)(2”) + (68”)(0.5625”) + (22”)(2”) = 110.25 in2 ⎛ ⎛ 490pcf ⎞ ⎞ ⎟ ⎟ (1.15 ) = 431.4 lbs per girder Wsection1 = ⎜110.3 in 2 ⎜ ft ⎜ (12 in )2 ⎟ ⎟ ⎜ ft ⎝ ⎠⎠ ⎝ (c) Section 3 A = (14”)(2”) + (68”)(0.5625”) + (22”)(2.375”) = 118.5 in2 ⎛ ⎛ 490pcf ⎞ ⎞ ⎟ ⎟ (1.15 ) = 463.7 lbs per girder Wsection1 = ⎜118.5 in 2 ⎜ ft ⎜ (12 in )2 ⎟ ⎟ ⎜ ft ⎝ ⎠⎠ ⎝ (d) Average Girder Self Weight Wave = ( 2 )( 40.17 ') ( 373.7 lbs ) + ( 2 )(18.0 ') ( 431.4 lbs ) + ( 50.0 ') ( 463.7 lbs ) ft ft ft 166.3' = 413.3 lbs ft Deck Self-Weight (DC1): ⎛ ( 9.5'')( 64.0 ') ⎞ ⎛ 150pcf ⎞ WDeck = ⎜ ⎟ = 1,267 lbs per girder ⎟⎜ ft 6 Girders ⎠ ⎜ (12 in ) ⎟ ⎝ ft ⎠ ⎝ Single-Span Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Created July2007: Page 4 of 23 -- 66 --
  72. 72. Haunch Self-Weight (DC1): Average width of haunch: 14’’ ⎛ 150pcf ⎞ ⎟ = 94.33 lbs per girder Whaunch = (14 )( 4 ) + 2 ( ( 1 ) ( 9 '')( 4 '') ) ⎜ 2 ft ⎜ (12 in )2 ⎟ ft ⎝ ⎠ ( ) Barrier Walls (DC2): Wbarriers ⎛ ( 2 each ) ( 640plf ) ⎞ ⎟ = 213.3 lbs per girder =⎜ ft ⎜ ⎟ 6 girders ⎝ ⎠ Wearing Surface (DW): Wwearing_surface = ( 61.0') ( 60psf ) 6 Girders = 610.0 lbs per girder ft The moment effect due to dead loads was found using an FE model composed of six frame elements to model the bridge (a node was placed at mid-span). This data was input into Excel to be combined with data from moving live load analyses performed in SAP 2000. DC1 dead loads were applied to the noncomposite section (bare steel). All live loads were applied to the short-term composite section (1n = 8). DW (barriers) and DC2 (wearing surface) dead loads were applied to the long-term composite section (3n = 24). The maximum moments at mid-span are easily computed since the bridge is statically determinate. M DC1,Steel 2 ⎛ wL2 ⎞ ⎡ ( 413.3 lbs ) (166.3') ⎤ ft ⎥ = 1, 429k-ft =⎜ ⎟=⎢ 8 ⎠ ⎢ 8 ⎥ ⎝ ⎣ ⎦ M DC1, Deck 2 ⎛ wL2 ⎞ ⎡ (1, 267 lbs ) (166.3' ) ⎤ ft ⎥ = 4,379k-ft =⎜ ⎟=⎢ 8 ⎥ ⎝ 8 ⎠ ⎢ ⎣ ⎦ M DC 2, Barriers M DW 2 ⎛ wL2 ⎞ ⎡ ( 213.3 lbs ) (166.3' ) ⎤ ft ⎥ = 737.4k-ft =⎜ ⎟=⎢ 8 ⎥ ⎝ 8 ⎠ ⎢ ⎣ ⎦ 2 ⎛ wL2 ⎞ ⎡ ( 610.0 lbs ) (166.3' ) ⎤ ft ⎥ = 2,109k-ft =⎜ ⎟=⎢ 8 ⎠ ⎢ 8 ⎥ ⎝ ⎣ ⎦ Single-Span Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Created July2007: Page 5 of 23 -- 67 --
  73. 73. The maximum shear forces at the ends of the girder are also easily computed. lbs ⎛ wL ⎞ ⎡ ( 413.3 ft ) (166.3' ) ⎤ kip VDC1,Steel = ⎜ =⎢ ⎥ = 34.37 ⎟ 2 ⎠ ⎢ 2 ⎝ ⎥ ⎣ ⎦ VDC1, Deck lbs ⎛ wL ⎞ ⎡ (1, 267 ft ) (166.3' ) ⎤ kipt =⎜ ⎥ = 105.4 ⎟=⎢ 2 ⎝ 2 ⎠ ⎢ ⎥ ⎣ ⎦ lbs ⎛ wL ⎞ ⎡ ( 213.3 ft ) (166.3' ) ⎤ kip VDC 2, Barriers = ⎜ =⎢ ⎥ = 17.74 ⎟ 2 ⎝ 2 ⎠ ⎢ ⎥ ⎣ ⎦ lbs ⎛ wL ⎞ ⎡ ( 610.0 ft ) (166.3' ) ⎤ kip VDW = ⎜ =⎢ ⎥ = 50.72 ⎟ 2 ⎠ ⎢ 2 ⎝ ⎥ ⎣ ⎦ Single-Span Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Created July2007: Page 6 of 23 -- 68 --
  74. 74. 2b. Live Load Calculations The following design vehicular live load cases described in AASHTO-LRFD are considered: 1) The effect of a design tandem combined with the effect of the lane loading. The design tandem consists of two 25kip axles spaced 4.0’ apart. The lane loading consists of a 0.64klf uniform load on all spans of the bridge. (HL-93M in SAP) 2) The effect of one design truck with variable axle spacing combined with the effect of the 0.64klf lane loading. (HL-93K in SAP) 3) For negative moment between points of contraflexure only: 90% of the effect of a truck-train combined with 90% of the effect of the lane loading. The truck train consists of two design trucks (shown below) spaced a minimum of 50’ between the lead axle of one truck and the rear axle of the other truck. The distance between the two 32kip axles should be taken as 14’ for each truck. The points of contraflexure were taken as the field splices at 132’ and 198’ from the left end of the bridge. (HL-93S in SAP) All live load calculations were performed in SAP 2000 using a beam line analysis. The nominal moment data from SAP was then input into Excel. An Impact Factor of 1.33 was applied to the truck and tandem loads within SAP. Single-Span Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Created July2007: Page 7 of 23 -- 69 --
  75. 75. Unfactored HL-93 Moment Envelopes from SAP 6,000 Single Truck 4,000 Tandem Moment (kip-ft) 2,000 0 -2,000 -4,000 -6,000 0 30 60 90 120 150 Station (ft) The following results were obtained from the SAP analysis: • The maximum positive live-load moments occur at stations 83.15’ HL-93M HL-93K HL-93S Station 40.16’- Section 1 3,614k-ft 4,322k-ft N/A Station 58.15’- Section 2 4,481k-ft 5,238k-ft N/A Station 83.15’- Section 3 4,911k-ft 5,821k-ft N/A Before proceeding, these live-load moments will be confirmed with an influence line analysis. Single-Span Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Created July2007: Page 8 of 23 -- 70 --
  76. 76. 2c. Verify the Maximum Positive Live-Load Moment at Station 83.15’: kip kip 25 25 Tandem: kip 32 8 kip 32 kip : Single Truck 0.640kip/ft Moment (k-ft / kip) Lane 45 40 35 30 25 20 15 10 5 0 0 15 30 45 60 75 90 105 120 135 150 165 Station (ft) Tandem: Single Truck: Lane Load: ( 25 ) ( 41.58 ) + ( 25 ) ( 39.58 ) = 2, 029 (8 ) ( 34.57 ) + ( 32 ) ( 41.58 ) + ( 32 ) ( 34.57 ) = 2, 713 kip kip k-ft kip kip kip k-ft kip k-ft kip ( 0.640 )(3, 457 ) = 2, 212 k-ft kip k-ft kip k-ft k-ft kip kip k-ft kip k-ft k-ft (IM)(Tandem) + Lane: (1.33 ) ( 2, 029 k-ft ) + 2, 212k-ft = 4, 911k-ft kip (IM)(Single Truck) + Lane: (1.33 ) ( 2, 713 k-ft ) + 2, 212 k-ft = 5,821k-ft kip GOVERNS The case of two trucks is not considered here because it is only used when computing negative moments. Single-Span Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Created July2007: Page 9 of 23 -- 71 --
  77. 77. Based on the influence line analysis, we can say that the moments obtained from SAP appear to be reasonable and will be used for design. Before these Service moments can be factored and combined, we must compute the distribution factors. Since the distribution factors are a function of Kg, the longitudinal stiffness parameter, we must first compute the sections properties of the girders. 3. SECTION PROPERTIES AND CALCULATIONS: 3a. Effective Flange Width, bs: For an interior beam, bs is the lesser of: bf ⎧ 14" = (12 )( 8.5") + = 109 '' ⎪•12ts + 2 2 ⎪ ⎪ ⎨•S = (11.33')(12 in ft ) = 135.96 '' ⎪ L ⎪• eff = 166.3' = 41.58' = 498.9 '' ⎪ 4 4 ⎩ Therefore, bs = 109” For computing the section properties shown on the two pages that follow, reinforcing steel in the deck was ignored for short-term and long-term composite calculations but was included for the cracked section. Note: At this point one should also check the effective of the outside girders as well. For this example, however, I will proceed sing the effective width for the interior girders. Single-Span Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Created July2007: Page 10 of 23 -- 72 --
  78. 78. 3b. Section 1 Flexural Properties Single Span Bridge Example - Section 1 Bare Steel t b A y Ay d Ix Ad2 IX Top Flange 1.1250 14.00 15.75 70.06 1,103.48 1.66 -40.87 26,308 26,310 Web Bot Flange 0.6875 1.5000 68.00 22.00 46.75 33.00 35.50 0.75 1,659.63 18,014.33 24.75 6.19 -6.31 28.44 1,860 26,696 19,874 26,702 2,787.86 ITotal = 72,886 29.19 SBS1,top = 1,759 SBS1,bot = 2,497 95.50 Y= Short-Term Composite (N=8) t b A y Ay d Ix Ad2 IX Slab 8.5000 109.00 115.81 74.88 8,671.46 697.29 -20.65 49,365 50,062 Haunch Top Flange 0.0000 1.1250 14.0000 14.0000 0.00 15.75 70.63 70.06 0.00 1,103.48 0.00 1.66 -16.40 -15.83 0 3,948 0 3,950 Web Bot Flange 0.6875 1.5000 68.0000 22.0000 46.75 33.00 35.50 0.75 1,659.63 18,014.33 24.75 6.19 18.73 53.48 16,399 94,381 34,414 94,387 11,459.32 ITotal = 182,813 54.23 SST1,top = 11,150 SST1,bot = 3,371 211.31 n= 8.00 Y= Long-Term Composite (N=24) t b A y Ay d Ix Ad2 IX Slab 8.5000 109.00 38.60 74.88 2,890.49 232.43 -32.53 40,856 41,089 Haunch 0.0000 14.00 0.00 70.63 0.00 0.00 -28.28 0 0 Top Flange 1.1250 14.0000 15.75 70.06 1,103.48 1.66 -27.72 12,102 12,104 Web Bot Flange 0.6875 1.5000 68.0000 22.0000 46.75 33.00 35.50 0.75 1,659.63 18,014.33 24.75 6.19 6.84 41.59 2,189 57,089 20,203 57,095 5,678.35 ITotal = 130,491 42.34 SLT1,top = 4,614 SLT1,bot = 3,082 134.10 n= 24.00 Y= Single-Span Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Created July2007: Page 11 of 23 -- 73 --
  79. 79. 3c. Section 2 Flexural Properties Single Span Bridge Example - Section 2 Bare Steel t b A y Ay d Ix Ad2 IX Top Flange 2.0000 14.00 28.00 71.00 1,988.00 9.33 -40.08 44,978 44,987 Web Bot Flange 0.5625 2.0000 68.00 22.00 38.25 44.00 36.00 1.00 1,377.00 14,739.00 44.00 14.67 -5.08 29.92 987 39,391 15,726 39,405 3,409.00 ITotal = 100,119 30.92 SBS1,top = 2,437 SBS1,bot = 3,238 110.25 Y= Short-Term Composite (N=8) t b A y Ay d Ix Ad2 IX Slab 8.5000 109.00 115.81 76.25 8,830.70 697.29 -22.11 56,600 57,297 Haunch Top Flange 0.0000 2.0000 14.0000 14.0000 0.00 28.00 72.00 71.00 0.00 1,988.00 0.00 9.33 -17.86 -16.86 0 7,956 0 7,966 Web Bot Flange 0.5625 2.0000 68.0000 22.0000 38.25 44.00 36.00 1.00 1,377.00 14,739.00 44.00 14.67 18.14 53.14 12,591 124,264 27,330 124,279 12,239.70 ITotal = 216,871 54.14 SST1,top = 12,145 SST1,bot = 4,006 226.06 n= 8.00 Y= Long-Term Composite (N=24) t b A y Ay d Ix Ad2 IX Slab 8.5000 109.00 38.60 76.25 2,943.57 232.43 -33.57 43,514 43,746 Haunch 0.0000 14.00 0.00 72.00 0.00 0.00 -29.32 0 0 Top Flange 2.0000 14.0000 28.00 71.00 1,988.00 9.33 -28.32 22,462 22,472 Web Bot Flange 0.5625 2.0000 68.0000 22.0000 38.25 44.00 36.00 1.00 1,377.00 14,739.00 44.00 14.67 6.68 41.68 1,705 76,425 16,444 76,439 6,352.57 ITotal = 159,101 42.68 SLT1,top = 5,426 SLT1,bot = 3,728 148.85 n= 24.00 Y= Single-Span Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Created July2007: Page 12 of 23 -- 74 --
  80. 80. 3d. Section 3 Flexural Properties Single Span Bridge Example - Section 3 Bare Steel t b A y Ay d Ix Ad2 IX Top Flange 2.0000 14.00 28.00 71.38 1,998.50 9.33 -42.25 49,970 49,980 Web Bot Flange 0.5625 2.3750 68.00 22.00 38.25 52.25 36.38 1.19 1,391.34 14,739.00 62.05 24.56 -7.25 27.94 2,008 40,796 16,747 40,820 3,451.89 ITotal = 107,546 29.13 SBS1,top = 2,487 SBS1,bot = 3,692 118.50 Y= Short-Term Composite (N=8) t b A y Ay d Ix Ad2 IX Slab 8.5000 109.00 115.81 76.63 8,874.13 697.29 -24.02 66,819 67,516 Haunch Top Flange 0.0000 2.0000 14.0000 14.0000 0.00 28.00 72.38 71.38 0.00 1,998.50 0.00 9.33 -19.77 -18.77 0 9,865 0 9,874 Web Bot Flange 0.5625 2.3750 68.0000 22.0000 38.25 52.25 36.38 1.19 1,391.34 14,739.00 62.05 24.56 16.23 51.42 10,076 138,137 24,815 138,161 12,326.02 ITotal = 240,366 52.61 SST1,top = 12,158 SST1,bot = 4,569 234.31 n= 8.00 Y= Long-Term Composite (N=24) t b A y Ay d Ix Ad2 IX Slab 8.5000 109.00 38.60 76.63 2,958.04 232.43 -35.82 49,544 49,777 Haunch 0.0000 14.00 0.00 72.38 0.00 0.00 -31.57 0 0 Top Flange 2.0000 14.0000 28.00 71.38 1,998.50 9.33 -30.57 26,174 26,184 Web Bot Flange 0.5625 2.3750 68.0000 22.0000 38.25 52.25 36.38 1.19 1,391.34 14,739.00 62.05 24.56 4.43 39.61 749 81,990 15,488 82,015 6,409.93 ITotal = 173,463 40.80 SLT1,top = 5,494 SLT1,bot = 4,251 157.10 n= 24.00 Y= Single-Span Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Created July2007: Page 13 of 23 -- 75 --
  81. 81. 4. DISTRIBUTION FACTOR FOR MOMENT 4a. Section 1: Interior Girder - One Lane Loaded: DFM1,Int,Sec1 0.4 0.3 ⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎞ = 0.06 + ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ ⎝ 14 ⎠ ⎝ L ⎠ ⎝ 12 LtS ⎠ 0.1 2 K g = n ( I + Aeg ) ( K g = (8) 72,890 in 4 + ( 95.5 in 2 ) ( 49.06") 2 ) K g = 2, 422, 000 in 4 0.4 DFM1,Int,Sec1 ⎛ 11.33' ⎞ ⎛ 11.33' ⎞ = 0.06 + ⎜ ⎟ ⎜ ⎟ ⎝ 14 ⎠ ⎝ 166.3' ⎠ 0.3 ⎛ 2, 422, 000 in 4 ⎞ ⎜ ⎟ ⎜ (12 )(166.3' )( 8.5")3 ⎟ ⎝ ⎠ 0.1 DFM1,Int,Sec1 = 0.4994 Interior Girder - Two or More Lanes Loaded: DFM2,Int,Sec1 0.4 0.3 ⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎞ = 0.075 + ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ ⎝ 9.5 ⎠ ⎝ L ⎠ ⎝ 12 LtS ⎠ DFM2,Int,Sec1 ⎛ 11.33' ⎞ ⎛ 11.33' ⎞ = 0.075 + ⎜ ⎟ ⎜ ⎟ ⎝ 9.5 ⎠ ⎝ 166.3' ⎠ 0.6 0.2 0.1 ⎛ 2, 422, 000 in 4 ⎞ ⎜ ⎟ ⎜ (12 )(166.3')( 8.5")3 ⎟ ⎝ ⎠ 0.1 DFM2,Int,Sec1 = 0.7703 Single-Span Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Created July2007: Page 14 of 23 -- 76 --
  82. 82. Exterior Girder – One Lane Loaded: The lever rule is applied by assuming that a hinge forms over the first interior girder as a truck load is applied near the parapet. The resulting reaction in the exterior girder is the distribution factor. DFM 1,Ext,Sec1 = 8.5' = 0.7500 11.33' Multiple Presence: DFM1,Ext,Sec1 = (1.2) (0.7500) = 0.9000 Exterior Girder - Two or More Lanes Loaded: DFM2,Ext,Sec1 = e DFM2,Int,Sec1 de 9.1 2.167 ' e = 0.77 + = 1.008 9.1 e = 0.77 + DFM2,Ext+ = (1.008) (0.7703) = 0.7765 4b. Section 2: Interior Girder – One Lane Loaded: 0.4 0.3 ⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎞ DFM1,Int,Sec2 = 0.06 + ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ ⎝ 14 ⎠ ⎝ L ⎠ ⎝ 12 LtS ⎠ 0.1 2 K g = n ( I + Aeg ) ( K g = (8) 100,100 in 4 + (110.3 in 2 ) ( 47.83") 2 ) K g = 2,819, 000 in 4 0.4 DFM1,Int,Sec2 ⎛ 11.33' ⎞ ⎛ 11.33' ⎞ = 0.06 + ⎜ ⎟ ⎜ ⎟ ⎝ 14 ⎠ ⎝ 166.3' ⎠ 0.3 ⎛ 2,819, 000 in 4 ⎞ ⎜ ⎟ ⎜ (12 )(166.3' ) ( 8.5 ")3 ⎟ ⎝ ⎠ 0.1 DFM1,Int,Sec2 = 0.5061 Single-Span Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Created July2007: Page 15 of 23 -- 77 --
  83. 83. Interior Girder – Two or More Lanes Loaded: 0.4 0.3 ⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎞ DFM2,Int,Sec2 = 0.075 + ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ ⎝ 9.5 ⎠ ⎝ L ⎠ ⎝ 12 LtS ⎠ DFM2,Int,Sec2 DFM2,Int,Sec2 0.1 0.6 0.2 4 ⎞ ⎛ 11.33' ⎞ ⎛ 11.33' ⎞ ⎛ 2,819, 000 in = 0.075 + ⎜ ⎜ ⎟ ⎟ ⎜ ⎟ ⎜ 3 ⎝ 9.5 ⎠ ⎝ 166.3' ⎠ ⎝ (12 )(166.3')( 8.5") ⎟ ⎠ = 0.7809 0.1 Exterior Girder - One Lane Loaded: Same as for the positive moment section: DFM1,Ext,Sec2 = 0.9000 Exterior Girder - Two or More Lanes Loaded: DFM2,Ext,Sec2 = e DFM2,Int,Sec2 e = 1.008 (same as before) DFM2,Ext,Sec2 =(1.008) (0.7809) = 0.7871 4c. Section 3: Interior Girder – One Lane Loaded: DFM 1, Int ,Sec 3 ( ⎛S ⎞ = 0.06 + ⎜ ⎟ ⎝ 14 ⎠ 2 K g = n I + Aeg ( 0.4 ⎛S⎞ ⎜ ⎟ ⎝L⎠ 0.3 ) ( ⎛ Kg ⎞ ⎜ 3 ⎟ ⎝ 12 LtS ⎠ ) 0.1 K g = ( 8 ) 107,500 in 4 + 118.5 in 2 ( 50.00") 2 ) K g = 3, 230, 000 in 4 DFM 1, Int , Sec 3 ⎛ 11.33' ⎞ = 0.06 + ⎜ ⎟ ⎝ 14 ⎠ 0.4 ⎛ 11.33' ⎞ ⎜ ⎟ ⎝ 166.3' ⎠ 0.3 ⎛ ⎞ ⎜ ⎟ 3 ⎟ ⎜ (12 )(166.3')( 8.5") ⎝ ⎠ 3, 230, 000 in 4 0.1 DFM 1, Int ,Sec 3 = 0.5122 Single-Span Bridge Example ODOT LRFD Short Course - Steel AASHTO-LRFD 2007 Created July2007: Page 16 of 23 -- 78 --

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